Past JEE Main Entrance Papers

Hint

(b) \(x^2-x-2=0 \Rightarrow(x-2)(x+1)=0\)
\(
\Rightarrow x=2,-1 \Rightarrow \alpha=2
\)
\(\therefore \lim _{x \rightarrow 2^{+}} \frac{\sqrt{1-\cos \left(x^2-x-2\right)}}{x-2}=\lim _{x \rightarrow 2^{+}} \frac{\sqrt{2}\left|\sin \left(\frac{\left.x^2-x-2\right)}{2}\right)\right|}{x-2}\)
\(=\lim _{x \rightarrow 2^{+}} \frac{\sqrt{2} \sin \frac{\left(x^2-x-2\right)}{2}}{\left(\frac{x^2-x-2}{2}\right)} \times \frac{\left(x^2-x-2\right)}{2(x-2)}\)
\(=\frac{1}{\sqrt{2}} \lim _{x \rightarrow 2^{-}}\left(\frac{\sin \left(\frac{x^2-x-2}{2}\right)}{\frac{x^2-x-2}{2}}\right) \times \lim _{x \rightarrow 2^{+}} \frac{(x-2)(x+1)}{(x-2)}\)
\(
=\frac{1}{\sqrt{2}} \times 1 \times 3=\frac{3}{\sqrt{2}}
\)

Hint

(b) Let
\(L=\lim _{x \rightarrow 0} \frac{x\left(e^{\frac{\sqrt{1+x^2+x^4}-1}{x}-1}\right)}{\sqrt{1+x^2+x^4}-1}=\lim _{x \rightarrow 0} \frac{e^{\frac{\sqrt{1+x^2+x^4}-1}{x}}-1}{\frac{\sqrt{1+x^2+x^4}-1}{x}}\)
Put \(\frac{\sqrt{1+x^2+x^4}-1}{x}=t\) when \(x \rightarrow 0 \Rightarrow t \rightarrow 0\)
\(
\therefore L=\lim _{t \rightarrow 0} \frac{e^t-1}{t}=1
\)

Hint

(b) Given \(\lim _{x \rightarrow 0}\left|\frac{1-x+|x|}{\lambda-x+[x]}\right|=L\)
Here, L.H.L. \(=\lim _{h \rightarrow 0}\left|\frac{1+h+h}{\lambda+h-1}\right|=\left|\frac{1}{\lambda-1}\right|\)
\(
\text { R.H.L. }=\lim _{h \rightarrow 0}\left|\frac{1-h+h}{\lambda+h+0}\right|=\left|\frac{1}{\lambda}\right|
\)
Given that limit exists. Hence L.H.L. = R.H.L.
\(
\begin{aligned}
&\Rightarrow|\lambda-1|=|\lambda| \\
&\Rightarrow \lambda=\frac{1}{2} \text { and } L=\left|\frac{1}{\lambda}\right|=2
\end{aligned}
\)

Hint

(a) 

Maxima of \(\mathrm{f}(\mathrm{x})\) occured at \(\mathrm{x}=2\) i.e. \(\alpha=2\)
Minima of \(\mathrm{g}(\mathrm{x})\) occured at \(\mathrm{x}=-1\) i.e. \(\beta=-1\)
\(
\therefore \lim _{x \rightarrow 2} \frac{(x-1)(x-2)(x-3)}{(x-2)(x-4)}=\frac{1}{2}
\)

Hint

(c)

Step 1: Find a condition between \(a\) and \(b\) using indeterminate form of limit.
Given, \(\lim _{x \rightarrow 1} \frac{x^2-a x+b}{x-1}=5\)
(Since denominators become zero on putting limit means numerators must be zero when \(x \rightarrow 1\) otherwise limit does not exist.
It can shown that if \(\lim _{x \rightarrow c} g(x)=0\) and \(\lim _{x \rightarrow c} f(x) \neq 0\), then \(\lim _{x \rightarrow c} \frac{f(x)}{g(x)}\) does not exist.
Hence, in order for the limit to exist, it must be the case that \(\lim _{x \rightarrow c} f(x)=0\).)

\(
(1)^2-a(1)+b=0 \quad \rightarrow \quad b-a+1=0 \quad \cdots(1)
\)
Step 2: Use L’hospital rule to find \(a\).
\(
\begin{aligned}
& \lim _{x \rightarrow 1} \frac{x^2-a x+b}{x-1}=5 \\
\Rightarrow & \lim _{x \rightarrow 1} \frac{2 x-a}{1}=5 \quad\left[\because \lim _{x \rightarrow a} \frac{f(x)}{g(x)}=\lim _{x \rightarrow a} \frac{f^{\prime}(x)}{g^{\prime}(x)}\right] \\
\Rightarrow & 2-a =5 \\
\Rightarrow & a=-3
\end{aligned}
\)
Substitute a= -3 in equation (1), we get
\(
b=-4
\)
\(
\therefore a+b=-7
\)

Hint

(a) \(\lim _{x \rightarrow 0} \frac{\sin ^2 x}{\sqrt{2}-\sqrt{1+\cos x}}=\lim _{x \rightarrow 0} \frac{\sin ^2 x}{\sqrt{2}-\sqrt{2 \cos ^2 \frac{x}{2}}}\left[\frac{0}{0}\right]\)
\(
\begin{aligned}
&=\lim _{x \rightarrow 0} \frac{\sin ^2 x}{\sqrt{2}\left[1-\cos \frac{x}{2}\right]}=\lim _{x \rightarrow 0} \frac{\sin ^2 x}{2 \sqrt{2} \sin ^2 \frac{x}{4}} \\
&=\lim _{x \rightarrow 0} \frac{\left(\frac{\sin x}{x}\right)^2 \cdot 16}{2 \sqrt{2}\left(\frac{\sin \frac{x}{4}}{\frac{x}{4}}\right)^2}=\frac{16}{2 \sqrt{2}}=4 \sqrt{2}
\end{aligned}
\)

Hint

(d) \(\lim _{x \rightarrow \frac{\pi}{4}} \frac{\cot ^3 x-\tan x}{\cos \left(x+\frac{\pi}{4}\right)}=\lim _{x \rightarrow \frac{\pi}{4}} \frac{\cot ^3 x\left(1-\frac{\tan x}{\cos ^3 x}\right)}{\cos (x+\pi / 4)}\) \(=\lim _{x \rightarrow \frac{\pi}{4}} \frac{\left(1-\tan ^4 x\right)}{\tan ^3 x \cos (x+\pi / 4)}\)
\(
\begin{aligned}
&=\lim _{x \rightarrow \frac{\pi}{4}} \frac{\left(1+\tan ^2 x\right)(1-\tan x)(1+\tan x)}{\tan ^3 x\left(\frac{\cos x-\sin x}{\sqrt{2}}\right)}\\
&=\lim _{x \rightarrow \frac{\pi}{4}} \frac{\left(1+\tan ^2 x\right)(1+\tan x)(\cos x-\sin x)}{\frac{\sin ^3 x}{\cos ^2 x}\left(\frac{\cos x-\sin x}{\sqrt{2}}\right)}=\frac{(2)(2)}{(\sqrt{2})(\sqrt{2})}=8
\end{aligned}
\)

Hint

(d) \(\lim _{x \rightarrow 0} \frac{x \cot 4 x}{\sin ^2 x \cdot \cot ^2 2 x}=\lim _{x \rightarrow 0} \frac{x \cdot \tan ^2 2 x}{\sin ^2 x \cdot \tan 4 x}\)
\(
=\lim _{x \rightarrow 0}\left(\frac{x}{\sin x}\right)^2 \cdot\left(\frac{\tan 2 x}{2 x}\right)^2 \cdot\left(\frac{4 x}{\tan 4 x}\right) \cdot \frac{4}{2^2}=1
\)

Explanation:
Rewrite in sine and cosine using the identity \(\tan x=\frac{\sin x}{\cos x}\).
\(
\begin{aligned}
&=\lim _{x \rightarrow 0} \frac{\tan (4 x)}{x} \\
&=\lim _{x \rightarrow 0} \frac{\frac{\sin (4 x)}{\cos (4 x)}}{x} \\
&=\lim _{x \rightarrow 0} \frac{\sin (4 x)}{x \cos (4 x)}
\end{aligned}
\)
Rewrite so that that one expression is \(\frac{\sin (4 x)}{x}\).
\(
=\lim _{x \rightarrow 0} \frac{\sin (4 x)}{x} \times \frac{1}{\cos (4 x)}
\)
Use the well know limit that \(\lim _{x \rightarrow 0} \frac{\sin x}{x}=1\) to deduce the fact that \(\lim _{x \rightarrow 0} \frac{\sin (4 x)}{x}=4\)
\(
\begin{aligned}
&=4 \times \frac{1}{\cos (0)} \\
&=4 \times 1 \\
&=4
\end{aligned}
\)

Hint

(b) \(\lim _{x \rightarrow 1^{+}} \frac{(1-|x|+\sin (|1-x|)) \sin \left(\frac{\pi}{2}[1-x]\right)}{|1-x|[1-x]}\)
\(
=\lim _{h \rightarrow 0} \frac{(1-|1+h|+\sin (|1-1-h|)) \sin \left(\frac{\pi}{2}[1-1-h]\right)}{|1-1-h|[1-1-h]}
\)
\(
\begin{aligned}
&=\lim _{h \rightarrow 0} \frac{(1-1-h+\sinh ) \sin \left(\frac{\pi}{2}(-1)\right)}{h([0-h])} \\
&=\lim _{h \rightarrow 0} \frac{(-h+\sin h) \sin \left(-\frac{\pi}{2}\right)}{h(-1)}=0
\end{aligned}
\)

Hint

(a)

\(
L=\lim _{y \rightarrow 0} \frac{\sqrt{1+\sqrt{1+y^4}}-\sqrt{2}}{y^4}
\)
\(
\begin{aligned}
&=\lim _{y \rightarrow 0} \frac{\left(\sqrt{1+\sqrt{1+y^4}}-\sqrt{2}\right)\left(\sqrt{1+\sqrt{1+y^4}}+\sqrt{2}\right)}{y^4\left(\sqrt{1+\sqrt{1+y^4}}+\sqrt{2}\right)} \\
&=\lim _{y \rightarrow 0} \frac{1+\sqrt{1+y^4}-2}{y^4\left(\sqrt{1+\sqrt{1+y^4}}+\sqrt{2}\right)} \\
&=\lim _{y \rightarrow 0} \frac{\left(\sqrt{1+y^4}-1\right)\left(\sqrt{1+y^4+1}\right)}{y^4\left(\sqrt{1+\sqrt{1+y^4}+\sqrt{2}}\right)\left(\sqrt{1+y^4}+1\right)} \\
&=\lim _{y \rightarrow 0} \frac{1+y^4-1}{y^4\left(\sqrt{1+\sqrt{1+y^4}+\sqrt{2}}\right)\left(\sqrt{1+y^4}+1\right)} \\
&=\frac{1}{2 \sqrt{2} \times 2}=\frac{1}{4 \sqrt{2}}
\end{aligned}
\)

Hint

(a) \(\lim _{x \rightarrow 0^{-}} \frac{x([x]+|x|) \cdot \sin [x]}{|x|}\)
\(
\begin{aligned}
&=\lim _{h \rightarrow 0} \frac{(0-h)([0-h]+|0-h|) \cdot \sin [0-h]}{|0-h|} \\
&=\lim _{h \rightarrow 0} \frac{(-h)(-1+h) \sin (-1)}{h} \\
&=\lim _{h \rightarrow 0}(1-h) \sin (-1)=-\sin 1
\end{aligned}
\)

Hint

\(
\text { (d) Let, } L=\lim _{x \rightarrow 0} \frac{(x \tan 2 x-2 x \tan x)}{(1-\cos 2 x)^2}=\lim _{x \rightarrow 0} K \text { (say) }
\)
\(
\begin{aligned}
&\Rightarrow K=\frac{x\left[\frac{2 \tan x}{1-(\tan x)^2}\right]-2 x \tan x}{\left(1-\left(1-2 \sin ^2 x\right)\right)^2}\\
&=\frac{2 x \tan x-\left[2 x \tan x-2 x \tan ^3 x\right]}{4 \sin ^4 x \times\left(1-\tan ^2 x\right)}\\
&=\frac{2 x \tan ^3 x}{4 \sin ^4 x \times\left(1-\tan ^2 x\right)}=\frac{2 x \tan ^3 x}{4 \sin ^4 x \times\left(\frac{\cos ^2 x-\sin ^2 x}{\cos ^2 x}\right)}\\
&=\frac{2 x \frac{\sin ^3 x}{\cos ^3 x}}{4 \sin ^4 x \times\left(\frac{\cos ^2 x-\sin ^2 x}{\cos ^2 x}\right)}\\
&\Rightarrow K=\frac{x}{2 \sin x \times\left(\cos ^2 x-\sin ^2 x\right) \cos x}\\
&\therefore L=\lim _{x \rightarrow 0} \frac{x}{2 \sin x} \times \lim _{x \rightarrow 0} \frac{1}{\cos x\left(\cos ^2 x-\sin ^2 x\right)}\\
&=\lim _{x \rightarrow 0} \frac{x}{2 \sin x} \times \lim _{x \rightarrow 0} \frac{1}{\cos 0\left(\cos ^2 0-\sin ^2 0\right)}=\frac{1}{2}
\end{aligned}
\)

Hint

(c) \(\lim _{x \rightarrow \frac{\pi}{2}} \frac{\cot x(1-\sin x)}{-8\left(x-\frac{\pi}{2}\right)^3}=\lim _{x \rightarrow \frac{\pi}{2}} \frac{\cot x(1-\sin x)}{8\left(\frac{\pi}{2}-x\right)^3}\)
Put \(\frac{\pi}{2}-\mathrm{x}=\mathrm{t} \Rightarrow\) as \(\mathrm{x} \rightarrow \frac{\pi}{2} \Rightarrow \mathrm{t} \rightarrow 0\)
\(
\begin{aligned}
&=\lim _{t \rightarrow 0} \frac{\cot \left(\frac{\pi}{2}-t\right)\left(1-\sin \left(\frac{\pi}{2}-t\right)\right)}{8 t^3} \\
&=\lim _{t \rightarrow 0} \frac{\tan t(1-\cos t)}{8 t^3}=\lim _{t \rightarrow 0} \frac{\tan t}{8 t} \cdot \frac{1-\cos t}{t^2}=\frac{1}{8} \cdot 1 \cdot \frac{1}{2}=\frac{1}{16}
\end{aligned}
\)

Hint

(a) Multiply and divide by \(x\) in the given expression, we get
\(
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{(1-\cos 2 x)}{x^2} \frac{(3+\cos x)}{1} \cdot \frac{x}{\tan 4 x} \\
=& \lim _{x \rightarrow 0} \frac{2 \sin ^2 x}{x^2} \cdot \frac{3+\cos x}{1} \cdot \frac{x}{\tan 4 x} \\
=& 2 \lim _{x \rightarrow 0} \frac{\sin ^2 x}{x^2} \cdot \lim _{x \rightarrow 0} 3+\cos x \cdot \lim _{x \rightarrow 0} \frac{x}{\tan 4 x} \\
=& 2.4 \frac{1}{4} \lim _{x \rightarrow 0} \frac{4 x}{\tan 4 x}=2.4 . \frac{1}{4}=2
\end{aligned}
\)

Hint

(c) \(\lim _{x \rightarrow 0} \frac{2 x e^{x^2}+\sin x}{2 \sin x \cos x}\)
\(
\lim _{x \rightarrow 0}\left(\frac{x}{\sin x} e^{x^2}+\frac{1}{2}\right) \frac{1}{\cos x}=1+\frac{1}{2}=\frac{3}{2}
\)

Hint

(b) \(\lim _{x \rightarrow 0} \frac{\sin \left(\pi \cos ^2 x\right)}{x^2}=\lim _{x \rightarrow 0} \frac{\sin \left[\pi\left(1-\sin ^2 x\right)\right]}{x^2}\) \(=\lim _{x \rightarrow 0} \sin \frac{\left(\pi-\pi \sin ^2 x\right)}{x^2} \quad[\because \sin (\pi-\theta)=\sin \theta]\) \(=\lim _{x \rightarrow 0} \sin \frac{\left(\pi \sin ^2 x\right)}{\pi \sin ^2 x} \times \frac{\pi \sin ^2 x}{x^2}\)
\(
=\lim _{x \rightarrow 0} 1 \times \pi\left(\frac{\sin x}{x}\right)^2=\pi
\)

Hint

(d) Multiply and divide by \(x\) in the given expression, we get
\(
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{(1-\cos 2 x)}{x^2} \frac{(3+\cos x)}{1} \cdot \frac{x}{\tan 4 x} \\
=& \lim _{x \rightarrow 0} \frac{2 \sin ^2 x}{x^2} \cdot \frac{3+\cos x}{1} \cdot \frac{x}{\tan 4 x} \\
=& 2 \lim _{x \rightarrow 0} \frac{\sin ^2 x}{x^2} \cdot \lim _{x \rightarrow 0}(3+\cos x) \cdot \lim _{x \rightarrow 0} \frac{4 x}{\tan 4 x} \times \frac{1}{4}=2 \cdot 4 \cdot \frac{1}{4}=2
\end{aligned}
\)

Hint

(b) \((\sqrt[3]{1+a}-1) x^2+(\sqrt{1+a}-1) x+(\sqrt[6]{1+a}-1)=0\)
Let \(a+1=y\), then equation reduces to
\(
\left(y^{1 / 3}-1\right) x^2+\left(y^{1 / 2}-1\right) x+\left(y^{1 / 6}-1\right)=0
\)
On dividing both sides by \(y-1[latex], we get
[latex]
\left(\frac{y^{1 / 3}-1}{y-1}\right) x^2+\left(\frac{y^{1 / 2}-1}{y-1}\right) x+\left(\frac{y^{1 / 6}-1}{y-1}\right)=0
\)
On taking limit as \(y \rightarrow 1\) i.e. \(a \rightarrow 0\) on both sides, we get \(\frac{1}{3} x^2+\frac{1}{2} x+\frac{1}{6}=0 \Rightarrow 2 x^2+3 x+1=0\) \(\Rightarrow x=-1,-\frac{1}{2}\) (roots of the equation)
\(
\therefore \quad \lim _{a \rightarrow 0^{+}} \alpha(a)=-1, \lim _{a \rightarrow 0^{+}} \beta(a)=-\frac{1}{2}
\)

Hint

\(
\text { (b) Given: } \lim _{x \rightarrow \infty}\left(\frac{x^2+x+1}{x+1}-a x-b\right)=4
\)
\(
\begin{aligned}
&\Rightarrow \lim _{x \rightarrow \infty} \frac{x^2+x+1-a x^2-a x-b x-b}{x+1}=4 \\
&\Rightarrow \lim _{x \rightarrow \infty} \frac{(1-a) x^2+(1-a-b) x+(1-b)}{x+1}=4
\end{aligned}
\)
For this limit to be finite \(1-a=0 \Rightarrow a=1\) then the given limit reduces to
\(
\begin{aligned}
&\lim _{x \rightarrow \infty} \frac{-b x+(1-b)}{x+1}=4 \Rightarrow \lim _{x \rightarrow \infty} \frac{-b+\frac{(1-b)}{x}}{1+\frac{1}{x}}=4 \\
&\Rightarrow-b=4 \text { or } b=-4, \quad \therefore a=1, b=-4
\end{aligned}
\)

Hint

(d) \(\lim _{x \rightarrow 0} \frac{[(a-n) n x-\tan x] \sin n x}{x^2}=0\)
\(
\Rightarrow \lim _{x \rightarrow 0} n . \frac{\sin n x}{n x}\left[\left\{(a-n) n-\frac{\tan x}{x}\right\}\right]=0
\)
\(
\Rightarrow n .1[(a-n) n-1]=0 \Rightarrow a=\frac{1}{n}+n
\)
[where \(n\) is non zero real number]

Hint

(b) \(\lim _{x \rightarrow 0} \frac{\sin \left(\pi \cos ^2 x\right)}{x^2}=\lim _{x \rightarrow 0} \frac{\sin \left(\pi-\pi \sin ^2 x\right)}{x^2}\) \(=\lim _{x \rightarrow 0} \frac{\sin \left(\pi \sin ^2 x\right)}{\pi \sin ^2 x} \times \frac{\left(\pi \sin ^2 x\right)}{x^2}=\pi\)

Hint

(b) \(\lim _{n \rightarrow \infty}\left(\frac{1}{1-n^2}+\frac{2}{1-n^2}+\ldots+\frac{n}{1-n^2}\right)\)
\(
=\lim _{n \rightarrow \infty} \frac{1+2+3+\ldots+n}{1-n^2}=\lim _{n \rightarrow \infty} \frac{\Sigma n}{1-n^2}=\lim _{n \rightarrow \infty} \frac{n(n+1)}{2\left(1-n^2\right)}
\)
\(
=\lim _{n \rightarrow \infty} \frac{1+1 / n}{2\left[\frac{1}{n^2}-1\right]}=-1 / 2
\)

Hint

(d)
\(\lim _{x \rightarrow 1} \frac{G(x)-G(1)}{x-1}, \frac{0}{0}\) form
\(=\mathrm{G}^{\prime}(1)\)
Now G \({ }^{\prime}(x)=-\frac{1}{2 \sqrt{25-x^2}}(-2 x)\), differentiate using chain rule \(\Rightarrow \mathrm{G}^{\prime}(\mathrm{x})=\frac{\mathrm{x}}{\sqrt{25-\mathrm{x}^2}}\)
\(\therefore G^{\prime}(1)=\frac{1}{\sqrt{24}}\), which is required limit

Hint

(c) \(f(x)=\sqrt{\frac{x-\sin x}{x+\cos ^2 x}}\)
\(
\therefore \lim _{x \rightarrow \infty} f(x)=\lim _{x \rightarrow \infty} \sqrt{\frac{1-\frac{\sin x}{x}}{1+\frac{\cos ^2 x}{x}}}=\sqrt{\frac{1-0}{1+0}}=1
\)

Hint

\(
\begin{aligned}
&\lim _{x \rightarrow \frac{\pi}{2}} \frac{4 \sqrt{2} \cdot 2 \sin 2 x \cos x}{2 \sin 2 x \sin \frac{3 x}{2}+\left(\cos \frac{5 x}{2}-\cos \frac{3 x}{2}\right)-\sqrt{2}(1+\cos 2 x)} \\
&=\lim _{x \rightarrow \frac{\pi}{2}} \frac{8 \sqrt{2} \cdot 2 \sin x \cos x \cos x}{2 \sin 2 x \sin \frac{3 x}{2}-2 \sin 2 x \sin \frac{x}{2}-2 \sqrt{2} \cos ^2 x}
\end{aligned}
\)
\(
\begin{aligned}
&=\lim _{x \rightarrow \frac{\pi}{2}} \frac{16 \sqrt{2} \sin x \cos ^2 x}{2 \sin 2 x\left(\sin \frac{3 x}{2}-\sin \frac{x}{2}\right)-2 \sqrt{2} \cos ^2 x} \\
&=\lim _{x \rightarrow \frac{\pi}{2}} \frac{16 \sqrt{2} \sin x \cos ^2 x}{4 \sin x \cos x\left(2 \cos x \cdot \sin \frac{x}{2}\right)-2 \sqrt{2} \cos ^2 x} \\
&=\frac{16 \sqrt{2} \sin x \cos ^2 x}{2 \cos ^2 x\left(4 \sin x \sin \frac{x}{2}-\sqrt{2}\right)} \\
&=\lim _{x \rightarrow \frac{\pi}{2}} \frac{8 \sqrt{2} \sin x}{4 \sin x \cdot \sin \frac{x}{2}-\sqrt{2}}=8
\end{aligned}
\)

Hint

\(\lim _{\alpha \rightarrow 0} \frac{e^{\cos \alpha^n}-e}{\alpha^m}=\frac{-e}{2}\)
\(\Rightarrow \lim _{\alpha \rightarrow 0} \frac{e\left[e^{\cos \alpha^n-1}-1\right]}{\cos \alpha^n-1} \times \frac{\cos \alpha^n-1}{\alpha^m}=\frac{-e}{2}\)
\(
\begin{aligned}
&\Rightarrow e \lim _{\alpha \rightarrow 0} \frac{-2 \sin ^2 \frac{\alpha^n}{2}}{\left(\frac{\alpha^n}{2}\right)^2} \times \frac{\left(\frac{\alpha^n}{2}\right)^2}{\alpha^m}=\frac{-e}{2} \\
&\Rightarrow \frac{-e}{2} \alpha^{2 n-m}=\frac{-e}{2} \Rightarrow \frac{m}{n}=2
\end{aligned}
\)

Hint

\(
\begin{aligned}
&\lim _{x \rightarrow 1}\left\{\frac{-a x+\sin (x-1)+a}{x+\sin (x-1)-1}\right\}^{\frac{1-x}{1-\sqrt{x}}}=\frac{1}{4} \\
&\Rightarrow \lim _{x \rightarrow 1}\left\{\frac{a(1-x)+\sin (x-1)}{(x-1)+\sin (x-1)}\right\}^{1+\sqrt{x}} \\
&\Rightarrow \lim _{x \rightarrow 1}\left\{\frac{-a+\frac{\sin (x-1)}{x-1}}{1+\frac{\sin (x-1)}{x-1}}\right\}^{1+\sqrt{x}} \Rightarrow\left(\frac{-a+1}{2}\right)^2=\frac{1}{4} \\
&\Rightarrow a=0 \text { or } 2
\end{aligned}
\)
\(\therefore\) Largest value of a is 2.

Hint

\(
\begin{aligned}
&\lim _{x \rightarrow 0} \frac{1}{x^8}\left[\left(1-\cos \frac{x^2}{2}\right)-\cos \frac{x^2}{4}\left(1-\cos \frac{x^2}{2}\right)\right] \\
&=\lim _{x \rightarrow 0} \frac{1}{x^8}\left[\left(1-\cos \frac{x^2}{2}\right)\left(1-\cos \frac{x^2}{4}\right)\right] \\
&=\lim _{x \rightarrow 0} \frac{1}{x^8}\left[2 \sin ^2 \frac{x^2}{4} 2 \sin ^2 \frac{x^2}{8}\right] \\
&=\lim _{x \rightarrow 0} \frac{1}{256}\left(\frac{\sin \frac{x^2}{4}}{\left(\frac{x^2}{4}\right)}\right)^2\left(\frac{\sin \frac{x^2}{8}}{\left(\frac{x^2}{8}\right)}\right) \\
&=\frac{1}{256}=2^{-8} \\
&\therefore k=8
\end{aligned}
\)

Hint

\(
\text { Let } 3^x=t^2
\)
\(
\begin{aligned}
&\lim _{t \rightarrow 3} \frac{t^2+\frac{27}{t^2}-12}{\frac{1}{t}-\frac{3}{t^2}}=\lim _{t \rightarrow 3} \frac{t^4-12 t^2+27}{t-3} \\
&=\lim _{t \rightarrow 3} \frac{\left(t^2-3\right)(t+3)(t-3)}{t-3} \\
&=\left(3^2-3\right)(3+3)=36 .
\end{aligned}
\)

Hint

\(\begin{aligned}
&\lim _{x \rightarrow 0^{+}} \frac{(1-x)^{1 / x}-e^{-1}}{x^a}=\lim _{x \rightarrow 0^{+}} \frac{e^{\left(\frac{\ln (1-x)}{x}\right)}-\frac{1}{e}}{x^a}\\
&\left[\because(1-x)^{1 / x}=e^{1 / x(1-x)}\right]
\end{aligned}\)
\(
\lim _{x \rightarrow 0^{+}} \frac{1}{e} \frac{e^{\left(1+\frac{\ln (1-x)}{x}\right)}-1}{x^a}=\frac{1}{e} \lim _{x \rightarrow 0^{+}} \frac{\ln (1-x)+x}{x^{(a+1)}}
\)
\(
=\frac{1}{e} \lim _{x \rightarrow 0^{+}} \frac{\left(-x-\frac{x^2}{2}-\frac{x^3}{3}-\ldots\right)+x}{x^{a+1}} \quad \therefore a=1
\)

Hint

Given : \(f(9)=9, f^{\prime}(9)=4\)
\(
\begin{aligned}
& \therefore \quad \lim _{x \rightarrow 9} \frac{\sqrt{f(x)}-3}{\sqrt{x}-3}=\lim _{x \rightarrow 9} \frac{(\sqrt{f(x)}-3)(\sqrt{f(x)}+3)}{(\sqrt{x}-3)(\sqrt{x}+3)} \\
=& \lim _{x \rightarrow 9} \frac{\sqrt{x}+3}{\sqrt{f(x)}+3} \times \lim _{x \rightarrow 9} \frac{f(x)-9}{x-9} \\
=& {\left[\frac{3+3}{3+3}\right] \cdot f^{\prime}(9)=1 \times 4=4 }
\end{aligned}
\)

Hint

\(
\begin{aligned}
&\lim _{x \rightarrow-\infty}\left[\frac{x^4 \sin \left(\frac{1}{x}\right)+x^2}{\left(1+|x|^3\right)}\right]\\
&=\lim _{x \rightarrow-\infty} \frac{x^3}{1+|x|^3}\left[x \sin \left(\frac{1}{x}\right)+\frac{1}{x}\right]\\
&=\lim _{x \rightarrow-\infty} \frac{x^3}{|x|^3}\left[\frac{1}{1+\frac{1}{|x| x^2}}\right]\left[x \sin \left(\frac{1}{x}\right)+\frac{1}{x}\right]\\
&=\lim _{x \rightarrow-\infty} \frac{x^3}{|x|^3} .1=\lim _{x \rightarrow-\infty} \frac{x^3}{-x^3}=-1
\end{aligned}
\)

Hint

Given :
\(
f(x)=\left\{\begin{array}{cc}
\sin x, x \neq \pi, n=0, \pm 1, \pm 2, \ldots \\
2, & \text { otherwise }
\end{array}\right.
\)
And \(g(x)= \begin{cases}x^2+1, & x \neq 0,2 \\ 4, & x=0 \\ 5, & x=2\end{cases}\)
\(
\therefore \quad \lim _{x \rightarrow 0} g[f(x)]=\lim _{x \rightarrow 0} g(\sin x) \Rightarrow \lim _{x \rightarrow 0}\left(\sin ^2 x+1\right)=1
\)

Hint

The answer is False. \(f(x)=\frac{|x-a|}{x-a}\) and \(g(x)=\frac{x-a}{|x-a|}\) then \(\lim _{x \rightarrow a}(f(x) g(x))\) exists but neither \(\lim _{x \rightarrow a} f(x)\) nor \(\lim _{x \rightarrow a} g(x)\) exists.

Hint

Given:
\(f(x)=\frac{1-x(1+|1-x|)}{|1-x|} \cos \left(\frac{1}{1-x}\right)\) for \(x \neq 1\)
\(
\begin{aligned}
& \lim _{\mathrm{x} \rightarrow 1^{-}} \mathrm{f}(\mathrm{x})=\lim _{\mathrm{h} \rightarrow 0} \frac{1-(1-\mathrm{h})(1+\mathrm{h})}{\mathrm{h}} \cos \left(\frac{1}{\mathrm{~h}}\right) \\
=& \lim _{\mathrm{h} \rightarrow 0} \frac{1-1+\mathrm{h}^2}{\mathrm{~h}} \cos \left(\frac{1}{\mathrm{~h}}\right)=\lim _{\mathrm{h} \rightarrow 0} \mathrm{~h} \cos \left(\frac{1}{\mathrm{~h}}\right)=0 \\
& \lim _{\mathrm{x} \rightarrow 1^{+}} \mathrm{f}(\mathrm{x})=\lim _{\mathrm{h} \rightarrow 0} \frac{1-(1+\mathrm{h})(1+\mathrm{h})}{\mathrm{h}} \cos \left(\frac{1}{\mathrm{~h}}\right) \\
=& \lim _{\mathrm{h} \rightarrow 0} \frac{-2 \mathrm{~h}-\mathrm{h}^2}{\mathrm{~h}} \cos \left(\frac{1}{\mathrm{~h}}\right)=\lim _{\mathrm{h} \rightarrow 0}(-2-\mathrm{h}) \cos \left(\frac{1}{\mathrm{~h}}\right) \\
=&-2 \times(\text { Some value oscillating between }-1 \text { and } 1) \\
& \therefore \quad \lim _{x \rightarrow 1^{+}} f(x) \text { does not exist. }
\end{aligned}
\)

Hint

\((\mathbf{b}, \mathbf{d})\) Given:
\(
\lim _{n \rightarrow \infty} \frac{1^a+2^a+—+n^a}{(n+1)^{a-1}[(n a+1)+(n a+2)+—+(n a+n)]}=\frac{1}{60}
\)
\(
\Rightarrow \lim _{n \rightarrow \infty} \frac{n^a}{(n+1)^{a-1}} \frac{\left(\frac{1}{n}\right)^a+\left(\frac{2}{n}\right)^a+—+\left(\frac{n}{n}\right)^a}{n^2 a+\frac{n(n+1)}{2}}=\frac{1}{60}
\)
\(
\Rightarrow \lim _{n \rightarrow \infty} \frac{n^{a-1}}{(n+1)^{a-1}} \frac{\frac{1}{n} \sum_{r=1}^n\left(\frac{r}{n}\right)^a}{a+\frac{1}{2}\left(1+\frac{1}{n}\right)}=\frac{1}{60}
\)
\(
\Rightarrow \lim _{\mathrm{n} \rightarrow \infty}\left(\frac{1}{1+\frac{1}{\mathrm{n}}}\right)^{\mathrm{a}-1} \frac{\frac{1}{\mathrm{n}} \sum_{\mathrm{r}=1}^{\mathrm{n}}\left(\frac{\mathrm{r}}{\mathrm{n}}\right)^{\mathrm{a}}}{\mathrm{a}+\frac{1}{2}\left(1+\frac{1}{\mathrm{n}}\right)}=\frac{1}{60}
\)
\(
\begin{aligned}
&\Rightarrow \frac{\int_0^1 x^a d x}{a+\frac{1}{2}}=\frac{1}{60} \Rightarrow \frac{\left[x^{a+1}\right]_0^1}{(a+1)\left(a+\frac{1}{2}\right)}=\frac{1}{60} \\
&\Rightarrow \frac{1}{(a+1)\left(a+\frac{1}{2}\right)}=\frac{1}{60} \\
&\therefore 2 a^2+3 a-119=0 \Rightarrow(a-7)(2 a-17)=0 \\
&\therefore a=7 \text { or }-\frac{17}{2}
\end{aligned}
\)

Hint

\(
\begin{aligned}
&\frac{\sqrt{1-\cos [2(x-1)]}}{x-1}=\frac{\sqrt{2 \sin ^2(x-1)}}{x-1}\\
&=\sqrt{2} \cdot \frac{\sqrt{\sin ^2(x-1)}}{x-1}=\sqrt{2} \frac{|\sin (x-1)|}{x-1}\\
&\text { L.H.L. }=\sqrt{2} \cdot \lim _{x \rightarrow 1} \frac{|\sin (x-1)|}{x-1}=\sqrt{2} . \lim _{h \rightarrow 0} \frac{|\sin (-h)|}{-h}\\
&=\sqrt{2} \lim _{h \rightarrow 0} \frac{\sin \mathrm{h}}{-h}=-\sqrt{2}\\
&\text { R.H.L. }=\sqrt{2} \lim _{x \rightarrow 1^{+}} \frac{|\sin (x-1)|}{x-1}\\
&=\sqrt{2} \lim _{h \rightarrow 0} \frac{|\sin h|}{h}=\sqrt{2} \lim _{h \rightarrow 0} \frac{\sin h}{h}=\sqrt{2}\\
&\text { L.H.L. } \neq \text { R.H.L., } \quad \therefore \lim _{x \rightarrow 1} f(x) \text { does not exist. }
\end{aligned}
\)

Hint

\(
\begin{aligned}
&\lim _{x \rightarrow 0} \frac{\sqrt{\frac{1}{2}(1-\cos 2 x)}}{x} \\
&=\lim _{x \rightarrow 0} \frac{\sqrt{\left.\frac{1}{2} \cdot 2 \sin ^2 x\right)}}{x}=\lim _{x \rightarrow 0} \frac{|\sin x|}{x} \\
&\therefore \quad \text { L.H.L. }=\lim _{h \rightarrow 0} \frac{|\sin (0-h)|}{0-h}=\lim _{h \rightarrow 0} \frac{|-\sin h|}{-h} \\
&=\lim _{h \rightarrow 0} \frac{\sin h}{-h}=-1 \\
&\text { R.H.L. }=\lim _{h \rightarrow 0} \frac{|\sin (0+h)|}{0+h}=\lim _{h \rightarrow 0} \frac{\sin h}{h}=1
\end{aligned}
\)
Thus, L.H.L. \(\neq\) R.H.L.
Therefore, the given limit does not exist.

Hint

(a) \(f(x)=x^{2 / 3}\) for Property 1
\(
\lim _{h \rightarrow 0} \frac{h^{2 / 3}-0}{\sqrt{|h|}}=\lim _{h \rightarrow 0} \frac{|h|^{2 / 3}}{|h|^{1 / 2}}=\lim _{h \rightarrow 0}|h|^{1 / 6}=0
\)
\(\therefore\) option (a) is correct.
(b) \(\mathrm{f}(\mathrm{x})=\sin \mathrm{x}\) for Property 2
\(
\lim _{h \rightarrow 0} \frac{\sin h-\sin 0}{h^2}=\lim _{h \rightarrow 0} \frac{\sin h}{h} \times \frac{1}{h}
\)
when does not exist.
\(\therefore\) (b) is incorrect option.
(c) \(\mathrm{f}(\mathrm{x})=|\mathrm{x}|\) for Property 1
\(
\lim _{h \rightarrow 0} \frac{|h|-0}{\sqrt{|h|}}=\lim _{h \rightarrow 0} \sqrt{|h|}=0
\)
\(\therefore\) option (c) is correct
(d) \(\mathrm{f}(\mathrm{x})=\mathrm{x}|\mathrm{x}|\) for Property 2
\(
\lim _{h \rightarrow 0} \frac{h|h|-0}{h^2}=\lim _{h \rightarrow 0} \frac{|h|}{h}
\)
\(\mathrm{LHL}=-1\) and \(\mathrm{RHL}=1\)
\(\therefore \lim _{h \rightarrow 0} \frac{|h|}{h}\) does not exist
\(\therefore\) option (d) is incorrect.

Hint

\(\lim _{n \rightarrow \infty}\left[(n+1) \frac{2}{\pi} \cos ^{-1}\left(\frac{1}{n}\right)-n\right]\)
\(
=\lim _{n \rightarrow \infty} n\left[\left(1+\frac{1}{n}\right) \frac{2}{\pi} \cos ^{-1}\left(\frac{1}{n}\right)-1\right]=\lim _{n \rightarrow \infty} n f\left(\frac{1}{n}\right)
\)
where \(f(x)=\left[(1+x) \frac{2}{\pi} \cos ^{-1} x-1\right]\) such that
\(
f(0)=\left[(1+0) \frac{2}{\pi} \cos ^{-1} 0-1\right]=\frac{2}{\pi} \cdot \frac{\pi}{2}-1=0
\)
\(\therefore\) Using given relation \(\lim _{n \rightarrow \infty} n f\left(\frac{1}{n}\right)=f^{\prime}(0)\)
given limit becomes
\(
=f^{\prime}(0)=\left.\frac{d}{d x}\left[(1+x) \frac{2}{\pi} \cos ^{-1} x-1\right]\right|_{x=0}
\)
\(
=\left.\frac{2}{\pi}\left[\cos ^{-1} x-\frac{1-x}{\sqrt{1-x^2}}\right]\right|_{x=0}
\)
\(
=\frac{2}{\pi}\left[\frac{\pi}{2}-1\right]=1-\frac{2}{\pi}=\frac{\pi-2}{\pi} .
\)

Hint

\(
\begin{aligned}
&\lim _{x \rightarrow 0} \frac{2^x-1}{\sqrt{1+x}-1}=\lim _{x \rightarrow 0} \frac{2^x-1}{\sqrt{1+x}-1} \times \frac{\sqrt{1+x}+1}{\sqrt{1+x}+1}\\
&=\lim _{x \rightarrow 0} \frac{\left(2^x-1\right)(\sqrt{1+x}+1)}{1+x-1}\\
&=\lim _{x \rightarrow 0} \frac{2^x-1}{x} \cdot \lim _{x \rightarrow 0}(\sqrt{1+x}+1)\\
&=\ln 2 \cdot(1+1)=2 \ln 2 .
\end{aligned}
\)

Hint

\(\lim _{h \rightarrow 0} \frac{(a+h)^2 \sin (a+h)-a^2 \sin a}{h}\)
\(=\lim _{h \rightarrow 0} \frac{a^2[\sin (a+h)-\sin a]+2 a h \sin (a+h)+h^2 \sin (a+h)}{h}\)
\(
\begin{aligned}
&=\lim _{h \rightarrow 0} \frac{a^2\left[2 \cos \left(a+\frac{h}{2}\right) \sin \frac{h}{2}\right]}{2 \times \frac{h}{2}}+2 \mathrm{a} \sin (\mathrm{a}+\mathrm{h}) \quad+\mathrm{h} \sin (\mathrm{a}+\mathrm{h})\\
&=\mathrm{a}^2 \cos \mathrm{a}+2 \mathrm{a} \sin \mathrm{a}
\end{aligned}
\)

Hint

Given : \(f(x)=\int \frac{2 \sin x-\sin 2 x}{x^3} d x, x \neq 0\)
\(
\begin{aligned}
\therefore f^{\prime}(x) &=\frac{2 \sin x-\sin 2 x}{x^3}, x \neq 0 \\
\therefore \quad \lim _{x \rightarrow 0} f^{\prime}(x) &=\lim _{x \rightarrow 0} \frac{2 \sin x-\sin 2 x}{x^3} \\
&=\lim _{x \rightarrow 0} \frac{2 \sin x(1-\cos x)(1+\cos x)}{x^3(1+\cos x)}
\end{aligned}
\)
\(
=\lim _{x \rightarrow 0} 2 \cdot \frac{\sin ^3 x}{x^3} \cdot \frac{1}{1+\cos x}=2 \times(1)^3 \times \frac{1}{2}=1
\)

Hint

\(
\begin{aligned}
& \lim _{x \rightarrow a} \frac{\sqrt{a+2 x}-\sqrt{3 x}}{\sqrt{3 a+x}-2 \sqrt{x}} \\
=& \lim _{x \rightarrow a} \frac{(\sqrt{a+2 x}-\sqrt{3 x})(\sqrt{a+2 x}+\sqrt{3 x})(\sqrt{3 a+x}+2 \sqrt{x})}{(\sqrt{3 a+x}-2 \sqrt{x})(\sqrt{3 a+x}+2 \sqrt{x})(\sqrt{a+2 x}+\sqrt{3 x})} \\
=& \lim _{x \rightarrow a} \frac{(a-x)(\sqrt{3 a+x}+2 \sqrt{x})}{3(a-x)(\sqrt{a+2 x}+\sqrt{3 x})} \\
=& \lim _{x \rightarrow a} \frac{(\sqrt{3 a+x}+2 \sqrt{x})}{3(\sqrt{a+2 x}+\sqrt{3 x})}=\frac{\sqrt{3 a+a}+2 \sqrt{a}}{3(\sqrt{a+2 a}+\sqrt{3 a})} \\
=& \frac{4 \sqrt{a}}{3 \times 2 \sqrt{3 a}}=\frac{2}{3 \sqrt{3}}
\end{aligned}
\)

Hint

\(\lim _{x \rightarrow a} \frac{(a+2 x)^{\frac{1}{3}}-(3 x)^{\frac{1}{3}}}{(3 a+x)^{\frac{1}{3}}-(4 x)^{\frac{1}{3}}} \quad\left[\frac{0}{0}\right.\) case \(]\)
Apply L’Hospital rule
\(
\begin{aligned}
& \lim _{x \rightarrow a} \frac{\frac{1}{3}(a+2 x)^{-2 / 3} \cdot 2-\frac{1}{3} \cdot(3 x)^{-2 / 3} \cdot 3}{\frac{1}{3}(3 a+x)^{-2 / 3} \cdot-\frac{1}{3}(4 x)^{-2 / 3} \cdot 4} \\
=& \frac{\frac{1}{3}(3 a)^{-2 / 3} \cdot(2-3)}{\frac{1}{3}(4 a)^{-2 / 3} \cdot(1-4)}=\frac{3^{-2 / 3}}{4^{-2 / 3}} \cdot \frac{1}{3}=\frac{2^{4 / 3}}{9^{1 / 3}} \cdot \frac{1}{3}=\frac{2}{3} \cdot\left(\frac{2}{9}\right)^{1 / 3}
\end{aligned}
\)

Hint

\(\lim _{x \rightarrow 0}\left(\frac{1+\tan x}{1-\tan x}\right)^{1 / x}\)
\(
\Rightarrow e^{\lim _{x \rightarrow 0} \frac{1}{x}\left[\tan \left(\frac{\pi}{4}+x\right)-1\right]} \Rightarrow e^{\lim _{x \rightarrow 0} \frac{1}{x}\left(\frac{1+\tan x}{1-\tan x}-1\right)}
\)
\(
\Rightarrow e^{\lim _{x \rightarrow 0}\left(\frac{2 \tan x}{1-\tan x}\right) \frac{1}{x}}=e^{\lim _{x \rightarrow 0}\left(\frac{\tan x}{x}\right)\left(\frac{2}{1-\tan x}\right)}=e^2
\)

Hint

Let \(\mathrm{R}=\lim _{x \rightarrow 0}\left(\frac{3 x^2+2}{7 x^2+2}\right)^{\frac{1}{x^2}}=e^{\lim _{x \rightarrow 0} \frac{1}{x^2}\left\{\frac{3 x^2+2}{7 x^2+2}-1\right\}}\)
\(
=e^{\lim _{x \rightarrow 0} \frac{1}{x^2}\left\{\frac{-4 x^2}{7 x^2+2}\right\}}=e^{\frac{-4}{2}}=e^{-2}=\frac{1}{e^2}
\)

Hint

Using L’ Hospital rule, \(\lim _{x \rightarrow 0} \frac{x \sin (10 x)}{1}=0\)

Hint

Given equation is, \(375 x^2-25 x-2=0\)
Sum and product of the roots are,
\(
\begin{gathered}
a+b=\frac{25}{375} \text { and } a b=\frac{-2}{375} \\
\lim _{n \rightarrow \infty} \sum_{r=1}^n\left(\alpha^r+\beta^r\right) \\
=\left(\alpha+\alpha^2+\alpha^3+\ldots \infty\left(\beta+\beta^2+\beta^3 \ldots+\infty\right)\right. \\
=\frac{\alpha}{1-\alpha}+\frac{\beta}{1-\beta}=\frac{\alpha+\beta-2 \alpha \beta}{1-(\alpha+\beta)+\alpha \beta}
\end{gathered}
\)
\(
=\frac{\frac{25}{375}+\frac{4}{375}}{1-\frac{25}{375}-\frac{2}{375}}=\frac{29}{375-25-2}=\frac{29}{348}=\frac{1}{12}
\)

Hint

Since, \(\lim _{x \rightarrow 0^{+}} x\left(\left[\frac{1}{x}\right]+\left[\frac{2}{x}\right]+\ldots+\left[\frac{15}{x}\right]\right)\) \(=\lim _{x \rightarrow 0^{+}} x\left(\frac{1+2+3+\ldots+15}{x}\right)-\left(\left\{\frac{1}{x}\right\}+\left\{\frac{2}{x}\right\}+\ldots+\left\{\frac{15}{x}\right\}\right)\)
\(
\because 0 \leq\left\{\frac{r}{x}\right\}<1 \Rightarrow 0 \leq x\left\{\frac{r}{x}\right\}<x
\)
\(
\therefore \lim _{x \rightarrow 0^{+}} x\left(\frac{1+2+3+\ldots+15}{x}\right)=\frac{15 \times 16}{2}=120
\)

Hint

Let \(\mathrm{L}=\lim _{x \rightarrow 0} \frac{(27+x)^{\frac{1}{3}}-3}{9-(27+x)^{\frac{2}{3}}}\)
Here ‘ \(L\) ‘ is in the indeterminate form i.e., \(\frac{0}{0}\)
\(\therefore\) using the L’Hospital rule we get:
\(
\mathrm{L}=\lim _{x \rightarrow 0} \frac{\frac{1}{3}(27+x)^{\frac{-2}{3}}}{-\frac{2}{3}(27+x)^{\frac{-1}{3}}}=\frac{\frac{1}{3} \times(2)^{\frac{-2}{3}}}{\frac{-2}{3} \times 27^{\frac{-1}{3}}}=-\frac{1}{6}
\)

Hint

\(p=\lim _{x \rightarrow 0^{+}}\left(1+\tan ^2 \sqrt{x}\right)^{\frac{1}{2 x}}, 1^{\infty}\) form
\(
=\lim _{x \rightarrow 0^{+}}\left(1+\tan ^2 \sqrt{x}\right)^{\frac{1}{\tan ^2 \sqrt{x}}} \cdot \frac{\tan ^2 \sqrt{x}}{2 x}
\)
\(
=\left(\lim _{x \rightarrow 0^{-}}\left(1+\tan ^2 \sqrt{x}\right)^{\frac{1}{\tan ^2 \sqrt{x}}}\right)^{\lim _{x \rightarrow 0^{+}} \frac{\tan ^2 \sqrt{x}}{2 x}}
\)

We know

\(
\lim _{x \rightarrow 0}(1+x)^{\frac{1}{x}}=e
\)

\(
=e^{\lim _{x \rightarrow 0^{+}} \frac{1}{2}} \cdot\left(\frac{\tan \sqrt{x}}{\sqrt{x}}\right)^2
\)
\(
=\mathrm{e}^{\frac{1}{2} \cdot 1^2}=\mathrm{e}^{\frac{1}{2}}
\)
Therefore \(\ln p=\ln e^{1 / 2}=\frac{1}{2}\)

Hint

\(\lim _{x \rightarrow 0}\left[1+x \ell n\left(1+b^2\right)\right]^{\frac{1}{x}}=2 b \sin ^2 \theta\)
\(
\begin{gathered}
\Rightarrow \lim _{e^{x \rightarrow 0 x}} \frac{1}{\ln \left[1+x \ln \left(1+b^2\right)\right]}=2 b \sin ^2 \theta \\
\Rightarrow e_{x \rightarrow 0} \frac{\ln \left[1+x \ell \mathrm{n}\left(1+b^2\right)\right]}{x \ell n\left(1+b^2\right)} \times \ell n\left(1+b^2\right)=2 b \sin ^2 \theta \\
\Rightarrow e^{\ln \left(1+b^2\right)=2 \mathrm{~b} \sin ^2 \theta} \\
\Rightarrow 1+b^2=2 b \sin ^2 \theta \Rightarrow 2 \sin ^2 \theta=b+\frac{1}{b}
\end{gathered}
\)
We know that \(2 \sin ^2 \theta \leq 2\) and \(b+\frac{1}{b} \geq 2\) for \(b>0\)
\(
\therefore 2 \sin ^2 \theta=b+\frac{1}{b}=2 \Rightarrow \sin ^2 \theta=1
\)
\(\theta \in(-\pi, \pi], \quad \therefore \theta=\pm \frac{\pi}{2}\)

Hint

\(
\begin{aligned}
&\lim _{x \rightarrow \frac{\pi}{4}} \frac{\int_2^{\sec ^2 x} f(t) d t}{x^2-\frac{\pi^2}{16}}\left[\frac{0}{0}\right. \text { form] } \\
&=\lim _{x \rightarrow \frac{\pi}{4}} \frac{\frac{d}{d x}\left[\int_2^{\sec ^2 x} f(t) d t\right]}{\frac{d}{d x}\left(x^2-\frac{\pi^2}{16}\right)} \text { (using L’ Hospital rule) }
\end{aligned}
\)
\(
\begin{aligned}
&=\lim _{x \rightarrow \frac{\pi}{4}} \frac{f\left(\sec ^2 x\right) \cdot 2 \sec ^2 x \tan x}{2 x} \\
&{\left[\because \frac{d}{d x}\left[\int_{g(x)}^{h(x)} f(t) d t\right]=f(h(x)) h^{\prime}(x)-f(g(x)) \cdot g^{\prime}(x)\right]} \\
&=\frac{f(2) \times 2 \times 2 \times 1}{2 \times \frac{\pi}{4}}=\frac{8}{\pi} f(2)
\end{aligned}
\)

Hint

\(
\begin{aligned}
&\text { Limit }=\lim _{x \rightarrow 0}\left((\sin x)^{\frac{1}{x}}+\left(\frac{1}{x}\right)^{\sin x}\right)\\
&=e^{\lim _{x \rightarrow 0}\left(\frac{\log \sin x}{x}\right)}+e^{\lim _{x \rightarrow 0}\left(\frac{-\log x}{\operatorname{cosec} x}\right)}\\
&=e^{-\infty}+e^{\lim _{x \rightarrow 0}\left(\frac{-\log x}{\operatorname{cosec} x}\right)}\\
&\left(\because x \rightarrow 0^{+}, \log (\sin x) \rightarrow-\infty\right)\\
&=e^{-\infty}+e^{\lim _{x \rightarrow 0}\left(-\frac{\frac{1}{x}}{-\operatorname{cosec} x \cot x}\right)} \text { (using L’ Hospital rule) }\\
&=e^{-\infty}+e^{\lim _{x \rightarrow 0}\left(\frac{\tan x}{x} \times \sin x\right)}\\
&=e^{-\infty}+e^0\\
&=0+1\\
&=1
\end{aligned}
\)

Hint

\(
\begin{aligned}
&\text { Let } L=\lim _{x \rightarrow 0} \frac{f\left(x^2\right)-f(x)}{f(x)-f(0)} \quad \text { [using L.H. Rule] }\\
&\left[\begin{array}{l}
\because f^{\prime}(a)>0 \text { as } f \text { being } \\
\text { strictly increasing }
\end{array}\right]\\
&L=\lim _{x \rightarrow 0} \frac{f^{\prime}\left(x^2\right) \cdot 2 x-f^{\prime}(x)}{f^{\prime}(x)}=\lim _{x \rightarrow 0} \frac{f^{\prime}\left(x^2\right) \cdot 2 x}{f^{\prime}(x)}-1=0-1\\
&=-1
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \lim _{h \rightarrow 0} \frac{f\left(2 h+2+h^2\right)-f(2)}{f\left(h-h^2+1\right)-f(1)} \quad\left[\frac{0}{0} \text { form }\right]\\
&=\lim _{h \rightarrow 0} \frac{f^{\prime}\left(2 h+2+h^2\right) \cdot(2+2 h)}{f^{\prime}\left(h-h^2+1\right) \cdot(1-2 h)}\\
&=\frac{f^{\prime}(2) \cdot 2}{f^{\prime}(1) \cdot 1}=\frac{6 \times 2}{4 \times 1}=3
\end{aligned}
\)

Hint

Given \(f ; R \rightarrow R, f(1)=3\) and \(f^{\prime}(1)=6\) Then \(\lim _{x \rightarrow 0}\left[\frac{f(1+x)}{f(1)}\right]^{1 / x}\)
\(
\begin{aligned}
&=\lim _{e^{x \rightarrow 0}} \frac{1}{x}[\log f(1+x)-\log f(1)] \\
&=\lim _{e^{x \rightarrow 0}} \frac{\frac{1}{f(1+x)} f^{\prime}(1+x)}{1} \quad \text { [using L’Hospital rule] } \\
&=\frac{f^{\prime}(1)}{f(1)}=e^{6 / 3}=e^2
\end{aligned}
\)

Hint

\(\lim _{x \rightarrow 0} \frac{(\cos x-1)\left(\cos x-e^x\right)}{x^n}\)
\(
=\lim _{x \rightarrow 0} \frac{(1-\cos x)(1+\cos x)\left(e^x-\cos x\right)}{x^n(1+\cos x)}
\)
\(
=\lim _{x \rightarrow 0}\left(\frac{\sin ^2 x}{x^2}\right) \cdot\left(\frac{e^x-\cos x}{x^{n-2}}\right) \cdot\left(\frac{1}{1+\cos x}\right)
\)
\(
=1^2 \cdot \frac{1}{2} \lim _{x \rightarrow 0} \frac{e^x-\cos x}{x^{n-2}}
\)
\(=\frac{1}{2} \lim _{x \rightarrow 0} \frac{e^x+\sin x}{(n-2) x^{n-3}} \quad\) [using LH rule] For this limit to be finite, \(n-3=0 \Rightarrow n=3\)

Hint

\(
\lim _{x \rightarrow 0} \frac{x^2 \sin \beta x}{\alpha x-\sin x}=1
\)
Therefore,
\(
\begin{aligned}
&\lim _{x \rightarrow 0} \frac{x^2 \sin \beta x}{\alpha x-\sin x}=1 \\
&\Rightarrow \lim _{x \rightarrow 0} \frac{x^2\left(\frac{\sin \beta x}{\beta x}\right) \beta x}{\alpha x-\sin x}=1 \\
&\Rightarrow \beta \lim _{x \rightarrow 0} \frac{x^3}{\alpha x-\left(x-\frac{x^3}{3 !}+\frac{x^5}{5 !}-\cdots\right)}=1 \\
&\Rightarrow \beta \lim _{x \rightarrow 0} \frac{x^3}{x(\alpha-1)+\frac{x^3}{3 !}-\frac{x^5}{5 !}+\cdots}=1
\end{aligned}
\)
For finite limit \(\alpha=1\),
\(
3 ! \times \beta=1 \Rightarrow \beta=\frac{1}{6}
\)
Then,
\(
6(\alpha+\beta)=6\left(1+\frac{1}{6}\right)=6+1=7
\)

Hint

\(\lim _{x \rightarrow 1} \frac{x+x^2+x^3+\ldots .+x^n-n}{x-1}=820\left(\frac{0}{0}\right.\) case \()\) \(\lim _{x \rightarrow 1} \frac{1+2 x+3 x^2+\ldots .+n x^{n-1}}{1}=820\)
(Using L’ Hospital rule) \(\Rightarrow 1+2+3+\ldots+n=820\)
\(
\begin{aligned}
&\Rightarrow \frac{n(n+1)}{2}=820 \Rightarrow n^2+n-1640=0 \\
&\Rightarrow n=40, n \in \mathrm{N}
\end{aligned}
\)

Hint

\(
\begin{aligned}
&f(x)^{g(x)}=e^{\log f(x)^{g(x)}}=e^{g(x) \log f(x)}\\
&\Rightarrow \quad \lim _{x \rightarrow 0}[f(x)]^{g(x)}=e^{\lim _{x \rightarrow 0} g(x) \log f(x)}\\
&\therefore \lim _{x \rightarrow 0}\left(\frac{1+5 x^2}{1+3 x^2}\right)^{1 / x^2}=e^{\lim _{x \rightarrow 0} \frac{1}{x^2} \log \left[\frac{1+5 x^2}{1+3 x^2}\right]}\\
&=e^{\lim _{x \rightarrow 0}\left[5 \cdot \frac{\log \left(1+5 x^2\right)}{5 x^2}-3 \cdot \frac{\log \left(1+3 x^2\right)}{3 x^2}\right]}=e^{5-3}=e^2
\end{aligned}
\)

Hint

\(
\begin{aligned}
&\lim _{x \rightarrow \infty}\left(\frac{x+6}{x+1}\right)^{x+4}=\lim _{x \rightarrow \infty}\left(1+\frac{5}{x+1}\right)^{x+4}\left[1^{\infty}\right. \text { form] } \\
&=e^{\lim _{x \rightarrow \infty} \frac{5(x+4)}{x+1}}=e^5
\end{aligned}
\)

Note: \(
\left[\lim _{x \rightarrow \infty}\left(1+\frac{1}{x}\right)^x=e\right]
\)

Hint

\(\lim _{x \rightarrow 0}\left\{\tan \left(\frac{\pi}{4}\right)+x\right\}^{\frac{1}{x}}=\lim _{x \rightarrow 0} \ln \left\{\tan \left(\frac{\pi}{4}+x\right)\right\}^{\frac{1}{x}}\)
\(
=\lim _{e^{x \rightarrow 0}} \frac{\ln \tan \left(\frac{\pi}{4}+x\right)}{x} \quad\left[\frac{0}{0} \text { form }\right]
\)
\(
\begin{gathered}
=e^{\lim _{x \rightarrow 0}\left[\frac{\sec ^2\left(\frac{\pi}{4}+x\right)}{\tan \left(\frac{\pi}{4}\right)+x}\right]} \\
=e^{\frac{2}{1}}=e^2
\end{gathered}
\)

Note: LH rule was used to solve this.

Hint

\(f(x)\) has extremum values at \(x=1\) and \(x=2\)
\(f^{\prime}(1)=0\) and \(f^{\prime}(2)=0\)
As, \(f(x)\) is a polynomial of degree 4 .
Suppose \(f(x)=A x^4+B x^3+C x^2+D x+E\)
\(
\begin{aligned}
&\lim _{x \rightarrow 0}\left(\frac{f(x)}{x^2}+1\right)=3 \\
&\Rightarrow \quad \lim _{x \rightarrow 0}\left(\frac{A x^4+B x^3+C x^2+D x+E}{x^2}+1\right)=3 \\
&\Rightarrow \quad \lim _{x \rightarrow 0}\left(A x^2+B x+C+\frac{D}{x}+\frac{E}{x^2}+1\right)=3
\end{aligned}
\)
As limit has finite value, so \(D=0\) and \(E=0\)
Now \(A(0)^2+B(0)+C+0+0+1=3\)
\(
\begin{aligned}
&\Rightarrow c+1=3 \Rightarrow c=2 \\
&f^{\prime}(x)=4 A x^3+3 B x^2+2 C x+D \\
&f^{\prime}(1)=0 \Rightarrow 4 A(1)+3 B(1)+2 C(1)+D=0 \\
&\Rightarrow 4 A+3 B=-4 \\
&f^{\prime}(2)=0 \Rightarrow 4 A(8)+3 B(4)+2 C(2)+D=0 \\
&\Rightarrow 8 A+3 B=-2
\end{aligned}
\)
From equations (i) and (ii), we get
\(
A=\frac{1}{2} \text { and } B=-2 ; \text { So, } f(x)=\frac{x^4}{2}-2 x^3+2 x^2
\)
Therefore, \(f(-1)=\frac{(-1)^4}{2}-2(-1)^3+2(-1)^2\) \(=\frac{1}{2}+2+2=\frac{9}{2}\). Hence \(f(-1)=\frac{9}{2}\)

Hint

\(\lim _{x \rightarrow 0}\left[1+\frac{f(x)}{x^2}\right]=3 \Rightarrow \lim _{x \rightarrow 0} \frac{f(x)}{x^2}=2\)
So, \(f(x)\) contain terms in \(x^2, x^3\) and \(x^4\).
Let \(f(x)=a_1 x^2+a_2 x^3+a_3 x^4\)
Since \(\lim _{x \rightarrow 0} \frac{f(x)}{x^2}=2 \Rightarrow a_1=2\)
Hence, \(f(x)=2 x^2+a_2 x^3+a_3 x^4\)
\(
f^{\prime}(x)=4 x+3 a_2 x^2+4 a_3 x^3
\)
As given : \(\mathrm{f}^{\prime}(1)=0 \operatorname{and} \mathrm{f}^{\prime}(2)=0\)
Hence, \(4+3 \mathrm{a}_2+4 \mathrm{a}_3=0 \quad\)…(i)
and \(8+12 \mathrm{a}_2+32 \mathrm{a}_3=0\)…(ii)
By 4 x (i) – (ii), we get
\(
\begin{gathered}
16+12 \mathrm{a}_2+16 \mathrm{a}_3-\left(8+12 \mathrm{a}_2+32 \mathrm{a}_3\right)=0 \\
\Rightarrow 8-16 \mathrm{a}_3=0 \Rightarrow \mathrm{a}_3=1 / 2
\end{gathered}
\)
and by eqn. (i), \(4+3 \mathrm{a}_2+4 / 2=0 \Rightarrow \mathrm{a}_2=-2\)
\(
\Rightarrow \mathrm{f}(\mathrm{x})=2 \mathrm{x}^2-2 \mathrm{x}^3+\frac{1}{2} \mathrm{x}^4
\)
\(
f(2)=2 \times 4-2 \times 8+\frac{1}{2} \times 16=0
\)

Hint

\(
\text { Given } f(1)=-2 \text { and } f^{\prime}(x) \geq 4.2 \text { for } 1 \leq x \leq 6
\)
Consider \(f^{\prime}(x)=\frac{f(x+h)-f(x)}{h}\) \(\Rightarrow f(x+h)-f(x)=f^{\prime}(x) \cdot h \geq(4.2) h\)
So, \(f(x+h) \geq f(x)+(4.2) h\)
put \(x=1\) and \(h=5\), we get
\(
f(6) \geq f(1)+5(4.2) \Rightarrow f \quad(6) \geq 19
\)
Hence \(f(6)\) lies in \([19, \infty)\)

Hint

Let \(F(x)=\sin \left(x^2+1\right)\), then, \(f(x+h)=\sin \left[(x+h)^2+1\right]\)
\(
\therefore \lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}
\)
\(
=\lim _{h \rightarrow 0} \frac{\sin \left[(\mathrm{x}+\mathrm{h})^2+1\right]-\sin \left[\mathrm{x}^2+1\right]}{\mathrm{h}}
\)
\(
\Rightarrow f^{\prime}(x)=\lim _{h \rightarrow 0} 2 \cos \left(\frac{2 x^2+h^2+2 x h+2}{2}\right) \times \frac{\sin \left(\frac{h^2+2 \times h}{2}\right)}{h}
\)
\(
=2 \cos \left(x^2+1\right) \lim _{h \rightarrow 0} \frac{\sin \left[\frac{h^2+2 x h}{2}\right]}{h\left[\frac{h+2 x}{2}\right]}\left(\frac{h+2 x}{2}\right)
\)
\(
=2 x \cos \left(x^2+1\right)
\)

Hint

Finding right hand limit
\(
\begin{aligned}
& \lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} f(0+h) \\
& =\lim _{h \rightarrow 0} f(h) \\
& =\lim _{h \rightarrow 0} \frac{\cos ^{-1}\left(1-h^2\right) \sin ^{-1}(1-h)}{h\left(1-h^2\right)} \\
& =\lim _{h \rightarrow 0} \frac{\cos ^{-1}\left(1-h^2\right)}{h}\left(\frac{\sin ^{-1} 1}{1}\right)
\end{aligned}
\)
Let \(\cos ^{-1}\left(1-h^2\right)=\theta \Rightarrow \cos \theta=1-h^2\)
\(
\begin{aligned}
& =\frac{\pi}{2} \lim _{\theta \rightarrow 0} \frac{\theta}{\sqrt{1-\cos \theta}} \\
& =\frac{\pi}{2} \lim _{\theta \rightarrow 0} \frac{1}{\sqrt{\frac{1-\cos \theta}{\theta^2}}} \\
& =\frac{\pi}{2} \frac{1}{\sqrt{1 / 2}} \\
& R=\frac{\pi}{\sqrt{2}}
\end{aligned}
\)
Now finding left hand limit
\(
\begin{aligned}
& L=\lim _{x \rightarrow 0^{-}} f(x) \\
& =\lim _{h \rightarrow 0} f(-h) \\
& =\lim _{h \rightarrow 0} \frac{\cos ^{-1}\left(1-\{-h\}^2\right) \sin ^{-1}(1-\{-h\})}{\{-h\}-\{-h\}^3} \\
& =\lim _{h \rightarrow 0} \frac{\cos ^{-1}\left(1-(-h+1)^2\right) \sin ^{-1}(1-(-h+1))}{(-h+1)-(-h+1)^3} \\
& =\lim _{h \rightarrow 0} \frac{\cos ^{-1}\left(-h^2+2 h\right) \sin ^{-1} h}{(1-h)\left(1-(1-h)^2\right)} \\
& =\lim _{h \rightarrow 0}\left(\frac{\pi}{2}\right) \frac{\sin ^{-1} h}{\left(1-(1-h)^2\right)} \\
& =\frac{\pi}{2} \lim _{ h \rightarrow 0}\left(\frac{\sin ^{-1} h }{- h ^2+2 h }\right) \\
& =\frac{\pi}{2} \lim _{ h \rightarrow 0}\left(\frac{\sin ^{-1} h }{ h }\right)\left(\frac{1}{- h +2}\right) \\
& L =\frac{\pi}{4} \\
& \frac{32}{\pi^2}\left( L ^2+ R ^2\right)=\frac{32}{\pi^2}\left(\frac{\pi^2}{2}+\frac{\pi^2}{16}\right) \\
& =16+2 \\
& =18
\end{aligned}
\)

Hint

\(
\begin{aligned}
& f( x )=\left\{\begin{array}{cc}
x -1 ; & x =\text { even } \\
2 x ; & x =\text { odd }
\end{array}\right. \\
& f(f(f( a )))=21
\end{aligned}
\)
C-1: If \(a=\) even
\(
\begin{aligned}
& f( a )= a -1=\text { odd } \\
& f ( f ( a ))=2( a -1)=\text { even } \\
& f(f(f( a )))=2 a -3=21 \Rightarrow a =12
\end{aligned}
\)
C-2: If \(a =\) odd
\(
\begin{aligned}
& f( a )=2 a =\text { even } \\
& f(f( a ))=2 a -1=\text { odd } \\
& f(f(f( a )))=4 a -2=21 \text { (Not possible) }
\end{aligned}
\)
\(
\text { Hence } a =12
\)
Now
\(
\begin{aligned}
& \lim _{x \rightarrow 12^{-}}\left(\frac{|x|^3}{2}-\left[\frac{x}{12}\right]\right) \\
& =\lim _{x \rightarrow 12^{-}} \frac{|x|^3}{12}-\lim _{x \rightarrow 12^{-}}\left[\frac{x}{12}\right] \\
& =144-0=144 .
\end{aligned}
\)

Hint

\(
a=\lim _{x \rightarrow 0} \frac{\sqrt{1+\sqrt{1+x^4}}-\sqrt{2}}{x^4}
\)
\(
=\lim _{x \rightarrow 0} \frac{\sqrt{1+x^4}-1}{x^4\left(\sqrt{1+\sqrt{1+x^4}}+\sqrt{2}\right)}
\)
\(
=\lim _{x \rightarrow 0} \frac{x^4}{x^4\left(\sqrt{1+\sqrt{1+x^4}}+\sqrt{2}\right)\left(\sqrt{1+x^4}+1\right)}
\)
\(
\begin{aligned}
& \text { Applying limit } a=\frac{1}{4 \sqrt{2}} \\
& b=\lim _{x \rightarrow 0} \frac{\sin ^2 x}{\sqrt{2}-\sqrt{1+\cos x}} \\
& =\lim _{x \rightarrow 0} \frac{\left(1-\cos ^2 x\right)(\sqrt{2}+\sqrt{1+\cos x})}{2-(1+\cos x)} \\
& b=\lim _{x \rightarrow 0}(1+\cos x)(\sqrt{2}+\sqrt{1+\cos x})
\end{aligned}
\)
Applying limits \(b =2(\sqrt{2}+\sqrt{2})=4 \sqrt{2}\)
Now, \(ab ^3=\frac{1}{4 \sqrt{2}} \times(4 \sqrt{2})^3=32\)

Hint

\(
\lim _{x \rightarrow 0} \frac{3+\alpha \sin x+\beta \cos x+\log _e(1-x)}{3 \tan ^2 x}=\frac{1}{3}
\)
\(
\Rightarrow \lim _{x \rightarrow 0} \frac{3+\alpha\left[x-\frac{x^3}{3!}+\ldots\right]+\beta\left[1-\frac{x^2}{2!}+\frac{x^4}{4!} \ldots\right]+\left(-x-\frac{x^2}{2}-\frac{x^3}{3} \ldots\right)}{3 \tan ^2 x}=\frac{1}{3}
\)
\(
\Rightarrow \lim _{x \rightarrow 0} \frac{(3+\beta)+(\alpha-1) x+\left(-\frac{1}{2}-\frac{\beta}{2}\right) x^2+\ldots}{3 x^2} \times \frac{x^2}{\tan ^2 x}=\frac{1}{3}
\)
\(
\begin{aligned}
& \Rightarrow \beta+3=0, \alpha-1=0 \text { and } \frac{-\frac{1}{2}-\frac{\beta}{2}}{3}=\frac{1}{3} \\
& \Rightarrow \beta=-3, \alpha=1 \\
& \Rightarrow 2 \alpha-\beta=2+3=5
\end{aligned}
\)

Hint

Using L’hopital rule
\(
\begin{aligned}
& =\lim _{x \rightarrow \frac{\pi^{-}}{2}} \frac{0-\cos x \times 3 x^2}{2\left(x-\frac{\pi}{2}\right)} \\
& =\lim _{x \rightarrow \frac{\pi^{-}}{2}} \frac{\sin \left(x-\frac{\pi}{2}\right)}{2\left(x-\frac{\pi}{2}\right)} \times \frac{3 \pi^2}{4} \\
& =\frac{3 \pi^2}{8}
\end{aligned}
\)

Hint

According to the question,
\(
\begin{aligned}
& 27 r_1+\frac{9 r_2}{2}=-9 \\
& \lim _{x \rightarrow 3} \frac{\int_3^x 8 t^2 d t}{\frac{\beta r_2 x}{2}-r_2 x^2-r_1 x^3-3 x} \\
& =\lim _{x \rightarrow 3} \frac{8 x^2}{\frac{3 r_2^2}{2}-2 r_2 x-3 r_1 x^2-3} \text { (using LH’ Rule) } \\
& =\frac{72}{\frac{3 r_2}{2}-6 r_2-27 r_1-3} \\
& =\frac{72}{-\frac{9 r_2}{2}-27 r_1-3} \\
& =\frac{72}{9-3}=12 \\
&
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{x \int_0^x f(t) d t}{\left(\frac{x^2-1}{x^2}\right) \times x^2} \\
& \lim _{x \rightarrow 0} \frac{\int_0^x f(t) d t}{x} \quad\left(\lim _{x \rightarrow 0} \frac{e^{x^2}-1}{x^2}=1\right) \\
& =\lim _{x \rightarrow 0} \frac{f(x)}{1} \quad \text { (using } L \text { Hospital) }
\end{aligned}
\)
\(
\begin{aligned}
& f (0)=\frac{1}{2} \\
& \alpha=\frac{1}{2} \\
& 8 \alpha^2=2
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \text { Put } x=1 \\
& \therefore a=1
\end{aligned}
\)
\(
b=\lim _{x \rightarrow 0} \frac{\int_0^x \frac{\ln (1+t)}{1+t^{2024}} d t}{x^2}
\)
Using L’ HOPITAL Rule
\(
b=\lim _{x \rightarrow 0} \frac{\ln (1+x)}{\left(1+x^{2024}\right)} \times \frac{1}{2 x}=\frac{1}{2}
\)
\(
\text { Now, } cx ^2+ dx + e =0, x ^2+ x +4=0
\)
\(
\begin{aligned}
& ( D <0) \\
& \therefore \frac{c}{1}=\frac{d}{1}=\frac{e}{4}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{e^{2|\sin x|}-2|\sin x|-1}{x^2} \\
& \lim _{x \rightarrow 0} \frac{e^{2|\sin x|}-2|\sin x|-1}{|\sin x|^2} \times \frac{\sin ^2 x}{x^2}
\end{aligned}
\)
Let \(|\sin x |= t\)
\(
\begin{aligned}
& \lim _{t \rightarrow 0} \frac{e^{2 t}-2 t-1}{t^2} \times \lim _{x \rightarrow 0} \frac{\sin ^2 x}{x^2} \\
& =\lim _{t \rightarrow 0} \frac{2 e^{2 t}-2}{2 t} \times 1=2 \times 1=2
\end{aligned}
\)

Hint

\(
\begin{aligned}
& f: R \rightarrow(0, \infty) \\
& \lim _{x \rightarrow \infty} \frac{f(7 x)}{f(x)}=1
\end{aligned}
\)
\(\because f\) is increasing
\(
\begin{aligned}
& \therefore f(x)<f(5 x)<f(7 x) \\
& \because \frac{f(x)}{f(x)}<\frac{f(5 x)}{f(x)}<\frac{f(7 x)}{f(x)} \\
& 1<\lim _{x \rightarrow \infty} \frac{f(5 x)}{f(x)}<1 \\
& \therefore\left[\frac{f(5 x)}{f(x)}-1\right] \\
& \Rightarrow 1-1=0
\end{aligned}
\)

Hint

\(
\lim _{x \rightarrow 0} \frac{a x^2\left(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\ldots .\right)-b\left(x-\frac{x^2}{2}+\frac{x^3}{3}-\ldots . .\right)+c x\left(1-x+\frac{x^2}{x!}-\frac{x^3}{3!}+\ldots . .\right)}{x^3 \cdot \frac{\sin x}{x}}
\)
\(
=\lim _{x \rightarrow \infty} \frac{(c-b) x+\left(\frac{b}{2}-c+a\right) x^2+\left(a-\frac{b}{3}+\frac{c}{2}\right) x^3+\ldots \ldots}{x^3}=1
\)
\(
\begin{aligned}
& c – b =0, \quad \frac{ b }{2}- c + a =0 \\
& a -\frac{ b }{3}+\frac{ c }{2}=1 \quad a =\frac{3}{4} \quad b = c =\frac{3}{2} \\
& a ^2+ b ^2+ c ^2=\frac{9}{16}+\frac{9}{4}+\frac{9}{4} \\
& 16\left( a ^2+ b ^2+ c ^2\right)=81
\end{aligned}
\)

Hint

\(
\lim _{x \rightarrow 0} \frac{e-(1+2 x)^{\frac{1}{2 x}}}{x}
\)
Using expansion
\(
\begin{aligned}
& =\lim _{x \rightarrow 0} \frac{e-e\left[1-\frac{2 x}{2}+\frac{11 \times 4 x^2}{24}+\ldots\right]}{x} \\
& =\lim _{x \rightarrow 0}\left(e-\frac{11 x}{6} e+\ldots\right)=e
\end{aligned}
\)

Alternate:

\(
\operatorname{Lim}_{x \rightarrow 0} \frac{e-e^{\frac{1}{2 z} \ln (1+2 x)}}{x}
\)
\(
\begin{aligned}
& =\operatorname{Lim}_{x \rightarrow 0}(-e) \frac{\left(e^{\frac{\ln (1+2 x)}{2 x}-1}-1\right)}{x} \\
& =\operatorname{Lim}_{x \rightarrow 0}(-e) \frac{\ln (1+2 x)-2 x}{2 x^2} \\
& =(-e) \times(-1) \frac{4}{2 \times 2}=e
\end{aligned}
\)

Hint

\(
f(x)=\left\{\begin{array}{cc}
\frac{\tan ((a+1) x)+b \tan x}{x}, & x<0 \\
3 & x=0 \\
\frac{\sqrt{a x+b^2 x^2}-\sqrt{a x}}{b \sqrt{a} x \sqrt{x}}, & x>0
\end{array}\right.
\)
\(
\begin{aligned}
& f(x) \text { is continuous at } x=0 \\
& \Rightarrow \lim _{x \rightarrow 0^{-}} f(x)=f(0)=\lim _{x \rightarrow 0^{+}} f(x) \\
& \lim _{x \rightarrow 0^{-}} f(x)=3 \\
& \Rightarrow \lim _{x \rightarrow 0^{-}} \frac{\tan ((a+1) x)+b+a x}{x}=3 \\
& \Rightarrow a+1+b=3 \\
& \Rightarrow a+b=2 \quad \ldots . .(1)
\end{aligned}
\)
also, \(\lim _{x \rightarrow 0^{+}} \frac{\sqrt{a x+b^2 x^2}-\sqrt{a x}}{b \sqrt{a} x \sqrt{x}}=3\)
\(
\begin{aligned}
& =\lim _{x \rightarrow 0^{+}} \frac{\sqrt{a h+b^2 h^2}-\sqrt{a h}}{b \sqrt{a} \times h \sqrt{h}}=3 \\
& =\lim _{h \rightarrow 0} \frac{\sqrt{a+b^2 h^2}-\sqrt{a}}{b \sqrt{a} h} \times \frac{\sqrt{a+b^2 h}+\sqrt{a}}{\sqrt{a+b^2 h}+\sqrt{a}}=3 \\
& =\lim _{h \rightarrow 0} \frac{a+b^2 h-a}{b \sqrt{a} h\left(\sqrt{a+b^2 h}+\sqrt{a}\right)}=3 \\
& \Rightarrow \frac{b^2}{b \sqrt{a}(2 \sqrt{a})}=3 \\
& \Rightarrow \frac{b}{2 a}=3 \\
& \Rightarrow \frac{b}{a}=6
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{\left(1^2-1\right)(n-1)+\left(2^2-2\right)(n-2)+\ldots+\left((n-1)^2-(n-1)\right) \times 1}{\left(1^3+2^3+\ldots+n^3\right)-\left(1^2+2^2+\ldots+n^2\right)} \\
& \text { Numerator }=\sum_{r=1}^{n-1}\left((r-1)^2-(r-1)\right)(n-r) \\
& =\sum_{r=1}^{n-1}(r-1)-(r-2)(n-r) \\
& =\sum_{r=1}^{n-1}-r^3+(n+3) r^2-(2+3 n) r+2 n
\end{aligned}
\)
We will take term with the greatest power of \(n\)
\(
\begin{aligned}
& =\frac{-1}{4} n^4+\frac{1}{3} n^4=\frac{1}{12} n^4 \\
& \text { Denominator }=\sum_{r=1}^n r^3-\sum_{r=1}^n r^2 \\
& =\left(\frac{n(n+1)}{2}\right)^2-\left(\frac{n(n+1)(2 n+1)}{6}\right)
\end{aligned}
\)
Greatest power of \(n\) is \(\frac{n^4}{4}\)
\(
\lim _{n \rightarrow \infty} \frac{\frac{1}{12} n^4}{\frac{n^4}{4}}=\frac{1}{3}
\)

Alternate:

\(
\lim _{n \rightarrow \infty} \frac{\sum_{r=1}^{n-1}\left(r^2-r\right)(n-r)}{\sum_{r=1}^n r^3-\sum_{r=1}^n r^2}
\)
\(
\lim _{n \rightarrow \infty} \frac{\sum_{r=1}^{n-1}\left(-r^3+r^2(n+1)-n r\right)}{\left(\frac{n(n+1)}{2}\right)^2-\frac{n(n+1)(2 n+1)}{6}}
\)
\(
\lim _{n \rightarrow \infty} \frac{\left(\frac{((n-1) n)}{2}\right)^2+\frac{(n+1)(n-1) n(2 n-1)}{6}-\frac{n^2(n-1)}{2}}{\frac{n(n+1)}{2}\left(\frac{n(n+1)}{2}-\frac{2 n+1}{3}\right)}
\)
\(
\lim _{n \rightarrow \infty} \frac{\frac{n(n-1)}{2}\left(\frac{-n(n-1)}{2}+\frac{(n+1)(2 n-1)}{3}-n\right)}{\frac{n(n+1)}{2} \frac{3 n^2+3 n-4 n-2}{6}}
\)
\(
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{(n-1)\left(-3 n^2+3 n+2\left(2 n^2+n-1\right)-6\right)}{(n+1)\left(3 n^2-n-2\right)} \\
& \lim _{n \rightarrow \infty} \frac{(n-1)\left(n^2+5 n-8\right)}{(n+1)\left(3 n^2-n-2\right)}=\frac{1}{3}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& f(x)=2 x^2+x+\left[x^2\right]-[x]=2 x^2+\left[x^2\right]+[x] \\
& f(-1)=2+1+0=3 \\
& f\left(-1^{+}\right)=2+0+0=2 \\
& f\left(0^{-}\right)=0+1=1 \\
& f\left(0^{+}\right)=0+0+0=0 \\
& f\left(1^{+}\right)=2+1+0=3 \\
& f\left(1^{-}\right)=2+0+1=3 \\
& f\left(2^{-}\right)=8+3+1=12 \\
& f\left(2^{+}\right)=8+4+0=12
\end{aligned}
[latex]
[latex]\therefore\) discontinuous at \(x=0, \sqrt{2}, \sqrt{3},-1\)

Alternate:

Doubtful points : \(-1,0,1, \sqrt{2}, \sqrt{3}, 2\) at \(x=\sqrt{2}, \sqrt{3}\)
\(
f(x)=\left(2 x^2+x-[x]\right)+\left[x^2\right]=\text { Discount }
\)
\(
\quad \quad  \quad \quad  \downarrow \text { (Cont.) } \quad \quad \downarrow \text { (Cont.) }
\)
\(
\text { at } x=-1 \text { : }
\)
\(
\text { RHL } \left.\Rightarrow \begin{array}{l}
f(x)=(2-1-(-1))+0=2 \\
f(-1)=2-1-(-1)+1=3
\end{array}\right\} \text { Dis. }
\)
\(
\text { at } x=2 \text { : }
\)
\(
\left.\begin{array}{rl}
LHL \Rightarrow f(x) & =8+2-1+3=12 \\
f(2) & =8+2-2+4=12
\end{array}\right\} \text { Cont. }
\)
\(
\text { at } x=0 \text { : }
\)
\(
\left.\begin{array}{rl}
LHL \Rightarrow 0+0-(-1)+0=1 \\
f(0=0
\end{array}\right\} \text { Dis. }
\)
\(
\text { at } x =1:
\)
\(
\left.\begin{array}{rl}
LHL \Rightarrow 2+1-0+0=3 \\
f(1) =3-1+1=3 \\
RHL \Rightarrow 2+1-1+1=3
\end{array}\right\} \text { Cont. }
\)

Hint

\(
\begin{aligned}
& \lim _{x \rightarrow 0} f(x)=f(0) \quad \text { (continuous at } x=0 \text { ) } \\
& \lim _{x \rightarrow 0} \frac{\sin 3 x+\alpha \sin x-\beta \cos 3 x}{x^3}
\end{aligned}
\)
For limit to exist \(\beta=0\)
\(
\begin{aligned}
& \Rightarrow \lim _{x \rightarrow 0} \frac{\sin 3 x+\alpha \sin x}{x^3} \\
& \Rightarrow \lim _{x \rightarrow 0} \frac{(3+\alpha) \sin x-4 \sin ^3 x}{x^3}
\end{aligned}
\)
For limit to exist \(\alpha+3=0 \Rightarrow \alpha=-3\)
\(
\Rightarrow \lim _{x \rightarrow 0} \frac{-4 \sin ^3 x}{x^3}=-4=f(0)
\)

Alternate: 

Hint

\(
\begin{aligned}
& f(x)=\left\{\begin{array}{l}
\frac{72^x-9^x-8^x+1}{\sqrt{2}-\sqrt{1+\cos x}}, x \neq 0 \\
a \log _e 2 \log _e 3 \quad, x=0
\end{array}\right. \\
& \because f(x) \text { is continuous at } x=0 \\
& \Rightarrow \lim _{x \rightarrow 0} \frac{72^x-9^x-8^x+1}{\sqrt{2}-\sqrt{1+\cos x}} \\
& \lim _{x \rightarrow 0} \frac{\left(9^x-1\right)\left(8^x-1\right)(\sqrt{2}+\sqrt{1+\cos x})}{\frac{(1-\cos x)}{x^2} \times x^2} \\
& =(\ln 9 \cdot \ln 8)(2 \sqrt{2}) \times 2 \\
& =4 \sqrt{2} \times 2 \times 3 \ln 2 \cdot \ln 3 \\
& 24 \sqrt{2} \cdot \ln 2 \cdot \ln 3 \\
& \Rightarrow \quad a=24 \sqrt{2} \\
& a^2=1152
\end{aligned}
\)

Alternate:

\(
\begin{aligned}
& \lim _{x \rightarrow 0} f(x)=a \ln 2 \ln 3 \\
& \lim _{n \rightarrow 0} \frac{72^x-9^x-8^x+1}{\sqrt{2}-\sqrt{1+\cos x}}=\lim _{x \rightarrow 0} \frac{\left(8^x-1\right)\left(9^x-1\right)}{\sqrt{2}-\sqrt{1+\cos x}} \\
& \lim _{n \rightarrow 0}\left(\frac{8^x-1}{x}\right)\left(\frac{9^x-1}{x}\right)\left(\frac{x^2}{1-\cos x}\right)(\sqrt{2}+\sqrt{1+\cos x}) \\
& \therefore \ln 8 \times \ln 9 \times 2 \times 2 \sqrt{2}=24 \sqrt{2} \ln 2 \ln 3 \\
& \therefore a=24 \sqrt{2}, a^2=576 \times 2=1152
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \lim _{x \rightarrow 1} \frac{\frac{1}{3}(5 x+1)^{-2 / 3} 5-\frac{1}{3}(x+5)^{-2 / 3}}{\frac{1}{2}(2 x+3)^{-1 / 2} \cdot 2-\frac{1}{2}(x+4)^{-1 / 2}} \\
& =\frac{8}{3} \frac{\sqrt{5}}{6^{2 / 3}} m=8 \\
& 8 m+12 n=100
\end{aligned}
\)

Alternate:

\(I=\lim _{x \rightarrow 1} \frac{(5 x+1)^{1 / 3}-(x+5)^{1 /3}}{(2 x+3)^{1 / 2}-(x+4)^{1 / 2}}\)
From: \(\frac{0}{0}\), using \(L – H\) rule
\(
\begin{aligned}
I & =\lim _{x \rightarrow 1} \frac{\frac{1}{3} \times 5(5 x+1)^{-2 / 3}–(+5)}{\frac{1}{2} \times 2(2 x+3)^{-1 / 2}-\frac{1}{2}(x+4)^{-1 / 2}} \\
& =\frac{\left(\left.\frac{5}{3}-\frac{1}{3} \right\rvert\, 6^{-2 / 3}\right.}{\frac{1}{2} 5^{-1 / 2}}=\frac{8}{3} \times \frac{5^{1 / 2}}{6^{2 / 3}}=\frac{m \sqrt{5}}{n(2 n)^{2 / 3}} \\
& \Rightarrow m=8, n=3 \\
& \Rightarrow 8 m+12 n=100
\end{aligned}
\)

Hint

\(
\lim _{x \rightarrow 0} \frac{f(x)}{x^3}
\)
Using L Hopital Rule.
\(
\lim _{x \rightarrow 0} \frac{f^{\prime}(x)}{3 x^2}=\lim _{x \rightarrow 0} \frac{x+\sin \left(1-e^x\right)}{3 x^2} \text { (Again L Hopital) }
\)
Using L.H. Rule
\(
\begin{aligned}
& =\lim _{x \rightarrow 0} \frac{-\left[\sin \left(1-e^x\right)\left(-e^x\right) \cdot e^x+\cos \left(1-e^x\right) \cdot e^x\right]}{6} \\
& =-\frac{1}{6}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \lim _{t \rightarrow x} \frac{t^2 f(x)-x^2 f(t)}{t-x}=1 \\
& \lim _{t \rightarrow x} \frac{2 t . f(x)-x^2 f^{\prime}(x)}{1}=1 \\
& 2 x \cdot f(x)-x 2 f^{\prime}(x)=1 \\
& \frac{d y}{d x}-\frac{2}{x} \cdot y=\frac{-1}{x^2} \\
& \text { I.f. }=e^{\int-\frac{2}{x} d x}=\frac{1}{x^2} \\
& \therefore \frac{y}{x^2}=\int-\frac{1}{x^4} d x+C \\
& \frac{y}{x^2}=\frac{1}{3 x^3}+C \\
& \text { Put } f(1)=1
\end{aligned}
\)
\(
\begin{aligned}
& C=\frac{2}{3} \\
& y=\frac{1}{3 x}+\frac{2 x^2}{3} \\
& y=\frac{2 x^3+1}{3 x} \\
& f(2)=\frac{17}{6} \\
& f(3)=\frac{55}{9} \\
& 2 f(2)+3 f(3)=\frac{17}{3}+\frac{55}{3}=\frac{72}{3}=24
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \lim _{x \rightarrow \frac{1}{a}} 16 \cdot \frac{\left(1-\cos 2\left(x-\frac{1}{a}\right)\left(x-\frac{1}{b}\right)\right)}{4\left(x-\frac{1}{b}\right)^2} \times \frac{4\left(x-\frac{1}{b}\right)^2}{a^2\left(x-\frac{1}{a}\right)^2} \\
& =16 \times \frac{2}{a^2}\left(\frac{1}{a}-\frac{1}{b}\right)^2 \\
& =\frac{32}{a^2}\left(\frac{17}{4}\right)=\frac{17.8}{a^2}=\frac{17 \times 8 \times 16}{(-1+\sqrt{117})^2} \\
& =\frac{136.16}{18.2 \sqrt{7}} \times \frac{18+2 \sqrt{7}}{18+2 \sqrt{7}} \\
& =\frac{136}{256}(18+2 \sqrt{7}) \cdot 16 \\
& =153+17 \sqrt{17}=\alpha+\beta \sqrt{17} \\
& \alpha+\beta=153+17=170 \\
&
\end{aligned}
\)

Hint

\(
\begin{aligned}
& f:(-\infty, \infty)-\{0\} \rightarrow R \\
& f^{\prime}(1)=\lim _{a \rightarrow \infty} a^2 f\left(\frac{1}{a}\right) \\
& \lim _{a \rightarrow \infty} \frac{a(a+1)}{2} \tan ^{-1}\left(\frac{1}{a}\right)+a^2-2 \ln (a) \\
& \lim _{a \rightarrow \infty} a^2\left(\frac{\left(1+\frac{1}{a}\right)}{2} \tan ^{-1}\left(\frac{1}{a}\right)+1-\frac{2}{a^2} \ln (a)\right) \\
& f(x)=\frac{1}{2}(1+x) \tan ^{-1}(x)+1-2 x^2 \ln (x) \\
& f ^{\prime}( x )=\frac{1}{2}\left(\frac{1+ x }{1+ x ^2}+\tan ^{-1}( x )+4 x \ell( x )\right)+2 x \\
& f ^{\prime}(1)=\frac{1}{2}\left(1+\frac{\pi}{4}\right)+2 \\
& f ^{\prime}(1)=\frac{5}{2}+\frac{\pi}{8}
\end{aligned}
\)

Hint

\(
\lim _{x \rightarrow 0} 2\left(\frac{1-\cos x(\cos 2 x)^{\frac{1}{2}}(\cos 3 x)^{\frac{1}{3}} \ldots(\cos 10 x)^{\frac{1}{10}}}{x^2}\right)\left(\frac{0}{0} \text { form }\right)
\)
Using L’ hospital
\(
\begin{aligned}
& 2 \lim _{x \rightarrow 0} \frac{\sin x(\cos 2 x)^{\frac{1}{2}} \ldots(\cos 10 x)^{\frac{1}{10}} \ldots(\sin 2 x)(\cos x)(\cos 3 x)^{\frac{1}{3}}+\ldots}{2 x} \\
& \Rightarrow \lim _{x \rightarrow 0}\left(\frac{\sin x}{x}+\frac{\sin 2 x}{x}+\ldots+\frac{\sin 10 x}{x}\right) \\
& \quad=1+2+\ldots+10=55
\end{aligned}
\)

Alternate:
\(
\lim _{x \rightarrow 0} 2\left(\frac{1-\left(1-\frac{z^2}{2!}\right)\left(1-\frac{4 x^2}{2!}\right)\left(1-\frac{1 x^2}{2}\right) \ldots .\left(1-\frac{100 z^2}{2!}\right)}{x^2}\right)
\)
By expansion
\(
\begin{aligned}
& \left.\lim _{x \rightarrow 0} \frac{2\left(1-\left(1-\frac{x^2}{2}\right)\right)\left(1-\frac{1}{2} \cdot \frac{4 x^2}{2}\right)\left(1-\frac{1}{3} \cdot \frac{9 x^2}{2}\right) \ldots \ldots\left(1-\frac{1}{10} \cdot \frac{100 x^2}{2}\right)}{x^2}\right) \\
& \lim _{x \rightarrow 0} 2\left(\frac{\left.1-\left(1-\frac{x^2}{2}\right)\left(1-\frac{2 x^2}{2}\right)\left(1-\frac{3 x^2}{2}\right) \ldots \ldots\left(1-\frac{10 x^2}{2}\right)\right)}{x^2}\right) \\
& \lim _{x \rightarrow 0} \frac{2\left(1-1+x^2\left(\frac{1}{2}+\frac{2}{2}+\frac{3}{2}+\ldots+\frac{10}{2}\right)\right)}{x^2} \\
& 2\left(\frac{1}{2}+\frac{2}{2}+\frac{3}{2}+\ldots .+\frac{10}{2}\right) \\
& 1+2+\ldots \ldots+10=\frac{10 \times 11}{2}=55 \\
&
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \alpha=\lim _{x \rightarrow 0^{+}} \frac{e^{\sqrt{\tan x}}-e^{\sqrt{x}}}{(\sqrt{\tan x}-\sqrt{x})} \\
& =\lim _{x \rightarrow 0} \frac{e^{\sqrt{x}}\left(e^{\sqrt{\tan x}-\sqrt{x}}-1\right)}{(\sqrt{\tan x}-\sqrt{x})}=1 \\
& \beta=\lim _{x \rightarrow 0}(1+\sin x)^{\frac{1}{2} \cot x}=\lim _{x \rightarrow 0} e^{(\sin x)\left(\frac{1}{2} \cot x\right)} \\
& =\lim _{x \rightarrow 0} e^{\frac{1}{2} \cos x}=e^{1 / 2} \\
& \text { Product of roots }=\sqrt{e}=\frac{-\sqrt{e}}{a} \Rightarrow a=-1 \\
& \text { Sum of roots }=\frac{-b}{a}=1+\sqrt{e} \\
& =b=\sqrt{e}+1 \\
& \Rightarrow 12 \ln (a+b)=12 \ln (\sqrt{e}+1-1)=12 \ln \left(e^{1 / 2}\right)=6 \\
&
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \sum_{r=1}^{\infty} \frac{n}{\sqrt{n^4+r^4}}-\frac{2 r^2}{\left(n^2+r^2\right) \sqrt{n^4+r^4}} \\
& \sum_{r=1}^{\infty} \frac{\frac{1}{n}}{\sqrt{1+\left(\frac{r}{n}\right)^4}}-\frac{2\left(\frac{1}{n}\right)\left(\frac{r}{n}\right)^2}{\left(1+\left(\frac{r}{n}\right)^2\right) \sqrt{1+\left(\frac{r}{n}\right)^4}}
\end{aligned}
\)
\(
\begin{aligned}
& \Rightarrow \int_0^1 \frac{ dx }{\sqrt{1+ x ^4}}-\frac{2 x ^2 dx }{\left(1+ x ^2\right) \sqrt{1+ x ^4}} \\
& \Rightarrow \int_0^1 \frac{1- x ^2}{\left(1+ x ^2\right) \sqrt{1+ x ^4}} d x \\
& \Rightarrow \int_0^1 \frac{\frac{1}{x^2}-1}{\left(x+\frac{1}{x}\right) \sqrt{x^2+\frac{1}{x^2}}} d x \\
& \Rightarrow-\int_0^1 \frac{1-\frac{1}{x^2}}{\left(x+\frac{1}{x}\right) \sqrt{\left(x+\frac{1}{x}\right)^2-2}} d x \\
&
\end{aligned}
\)
\(
\begin{aligned}
& x+\frac{1}{x}=t \Rightarrow 1-\frac{1}{x^2} d x=d t \\
& \Rightarrow-\int_{\infty}^2 \frac{ dt }{ t \sqrt{ t ^2-2}} \\
& \Rightarrow-\int_{\infty}^2 \frac{t d t}{t^2 \sqrt{t^2-2}} \\
& \text { take } t ^2-2=\alpha^2 \\
& tdt =\alpha d \alpha \\
& \Rightarrow-\int_{\infty}^{\sqrt{2}} \frac{\alpha d \alpha}{\left(\alpha^2+2\right) \alpha} \\
& \Rightarrow-\int_{\infty}^{\sqrt{2}} \frac{ d \alpha}{\alpha^2+2} \\
& \left.\Rightarrow \frac{-1}{\sqrt{2}} \tan ^{-1} \frac{\alpha}{\sqrt{2}}\right]_{\infty}^{\sqrt{2}} \\
& \Rightarrow \frac{-1}{\sqrt{2}}\left\{\tan ^{-1} 1\right\}+\frac{1}{\sqrt{2}} \tan ^{-1} \infty \\
& \Rightarrow \frac{1}{\sqrt{2}}\left\{\frac{\pi}{2}-\frac{\pi}{4}\right\} \\
& \Rightarrow \frac{\pi}{4 \sqrt{2}}=\frac{\pi}{K} \\
&
\end{aligned}
\)
So \(K=4 \sqrt{2}\)
\(
K ^2=32
\)

Hint

\(
\begin{aligned}
& \lim _{x \rightarrow \frac{\pi}{2}} \frac{0-\{\sin (2 x)+\cos (x)\} \cdot 3 x^2}{2\left(x-\frac{\pi}{2}\right)} \\
& =\lim _{x \rightarrow \frac{\pi}{2}} \frac{-\{2 \sin x \cos x+\cos x\} 3 x^2}{2\left(x-\frac{\pi}{2}\right)} \\
& =\lim _{x \rightarrow \frac{\pi}{2}}\left\{\frac{2 \sin x \sin \left(\frac{\pi}{2}-x\right)}{2\left(x-\frac{\pi}{2}\right)}+\frac{\sin \left(\frac{\pi}{2}-x\right)}{2\left(\frac{\pi}{2}-x\right)}\right\} 3 x^2 \\
& =\left(1(1)+\frac{1}{2}\right) 3\left(\frac{\pi}{2}\right)^2 \\
& =\frac{9 \pi^2}{8}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& y=\frac{(\sqrt{x}+1)\left(x^2-\sqrt{x}\right)}{x \sqrt{x}+x+\sqrt{x}}+\frac{1}{15}\left(3 \cos ^2 x-5\right) \cos ^3 x \\
& y=\frac{(\sqrt{x}+1)(\sqrt{x})\left((\sqrt{x})^3-1\right)}{(\sqrt{x})\left((\sqrt{x})^2+(\sqrt{x})+1\right)}+\frac{1}{5} \cos ^5 x-\frac{1}{3} \cos ^3 x \\
& y=(\sqrt{x}+1)(\sqrt{x}-1)+\frac{1}{5} \cos ^5 x-\frac{1}{3} \cos ^3 x \\
& y^{\prime}=1-\cos ^4 x \cdot(\sin x)+\cos ^2 x(\sin x) \\
& y^{\prime}\left(\frac{\pi}{6}\right)=1-\frac{9}{16} \times \frac{1}{2}+\frac{3}{4} \times \frac{1}{2} \\
& =\frac{32-9+12}{32}=\frac{35}{32} \\
& =96 y^{\prime}\left(\frac{\pi}{6}\right)=105
\end{aligned}
\)

Hint

\(
\begin{aligned}
& f(x)-f(y) \geq \ln x-\ln y+x-y \\
& \frac{ f ( x )- f ( y )}{ x – y } \geq \frac{\ln x -\ln y }{ x – y }+1 \\
& \text { Let } x>y \\
& \lim _{y \rightarrow x} f^{\prime}\left(x^{-}\right) \geq \frac{1}{x}+1 \ldots(1) \\
& \text { Let } x < y \\
& \lim _{y \rightarrow x} f^{\prime}\left(x^{+}\right) \leq \frac{1}{x}+1 \ldots(2) \\
& f ^1\left( x ^{-}\right)= f ^1\left( x ^{+}\right) \\
& f ^1( x )=\frac{1}{ x }+1 \\
& f ^{\prime}\left(\frac{1}{ x ^2}\right)= x ^2+1 \\
& \sum_{ x =1}^{20}\left( x ^2+1\right)=\sum_{ x =1}^{20} x ^2+20 \\
& =\frac{20 \times 21 \times 41}{6}+20 \\
& =2890
\end{aligned}
\)

Hint

\(
\begin{aligned}
& f(x)=x^3+x^2 \cdot f^{\prime}(1)+x \cdot f^{\prime \prime}(2)+f^{\prime \prime \prime}(3) \\
& f^{\prime}(x)=3 x^2+2 x f^{\prime}(1)+f^{\prime \prime}(2) \\
& f^{\prime \prime}(x)=6 x+2 f^{\prime}(1) \\
& f^{\prime \prime \prime}(x)=6 \\
& f^{\prime}(1)=-5, f^{\prime \prime}(2)=2, f^{\prime \prime \prime}(3)=6 \\
& f(x)=x^3+x^2 \cdot(-5)+x \cdot(2)+6 \\
& f^{\prime}(x)=3 x^2-10 x+2 \\
& f^{\prime}(10)=300-100+2=202
\end{aligned}
\)

Hint

\(
\begin{aligned}
& f^{\prime}(0)=\lim _{h \rightarrow 0} \frac{f(h)-f(0)}{h} \\
& =\lim _{h \rightarrow 0} \frac{\left(2^h+2^{-h}\right) \tan h \sqrt{\tan ^{-1}\left(h^2-h+1\right)}-0}{\left(7 h^2+3 h+1\right)^3 h} \\
& =\sqrt{\pi}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& y=\log _e\left(\frac{1-x^2}{1+x^2}\right) \\
& \frac{d y}{d x}=y^{\prime}=\frac{-4 x}{1-x^4}
\end{aligned}
\)
Again,
\(
\frac{d^2 y}{d x^2}=y^{\prime \prime}=\frac{-4\left(1+3 x^4\right)}{\left(1-x^4\right)^2}
\)
Again
\(
\begin{aligned}
& y^{\prime}-y^{\prime \prime}=\frac{-4 x}{1-x^4}+\frac{4\left(1+3 x^4\right)}{\left(1-x^4\right)^2} \\
& \text { at } x =\frac{1}{2} \text {, } \\
& y^{\prime}-y^{\prime \prime}=\frac{736}{225}
\end{aligned}
\)
Thus \(225\left(y^{\prime}-y^{\prime \prime}\right)=225 \times \frac{736}{225}=736\)

Hint

\(
\begin{aligned}
& f(x)=2 \\
& \text { when } x=0 \\
& \because g^{\prime}(f(x)) f^{\prime}(x)=1 \\
& g^{\prime}(2)=\frac{1}{f^{\prime}(0)} \\
& \because f^{\prime}(x)=5 x^4+\frac{2}{4} e^{x / 4} \\
& g^{\prime}(2)=2 \\
& 8 g^{\prime}(2)=16
\end{aligned}
\)

Hint

\(
\begin{aligned}
& y=\frac{2 \cos \theta+2 \cos ^2 \theta-1}{4 \cos ^3 \theta-3 \cos \theta+8 \cos ^2 \theta-4+5 \cos \theta+2} \\
& y=\frac{\left(2 \cos ^2 \theta+2 \cos \theta-1\right)}{\left(2 \cos ^2 \theta+2 \cos \theta-1\right)(2 \cos \theta+2)} \\
& y=\frac{1}{2}\left(\frac{1}{1+\cos \theta}\right) \\
& \Rightarrow \theta=\frac{\pi}{2} \quad y=\frac{1}{2} \\
& y^{\prime}=\frac{1}{2}\left(\frac{-1}{(1+\cos \theta)^2} \times(-\sin \theta)\right) \\
& \Rightarrow \theta=\frac{\pi}{2} \quad y=\frac{1}{2} \\
& y^{\prime \prime}=\frac{1}{2}\left[\frac{\cos \theta(1+\cos \theta)^2-\sin \theta(2)(1+\cos \theta)(-\sin \theta)}{(1+\cos \theta)^4}\right] \\
& \Rightarrow \theta=\frac{\pi}{2} \quad y=1 \\
&
\end{aligned}
\)
\(
y^{\prime \prime}+y^{\prime}+y=2
\)

Hint

\(
\begin{aligned}
& f^{\prime}(x)=3 x^2 \sin \left(\frac{1}{x}\right)-x \cos \left(\frac{1}{x}\right) \\
& f^{\prime \prime}(x)=6 x \sin \left(\frac{1}{x}\right)-3 \cos \left(\frac{1}{x}\right)-\cos \left(\frac{1}{x}\right)-\frac{\sin \left(\frac{1}{x}\right)}{x} \\
& f^{\prime \prime}\left(\frac{2}{\pi}\right)=\frac{12}{\pi}-\frac{\pi}{2}=\frac{24-\pi^2}{2 \pi}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& g(x)=h\left(e^x\right) \cdot e^{h(x)} \\
& g^{\prime}(x)=h\left(e^x\right) \cdot e^{h(x)} \cdot h^{\prime}(x)+e^{h(x)} h^{\prime}\left(e^x\right) \cdot e^x \\
& g^{\prime}(0)=h(1) e^{h(0)} h^{\prime}(0)+e^{h(0)} h^{\prime}(1) \\
& =2+2=4
\end{aligned}
\)

Hint

\(
\begin{aligned}
& f(x)=a x^3+b x^2+c x+41 \\
& f^{\prime}(x)=3 a x^2+2 b x+c x \\
& \Rightarrow f^{\prime}(1)=3 a+2 b+c=2 \ldots(1) \\
& f ^{\prime \prime}( x )=6 ax +2 b \\
& \Rightarrow f ^{\prime \prime}(1)=6 a +2 b =4 \\
& 3 a + b =2 \ldots \ldots \ldots .(2)
\end{aligned}
\)
\(
\begin{aligned}
& \text { (1) }-(2) \\
& b+c=0 \ldots(3) \\
& f (1)=40 \\
& a + b + c +41=40 \\
& \text { use (3) } \\
& a+41=40 \\
& \text { by (2) } \\
& -3+b=2 \Rightarrow b=5 \& c=-5 \\
& a^2+b^2+c^2=1+25+25=51 \\
&
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \ln (y)=3 \sin ^{-1} x \\
& \frac{1}{y} \cdot y^{\prime}=3\left(\frac{1}{\sqrt{1-x^2}}\right) \\
& \Rightarrow y^{\prime}=\frac{3 y}{\sqrt{1-x^2}} \text { at } x=\frac{1}{2} \\
& \Rightarrow y^{\prime}=\frac{3 e^{3\left(\frac{\pi}{6}\right)}}{\frac{\sqrt{3}}{2}}=2 \sqrt{3} e^{\frac{\pi}{2}} \\
& \Rightarrow y^{\prime \prime}=3\left(\frac{\sqrt{1-x^2} y^{\prime}-y \frac{1}{2 \sqrt{1-x^2}}(-2 x)}{\left(1-x^2\right)}\right) \\
& \Rightarrow\left(1-x^2\right) y^{\prime \prime}=3\left(3 y+\frac{x y}{\sqrt{1-x^2}}\right) \\
& \downarrow \text { at } x=\frac{1}{2}, y=e^{3 \sin ^{-1}\left(\frac{1}{2}\right)}= e ^{3\left(\frac{\pi}{6}\right)}= e ^{\frac{\pi}{2}}
\end{aligned}
\)
\(
\begin{aligned}
& \left.\left(1-x^2\right) y^{\prime \prime}\right|_{\text {at } x=\frac{1}{2}}=3\left(3 e ^{\frac{\pi}{2}}+\frac{\frac{1}{2}\left(e^{\frac{\pi}{2}}\right)}{\frac{\sqrt{3}}{2}}\right) \\
& =3 e ^{\frac{\pi}{2}}\left(3+\frac{1}{\sqrt{3}}\right) \\
& \left(1-x^2\right) y^{\prime \prime}-\left.x y^{\prime}\right|_{\text {atx } x=\frac{1}{2}} \\
& =3 e ^{\frac{\pi}{2}}\left(3+\frac{1}{\sqrt{3}}\right)-\frac{1}{2}\left(2 \sqrt{3} e ^{\frac{\pi}{2}}\right)=9 e ^{\frac{\pi}{2}}
\end{aligned}
\)

Hint

\(
\lim _{t \rightarrow 0}\left(1^{\operatorname{cosec}^2 t}+2^{\operatorname{cosec}^2 t}+\ldots \ldots . .+n^{\operatorname{cosec}^2 t}\right)^{\sin ^2 t}
\)
\(
\begin{aligned}
& =\lim _{t \rightarrow 0} n\left(\left(\frac{1}{n}\right)^{\operatorname{cosec}^2 t}+\left(\frac{2}{n}\right)^{\operatorname{cosecs}^2 t}+\ldots \ldots \ldots+1\right)^{\sin ^2 t} \\
& =n
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \lim _{x \rightarrow a}([x-5]-[2 x+2])=0 \\
& \lim _{x \rightarrow a}([x]-5-[2 x]-2)=0 \\
& \lim _{x \rightarrow a}([x]-[2 x])=7
\end{aligned}
\)
\(
\begin{aligned}
& {[a]-[2 a]=7} \\
& a \in I, \quad a=-7 \\
& a \notin I, \quad a=I+f \\
& \text { Now, }[a]-[2 a]=7 \\
& \quad-I-[2 f]=7
\end{aligned}
\)
\(
\begin{aligned}
& \text { Case-I: } f \in\left(0, \frac{1}{2}\right) \\
& \text { 2f } \in(0,1) \\
& – I =7 \\
& I =-7 \Rightarrow a \in(-7,-6.5)
\end{aligned}
\)
Case-II: \(f \in\left(\frac{1}{2}, 1\right)\)
\(
\begin{aligned}
& \text { 2f } \in(1,2) \\
& -I-I=7 \\
& I=-8 \Rightarrow a \in(-7.5,-7)
\end{aligned}
\)
Hence, \(a \in(-7.5,-6.5)\)

Hint

\(
\begin{aligned}
& LHL =\lim _{ k \rightarrow 0} g ( h (- k )) \quad, k >0 \text { mathongo } \\
& =\lim _{ k \rightarrow 0} g (-2+1) \text { mathon } \because f ( x )=-1 \forall x <0 \\
& = g (-1)=1 \\
& RHL =\lim _{ k \rightarrow 0} g ( h ( k )) \quad, k >0 \\
& =\lim _{ k \rightarrow 0} g (-1) \quad, \because f ( x )=1, \forall x >0 \\
& =1
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \lim _{x \rightarrow \infty} \frac{(\sqrt{3 x+1}+\sqrt{3 x-1})^6+(\sqrt{3 x+1}-\sqrt{3 x-1})^6}{\left(x+\sqrt{x^2-1}\right)^6+\left(x-\sqrt{x^2-1}\right)^6} x^3 \\
& \lim _{x \rightarrow \infty} x^3 \times\left\{\frac{x^3\left\{\left(\sqrt{3+\frac{1}{x}}+\sqrt{3-\frac{1}{x}}\right)^6+\left(\sqrt{3+\frac{1}{x}}-\sqrt{3-\frac{1}{x}}\right)^6\right\}}{x^6\left\{\left(1+\sqrt{1-\frac{1}{x^2}}\right)^6+\left(1-\sqrt{1-\frac{1}{x^2}}\right)^6\right\}}\right\} \\
& =\frac{(2 \sqrt{3})^6+0}{2^6+0}=3^3=(27)
\end{aligned}
\)

Hint

Let,
\(
L=\lim _{n \rightarrow \infty}\left\{\left(2^{\frac{1}{2}}-2^{\frac{1}{3}}\right)\left(2^{\frac{1}{2}}-2^{\frac{1}{5}}\right) \ldots\left(2^{\frac{1}{2}}-2^{\frac{1}{2 n+1}}\right)\right\}
\)
Now,
\(
\begin{gathered}
2^{\frac{1}{2}}-2^{\frac{1}{3}}<1 \\
2^{\frac{1}{2}}-2^{\frac{1}{5}}<1
\end{gathered}
\)
\(
2^{\frac{1}{2}}-2^{\frac{1}{2 n+1}}<1 \quad \forall n \in N
\)
And \(\left(2^{\frac{1}{2}}-2^{\frac{1}{3}}\right)^n<\left(2^{\frac{1}{2}}-2^{\frac{1}{3}}\right)\left(2^{\frac{1}{2}}-2^{\frac{1}{5}}\right) \ldots\left(2^{\frac{1}{2}}-2^{\frac{1}{2 n+1}}\right)<\left(2^{\frac{1}{2}}-2^{\frac{1}{2 n+1}}\right)^n\)
\(
\begin{aligned}
& \Rightarrow \lim _{n \rightarrow \infty}\left(2^{\frac{1}{2}}-2^{\frac{1}{3}}\right)^n<\lim _{n \rightarrow \infty}\left\{\left(2^{\frac{1}{2}}-2^{\frac{1}{3}}\right)\left(2^{\frac{1}{2}}-2^{\frac{1}{5}}\right) \ldots\left(2^{\frac{1}{2}}-2^{\frac{1}{2 n+1}}\right)\right\}<\lim _{n \rightarrow \infty}\left(2^{\frac{1}{2}}-2^{\frac{1}{2 n+1}}\right)^n \\
& \Rightarrow \lim _{n \rightarrow \infty}\left(2^{\frac{1}{2}}-2^{\frac{1}{3}}\right)^n<L<\lim _{n \rightarrow \infty}\left(2^{\frac{1}{2}}-2^{\frac{1}{2 n+1}}\right)^n
\end{aligned}
\)
And \(\lim _{n \rightarrow \infty}\left(2^{\frac{1}{2}}-2^{\frac{1}{3}}\right)^n=0 \& \lim _{n \rightarrow \infty}\left(2^{\frac{1}{2}}-2^{\frac{1}{2 n+1}}\right)^n=0\)
\(
\left\{\text { as } 2^{\frac{1}{2}}-2^{\frac{1}{3}}<1 \& 2^{\frac{1}{2}}-2^{\frac{1}{2 n+1}}<1\right\}
\)
Hence, \(\lim _{n \rightarrow \infty}\left\{\left(2^{\frac{1}{2}}-2^{\frac{1}{3}}\right)\left(2^{\frac{1}{2}}-2^{\frac{1}{5}}\right) \ldots\left(2^{\frac{1}{2}}-2^{\frac{1}{2 n+1}}\right)\right\}=0\)

Hint

Given,
\(
\lim _{x \rightarrow 0}\left(\left(\frac{1-\cos ^2(3 x)}{\cos ^3(4 x)}\right)\left(\frac{\sin ^3(4 x)}{\left(\log _e(2 x+1)\right)^5}\right)\right)
\)
Now we now that,
\(
\lim _{x \rightarrow 0} \frac{\sin x}{x}=1, \lim _{x \rightarrow 0} \frac{1-\cos x}{x^2}=\frac{1}{2} \& \lim _{x \rightarrow 0} \frac{\log (1+x)}{x}=1
\)
Now using the above formula we get,
\(
\begin{aligned}
& \lim _{x \rightarrow 0}\left(\left(\frac{1-\cos ^2 3 x}{\cos ^3 4 x}\right)\left(\frac{\sin ^3 4 x}{\left(\log _e(2 x+1)\right)^5}\right)\right) \\
& =\lim _{x \rightarrow 0} \frac{(1-\cos 3 x)(1+\cos 3 x) 9 x^2}{\left(\cos ^3 4 x\right) 9 x^2} \frac{(\sin 4 x)^3(64 x)^3(2 x)^5}{\left(64 x^3\right)\left(\log _e(2 x+1)\right)^5(2 x)^5} \\
& =\lim _{x \rightarrow 0} \frac{\left(\frac{1-\cos 3 x}{9 x^2}\right)(1+\cos 3 x)}{\left(\cos ^3 4 x\right)} \frac{\left(\frac{\sin 4 x}{4 x}\right)^3}{\left(\frac{\log _e(2 x+1)}{2 x}\right)^5} \times \frac{9 \times 64}{32} \\
& =\frac{\left(\frac{1}{2}\right) \times(2)}{(1)} \frac{(1)^3}{(1)^5} \times \frac{9 \times 64}{32} \\
& =\frac{9 \times 2}{1}=18
\end{aligned}
\)

Hint

Since, \(\alpha\) and \(\beta\) are the roots of the equation \(a x^2+b x+1=0\), therefore \(\frac{1}{\alpha}\) and \(\frac{1}{\beta}\) would be the roots of \(x^2+b x+a=0\).
Let
\(
\begin{aligned}
& L=\lim _{x \rightarrow \frac{1}{\alpha}}\left[\frac{1-\cos \left(x^2+b x+a\right)}{2(1-\alpha x)^2}\right]^{\frac{1}{2}} \rightarrow \frac{0}{0} \\
& \Rightarrow L=\lim _{x \rightarrow \frac{1}{\alpha}}\left[\frac{2 \sin ^2\left(\frac{x^2+b x+\alpha}{2}\right)}{2(1-\alpha x)^2}\right]^{\frac{1}{2}} \\
& \Rightarrow L=\lim _{x \rightarrow \frac{1}{\alpha}}\left[\frac{\sin ^2\left(\frac{1}{2}\left(x-\frac{1}{\alpha}\right)\left(x-\frac{1}{\beta}\right)\right)}{\alpha^2\left(x-\frac{1}{\alpha}\right)^2}\right]^{\frac{1}{2}} \\
& \Rightarrow L=\lim _{x \rightarrow \frac{1}{\alpha}}\left[\frac{\left(x-\frac{1}{\beta}\right)^2}{4 \alpha^2} \times \frac{\left\{\sin \left(\frac{1}{2}\left(x-\frac{1}{\alpha}\right)\left(x-\frac{1}{\beta}\right)\right)\right\}^2}{\left(\frac{1}{2}\right)^2\left(x-\frac{1}{\alpha}\right)^2\left(x-\frac{1}{\beta}\right)^2}\right]^{\frac{1}{2}} \\
& \Rightarrow L=\frac{\left(\frac{1}{\alpha}-\frac{1}{\beta}\right)}{2 \alpha}
\end{aligned}
\)
On comparing this value with the value given in the question we get, \(k=2 \alpha\).

Hint

\(
\begin{aligned}
& S_1: \lim _{n \rightarrow \infty} \frac{1}{n^2}[2+4+6+\ldots+2 n] \\
& \lim _{n \rightarrow \infty} 2 \frac{n(n+1)}{2 n^2}=1 \\
& S_2: \lim _{n \rightarrow \infty} \frac{1}{n^{16}}\left(1^{15}+2^{15}+3^{15}+\ldots+n^{15}\right)
\end{aligned}
\)
This is limit of sum.
We know that \(\lim _{n \rightarrow \infty} \sum_{r=1}^n f\left(\frac{r}{n}\right) \frac{1}{n}=\int_0^1 f(x) d x\)
\(
\begin{aligned}
& \because \lim _{n \rightarrow \infty} \frac{\sum_{r=1}^n r^k}{n^{k+1}}=\lim _{n \rightarrow \infty} \frac{\sum_{r=1}^n\left(\frac{r}{n}\right)^k}{n} \\
& =\int_0^1 x^k d x=\frac{1}{k+1}
\end{aligned}
\)
Here \(k=15\)
\(
\therefore \lim _{n \rightarrow \infty} \frac{\sum_{r=1}^n r^{15}}{n^{16}}=\frac{1}{16}
\)
\(\therefore\) Both \(S 1\) and \(S 2\) are correct

Hint

Given that
\(
\Rightarrow \lim _{x \rightarrow 0} \frac{e^{a x}-\cos (b x)-\frac{c x}{2} e^{-c x}}{1-\cos 2 x}=17
\)
We know that \(e^{a x}=1+a x+\frac{(a x)^2}{2!}+\ldots\) and \(\cos (b x)=1-\frac{(b x)^2}{2!}+\ldots\)
\(
\begin{aligned}
& \Rightarrow \lim _{x \rightarrow 0} \frac{\left(1+a x+\frac{(a x)^2}{2!}+\ldots\right)-\left(1-\frac{(b x)^2}{2!}+\ldots\right)-\frac{c x}{2}\left(1-(c x)+\frac{(c x)^2}{2!}-\ldots\right)}{\left(\frac{1-\cos 2 x}{(2 x)^2}\right) \times 4 x^2}=17 \\
& \Rightarrow \lim _{x \rightarrow 0} \frac{\left(a-\frac{c}{2}\right) x+\left(\frac{a^2+b^2+c^2}{2}\right) x^2+\ldots}{\frac{1}{2} \times 4 x^2}=17
\end{aligned}
\)
Now for limit to exist, \(a-\frac{c}{2}=0 \Rightarrow c=2 a\)
\(
\begin{aligned}
& \Rightarrow \lim _{x \rightarrow 0} \frac{\left(\frac{a^2+b^2+c^2}{2}\right) x^2+\ldots}{\frac{1}{2} \times 4 x^2}=17 \\
& \Rightarrow \frac{a^2+b^2+c^2}{4}=17 \\
& \Rightarrow a^2+b^2+4 a^2=68 \\
& \Rightarrow 5 a^2+b^2=68
\end{aligned}
\)

Hint

\(
\begin{aligned}
& f(x)=x^3-x^2 f^{\prime}(1)+x f^{\prime \prime}(2)-f^{\prime \prime \prime}(3), x \in R \\
& \text { Let } f^{\prime}(1)=a, f^{\prime \prime}(2)=b, f^{\prime \prime \prime}(3)=c \\
& f( x )=x^3- a x^2 + b x- c \\
& f^{\prime}(x)=3 x^2-2 a x+b
\end{aligned}
\)
\(
\begin{aligned}
& f ^{\prime \prime}( x )=6 x -2 a \\
& f ^{\prime \prime \prime}( x )=6 \\
& c =6, a =3, b =6 \\
& f ( x )= x ^3-3 x ^2+6 x -6 \\
& f (1)=-2, f (2)=2, f (3)=12, f (0)=-6 \\
& 2 f (0)- f (1)+ f (3)=2= f (2)
\end{aligned}
\)

Hint

\(
\begin{aligned}
& y=\frac{1-x^{32}}{1-x} \Rightarrow y-x y=1-x^{32} \\
& y^{\prime}-x y^{\prime}-y=-32 x^{31} \\
& y^{\prime \prime}-x y^{\prime \prime}-y^{\prime}-y^{\prime}=-(32)(31) x^{30} \\
& \text { at } x=-1 \Rightarrow y^{\prime}-y^{\prime \prime}=496
\end{aligned}
\)

Hint

\(
\begin{aligned}
& f ( x + y )= f ( x )+ f ( y )-1 \\
& f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\
& f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(h)-f(0)}{h}=f^{\prime}(0)=2 \\
& f^{\prime}(x)=2 \Rightarrow d y=2 d x \\
& y =2 x + C \\
& x =0, y =1, c =1 \\
& y =2 x +1 \\
& |f(-2)|=|-4+1|=|-3|=3
\end{aligned}
\)

Hint

\(
\begin{aligned}
& f^{\prime \prime}(x)=g^{\prime \prime}(x)+6 x \ldots(1) \\
& f^{\prime}(1)=4 g^{\prime}(1)-3=9 \ldots(2) \\
& f(2)=3 g(2)=12 \ldots(3)
\end{aligned}
\)
By integrating (1)
\(
f ^{\prime}( x )= g ^{\prime}( x )+6 \frac{ x ^2}{2}+C
\)
At \(x =1\),
\(
\begin{aligned}
& f ^{\prime}(1)=g^{\prime}(1)+3+C \\
\Rightarrow & 9=3+3+C \Rightarrow C=3 \\
\therefore & f^{\prime}(x)=g^{\prime}(x)+3 x^2+3
\end{aligned}
\)
Again by integrating,
\(
f(x)=g(x)+\frac{3 x^3}{3}+3 x+D
\)
At \(x=2\),
\(
f(2)=g(2)+8+3(2)+D
\)
\(
\Rightarrow 12=4+8+6+D \Rightarrow D=-6
\)
So, \(f(x)=g(x)+x^3+3 x-6\)
\(
\Rightarrow f ( x )- g ( x )= x ^3+3 x -6
\)
At \(x=-2\),
\(
\Rightarrow g (-2)- f (-2)=20
\)
(Option (1) is true)
Now, for \(-1<x<2\)
\(
\begin{aligned}
& h(x)=f(x)-g(x)=x^3+3 x-6 \\
& \Rightarrow h^{\prime}(x)=3 x^2+3 \\
& \Rightarrow h(x) \uparrow
\end{aligned}
\)
\(
\begin{aligned}
& \text { So, } h(-1)<h(x)<h(2) \\
& \Rightarrow-10<h(x)<8 \\
& \Rightarrow|h(x)|<10
\end{aligned}
\)
(option (2) is NOT true)
Now, \(h ^{\prime}( x )= f ^{\prime}( x )- g ^{\prime}( x )=3 x ^2+3\)
\(
\begin{aligned}
& \text { If }\left| h ^{\prime}( x )\right|<6 \Rightarrow\left|3 x ^2+3\right|<6 \\
& \Rightarrow 3 x ^2+3<6 \\
& \Rightarrow x ^2<1
\end{aligned}
\)
\(\Rightarrow-1< x <1 \quad\) (option (3) is True)
If \(x \in(-1,1)\left|f^{\prime}(x)-g^{\prime}(x)\right|<6\)
option (3) is true and now to solve
\(
\begin{aligned}
& f ( x )- g ( x )=0 \\
& \Rightarrow x ^3+3 x -6=0 \\
& h ( x )= x ^3+3 x -6
\end{aligned}
\)
here, \(h(1)=-\) ve and \(h\left(\frac{3}{2}\right)=+\) ve
So there exists \(x_0 \in\left(1, \frac{3}{2}\right)\) such that \(f\left(x_0\right)=g\left(x_0\right)\) (option (4) is true)

Hint