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If \(f^{\prime}(x)=\phi(x)\) and \(\phi^{\prime}(x)=f(x)\) for all \(x\). Also \(f(3)=5\) and \(f^{\prime}(3) =4\). Then the value of \([f(10)]^2-[\phi(10)]^2\) is
(c)
\(
\begin{aligned}
& \frac{d}{d x}\left\{[f(x)]^2-[\phi(x)]^2\right\} \\
& =2\left[f(x) \cdot f^{\prime}(x)-\phi(x) \cdot \phi^{\prime}(x)\right] \\
& =2[f(x) \cdot \phi(x)-\phi(x) \cdot f(x)]\left[\because f^{\prime}(x)=\phi(x) \text { and } \phi^{\prime}(x)=f(x)\right] \\
& =0 \\
& {[f(x)]^2-[\phi(x)]^2=\text { constant }} \\
& {[f(10)]^2-[\phi(10)]^2=[f(3)]^2-[\phi(3)]^2=[f(3)]^2-\left[f^{\prime}(3)\right]^2=25} \\
& -16=9 .
\end{aligned}
\)
If \(y=f(x)\) is an odd differentiable function defined on \((-\infty\), \(\infty)\) such that \(f^{\prime}(3)=-2\), then \(\left|f^{\prime}(-3)\right|\) equals
(a)
(a) Since \(f(x)\) is odd. Therefore \(f(-x)=-f(x) \Rightarrow f^{\prime}(-x)(-1)= -f^{\prime}(x)\)
\(
\Rightarrow f^{\prime}(-x)=f^{\prime}(x) \therefore f^{\prime}(-3)=f^{\prime}(3)=-2 .
\)
If \(x^3+3 x^2-9 x+c\) is of the form \((x-\alpha)^2(x-\beta)\), then positive value of \(c\) is
(b) Here \(x=\alpha\) is a repeated root of the equation \(f(x)=0\)
hence \(x=\alpha\) is also a root of the equation \(f^{\prime}(x)=0\) i.e., \(3 x^2\)
\(+6 x-9=0\)
\(x^2+2 x-3=0\) or \((x+3)(x-1)=0\)
has the root \(\alpha\) once which can be either -3 , or 1.
If \(\alpha=1\), then \(f(x)=0\) gives \(c-5=0\) or \(c=5\)
If \(\alpha=-3\), then \(f(x)=0\) gives \(-27+27+27+c=0 \therefore c=-27\)
If graph of \(y=f(x)\) is symmetrical about the point \((5,0)\) and \(f^{\prime}(7)=3\), then the value of \(f^{\prime}(3)\) is
(d) We have \(f(5-x)=-f(5+x) \Rightarrow-f^{\prime}(5-x)=-f^{\prime}(5+x) \Rightarrow f^{\prime}(5-2)=f^{\prime}(5+2) \Rightarrow f^{\prime}(3)=f^{\prime}(7)=3\)
Let \(g(x)=f(x) \sin x\), where \(f(x)\) is a twice differentiable function on \((-\infty, \infty)\) such that \(f^{\prime}(-\pi)=1\). The value of \(\left|g^{\prime \prime}(-\pi)\right|\) equals
(a) We have \(g(x)=f(x) \sin \dot{x} \dots(1)\)
On differentiating equation (1) w.r.t. \(x\), we get
\(
g^{\prime}(x)=f(x) \cos x+f^{\prime}(x) \sin x \dots(2)
\)
Again differentiating equation (2) w.r.t. \(x\), we get
\(
g^{\prime \prime}(x)=f(x)(-\sin x)+f^{\prime}(x) \cos x+f^{\prime}(x) \cos x+f^{\prime \prime}(x) \sin x \dots(3)
\)
\(
g^{\prime \prime}(-\pi)=2 f^{\prime}(-\pi) \cos (-\pi)=2 \times 1 \times(-1)=-2
\)
Hence \(g^{\prime \prime}(-\pi)=-2\)
Let \(f(x)=(x-1)(x-2)(x-3) \cdots(x-n), n \in N\) and \(ff^{\prime}(n)=\) 5040 , then the value of \(n\) is
(c) \(\ln (f(x))=\ln (x-1)+\ln (x-2)+\cdots+\ln (x-n)\)
\(
\Rightarrow f^{\prime}(x)=f(x)\left[\frac{1}{x-1}+\frac{1}{x-2}+\cdots \frac{1}{x-n}\right]
\)
\(
\begin{aligned}
\Rightarrow & f^{\prime}(x)=(x-2)(x-3)(x-n)+(x-1)(x-3) \cdots(x-n)+\cdots+ \\
& (x-1)(x-2) \cdots(x-(n-1)) \\
\Rightarrow & f^{\prime}(n)=(n-1)(n-2)(n-3) \cdot 3 \cdot 2 \cdot 1 \text { (all other factors except } \\
& \text { the last vanishes when } x=n) \\
\Rightarrow & 5040=(n-1)! \\
\Rightarrow & n=8
\end{aligned}
\)
\(y=f(x)\), where \(f\) satisfies the relation \(f(x+y)=2 f(x)+x f(y)\)\(+y \sqrt{f(x)} \forall x, y \in R\) and \(f^{\prime}(0)=0\), then \(f(6)\) is equal to
(b)
\(
\begin{aligned}
& f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x+0)}{h} \\
& \lim _{h \rightarrow 0} \frac{2 f(x)+x f(h)+h \sqrt{f(x)}-2 f(x)-x f(0)-0 \sqrt{f(x)}}{h} \\
& \text { as } f(0)=0 \\
& \lim _{h \rightarrow 0} x\left(\frac{f(h)-f(0)}{h-0}\right)+\sqrt{f(x)}=f^{\prime}(0)+\sqrt{f(x)} \\
& f^{\prime}(x)=\sqrt{f(x)} \quad\left(\because f^{\prime}(0)=0\right) \\
& \int \frac{f^{\prime}(x)}{\sqrt{f(x)}} d x=\int d x \\
& 2 \sqrt{f(x)}=x+c \quad \\
& f(x)=\frac{x^2}{4} \quad \quad(\because f(0)=0)
\end{aligned}
\)
If function \(f\) satisfies the relation \(f(x) \times f^{\prime}(-x)=f(-x) \times f^{\prime}(x)\) for all \(x\), and \(f(0)=3\), now if \(f(3)=3\), then the value of \(f(-3)\) is
\(
\begin{aligned}
& \text { (c) } f(x) \times f^{\prime}(-x)=f(-x) \times f^{\prime}(x) \\
& \Rightarrow f^{\prime}(x) \times f(-x)-f(x) \times f^{\prime}(-x)=0 \\
& \Rightarrow \frac{d}{d x}[f(x) f(-x)]=0 \\
& \Rightarrow f(x) f(-x)=k \\
& \text { Given }(f(0))^2=k=9 \Rightarrow k=9 \\
& \text { Then } f(3) f(-3)=9 \Rightarrow f(-3)=3
\end{aligned}
\)
If \(y=\frac{a+b x^{3 / 2}}{x^{5 / 4}}\) and \(y^{\prime}=0\) at \(x=5\), then the value of \(a^2 / b^2\) is
(a)
\(
\begin{aligned}
&y=\frac{a+b x^{3 / 2}}{x^{5 / 4}} \Rightarrow y^{\prime}=\frac{\frac{3}{2} b x^{1 / 2} x^{5 / 4}-\frac{5}{4} x^{1 / 4}\left(a+b x^{3 / 2}\right)}{x^{5 / 2}}\\
&\text { According to the question, }\\
&0=\frac{\frac{3}{2} b 5^{1 / 2} 5^{5 / 4}-\frac{5}{4} 5^{1 / 4}\left(a+b 5^{3 / 2}\right)}{5^{5 / 2}}
\end{aligned}
\)
\(
\begin{aligned}
& \Rightarrow \frac{3 b}{2} 5^{7 / 4}-a \frac{5^{5 / 4}}{4}-5 b \frac{5^{7 / 4}}{4}=0 \\
& \Rightarrow b 5^{7 / 4}=a 5^{5 / 4} \\
& \Rightarrow b \sqrt{5}=a \\
& \Rightarrow a: b=\sqrt{5}: 1
\end{aligned}
\)
\(
a^2 / b^2=5:1
\)
Let \(y=\frac{2^{\log _{2^{1 / 4}} x}-3^{\log _{27}\left(x^2+1\right)^3}-2 x}{7^{4 \log _{49} x}-x-1}\) and \(\frac{d y}{d x}=a x+b\), then the value of \(a+b\) is
(c) The numerator terms are simplified using logarithm properties. The first term, \(2^{\log _{21 / 4} x}\), is simplified as follows:
\(
2^{\log _2 1 / 4 x}=2^{\frac{\log _2 x}{\log _2 2^{1 / 4}}}=2^{\frac{\log _2 x}{1 / 4}}=2^{4 \log _2 x}=2^{\log _2 x^4}=x^4 .
\)
\(
\text { The second term, } 3^{\log _{27}\left(x^2+1\right)^3} \text {, is simplified as follows: }
\)
\(
3^{\log _{27}\left(x^2+1\right)^3}=3^{\log _{33}\left(x^2+1\right)^3}=3^{\frac{3 \log _3\left(x^2+1\right)}{3}}=3^{\log _3\left(x^2+1\right)}=x^2+1 .
\)
\(
\text { Therefore, the numerator becomes } x^4-\left(x^2+1\right)-2 x=x^4-x^2-2 x-1 \text {. }
\)
The denominator term is simplified using logarithm properties.
The term \(7^{4 \log _{49} x}\) is simplified as follows:
\(
7^{4 \log _{49} x}=7^{4 \log _7 2 x}=7^{4 \frac{\log _7 x}{2}}=7^{2 \log _7 x}=7^{\log _7 x^2}=x^2
\)
Therefore, the denominator becomes \(x^2-x-1\).
\(
\begin{aligned}
& y=\frac{x^4-\left(x^2+2 x+1\right)}{x^2-x-1}=x^2+x+1 \\
& \frac{d y}{d x}=2 x+1=\mathrm{a} x+b \\
& \text { hence } a=2 \text { and } b=1
\end{aligned}
\)
\(\operatorname{Lim}_{h \rightarrow 0} \frac{(e+h)^{\ln (e+h)}-e}{h}\) is
\(
\begin{aligned}
& \text { (b) Limit is } f^{\prime}(e) \text { where } f(x)=x^{\ln x}=e^{\ln ^2 x} \\
& \Rightarrow g^{\prime}(f(x)) f^{\prime}(x)=e^{\ln ^2 x} \cdot \frac{2 \ln x}{x} \\
& \Rightarrow f^{\prime}(e)=e \cdot \frac{2}{e}=2
\end{aligned}
\)
If the function \(f(x)=-4 e^{\frac{1-x}{2}}+1+x+\frac{x^2}{2}+\frac{x^3}{3}\) and \(g(x) =f^{-1}(x)\), then the reciprocal of \(g^{\prime}\left(\frac{-7}{6}\right)\) is
\(
\begin{aligned}
&\text { (a) We have }(g o f)(x)=x\\
&\begin{aligned}
\Rightarrow & g^{\prime}(f(x)) f^{\prime}(x)=1 \\
& \text { when } f(x)=-\frac{7}{6} \Rightarrow x=1 \\
\Rightarrow & g^{\prime}(f(x)) g^{\prime}\left(-\frac{7}{6}\right) f^{\prime}(1)=1 \\
& \text { Hence } g^{\prime}\left(-\frac{7}{6}\right)=\frac{1}{f^{\prime}(1)}=\frac{1}{5}
\end{aligned}
\end{aligned}
\)
Suppose that \(f(0)=0\) and \(f^{\prime}(0)=2\), and let \(g(x)=f(-x+ f(f(x)))\). The value of \(g^{\prime}(0)\) is equal to
(d)
\(
\begin{aligned}
& g(x)=f(-x+f(f(x))) ; \quad f(0)=0 ; \quad f^{\prime}(0)=2 \\
& g^{\prime}(x)=f^{\prime}(-x+f(f(x))) \cdot\left[-1+f^{\prime}(f(x)) \cdot f^{\prime}(x)\right] \\
& g^{\prime}(0)=f^{\prime}(f(0)) \cdot\left[-1+f^{\prime}(0) \cdot f^{\prime}(0)\right] \\
& =f^{\prime}(0)[-1+(2)(2)] \\
& =(2)(3)=6
\end{aligned}
\)
Suppose \(f(x)=e^{a x}+e^{b x}\), where \(a \neq b\), and that \(f^{\prime \prime}(x)- 2 f^{\prime}(x)-15 f(x)=0\) for all \(x\). Then the value of \(|a b| / 3\) is
\(
\begin{aligned}
& \text { (c) According to question }\left(a^2-2 a-15\right) e^{a x}+\left(b^2-2 b-15\right) e^{b x} \\
& =0 \\
& \Rightarrow\left(a^2-2 a-15\right)=0 \text { and } b^2-2 b-15=0 \\
& \Rightarrow(a-5)(a+3)=0 \text { and }(b-5)(b+3)=0 \\
& \Rightarrow a=5 \text { or }-3 \text { and } b=5 \text { or }-3 \\
& \therefore a \neq b \text { hence } a=5 \text { and } b=-3 \\
& \text { or } a=-3 \text { and } b=5 \\
& \Rightarrow a b=-15
\end{aligned}
\)
\(
|a b| / 3=5
\)
A non-zero polynomial with real coefficients has the property that \(f(x)=f^{\prime}(x) \cdot f^{\prime \prime}(x)\). If \(a\) is the leading coefficient of \(f(x)\), then the value of \(1 /(2 a)\) is
(d) Let degree of \(f(x)\) is \(n\); degree of \(f^{\prime}(x)=n-1\) degree of \(f^{\prime \prime}(x)\) is \((n-2)\)
Hence \(n=(n-1)+(n-2)=2 n-3\)
\(
n=3
\)
Hence \(f(x)=a x^3+b x^2+c x+d \quad(a \neq 0)\)
\(
\begin{aligned}
& f^{\prime}(x)=3 a x^2+2 b x+c \\
& f^{\prime \prime}(x)=6 a x+2 b \\
& a x^3+b x^2+c x+d=\left(3 a x^2+2 b x+c\right)(6 a x+2 b) \\
& 18 a^2=a \Rightarrow a=\frac{1}{18}
\end{aligned}
\)
\(
1 /(2 a)=9
\)
A function is represented parametrically by the equations \(x=\frac{1+t}{t^3} ; y=\frac{3}{2 t^2}+\frac{2}{t}\), then the value of \(\left|\frac{d y}{d x}-x\left(\frac{d y}{d x}\right)^3\right|\) is
(a)
\(
\begin{aligned}
& \frac{d x}{d t}=-\frac{3}{t^4}-\frac{2}{t^3}=-\left(\frac{3+2 t}{t^4}\right) \\
& \frac{d y}{d t}=-\left(\frac{3}{t^3}+\frac{2}{t^2}\right)=-\left(\frac{3+2 t}{t^3}\right) \\
& \frac{d y}{d x}=t \\
& \frac{d y}{d x}-x\left(\frac{d y}{d x}\right)^3=t-\left(\frac{1+t}{t^3}\right) \cdot t^3=-1
\end{aligned}
\)
Let \(z=(\cos x)^5\) and \(y=\sin x\). Then the value of \(2 \frac{d^2 z}{d y^2}\) at \(x=\frac{2 \pi}{9}\) is
(d)
\(
\begin{aligned}
& z=(\cos x)^5 ; y=\sin x \\
& \frac{d z}{d x}=-5 \cos ^4 x \cdot \sin x ; \frac{d y}{d x}=\cos x \\
& \frac{d z}{d y}=-5 \cos ^3 x \cdot \sin x \\
& \text { Now } \frac{d^2 z}{d y^2}=\frac{d}{d x}\left(\frac{d z}{d y}\right) \cdot \frac{d x}{d y} \\
& =-5 \frac{d}{d x}\left[\cos ^3 x \cdot \sin x\right] \frac{1}{\cos x} \\
& =-5\left[\cos ^4 x-3 \sin ^2 x \cdot \cos ^2 x\right] \frac{1}{\cos x} \\
& =-5\left(\cos ^3 x-3 \sin ^2 x \cdot \cos ^x\right) \\
& =-5\left(\cos ^3 x-3 \cos x\left(1-\cos ^2 x\right)\right) \\
& =-5\left(4 \cos ^3 x-3 \cos x\right) \\
& =-5 \cos 3 x \\
& \left.\frac{d^2 z}{d y^2}\right|_{x=\frac{2 \pi}{9}}=-5 \cos 120^{\circ}=\frac{5}{2}
\end{aligned}
\)
\(
2 \frac{d^2 z}{d y^2}=5
\)
Let \(g(x)=\left\{\begin{array}{cc}\frac{x^2+x \tan x-x \tan 2 x}{a x+\tan x-\tan 3 x} ; & x \neq 0 \\ 0 ; & x=0\end{array}\right.\). If \(g^{\prime}(0)\) exists and is equal to non-zero value \(b\), then \(52 \frac{b}{a}\) is equal to
(d)
\(
\begin{aligned}
& g^{\prime}(0)=b=\lim _{x \rightarrow 0} \frac{x^2+x \tan x-x \tan 2 x}{x(a x+\tan x-\tan 3 x)} \\
& =\lim _{x \rightarrow 0} \frac{x+\tan x-\tan 2 x}{a x+\tan x-\tan 3 x} \\
& x+\left(x+\frac{x^3}{3}+\frac{2}{15} x^5+\cdots \infty\right) \\
& =\lim _{x \rightarrow 0} \frac{-\left(2 x+\frac{8 x^3}{3}+\frac{2}{15} \cdot 32 x^5+\cdots \infty\right)}{a x+\left(x+\frac{x^3}{3}+\frac{2}{15} x^5+\cdots \infty\right)} \\
& \quad-\left(3 x+\frac{27 x^3}{3}+\frac{2}{15} 243 x^5+\cdots \infty\right)
\end{aligned}
\)
\(
\begin{aligned}
&=\lim _{x \rightarrow 0} \frac{x^3\left(-\frac{7}{3}+\frac{-62}{15} x^2+\cdots \infty\right)}{(a+1-3) x+\left(\frac{1}{3}-9\right) x^3+\frac{2}{15}(-242) x^5+\cdots \infty}\\
&b \text { can be finite if } a+1-3=0\\
&\therefore a=2 \text { and } b=\frac{-\frac{7}{3}}{\frac{1}{3}-9}=\left(\frac{-7}{3}\right)\left(\frac{3}{-26}\right)=\frac{7}{26} \Rightarrow 52 \frac{b}{a}=7
\end{aligned}
\)
The reciprocal of the value of \(\lim _{n \rightarrow \infty}\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right) \cdots\left(1-\frac{1}{n^2}\right)\) is
(b)
\(
\begin{aligned}
&\text { We have }\\
&L=\lim _{n \rightarrow \infty} \prod_{n=2}^n \frac{n^2-1}{n^2}
\end{aligned}
\)
\(
\begin{aligned}
& =\lim _{n \rightarrow \infty} \prod_{n=2}^n \frac{n-1}{n} \cdot \prod_{n=2}^n \frac{n+1}{n} \\
& =\lim _{n \rightarrow \infty}\left(\frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdots \frac{n-1}{n}\right)\left(\frac{3}{2} \cdot \frac{4}{3} \cdot \frac{5}{4} \cdots \frac{n+1}{n}\right) \\
& =\lim _{n \rightarrow \infty} \frac{1}{n} \cdot \frac{n+1}{2}=\frac{1}{2}
\end{aligned}
\)
If \(f(x)=\left\{\begin{array}{cc}x^2+2 & x \geq 2 \\ 1-x & x<2\end{array}\right.\) and \(g(x)=\left\{\begin{array}{cc}2 x & x>1 \\ 3-x & x \leq 1\end{array}\right.\), then the value of \(\lim _{x \rightarrow 1} f(g(x))\) is
(c) \(\lim _{x \rightarrow 1^{+}} f(g(x))=f\left(g\left(1^{+}\right)\right)=f\left(\left(2^{+}\right)=2^2+2=6\right.\) and \(\lim _{x \rightarrow 1^{-}} f(g(x))=f\left(g\left(1^{-}\right)\right)=f\left(3-1^{-}\right)=f\left(2^{+}\right)=2^2+2=6\) Hence \(\lim _{x \rightarrow 1} f(g(x))=6\)
If \(\lim _{x \rightarrow 1}\left(1+a x+b x^2\right)^{\frac{c}{x-1}}=e^3\), then the value of \(b c\) is
(a) \(\lim _{x \rightarrow 1}\left(1+a x+b x^2\right)^{\frac{c}{x-1}}=e^3\)
\(
\begin{aligned}
& \Rightarrow e^{\lim _{x \rightarrow 1}\left(1+a x+b x^2-1\right)^{\frac{c}{x-1}}}=e^3 \\
& \Rightarrow e^{\lim _{x \rightarrow 1} \frac{c\left(a x+b x^2\right)}{x-1}}=e^3 \\
& \Rightarrow \lim _{x \rightarrow 1} \frac{c\left(a x+b x^2\right)}{x-1}=3 \\
& \Rightarrow \lim _{h \rightarrow 0} \frac{c\left(a(1+h)+b(1+h)^2\right)}{1+h-1}=3 \\
& \Rightarrow \lim _{h \rightarrow 0} \frac{(c a+b)+(a c+2 b) h+b h^2}{h}=3 \\
& \Rightarrow c a+b=0 \text { and } a c+2 b=3 \\
& \Rightarrow b=3 \text { and } a c=-3
\end{aligned}
\)
Also the form must be \(1^{\infty}\) for which \(a+b=0 \Rightarrow a=-3\) and \(c=1\)
The value of \(\lim _{n \rightarrow \infty}\left[\sqrt[3]{(n+1)^2}-\sqrt[3]{(n-1)^2}\right]\) is
(d)
\(
\begin{aligned}
& \lim _{n \rightarrow \infty}\left[\sqrt[3]{(n+1)^2}-\sqrt[3]{(n-1)^2}\right] \\
= & \lim _{n \rightarrow \infty} n^{2 / 3}\left[\left(1+\frac{1}{n}\right)^{2 / 3}-\left(1-\frac{1}{n}\right)^{2 / 3}\right] \\
= & \lim _{n \rightarrow \infty} n^{2 / 3}\left[\left(1+\frac{2}{3} \cdot \frac{1}{n}+\frac{\frac{2}{3}\left(\frac{2}{2}-1\right)}{2!} \frac{1}{n^2} \cdots\right)\right. \\
& \left.\quad-\left(1-\frac{2}{3} \cdot \frac{1}{n}+\frac{\frac{2}{3}\left(\frac{2}{3}-1\right)}{2!} \frac{1}{n^2} \cdots\right)\right] \\
= & \lim _{n \rightarrow \infty} n^{2 / 3}\left[\frac{4}{3} \cdot \frac{1}{n}+\frac{8}{81} \cdot \frac{1}{n^3}+\cdots\right] \\
= & \lim _{n \rightarrow \infty}\left[\frac{4}{3} \cdot \frac{1}{n^{1 / 3}}+\frac{8}{81} \cdot \frac{1}{n^{7 / 3}}+\cdots\right]=0
\end{aligned}
\)
If \(\lim _{x \rightarrow 0}\left[1+x+\frac{f(x)}{x}\right]^{1 / x}=e^3\), then the value of \(\ln \left(\lim _{x \rightarrow 0}\left[1+\frac{f(x)}{x}\right]^{1 / x}\right)\) is
(b) \(\lim _{x \rightarrow 0}\left[1+x+\frac{f(x)}{x}\right]^{1 / x}=e^3\)
\(
\begin{aligned}
& \Rightarrow \lim _{x \rightarrow 0} e^{\lim _{x \rightarrow 0}\left[1+x+\frac{f(x)}{x}\right] \frac{1}{x}}=e^3 \\
& \Rightarrow \lim _{x \rightarrow 0} e^{\lim _{x \rightarrow 0}\left[1+\frac{f(x)}{x^2}\right]}=e^3 \\
& \Rightarrow \lim _{x \rightarrow 0} \frac{f(x)}{x^2}=2
\end{aligned}
\)
\(
\text { Now } \lim _{x \rightarrow 0}\left[1+\frac{f(x)}{x}\right]^{1 / x}=e^{\lim _{x \rightarrow 0}\left[1+\frac{f(x)}{x}-1\right] \frac{1}{x}}=e^{\lim _{x \rightarrow 0} \frac{f(x)}{x^2}}=e^2
\)
\(\lim _{x \rightarrow \infty} f(x) \text {, where } \frac{2 x-3}{x}<f(x)<\frac{2 x^2+5 x}{x^2} \text {, is }\)
\(
\text { (c) } \begin{aligned}
& \lim _{x \rightarrow \infty} \frac{2 x-3}{x}<\lim _{x \rightarrow \infty} f(x)<\lim _{x \rightarrow \infty} \frac{2 x^2+5 x}{x^2} \\
\Rightarrow & \lim _{x \rightarrow \infty} \frac{2-\frac{3}{x}}{1}<\lim _{x \rightarrow \infty} f(x)<\lim _{x \rightarrow \infty} \frac{2+\frac{5}{x^2}}{1} \\
\Rightarrow & \lim _{x \rightarrow \infty} f(x)=2
\end{aligned}
\)
If \(f(x)=\left\{\begin{array}{cc}x-1, & x \geq 1 \\ 2 x^2-2 & x<1\end{array}, g(x)=\left\{\begin{array}{cc}x+1, & x>0 \\ -x^2+1 & x \leq 0\end{array}\right.\right.\) and \(h(x) =|x|\), then find \(\lim _{x \rightarrow 0} f(g(h(x)))\)
(a)
\(
\begin{aligned}
&\begin{aligned}
& \lim _{x \rightarrow 0^{+}} f(g(h(x)))=f\left(g\left(0^{+}\right)\right)=f\left(1^{+}\right)=0 \\
& \lim _{x \rightarrow 0^{-}} f(g(h(x)))=f\left(g\left(0^{+}\right)\right)=f\left(1^{+}\right)=0
\end{aligned}\\
&\text { Hence } \lim _{x \rightarrow 0} f(g(h(x)))=0
\end{aligned}
\)
If \(\lim _{x \rightarrow \infty} f(x)\) exists and is finite and nonzero and if \(\lim _{x \rightarrow \infty}\left(f(x)+\frac{3 f(x)-1}{f^2(x)}\right)=3\), then the value of \(\lim _{x \rightarrow \infty} f(x)\) is
\(
\begin{aligned}
&\text { (d) } \lim _{x \rightarrow \infty}\left(f(x)+\frac{3 f(x)-1}{f^2(x)}\right)=3\\
&\begin{aligned}
& \Rightarrow\left(\lim _{x \rightarrow \infty} f(x)+\frac{3 \lim _{x \rightarrow \infty} f(x)-1}{\left(\lim _{x \rightarrow \infty} f(x)\right)^2}\right)=3 \\
& \Rightarrow\left(y+\frac{3 y-1}{y^2}\right)=3 \\
& \Rightarrow y^3-3 y^2+3 y-1=0 \\
& \Rightarrow(y-1)^3=0 \\
& \Rightarrow y=1
\end{aligned}
\end{aligned}
\)
If \(L=\lim _{x \rightarrow 0} \frac{e^{-x^2 / 2}-\cos x}{x^3 \sin x}\) then the value of \(1 /(3 L)\) is
(b)
\(
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{e^{-x^2 / 2}-\cos x}{x^3 \sin x} \\
& =\lim _{x \rightarrow 0} \frac{\left(1-\frac{\left(x^2 / 2\right)}{1!}+\frac{\left(x^2 / 2\right)^2}{2!}\right)-\left(1-\frac{x^2}{2!}+\frac{x^4}{4!}\right)}{x^3\left(x-\frac{x^3}{3!}\right)}
\end{aligned}
\)
\(
=\lim _{x \rightarrow 0} \frac{\left(\frac{x^4}{8}\right)-\left(\frac{x^4}{24}\right)}{x^4\left(1-\frac{x^2}{3!}\right)}=\frac{1}{12}
\)
\(
1 /(3 L)=4
\)
If \(L=\lim _{x \rightarrow 2} \frac{(10-x)^{1 / 3}-2}{x-2}\), then the value of \(|1 /(4 L)|\) is
(c)
\(
\begin{aligned}
& \lim _{x \rightarrow 2} \frac{(10-x)^{1 / 3}-2}{x-2} \\
& =\lim _{h \rightarrow 0} \frac{(8-h)^{1 / 3}-2}{h} \quad(\text { Put } x=2+h) \\
& =\lim _{h \rightarrow 0} \frac{2\left(1-\frac{h}{8}\right)^{1 / 3}-2}{h} \\
& =2 \lim _{h \rightarrow 0} \frac{\left(1-\frac{h}{8}\right)^{1 / 3}-1}{h} \\
& =2 \lim _{h \rightarrow 0} \frac{1-\frac{1}{3} \frac{h}{8}-1}{h}=-\frac{1}{12}
\end{aligned}
\)
\(
|1 /(4 L)|=3
\)
The value of \(\lim _{x \rightarrow \infty} \frac{\log _e\left(\log _e x\right)}{e^{\sqrt{x}}}\) is
(d)
\(
\begin{aligned}
& \text { Let } L=\lim _{x \rightarrow \infty} \frac{\log _e\left(\log _e x\right)}{e^{\sqrt{x}}}=\left(\frac{\infty}{\infty} \text { form }\right) \\
& =\lim _{x \rightarrow \infty} \frac{\frac{1}{x \log _e x}}{e^{\sqrt{x}} \frac{1}{2 \sqrt{x}}} \\
& =\lim _{x \rightarrow \infty} \frac{2 \sqrt{x}}{e^{\sqrt{x}} x \log _e x} \\
& =\lim _{x \rightarrow \infty} \frac{2}{e^{\sqrt{x}} \sqrt{x} \log _e x} \\
& =0
\end{aligned}
\)
If \(L=\lim _{n \rightarrow \infty}\left(2 \cdot 3^2 \cdot 2^3 \cdot 3^4 \cdots \cdot 2^{n-1} \cdot 3^n\right)^{\frac{1}{\left(n^2+1\right)}}\), then the value of \(L^4\) is
(d)
\(
\begin{aligned}
&\text { It is obvious } n \text { is even, then }\\
&\begin{aligned}
& \lim _{n \rightarrow \infty}\left(2^{1+3+5+\cdots+n / 2 \text { terms }} \cdot 3^{2+4+6+\cdots+n / 2 \text { terms })^{\frac{1}{\left(n^2+1\right)}}}\right. \\
& =\lim _{n \rightarrow \infty}\left(2^{\frac{n^2}{4}} \cdot 3^{\frac{n(n+2)}{4}}\right)^{\frac{1}{\left(n^2+1\right)}} \\
& =\lim _{n \rightarrow \infty} 2^{\frac{n^2}{4\left(n^2+1\right)}} \cdot 3^{\frac{n(n+2)}{4\left(n^2+1\right)}} \\
& =2^{\lim _{n \rightarrow \infty} \frac{1}{4\left(1+\frac{1}{n^2}\right)}} \cdot 3^{\lim _{n \rightarrow \infty} \frac{\left(1+\frac{2}{n}\right)}{4\left(1+\frac{1}{n^2}\right)}} \\
& =2^{\frac{1}{4}} 3^{\frac{1}{4}}=(6)^{\frac{1}{4}}
\end{aligned}
\end{aligned}
\)
If \(\lim _{x \rightarrow 1} \frac{a \sin (x-1)+b \cos (x-1)+4}{x^2-1}=-2\), then \(|a+b|\) is
\(
\begin{aligned}
&\text { (b) Since RHS is finite quantity }\\
&\begin{aligned}
\therefore & \text { At } x \rightarrow 1, \text { Numerator must be }=0 \\
\therefore & 0+b+4=0 \\
\therefore & b=-4 \\
& \text { Then } \lim _{x \rightarrow 1} \frac{a \sin (x-1)-4 \cos (x-1)+4}{\left(x^2-1\right)}=-2 \\
& \text { Put } x=1+h, \text { Then } \lim _{h \rightarrow 0} \frac{a \sinh +4(1-\cosh )}{h(2+h)}=-2 \\
\Rightarrow & \lim _{h \rightarrow 0} \frac{a\left(\frac{\sinh }{h}\right)+4\left(\frac{1-\cosh }{h}\right)}{2+h}=-2 \\
\Rightarrow & \frac{a(1)+0}{2}=-2 \\
\Rightarrow & a=-4 \\
\Rightarrow & |a+b|=8
\end{aligned}
\end{aligned}
\)
Let \(\lim _{x \rightarrow 1} \frac{x^a-a x+a-1}{(x-1)^2}=f(a)\). Then the value of \(f(4)\) is
\(
\begin{aligned}
&\text { (c) Put } x=1+h\\
&\begin{aligned}
\text { Then } f(a) & =\lim _{h \rightarrow 0} \frac{(1+h)^a-a(1+h)+a-1}{h^2} \\
& =\lim _{h \rightarrow 0} \frac{\left(1+a h+\frac{a(a-1)}{2!} h^2+\cdots\right)-a-a h+a-1}{h^2} \\
\therefore \quad f(a) & =\frac{a(a-1)}{2} \\
\therefore \quad f(4) & =6
\end{aligned}
\end{aligned}
\)
The integer \(n\), for which \(\lim _{x \rightarrow 0} \frac{(\cos x-1)\left(\cos x-e^x\right)}{x^n}\) is a finite non-zero number, is
(b)
\(
\begin{aligned}
L & =\lim _{x \rightarrow 0} \frac{(\cos x-1)\left(\cos x-e^x\right)}{x^n} \\
& =-\lim _{x \rightarrow 0} \frac{(1-\cos x)(1+\cos x)\left(\cos x-e^x\right)}{(1+\cos x) x^n} \\
& =\lim _{x \rightarrow 0} \frac{\left(\frac{\sin x}{x}\right)^2\left(\frac{1-\cos x}{x}+\frac{e^x-1}{x}\right)}{x^{n-3}} \frac{1}{1+\cos x}
\end{aligned}
\)
If \(L\) is finite non-zero, then \(n=3\) (as for \(n=1,2, L=0\) and for \(n=4, L=\infty\) )
If \(\lim _{x \rightarrow 0} \frac{1-\sqrt{\cos 2 x} \cdot \sqrt[3]{\cos 3 x} \cdot \sqrt[4]{\cos 4 x} \cdots \sqrt[n]{\cos n x}}{x^2}\) has the value equal to 10, then the value of \(n\) equals
(a) \(L=\lim _{x \rightarrow 0}=-\lim _{x \rightarrow 0} \frac{D \prod_{r=2}^n(\cos r x)^{1 / r}}{2 x} \text { (Using L’Hospital’s rule) }\)
\(
\begin{aligned}
& \text { let } y=\prod_{r=2}^n(\cos r x)^{1 / r} \\
\Rightarrow & \ln y=\sum_{r=2}^n\left(\frac{1}{r} \ln (\cos r x)\right) \\
\Rightarrow & \frac{1}{y} \frac{d y}{d x}=-\sum_{r=2}^n \tan (r x) \\
\Rightarrow & -D y=y \sum_{r=2}^n \tan (r x)
\end{aligned}
\)
\(
\begin{aligned}
\Rightarrow D & \prod_{r=2}^n(\cos r x)^{1 / r}=-y \sum_{r=2}^n \tan (r x) \\
\Rightarrow L & =\lim _{x \rightarrow 0} \frac{y \cdot \sum_{r=2}^n \tan (r x)}{2 x} \\
& =\frac{1}{2}[2+3+4+\cdots+n] \\
& =\frac{1}{2}\left[\frac{n(n+1)}{2}-1\right] \\
& =\frac{n^2+n-2}{4} \\
\Rightarrow & \frac{n^2+n-2}{4}=10 \\
\Rightarrow & n^2+n-42=0 \\
\Rightarrow & (n+7)(n-6)=0 \\
\Rightarrow & n=6
\end{aligned}
\)
\(f(x)=\frac{3 x^2+a x+a+1}{x^2+x-2}\) and \(\lim _{x \rightarrow-2} f(x)\) exists, then the value of \((a-4)\) is
(c)
\(
\begin{aligned}
&f(x)=\frac{3 x^2+a x+a+1}{(x+2)(x-1)}\\
&\text { as } x \rightarrow-2, D^r \rightarrow 0 \text {, hence as } x \rightarrow-2, N^r \rightarrow 0\\
&12-2 a+a+1=0 \Rightarrow a=13
\end{aligned}
\)
\(
(a-4)=9
\)
If \(L=\lim _{x \rightarrow \infty}\left(x-x^2 \log _e\left(1+\frac{1}{x}\right)\right)\), then the value of \(8 L\) is
(b) \(\text { Let } x=1 / y\)
\(
\begin{array}{r}
\Rightarrow \lim _{x \rightarrow \infty}\left(x-x^2 \log _e\left(1+\frac{1}{x}\right)\right) \\
=\lim _{y \rightarrow 0}\left(\frac{1}{y}-\frac{\log _e(1+y)}{y^2}\right) \\
=\lim _{y \rightarrow 0}\left(\frac{y-\log _e(1+y)}{y^2}\right)
\end{array}
\)
\(
=\lim _{y \rightarrow 0}\left(\frac{y-\left(y-\frac{y^2}{2}\right)}{y^2}\right)=1 / 2
\)
Let \(S_n=1+2+3+\cdots+n\) and \(P_n=\frac{S_2}{S_2-1} \cdot \frac{S_3}{S_3-1} \cdot \frac{S_4}{S_4-1} \cdot \cdots \frac{S_n}{S_n-1}\), where \(n \in N(n \geq 2)\). Then \(\lim _{n \rightarrow \infty} P_n=\)
\(
\text { (c) } S_n=\frac{n(n+1)}{2} \text { and } S_n-1=\frac{(n+2)(n-1)}{2}
\)
\(
\begin{aligned}
& \therefore \frac{S_n}{S_n-1}=\frac{n(n+1)}{2} \cdot \frac{2}{(n+2)(n-1)} \\
& \Rightarrow \frac{S_n}{S_n-1}=\left(\frac{n}{n-1}\right)\left(\frac{n+1}{n+2}\right) \\
& \Rightarrow P_n=\left(\frac{2}{1} \cdot \frac{3}{2} \cdot \frac{4}{3} \cdot \frac{5}{4} \cdots \cdot \frac{n}{n-1}\right)\left(\frac{3}{4} \cdot \frac{4}{5} \cdot \frac{5}{6} \cdots \cdot \frac{n+1}{n+2}\right) \\
& \Rightarrow P_n=\left(\frac{n}{1}\right)\left(\frac{3}{n+2}\right) \\
& \Rightarrow \lim _{n \rightarrow \infty} P_n=3
\end{aligned}
\)
Let \(f^{\prime \prime}(x)\) be continuous at \(x=0\).
If \(\lim _{x \rightarrow 0} \frac{2 f(x)-3 a f(2 x)+b f(8 x)}{\sin ^2 x}\) exists and \(f(0) \neq 0\),
\(f^{\prime}(0) \neq 0\), then the value of \(3 a / b\) is
(a) We have,
\(
L=\lim _{x \rightarrow 0} \frac{2 f(x)-3 a f(2 x)+b f(8 x)}{\sin ^2 x}
\)
For the limit to exist, we have \(2 f(0)-3 a f(0)+b f(0)=0\)
\(
\Rightarrow 3 a-b=2[\because f(0) \neq 0, \text { given }] \dots(1)
\)
\(
\Rightarrow L=\lim _{x \rightarrow 0} \frac{2 f^{\prime}(x)-6 a f^{\prime}(2 x)+8 b f^{\prime}(8 x)}{2 x}
\)
For the limit to exist, we have \(2 f^{\prime}(0)-6 a f^{\prime}(0)+8 b f^{\prime}(0)=0\)
\(
\Rightarrow 3 a-4 b=1\left[\because f^{\prime}(0) \neq 0, \text { given }\right] \dots(2)
\)
Solving equations (1) and (2), we have \(a=7 / 9\) and \(b=1 / 3\).
\(
3 a / b=7
\)
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