Past JEE Main Entrance Paper

Hint

\(
\begin{aligned}
& \text { (b) } \because|z|-\operatorname{Re}(z) \leq 1 \quad(\because z=x+i y) \\
& \Rightarrow \sqrt{x^2+y^2}-x \leq 1 \Rightarrow \sqrt{x^2+y^2} \leq 1+x \\
& \Rightarrow x^2+y^2 \leq 1+x^2+2 x \\
& \Rightarrow y^2 \leq 1+2 x \Rightarrow y^2 \leq 2\left(x+\frac{1}{2}\right)
\end{aligned}
\)

Hint

(a) Given sets \(A=\left\{m \in R\right.\) : both the roots of \(x^2-(m+1) x+m+4=0\) are real \(\}\) and \(B=[-3,5)\) \(\because\) Roots of \(x^2-(m+1) x+m+4=0\) are real, \(m \in R\)
\(\therefore \quad D \geq 0 \Rightarrow(m+1)^2-4(m+4) \geq 0\)
\(\Rightarrow \quad m^2-2 m-15 \geq 0\)
\(\Rightarrow \quad m^2-5 m+3 m-15 \geq 0\)
\(\Rightarrow \quad m(m-5)+3(m-5) \geq 0\)
\(\Rightarrow \quad(m+3)(m-5) \geq 0\)
\(\Rightarrow \quad m \in(-\infty,-3] \cup[5, \infty) \quad[\because A=(-\infty,-3] \cup[5, \infty)]\)
\(\therefore \quad A-B=(-\infty,-3) \cup[5, \infty)\)
\(A \cap B=\{-3\}, B-A=(-3,5)\) and \(A \cup B=R\)
Hence, option (a) is correct.

Hint

(b) \(A=\{x: x \in(-2,2)\}\)
\(
\begin{aligned}
& B=\{x: x \in(-\infty,-1] \cup[5, \infty)\} \\
& A \cap B=\{x: x \in(-2,-1]\} \\
& A \cup B=\{x: x \in(-\infty, 2) \cup[5, \infty)\} \\
& A-B=\{x: x \in(-1,2)\} \\
& B-A=\{x: x \in(-\infty,-2] \cup[5, \infty)\}
\end{aligned}
\)

Hint

(b)
Let \(3^x=y\)
\(
\therefore \quad y(y-1)+2=|y-1|+|y-2|
\)
Case 1: when \(y>2\)
\(
\begin{aligned}
& y^2-y+2=y-1+y-2 \\
& y^2-3 y+5=0
\end{aligned}
\)
\(D<0[\therefore\) Equation not satisfy. \(]\)
Case 2: when \(1 \leq y \leq 2\)
\(
\begin{aligned}
& y^2-y^2+2=y-1-y+2 \\
& y^2-y+1=0 \\
&  \quad D<0[\therefore \text { Equation not satisfy. }]
\end{aligned}
\)
Case 3: when \(y \leq 1\)
\(
\begin{aligned}
& y^2-y+2=-y+1-y+2 \\
& y^2+y-1=0 \\
& \therefore y=\frac{-1+\sqrt{5}}{2} \\
& =\frac{-1-\sqrt{5}}{2}[\therefore \text { Equation not Satisfy }]
\end{aligned}
\)
\(\therefore\) Only one \(-1+\frac{\sqrt{5}}{2}\) satisfy equation

Hint

(d) Given inequality is,
\(
\begin{aligned}
& 2 \sqrt{\sin ^2 x-2 \sin x+5} \leq 2^{2 \sin ^2 y} \\
& \Rightarrow \sqrt{\sin ^2 x-2 \sin x+5} \leq 2 \sin ^2 y \\
& \Rightarrow \sqrt{(\sin x-1)^2+4} \leq 2 \sin ^2 y
\end{aligned}
\)
We know that for \(\sin ^2 y \leqslant 1\)
max value is \(\left|\sin ^2 y\right|=1\)
Now max value for \(\sqrt{(\sin x-1)^2+4}=2\) is
\(
\begin{aligned}
& (\sin x-1)^2+4=4 \\
& \sin x-1=0 \\
& \sin x=1 \\
\therefore & \sin x=|\sin y|
\end{aligned}
\)
\(\Rightarrow\) all the pairs that satisfy the inequality along with the equation \(
\sin x=|\sin y|\)

Hint

\(
\begin{aligned}
& (1+2 m) x^2-2(1+3 m) x+4(1+m)>0 \\
& \therefore D<0 \\
& 4(1+3 m)^2-16(1+2 m)(1+m)<0 \\
& \Rightarrow 1+9 m^2+6 m-4-12 m-8 m^2<0 \\
& \Rightarrow m^2-6 m-3<0 \\
& \Rightarrow(m-3)^2<12 \\
& \Rightarrow-2 \sqrt{3}<m-3<2 \sqrt{3} \\
& \Rightarrow 3-2 \sqrt{3}<m<3+2 \sqrt{3}
\end{aligned}
\)
Also
\(
\begin{aligned}
& 2 m+1>0 \\
& \Rightarrow m>-\frac{1}{2}
\end{aligned}
\)
Possible integral values of \(m\) are \(0,1,2,3,4,5,6\)
Hence, number of integral values of \(m\) is 7.

Hint

(b) \(f(x)=\left(\frac{3}{5}\right)^x+\left(\frac{4}{5}\right)^x-1\)
Put \(f(x)=0\)
\(
\begin{gathered}
\Rightarrow \quad 0=\left(\frac{3}{5}\right)^x+\left(\frac{4}{5}\right)^x-1 \\
\Rightarrow \quad\left(\frac{3}{5}\right)^x+\left(\frac{4}{5}\right)^x=1 \\
\Rightarrow 3^x+4^x=5^x \dots(i)
\end{gathered}
\)
For \(x=1\)
\(
3^1+4^1>5^1
\)
For \(x=3\)
\(
3^3+4^3=91<5^3
\)
Only for \(x=2\), equation (i) Satisfy
So, only one solution \((x=2)\)

Hint

(a) Let \(a=\sqrt{x^2+x}\) and \(b=\frac{\tan ^2 \alpha}{\sqrt{x^2+x}}\)
\(
\begin{array}{ll}
\quad & \mathrm{AM} \geq \mathrm{GM}, \quad \therefore \quad \frac{a+b}{2} \geq \sqrt{a b} \\
\Rightarrow & a+b \geq 2 \sqrt{a b} \\
\Rightarrow & \sqrt{x^2+x}+\frac{\tan ^2 \alpha}{\sqrt{x^2+x}} \geq 2 \sqrt{\tan ^2 \alpha}=2 \tan \alpha
\end{array}
\)
\(
[\because \alpha \in(0, \pi / 2)]
\)

Hint

(b) For \(x<-2,|x+2|=-(x+2)\)
\(
\begin{array}{ll}
\therefore & x^2-|x+2|+x>0 \\
\Rightarrow & x^2+x+2+x>0 \quad \Rightarrow(x+1)^2+1>0,
\end{array}
\)
which is valid \(\forall x \in R\)
But \(x<-2, \quad \therefore x \in(-\infty,-2) \dots(1)\)
For \(x \geq 2,|x+2|=x+2\)
\(
\begin{aligned}
& \therefore x^2-|x+2|+x>0 \Rightarrow x^2-x-2+x>0 \\
& \Rightarrow x^2>2 \Rightarrow x>\sqrt{2} \text { or } x<-\sqrt{2}
\end{aligned}
\)
i.e., \(x \in(-\infty,-\sqrt{2)} \cup(\sqrt{2, \infty})\)
But \(x \geq-2 \Rightarrow x \in[-2,-\sqrt{2}) \cup(\sqrt{2, \infty}) \dots(ii)\)
From (i) and (ii),
\(
\begin{aligned}
& x \in(-\infty,-2) \cup[-2,-\sqrt{2}) \cup(\sqrt{2}, \infty) \\
\Rightarrow \quad & x \in(-\infty,-\sqrt{2}) \cup(\sqrt{2}, \infty)
\end{aligned}
\)

Hint

(a) Since for positive real numbers,
\(
\begin{aligned}
& \text { A.M. } \geq \text { G.M. } \\
& \therefore \quad \frac{\left(a_1+a_2+\ldots+a_{n-1}+2 a_n\right)}{n} \geq\left(a_1 a_2 \ldots a_{n-1} 2 a_n\right)^{1 / n} \\
& \Rightarrow \quad a_1+a_2+a_3+\ldots .+a_{n-1}+2 a_n \geq n(2 c)^{1 / n}
\end{aligned}
\)
Hence, required minimum value \(=n(2 c)^{1 / n}\)

Hint

(a) Since, for positive real numbers,
A.M. \(\geq\) G.M.
\(
\begin{aligned}
& \quad \frac{(a+b)+(c+d)}{2} \geq \sqrt{(a+b)(c+d)} \Rightarrow M \leq 1 \\
& \text { Also }(\mathrm{a}+\mathrm{b})(\mathrm{c}+\mathrm{d})>0 \\
& \therefore \quad 0 \leq M \leq 1
\end{aligned}
\)

Hint

(b) For real roots \(q^2-4 p r \geq 0\)
\(
\begin{aligned}
& \Rightarrow\left(\frac{p+r}{2}\right)^2-4 p r \geq 0 \quad(\because p, q, r \text { are in A.P. }) \\
& \Rightarrow \quad p^2+r^2-14 p r \geq 0 \Rightarrow \frac{p^2}{r^2}-14 \frac{p}{r}+1 \geq 0 \\
& \Rightarrow \quad\left(\frac{p}{r}-7\right)^2-48 \geq 0 \Rightarrow\left|\frac{p}{r}-7\right| \geq 4 \sqrt{3}
\end{aligned}
\)

Hint

(a) Given: \(y=5 x^2+2 x+3=5\left(x+\frac{1}{5}\right)^2+\frac{14}{5}>2, \forall x \in R\) While \(y=2 \sin x \leq 2, \forall x \in R\)
\(\therefore\) The two curves do not meet at all.

Alternate:

We have
\(
\begin{aligned}
& y=5 x^2+2 x+3=5\left[x^2+\frac{2}{5} x+\frac{3}{5}\right] \\
& =5\left[\left(x+\frac{1}{5}\right)^2+\frac{3}{5}-\frac{1}{25}\right] \\
& =5\left[\left(x+\frac{1}{5}\right)^2+\frac{14}{25}\right]=5\left(x+\frac{1}{5}\right)^2+\frac{14}{25}>2
\end{aligned}
\)
Since \(y=2 \sin x \leq 2[latex]
[latex]\therefore\) we cannot have any point of intersection.
Hence number of points of intersection is \({ }^{\prime} 0^{\prime}\)

Hint

(a) Given : \(\sin \left(e^x\right)=5^x+5^{-x}\)
We know \(5^x\) and \(5^{-x}\) both are +ve real numbers.
By using \(\mathrm{AM} \geq \mathrm{GM}, 5^x+\frac{1}{5^x} \geq 2 \Rightarrow 5^x+5^{-x} \geq 2\)
\(\therefore \quad\) R.H.S. of given equation \(\geq 2\)
While \(\sin e^x \in[-1,1]\) i.e. LHS \(\in[-1,1]\)
\(\therefore\) The equation is not possible for any real value of \(x\).

Hint

(a) First of all for \(\log (x-1)\) to be defined, \(x-1>0\)
\(
\Rightarrow x>1
\)
Now, \(\log _{0.3}(x-1)<\log _{0.09}(x-1)\)
\(
\begin{aligned}
& \Rightarrow \log _{0.3}(x-1)<\log _{(0.3)}(x-1) \\
& \Rightarrow \log _{0.3}(x-1)<\frac{1}{2} \log _{0.3}(x-1) \\
& \Rightarrow 2 \log _{0.3}(x-1)<\log _{0.3}(x-1) \\
& \Rightarrow \log _{0.3}(x-1)^2<\log _{0.3}(x-1) \\
& \Rightarrow(x-1)^2>(x-1)
\end{aligned}
\)
[Here inequality is reversed because base lies between 0 and 1]
\(
\Rightarrow(x-1)^2-(x-1)>0 \Rightarrow(x-1)(x-2)>0
\)
On combining (i) and (ii), we get \(x>2\)
\(
\therefore \quad x \in(2, \infty)
\)

Hint

(c) Given : \(a^2+b^2+c^2=1 \ldots .(\mathrm{i})\)
We know \((a+b+c)^2 \geq 0\)
\(
\begin{aligned}
& \Rightarrow \quad a^2+b^2+c^2+2 a b+2 b c+2 c a \geq 0 \\
& \Rightarrow \quad 2(a b+b c+c a) \geq-1 \quad \text { [using (i)] } \\
& \Rightarrow \quad a b+b c+c a \geq-1 / 2 \dots(ii)
\end{aligned}
\)
Also we know that
\(
\begin{aligned}
& \frac{1}{2}\left[(a-b)^2+(b-c)^2+(c-a)^2\right] \geq 0 \\
\Rightarrow & a^2+b^2+c^2-a b-b c-c a \geq 0 \\
\Rightarrow & a b+b c+c a \leq 1 \quad \text { [using (i) }]
\end{aligned}
\)
On combining (ii) and (iii), we get
\(
-1 / 2 \leq a b+b c+c a \leq 1, \therefore a b+b c+c a \in[-1 / 2,1]
\)

Hint

\(
\text { (d) Given expression } x^{12}-x^9+x^4-x+1=f(x) \text { (let) }
\)
For \(\mathrm{x}<0\), put \(x=-y\) where \(y>0\) then we get
\(
f(x)=y^{12}+y^9+y^4+y+1>0 \text { for } y>0
\)
For \(0<x<1, \quad x^9<x^4 \Rightarrow-x^9+x^4>0\)
Also \(1-x>0\) and \(x^{12}>0\)
\(
\Rightarrow x^{12}-x^9+x^4+1-x>0 \quad \Rightarrow f(x)>0
\)
For \(x>1 ; f(x)=x\left(x^3-1\right)\left(x^8+1\right)+1>0\)
So, \(f(x)>0\) for \(-\infty<x<\infty\).

Hint

(b) If \(p=5, q=3, r=2 ; \max (p, q)=5 ; \max (p, q, r)=5\) \(\Rightarrow \max (p, q)=\max (p, q, r)\)
\(\therefore \quad\) (a) is not true. Similarly we can show that (c) is not true.
Also \(\min (p, q)=\frac{1}{2}(p+q-|p-q|)\)
Let \(p<q\), then \(\mathrm{LHS}=p\)
and R.H.S. \(=\frac{1}{2}(p+q-q+p)=p\)
Similarly, we can prove that \((b)\) is true for \(q<p\) too.

Hint

(c) Let the distance of school from \(A=x\)
\(\therefore \quad\) The distance of the school form \(B=60-x\) Total distance covered by 200 students
\(
=2[150 x+50(60-x)]=2[100 x+3000] \text {, }
\)
Which is minimum, when \(x=0\)
\(\therefore \quad\) School should be built at town \(A\).

Hint

(a) \(|x|^2-3|x|+2=0\)
Case I : \(x<0\), then \(|x|=-x\)
\(\Rightarrow x^2+3 x+2=0 \Rightarrow(x+1)(x+2)=0\)
\(x=-1,-2(\) both acceptable as \(x<0)\)
Case II : \(x>0\), then \(|x|=x\)
\(\Rightarrow x^2-3 x+2=0 \Rightarrow(x-1)(x-2)=0\)
\(x=1,2\) ( both acceptable as \(x>0\) )
Hence, there are 4 real solutions.

Hint

\(
\text { (b) Let } y=2 \log _{10} x-\log _x 0.01
\)
\(
\begin{aligned}
& =2 \log _{10} x-\frac{\log _{10} 0.01}{\log _{10} x}=2 \log _{10} x+\frac{2}{\log _{10} x} \\
& =2\left[\log _{10} x+\frac{2}{\log _{10} x}\right]
\end{aligned}
\)
Here \(x>1 \Rightarrow \log _{10} x>0\)
Also the sum of a real positive number and its reciprocal is always greater than or equal to 2.
\(\therefore \quad y \geq 2 \times 2 \Rightarrow y \geq 4, \therefore\) Least value of \(y\) is 4.

Hint

(c) Since, \(a, b, c>0\); therefore \(a, b, c\) should be real because order relation is not defined in the set of complex numbers.
\(\therefore \quad\) Roots of equation are either real or complex conjugate.
Let \(\alpha, \beta\) be the roots of \(a x^2+b x+c=0\), then
\(
\alpha+\beta=-\frac{b}{a}=-v e, \quad \alpha \beta=\frac{c}{a}=+v e
\)
\(\Rightarrow\) Either both \(\alpha, \beta\) are – ve, if roots are real or both \(\alpha, \beta\) have \(–\) ve real parts, if roots are complex conjugate.

Hint

\(
\begin{aligned}
& \text { (a) } u=x^2+4 y^2+9 z^2-6 y x-3 z x-2 x y \\
& \quad=\frac{1}{2}\left[2 x^2+8 y^2+18 z^2-12 y z-6 z x-4 x y\right] \\
& =\frac{1}{2}\left[\left(x^2-4 x y+4 y^2\right)+\left(4 y^2+9 z^2-12 y z\right)\right. \\
& \left.+\left(x^2+9 z^2-6 z x\right)\right] \\
& =\frac{1}{2}\left[(x-2 y)^2+(2 y-3 z)^2+(3 z-x)^2\right] \geq 0
\end{aligned}
\)
Hence, \(\)u\(\) is always non-negative.

Hint

(d) The given equations are
\(
x+2 y+2 z=1 \dots(i)
\)
and \(2 x+4 y+4 z=9 \dots(ii)\)
On multiplying (i), by 2 and then subtracting from (ii), we get \(0=7\), which is not possible
\(\therefore \quad\) No solution.

Hint

(2) Given : \(x^4-4 x^3+12 x^2+x-1=0\)
\(
\Rightarrow x^4-4 x^3+6 x^2-4 x+1+6 x^2+5 x-2=0
\)
\(
\begin{aligned}
& \Rightarrow(x-1)^4+6 x^2+5 x-2=0 \\
& \Rightarrow(x-1)^4=-6 x^2-5 x+2
\end{aligned}
\)
The solution of the above polynomial is the intersection points of the curves \(y=(x-1)^4\) and
\(
y=-6 x^2-5 x+2 \text { or } y=(x-1)^4 \text { and }\left(x+\frac{5}{12}\right)^2=-\frac{1}{6}\left(y-\frac{73}{24}\right)
\)
Clearly, the two curves have two points of intersection. Hence the given polynomial has two real roots.

Hint

(8) \(\because a>0, \therefore a^{-5}, a^{-4}, 3 a^{-3}, 1, a^8, a^{10}>0\)
Using AM \(\geq\) GM for positive real numbers, we get
\(
\begin{aligned}
& \frac{\frac{1}{a^5}+\frac{1}{a^4}+\frac{1}{a^3}+\frac{1}{a^3}+\frac{1}{a^3}+1+a^8+a^{10}}{8} \geq \\
& \left(\frac{1}{a^5} \cdot \frac{1}{a^4} \cdot \frac{1}{a^3} \cdot \frac{1}{a^3} \cdot \frac{1}{a^3} \cdot 1 \cdot a^8 \cdot a^{10}\right)^{\frac{1}{8}} \\
& \Rightarrow \frac{1}{a^5}+\frac{1}{a^4}+\frac{3}{a^3}+1+a^8+a^{10} \geq 8(1)^{\frac{1}{8}}
\end{aligned}
\)

Hint

(7) Given system of equations:
\(
\begin{aligned}
& 3 x-y-z=0 \\
& -3 x+z=0
\end{aligned}
\)
and \(-3 x+2 y+z=0\)
Let \(x=p\), where \(p\) is an integer, then \(y=0\) and \(z=3 p\)
But \(x^2+y^2+z^2 \leq 100 \Rightarrow p^2+9 p^2 \leq 100\)
\(
\Rightarrow p^2 \leq 10 \Rightarrow p=0, \pm 1, \pm 2 \pm 3
\)
i.e. \(p\) can take 7 different values.
\(\therefore \quad\) Number of points \((x, y, z)\) are 7.

Hint

\(|x-2|^2+|x-2|-2=0\)
Case 1: \(x \geq 2\)
\(
\begin{aligned}
& \Rightarrow \quad(x-2)^2+(x-2)-2=0 \\
& \Rightarrow \quad x^2-3 x=0 \Rightarrow x(x-3)=0 \\
& \Rightarrow \quad x=0,3
\end{aligned}
\)
But 0 is rejected as \(x \geq 2\)
\(
\therefore \quad x=3
\)
Case 2: \(x<2\)
\(
\begin{aligned}
& \{-(x-2)\}^2-(x-2)-2=0 \\
& \Rightarrow x^2+4-4 x-x=0 \Rightarrow(x-1)(x-4)=0 \\
& \Rightarrow x=1,4
\end{aligned}
\)
But 4 is rejected as \(x<2\)
\(
\therefore \quad x=1
\)
\(\therefore\) the sum of the roots is \(3+1=4\).

Hint

The number of solution of \(x_1+x_2+\ldots+x_k=n\) \(=\) Coefficient of \(t^n\) in \(\left(t+t^2+t^3+\ldots\right)\left(t^2+t^3+\ldots\right)\left(t^k+t^{k+1}+\ldots\right)\) \(=\) Coefficient of \(t^n\) in \(t^{1+2+\ldots+k}\left(1+t+t^2+\ldots\right)^k\)
Now, \(1+2+\ldots+k=\frac{k(k+1)}{2}=p\) [say]
and \(1+t+t^2+\ldots=\frac{1}{1-t}\)
Thus, the number of required solutions
\(=\) Coefficient of \(t^{n-p}\) in \((1-t)^{-k}\)
\(=\) Coefficient of \(t^{n-p}\) in \(\left[1+{ }^k C_1 t+{ }^{k+1} C_2 t^2+{ }^{k+2} C_3 t^3+\ldots\right]\)
\(={ }^{k+n-p-1} C_{n-p}={ }^r C_{n-p}\)
where, \(r=k+n-p-1=k+n-1-\frac{1}{2} k(k+1)\)
\(
=\frac{1}{2}\left(2 k+2 n-2+k^2-k\right)=\frac{1}{2}\left(2 n-k^2+k-2\right)
\)

Hint

Given \(x<0, \quad y<0\),
\(
x+y+\frac{x}{y}=\frac{1}{2} \quad \text { and } \quad(x+y) \cdot \frac{x}{y}=-\frac{1}{2}
\)
Let \(x+y=a\) and \(\frac{x}{y}=b\)
\(
\therefore \quad a+b=\frac{1}{2} \text { and } a b=-\frac{1}{2}
\)
On solving the above two equations, we get \(a+\left(-\frac{1}{2 a}\right)=\frac{1}{2}\)
\(
\Rightarrow 2 a^2-a-1=0 \Rightarrow a=1,-1 / 2 \Rightarrow b=-1 / 2,1
\)
Hence, from equation (i), \(x+y=1\) and \(\frac{x}{y}=-\frac{1}{2}\)
or \(x+y=\frac{-1}{2}\) and \(\frac{x}{y}=1\)
But \(x, y<0\)
\(
\therefore x+y<0 \Rightarrow x+y=\frac{-1}{2} \text { and } \frac{x}{y}=1
\)
On solving the above equations, we get \(x=-1 / 4\) and \(y=-1 / 4\).

Hint

\(
\begin{aligned}
& \log _7 \log _5(\sqrt{x+5}+\sqrt{x})=0 \\
& \Rightarrow \quad \log _5(\sqrt{x+5}+\sqrt{x})=1 \\
& \Rightarrow \quad \sqrt{x+5}+\sqrt{x}=5 \Rightarrow x+5=25+x-10 \sqrt{x} \\
& \Rightarrow \quad 2=\sqrt{x} \Rightarrow x=4, \text { which satisfies the given equation. }
\end{aligned}
\)

Hint

\(\mathrm{x} \& \mathrm{y}\) are positive real numbers and \(\mathrm{m} \& \mathrm{n}\) are positive integers Since for two +ve numbers A.M. \(\geq\) G.M.
\(
\begin{aligned}
& \therefore \quad \frac{1+x^{2 n}}{2} \geq\left(1 \times x^{2 n}\right)^{1 / 2} \text { and } \frac{1+y^{2 m}}{2} \geq\left(1 \times y^{2 m}\right)^{1 / 2} \\
& \Rightarrow\left(\frac{1+x^{2 n}}{2}\right) \geq x^n \dots(i)
\end{aligned}
\)
and \(\left(\frac{1+y^{2 m}}{2}\right) \geq y^m\) \dots(ii)
Multiplying (i) and (ii), we get
\(
\frac{\left(1+x^{2 n}\right)\left(1+y^{2 m}\right)}{4} \geq x^n y^m \Rightarrow \frac{1}{4} \geq \frac{x^n y^m}{\left(1+x^{2 n}\right)\left(1+y^{2 m}\right)}
\)
Hence, the statement is false.

Hint

Consider \(N=n_1+n_2+n_3+\ldots+n_p\), where \(N\) is an even number.
Let \(k\) numbers among these \(p\) numbers be odd, then \((p-k)\) numbers are even numbers. Now sum of \((p-k)\) even numbers is even and for \(N\) to be an even number, sum of \(k\) odd numbers must be even which is possible only when \(k\) is even.
\(\therefore \quad\) The given statement is false.

Hint

Consider \(n\) numbers, namely \(1,2,3,4, \ldots, n\). Since, A.M. > G.M.
\(
\begin{aligned}
& \therefore \quad \frac{1+2+3+4 \ldots+n}{n}>(1.2 .3 .4 \ldots n)^{1 / n} \\
& \Rightarrow \frac{n(n+1)}{2 n}>(n !)^{1 / n} \Rightarrow(n !)^{1 / n}<\frac{n+1}{2}, \text { True }
\end{aligned}
\)

Hint

\(
\begin{aligned}
& (\mathbf{a}, \mathbf{b}, \mathbf{c}) \\
& 3^{\mathrm{x}}=4^{\mathrm{x}-1} \Rightarrow \mathrm{x} \log 3=2(\mathrm{x}-1) \log 2 \\
& \quad \Rightarrow \mathrm{x}=\frac{2 \log 2}{2 \log 2-\log 3} \Rightarrow \mathrm{x}=\frac{2 \log _3 2}{2 \log _3 2-1}=\frac{2}{2-\log _2 3}
\end{aligned}
\)

\(
\text { Also } x=\frac{1}{1-\frac{1}{2} \log _2 3}=\frac{1}{1-\log _4 3}
\)

Hint

We can write 240 as \(2^4 .3^1 .5^1\)
But we want divisors of the form \(4 \mathrm{n}+2\)
That is, we want even divisors of the form \(2(2 n+1)\) and \(n \geqslant 0\)
The divisor is 2 ( when \(\mathrm{n}=0\) ) or divisors are odd multiples of 2
So the divisors are \(2,6,10\) and 30
Hence the answer is 4

Hint

(d) Let \(x_1, x_2, \ldots, x_n\) be the \(n\) +ve numbers
According to the question,
\(
x_1 x_2 x_3 \ldots x_n=1 \dots(i)
\)
We know that for positive numbers A.M. \(\geq\) G.M.
\(
\begin{array}{ll}
\therefore & \frac{x_1+x_2+\ldots+x_n}{n} \geq \sqrt[n]{x_1 x_2 \ldots x_n} \\
\Rightarrow & \frac{x_1+x_2+\ldots+x_n}{n} \geq 1 \\
& \text { [using eq. (i)] } \\
\Rightarrow x_1+x_2+\ldots+x_n \geq n
\end{array}
\)

Hint

(a, b, c) Given equation: \(\frac{3}{x^4}\left(\log _2 x\right)^2+\log _2 x-\frac{5}{4}=\sqrt{2}\) For \(x>0\), taking log on both sides to the base \(x\), we get
\(
\frac{3}{4}\left(\log _2 x\right)^2+\left(\log _2 x\right)-\frac{5}{4}=\log _x \sqrt{2}=\frac{1}{2} \log _x 2
\)
Let \(\log _2 x=y\), then we get, \(\frac{3}{4} y^2+y-\frac{5}{4}=\frac{1}{2 y}\)
\(
\begin{aligned}
& \Rightarrow \quad 3 y^3+4 y^2-5 y-2=0 \\
& \Rightarrow \quad(y-1)(y+2)(3 y+1)=0 \Rightarrow y=1,-2,-1 / 3 \\
& \Rightarrow \quad \log _2 x=1,-2,-1 / 3 \Rightarrow x=2,2^{-2}, 2^{-1 / 3} \\
& \Rightarrow \quad x=2, \frac{1}{4}, \frac{1}{2^{1 / 3}}, \text { all are possible because they are positive }
\end{aligned}
\)

Hint

Since, as AM \(>\) GM
\(
\begin{aligned}
& \Rightarrow \frac{(b+c-a)+(c+a-b)}{2}>((b+c-a)(c+a-b))^{1 / 2} \\
& \Rightarrow \quad \quad c>((b+c-a)(c+a-b))^{1 / 2} \dots(i)
\end{aligned}
\)
Similarly, \(b>((a+b-c)(b+c-a))^{1 / 2} \dots(ii)\)
and \(\quad a>((a+b-c)(c+a-b))^{1 / 2} \dots(iii)\)
On multiplying Eqs. (i), (ii) and (iii), we get \(a b c>(a+b-c)(b+c-a)(c+a-b)\)
Hence, \(\quad(a+b-c)(b+c-a)(c+a-b)-a b c<0\)

Hint

(a, d) Wavy curve method:
Let \(f(x)=\left(x-\alpha_1\right)\left(x-\alpha_2\right) \ldots\left(x-\alpha_n\right)\)
To find sign of \(f(x)\), plot \(\alpha_1, \alpha_2, \ldots \alpha_n\) on number line in ascending order of magnitude. Starting from right extreme interval i.e., interval \(\left(\frac{1}{2}, \infty\right)\) , put \(+\) ve, \(–\) ve signs alternately in the intervals on the number line. \(f\) \((x)\) is positive in the intervals having + ve sign and negative in the intervals having -ve sign.
Using the wavy curve method for the function,
\(
f(x)=\frac{2 x-1}{2 x^3+3 x^2+x}=\frac{2 x-1}{x(2 x+1)(x+1)}
\)
We get

Clearly \(f(x)>0\), when \(x \in(-\infty,-1) \cup(-1 / 2,0) \cup(1 / 2, \infty)\)
Hence, \(S\) contains \((-\infty,-3 / 2)\) and \((1 / 2,3)\)

Hint

Given : \(\frac{2 x}{2 x^2+5 x+2}>\frac{1}{x+1}\)
\(
\begin{aligned}
& \Rightarrow \frac{2 x}{2 x^2+5 x+2}-\frac{1}{x+1}>0 \\
& \Rightarrow \frac{2 x^2+2 x-2 x^2-5 x-2}{\left(2 x^2+5 x+2\right)(x+1)}>0 \\
& \Rightarrow \frac{-3 x-2}{(2 x+1)(x+1)(x+2)}>0 \Rightarrow \frac{(3 x+2)}{(x+1)(x+2)(2 x+1)}<0 \\
& \Rightarrow \frac{(3 x+2)(x+1)(x+2)(2 x+1)}{(x+1)^2(x+2)^2(2 x+1)^2}<0 \\
& \Rightarrow(3 x+2)(x+1)(x+2)(2 x+1)<0
\end{aligned}
\)
Using wavey curve method, we get

\(
\text { Clearly Inequality (i) holds for, } x \in(-2,-1) \cup(-2 / 3,-1 / 2)
\)

Hint

Let \(\alpha, \beta\) be the roots of eq. \(a x^2+b x+c=0\)
According to the question, \(\beta=\alpha^n\)
\(
\begin{gathered}
\text { Also, } \alpha+\beta=-b / a ; \alpha \beta=c / a \\
\alpha \beta=\frac{c}{a} \Rightarrow \alpha \cdot \alpha^n=\frac{c}{a} \\
\Rightarrow \quad a^{n+1}=\frac{c}{a} \Rightarrow \alpha=\left(\frac{c}{a}\right)^{\frac{1}{n+1}}
\end{gathered}
\)
then
\(
\begin{aligned}
& \alpha+\beta=-b / a \Rightarrow \alpha+\alpha^n=\frac{-b}{a} \\
& \Rightarrow\left(\frac{c}{a}\right)^{\frac{1}{n+1}}+\left(\frac{c}{a}\right)^{\frac{n}{n+1}}=\frac{-b}{a} \\
& \Rightarrow \quad a \cdot\left(\frac{c}{a}\right)^{\frac{1}{n+1}}+a \cdot\left(\frac{c}{a}\right)^{\frac{n}{n+1}}+b=0 \\
& \Rightarrow \quad \frac{n}{n+1} c^{\frac{1}{n+1}}+a^{\frac{1}{n+1}} c^{\frac{n}{n+1}}+b=0 \\
& \Rightarrow \quad\left(a^n c\right)^{\frac{1}{n+1}}+\left(a c^n\right)^{\frac{1}{n+1}}+b=0
\end{aligned}
\)

Hint

\(
\text { For any square there can be at most } 4 \text {, neighbouring squares. }
\)

Let us consider a square \(d\) as shown in the figure. It has four neighbouring square \(p, q, r, s\) as shown.
According to the question,
\(
\begin{array}{ll}
& p+q+r+s=4 d \\
\Rightarrow \quad & (d-p)+(d-q)+(d-r)+(d-s)=0
\end{array}
\)
Sum of four tve numbers can be zero only if these are zero individually
\(
\begin{aligned}
& \therefore \quad d-p=0=d-q=d-r=d-s \\
& \Rightarrow \quad p=q=r=s=d
\end{aligned}
\)
Hence, all the numbers written are same.

Hint

\(e^{\sin x}-e^{-\sin x}-4=0\)
Let \(e^{\sin x}=y\) then \(e^{-\sin x}=1 / y\)
\(\therefore \quad\) Given equation becomes, \(y-\frac{1}{y}-4=0\)
\(
\Rightarrow y^2-4 y-1=0 \Rightarrow y=2+\sqrt{5}, 2-\sqrt{5}
\)
But \(\mathrm{y}\) is real \(+\mathrm{ve}\) number,
\(
\begin{aligned}
& \therefore \quad y \neq 2-\sqrt{5} \Rightarrow y=2+\sqrt{5} \\
& \Rightarrow \quad e^{\sin x}=2+\sqrt{5} \Rightarrow \sin x=\log _e(2+\sqrt{5})
\end{aligned}
\)
But \(2+\sqrt{5}>e \Rightarrow \log _e\left(2+\sqrt{5)}>\log _e e\right.\)
\(
\Rightarrow \log _e(2+\sqrt{5})>1
\)
Hence, \(\sin x>1\), which is not possible.
\(\therefore \quad\) Given equation has no real solution.

Hint

The given system is
\(
\begin{aligned}
& x+2 y+z=1 \dots(i)\\
& 2 x-3 y-\omega=2 \dots(ii)
\end{aligned}
\)
where \(x, y, z\), then \(\omega \geq 0\)
Multiplying eqn. (i) by 2 and subtracting from (ii), we get \(7 y+2 z+\omega=0 \Rightarrow \omega=-(7 y+2 z)\)
Now if \(y, z>0\), then \(\omega<0\) (not possible)
If \(y=0, z=0\), then \(x=1\) and \(\omega=0\).
\(\therefore \quad\) The only solution is \(x=1, y=0, z=0, \omega=0\).

Hint

The given equations are \(3 x+m y-m=0\) and \(2 x-5 y-20=0\)
By cross-multiplication method, we get
\(
\begin{aligned}
& \frac{x}{-20 m-5 m}=\frac{y}{-2 m+60}=\frac{1}{-15-2 m} \\
& \Rightarrow \quad x=\frac{25 m}{2 m+15}, y=\frac{2 m-60}{2 m+15}
\end{aligned}
\)
If \(x>0\), then \(\frac{25 m}{2 m+15}>0\)
\(
\Rightarrow m<-\frac{15}{2} \text { or } m>0 \dots(i)
\)
If \(y>0\), then \(\frac{2(m-30)}{2 m+15}>0\)
\(
\Rightarrow \mathrm{m}<-\frac{15}{2} \text { or } \mathrm{m}>30 \dots(ii)
\)
On combining (i) and (ii), we get the common values of \(\mathrm{m}\) as follows :
\(
\mathrm{m}<-\frac{15}{2} \text { or } \mathrm{m}>30 \quad \therefore \quad \mathrm{m} \in\left(-\infty, \frac{-15}{2}\right) \cup(30, \infty)
\)

Hint

\(y=\sqrt{\frac{(x+1)(x-3)}{(x-2)}}\)
\(\mathrm{y}\) will take all real values if \(\frac{(x+1)(x-3)}{(x-2)} \geq 0\)
By wavy curve method

\(
x \in[-1,2) \cup[3, \infty)
\)
[2 is not included as it makes the denominator zero, and hence \(\)y\(\) an undefined number.]

Hint

Given : \(n^4<10^n\) for a fixed \(+\) ve integer \(n \geq 2\).
To prove : \((n+1)^4<10^{n+1}\)
Proof : Since \(n^4<10^n \Rightarrow 10 n^4<10^{n+1} \dots(i)\)
So it is sufficient to prove that \((n+1)^4<10 n^4\)
Now \(\left(\frac{n+1}{n}\right)^4=\left(1+\frac{1}{n}\right)^4 \leq\left(1+\frac{1}{2}\right)^4[\because n \geq 2]\)
\(
\begin{aligned}
& =\frac{81}{16}<10 \\
\Rightarrow \quad(n+1)^4 & <10 n^4 \dots(ii)
\end{aligned}
\)
From (i) and (ii), \((n+1)^4<10^{n+1}\)

Hint

There are two parts of this question
\((5 x-1)<(x+1)^2\) and \((x+1)^2<(7 x-3)\)
Taking first part
\(
\begin{aligned}
& (5 x-1)<(x+1)^2 \Rightarrow 5 x-1<x^2+2 x+1 \\
& \Rightarrow x^2-3 x+2>0 \Rightarrow(x-1)(x-2)>0
\end{aligned}
\)
Using wavy curve method

\(
\Rightarrow x<1 \text { or } x>2 \dots(i)
\)

Taking second part
\(
\begin{aligned}
& (x+1)^2<(7 x-3) \Rightarrow x^2-5 x+4<0 \\
& \Rightarrow \quad(x-1)(x-4)<0
\end{aligned}
\)
Using wavy curve method

\(
\Rightarrow \quad 1<x<4 \dots(ii)
\)
On combining (i) and (ii) [taking common solution], we get \(2<\mathrm{x}<4\), but \(x\) is an integer therefore \(x=3\).

Hint

\(x^2+y^2-2 x \geq 0 \Rightarrow x^2-2 x+1+y^2 \geq 1\)
\(\Rightarrow(x-1)^2+y^2 \geq 1\), which represents the boundary and exterior region of the circle with centre at \((1,0)\) and radius as 1 .
For \(3 x-y \leq 12\), the corresponding equation is \(3 x-y=12\). Two points on it can be taken as \((4,0),(2,-6)\). Also putting \((0,0)\) in given inequation, we get \(0 \leq 12\) which is true.
\(\therefore\) given inequation represents that half plane region on one side of the line \(3 x-y=12\), which contains origin.
For \(y \leq x\), the corresponding equation is \(y=x\). Two points on it can be taken as \((0,0)\) and \((1,1)\). Also putting \((2,1)\) in the given inequation, we get \(1 \leq 2\), which is true. So \(y \leq x\) represents that half plane on one side of the line \(y=x\), which contains the points \((2,1) \cdot y \geq 0\) represents upper half cartesian plane.
Combining all we find the solution set as the shaded region in the graph.

Hint

\(
\begin{aligned}
& \text { Let } x=\frac{\sqrt{26-15 \sqrt{3}}}{5 \sqrt{2}-\sqrt{38+5 \sqrt{3}}} \\
& \Rightarrow \quad x^2=\frac{26-15 \sqrt{3}}{50+38+5 \sqrt{3}-10 \sqrt{76+10 \sqrt{3}}} \\
& \Rightarrow x^2=\frac{26-15 \sqrt{3}}{88+5 \sqrt{3}-10 \sqrt{75+1+10 \sqrt{3}}} \\
& \Rightarrow x^2=\frac{26-15 \sqrt{3}}{88+5 \sqrt{3}-10 \sqrt{\left(5 \sqrt{3)^2}+(1)^2+2 \times 5 \sqrt{3} \times 1\right.}} \\
& \Rightarrow \quad x^2=\frac{26-15 \sqrt{3}}{88+5 \sqrt{3}-10 \sqrt{(5 \sqrt{3}+1)^2}} \\
& =\frac{26-15 \sqrt{3}}{3(26-15 \sqrt{3})}=\frac{1}{3} \text {, which is a rational number. } \\
&
\end{aligned}
\)

Hint

\(
\mathrm{RHS}=(\mathrm{m}-1, \mathrm{n}+1)+\mathrm{x}^{\mathrm{m}-\mathrm{n}-1}(m-1, n)
\)
\(
\begin{aligned}
& =\frac{\left(1-x^{m-1}\right)\left(1-x^{m-2}\right) \ldots\left(1-x^{m-n-1}\right)}{(1-x)\left(1-x^2\right) \ldots\left(1-x^{n+1}\right)} \\
& +x^{m-n-1}\left[\frac{\left(1-x^{m-1}\right)\left(1-x^{m-2}\right) \ldots\left(1-x^{m-n}\right)}{(1-x)\left(1-x^2\right) \ldots\left(1-x^n\right)}\right] \\
& =\frac{\left(1-x^{m-1}\right)\left(1-x^{m-2}\right) \ldots\left(1-x^{m-n}\right)}{(1-x)\left(1-x^2\right) \ldots\left(1-x^n\right)} \\
& {\left[\frac{1-x^{m-n-1}}{1-x^{n+1}}+x^{m-n-1}\right]} \\
& {\left[\frac{1-x^{m-n-1}+x^{m-n-1}-x^m}{1-x^{n+1}}\right]} \\
& =\frac{\left(1-x^m\right)\left(1-x^{m-1}\right) \ldots\left(1-x^{m-n}\right)}{(1-x)\left(1-x^2\right) \ldots\left(1-x^n\right)\left(1-x^{n+1}\right)} \\
& =(m, n+1)=\text { L.H.S. } \\
&
\end{aligned}
\)

Hint

\(
\begin{aligned}
& 4^x-3^{x-1 / 2}=3^{x+1 / 2}-\frac{\left(2^2\right)^x}{2} \\
& \Rightarrow 4^x-\frac{3^x}{\sqrt{3}}=3^x \sqrt{3}-\frac{4^x}{2} \\
& \Rightarrow \frac{3}{2} \cdot 4^x=3^x\left(\sqrt{3}+\frac{1}{\sqrt{3}}\right) \Rightarrow \frac{3}{2} \cdot 4^x=3^x \frac{4}{\sqrt{3}} \\
& \Rightarrow \frac{4^{x-1}}{4^{1 / 2}}=\frac{3^{x-1}}{\sqrt{3}} \Rightarrow 4^{x-3 / 2}=3^{x-3 / 2} \\
& \Rightarrow\left(\frac{4}{3}\right)^{x-3 / 2}=1 \Rightarrow x-\frac{3}{2}=0 \Rightarrow x=3 / 2
\end{aligned}
\)

Hint

\(
\begin{aligned}
& {[\sqrt{1}]+[\sqrt{2}]+[\sqrt{3}]+\ldots+[120]} \\
& E=1+1+1+2+2+2+2+2+3+3+3+3+3 \\
& +3+3+4+4+\ldots \\
& E=3 \times 1+5 \times 2+7 \times 3+\ldots+19 \times 9+10 \times 21 \\
& =\sum_{r=1}^{10}(2 r+1) r=2\left[\frac{10 \times 11 \times 21}{6}\right]+\frac{10 \times 11}{2} \\
& =770+55 \\
& =825
\end{aligned}
\)

Hint

Firstly, we know that the given function
\(f(x)=e^{8 x}-e^{6 x}-3 e^{4 x}-e^{2 x}+1\) intersects the \(x\)-axis where \(f(x)=0\). Setting \(f(x)\) equal to zero gives us:
\(
e^{8 x}-e^{6 x}-3 e^{4 x}-e^{2 x}+1=0 \text {. }
\)
Let \(t=e^{2 x}\). The equation now becomes:
\(
t^4-t^3-3 t^2-t+1=0 \text {. }
\)
Dividing by \(t^2\) and rearranging the equation gives :
\(
t^2-t-3-\frac{1}{t}+\frac{1}{t^2}=0
\)
If we let \(y=t+\frac{1}{t}\), we get :
\(
y^2-y-5=0
\)
This quadratic equation in y can be solved using the quadratic formula to give two roots :
\(
y=\frac{1 \pm \sqrt{21}}{2} .
\)
Since \(y=t+\frac{1}{t}\), and \(t>0\) (as \(t=e^{2 x}\) ), we must choose the root where \(y>2\). Thus, we take \(y=\frac{1+\sqrt{21}}{2}\).
So, we have :
\(
t+\frac{1}{t}=\frac{1+\sqrt{21}}{2}
\)
Solving for \(t\) gives us :
\(
t^2-y t+1=0
\)
or \(t=\frac{y \pm \sqrt{y^2-4}}{2} .\)
Substituting \(y=\frac{1+\sqrt{21}}{2}\) into the formula gives :
\(
t=\frac{\frac{1+\sqrt{21}}{2} \pm \sqrt{\left(\frac{1+\sqrt{21}}{2}\right)^2-4}}{2} .
\)
This quadratic formula for \(t\) yields two possible values \((t=2.37\) and \(t=0.42)\). Both of them are positive, thus both are acceptable values for \(t=e^{2 x}\).
Finally, to get \(x\), take the natural \(\log\) of both sides and divide by 2 :
\(
x=\frac{1}{2} \ln (t)
\)
This gives us two solutions for \(x\), corresponding to the two positive solutions for \(t\). So, the number of points where the curve \(f(x)=e^{8 x}-e^{6 x}-3 e^{4 x}-e^{2 x}+1\) intersects the \(x\)-axis is 2 .

Hint

\(
x^2-7 x-1=0<_b^a
\)
By newton’s theorem
\(
\begin{aligned}
& S _{ n +2}-7 S _{ n +1}- S _{ n }=0 \\
& S _{21}-7 S _{20}- S _{19}=0 \\
& S _{20}-7 S _{19}- S _{18}=0 \\
& S _{19}-7 S _{18}- S _{17}=0 \\
& \frac{ S _{21}+ S _{17}}{ S _{19}}=\frac{ S _{21}+\left( S _{19}-7 S _{18}\right)}{ S _{19}} \\
& =\frac{ S _{21}+ S _{19}-7\left( S _{20}-7 S _{19}\right)}{ S _{19}} \\
& =\frac{50 S _{19}+\left( S _{21}-7 S _{20}\right)}{ S _{19}} \\
& =51 \cdot \frac{ S _{19}}{ S _{19}}=51
\end{aligned}
\)

Hint

\(
\begin{aligned}
& x^2-5 \alpha x+1=0 \dots(1)\\
& x^2-\alpha x-5=0 \dots(2)
\end{aligned}
\)
have a common root.
Subtracting (1) with (2) we’ll get \(x=\frac{6}{4 \alpha}\)
Substituting in (1)
\(
\begin{aligned}
& \frac{36}{16 \alpha^2}-\frac{30}{4}+1=0 \\
& \Rightarrow \alpha^2=\frac{9}{26} \\
& \alpha=\frac{3}{\sqrt{2 \times 13}} \\
& \therefore \beta=13
\end{aligned}
\)

Hint

\(
\begin{aligned}
& x^7+3 x^5-13 x^3-15 x=0 \\
& x\left(x^6+3 x^4-13 x^2-15\right)=0 \\
& x=0=\alpha_7 \\
& \text { Let } x^2=t \\
& t^3+3 t^2-13 t-15=0 \\
& (t+1)(t+5)(t-3)=0 \\
& t=x^2=-1,-5,3 \\
& x= \pm i, \pm \sqrt{5} i, \pm \sqrt{3} \\
& \alpha_1, \alpha_2= \pm \sqrt{5} i, \alpha_3, \alpha_4= \pm \sqrt{3}, \alpha_5, \alpha_6= \pm i \\
& \alpha_1 \alpha_2-\alpha_3 \alpha_4+\alpha_5 \alpha_6=5+3+1=9
\end{aligned}
\)

Hint

\(
\begin{aligned}
& x^2+60^{\frac{1}{4}} x+a=0 \\
& \therefore \alpha+\beta=-60^{\frac{1}{4}}, \alpha \beta=a \\
& \text { Now } \alpha^4+\beta^4=-30 \\
& \Rightarrow\left(\alpha^2+\beta^2\right)^2-2 a^2=-30 \\
& \Rightarrow\left[(\alpha+\beta)^2-2 a\right]^2-2 a^2=-30 \\
& \Rightarrow\left(60^{\frac{1}{2}}-2 a\right)^2-2 a^2=-30 \\
& \Rightarrow 60+4 a^2-4 \cdot 60^{\frac{1}{2}} a-2 a^2+30=0 \\
& \Rightarrow 2 a^2-8 \sqrt{15} a+90=0
\end{aligned}
\)
Product of value of \(a=45\)

Hint

\(
\begin{aligned}
& D \geq 0 \Rightarrow 4-4|\lambda-3| \geq 0 \\
& |\lambda-3| \leq 1 \\
& -1 \leq \lambda-3 \leq 1 \\
& 2 \leq \lambda \leq 4 \\
& |x|=\frac{2 \pm \sqrt{4-4|\lambda-3|}}{2} \\
& =1 \pm \sqrt{1-|\lambda-3|} \\
& x_{\text {largest }} \quad=1+1=2 \text {, when } \lambda=3
\end{aligned}
\)
Largest element of \(S=2+3=5\)

Hint

Case-I:
When \(x<-2\) then
\(
\begin{aligned}
& -x^2+5(x+2)+6=0 \\
& \Rightarrow x^2-5 x-16=0
\end{aligned}
\)
\(
\Rightarrow x=\frac{5 \pm \sqrt{25+64}}{2}
\)
\(\therefore x=\frac{5-\sqrt{89}}{2}\) is accepted

Case-II :
When \(-2 \leq x<0\) then
\(
\begin{aligned}
& -x^2-5(x+2)+6=0 \\
\Rightarrow & x^2+5 x+4=0 \\
\Rightarrow & (x+1)(x+4)=0
\end{aligned}
\)
\(x=-1\) is accepted

Case-III :
When \(x \geq 0\) then
\(
\begin{gathered}
x^2-5(x+2)+6=0 \\
\Rightarrow x^2-5 x-4=0 \\
x=\frac{5 \pm \sqrt{25+16}}{2} \\
=\frac{5 \pm \sqrt{41}}{2} \\
x=\frac{5-\sqrt{41}}{2} \text { is accepted }
\end{gathered}
\)
\(\therefore 3\) real roots are possible.

Hint

Find the roots of the quadratic equation:
The quadratic formula is \(x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\). For the quadratic equation \(x^2-\sqrt{2} x+2=0\), we have \(a=1, b=-\sqrt{2}, c=2\). Plugging these into the quadratic formula gives:
\(
x=\frac{\sqrt{2} \pm \sqrt{(-\sqrt{2})^2-4(1)(2)}}{2(1)}=\frac{\sqrt{2} \pm \sqrt{2-8}}{2}=\frac{\sqrt{2} \pm \sqrt{6} i}{2} .
\)
So, we have two roots, \(\alpha\) and \(\beta\), which are:
So, we have two roots, \(\alpha\) and \(\beta\), which are:
\(
\begin{aligned}
& \alpha=\frac{\sqrt{2}+\sqrt{6} i}{2}, \\
& \beta=\frac{\sqrt{2}-\sqrt{6} i}{2} .
\end{aligned}
\)
Express the roots in exponential form:
We can express complex numbers in the form \(r e^{i \theta}\). For \(\alpha\) and \(\beta\), we find the magnitude \(r=\sqrt{2}\) and the arguments \(\theta=\frac{\pi}{3},-\frac{\pi}{3}\) respectively. So, we have:
\(
\begin{aligned}
& \alpha=\sqrt{2} e^{i \frac{\pi}{3}} \\
& \beta=\sqrt{2} e^{-i \frac{\pi}{3}}
\end{aligned}
\)
Calculate the 14 th power of the roots:
To find \(\alpha^{14}\) and \(\beta^{14}\), we use the property of exponents which says that \(\left(a^m\right)^n=a^{m n}\). So, we have:
\(
\begin{aligned}
& \alpha^{14}=\left(\sqrt{2} e^{i \frac{\pi}{3}}\right)^{14}=2^7 e^{i \frac{14 \pi}{3}}=128 e^{i \frac{2 \pi}{3}}, \\
& \beta^{14}=\left(\sqrt{2} e^{-i \frac{\pi}{3}}\right)^{14}=2^7 e^{-i \frac{14 \pi}{3}}=128 e^{-i \frac{2 \pi}{3}} .
\end{aligned}
\)
Add the 14th powers of the roots:
We want to find the real part of \(\alpha^{14}+\beta^{14}\). To do this, we use the property that \(e^{i x}=\cos (x)+i \sin (x)\). We have:
\(
\alpha^{14}+\beta^{14}=128 e^{i \frac{2 \pi}{3}}+128 e^{-i \frac{2 \pi}{3}}=128(2) \cos \left(\frac{2 \pi}{3}\right)=-128
\)

Hint

\(
x|x-1|+|x+2|+a=0
\)

Case I: If \(x<-2\) then
\(
\begin{aligned}
& -x^2+x-x-2+a=0 \\
& a=x^2+2
\end{aligned}
\)
\(y=x^2+2\) is decreasing \(\forall x \in(-\infty,-2)\)
Case II: If \(-2 \leq x<1\) then
\(
\begin{aligned}
& -x^2+x+x+2+a=0 \\
& a=x^2-2 x-2
\end{aligned}
\)
\(y=x^2-2 x-2\) is decreasing \(\forall x \in[-2,1)\).
Case III: If \(x \geq 1\) then
\(
\begin{aligned}
& x^2-x+x+2+a=0 \\
& a=-\left(x^2+2\right)
\end{aligned}
\)
\(y=-\left(x^2+2\right)\) is decreasing \(\forall x \in[1, \infty)\)
\(\therefore\) Exactly one real root \(\forall x \in R\)

Hint

Given quadratic equation: \(x^2+\sqrt{6} x+3=0\)
We can find the roots using the quadratic formula: \(x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)
Here, \(a=1, b=\sqrt{6}[latex], and [latex]c=3\).
Substituting these values into the quadratic formula, we have:
\(
x=\frac{-\sqrt{6} \pm \sqrt{(\sqrt{6})^2-4(1)(3)}}{2(1)}
\)
Simplifying further :
\(
\begin{aligned}
& x=\frac{-\sqrt{6} \pm \sqrt{6-12}}{2} \\
& x=\frac{-\sqrt{6} \pm \sqrt{-6}}{2}
\end{aligned}
\)
Since the discriminant is negative, the roots are complex numbers. We can express them using the imaginary unit \(i\) :
\(
\begin{aligned}
& x=\frac{-\sqrt{6} \pm \sqrt{6} i}{2} \\
& \Rightarrow x=\frac{-1}{2} \sqrt{6} \pm \frac{1}{2} \sqrt{6} i \\
& \therefore \alpha, \beta=\sqrt{3} e ^{ \pm \frac{3 \pi i}{4}} .
\end{aligned}
\)
The required expression can be rewritten in terms of the argument of the exponential form of the roots, which simplifies the calculation:
\(
=\frac{(\sqrt{3})^{23}\left(2 \cos \frac{69 \pi}{4}\right)+(\sqrt{3})^{14}\left(2 \cos \frac{42 \pi}{4}\right)}{(\sqrt{3})^{15}\left(2 \cos \frac{45 \pi}{4}\right)+(\sqrt{3})^{10}\left(2 \cos \frac{30 \pi}{4}\right)}
\)
Here, the exponential power of \(\sqrt{3}\) in the numerator is larger by 8 compared to the denominator, so we can divide the numerator and denominator by \((\sqrt{3})^8=81\) to simplify:
\(
=\frac{(\sqrt{3})^{15}\left(2 \cos \frac{69 \pi}{4}\right)+(\sqrt{3})^6\left(2 \cos \frac{42 \pi}{4}\right)}{(\sqrt{3})^7\left(2 \cos \frac{45 \pi}{4}\right)+(\sqrt{3})^2\left(2 \cos \frac{30 \pi}{4}\right)}
\)
Since the cosine function has a period of \(2 \pi\), we can reduce the arguments of the cosine function in the numerator and denominator.
We have \(69 \pi / 4=\pi / 4+17 \pi=\pi / 4\),
\(
\begin{aligned}
& 42 \pi / 4=2 \pi / 4+10 \pi=\pi / 2, \\
& 45 \pi / 4=\pi / 4+11 \pi=\pi / 4, \text { and } \\
& 30 \pi / 4=2 \pi / 4+7 \pi=\pi / 2 .
\end{aligned}
\)
Therefore, the required expression simplifies to :
\(
\begin{aligned}
& =\frac{(\sqrt{3})^{15}\left(2 \cos \frac{\pi}{4}\right)+(\sqrt{3})^6\left(2 \cos \frac{\pi}{2}\right)}{(\sqrt{3})^7\left(2 \cos \frac{\pi}{4}\right)+(\sqrt{3})^2\left(2 \cos \frac{\pi}{2}\right)} \\
& =\frac{(\sqrt{3})^{15} \sqrt{2}+(\sqrt{3})^6 \cdot 0}{(\sqrt{3})^7 \sqrt{2}+(\sqrt{3})^2 \cdot 0} \\
& =\frac{(\sqrt{3})^{15} \sqrt{2}}{(\sqrt{3})^7 \sqrt{2}} \\
& =(\sqrt{3})^{15-7} \\
& =(\sqrt{3})^8 \\
& =81
\end{aligned}
\)

Hint

Concept:
For a cubic equation, \(a x^3+b x^2+c x+d=0\)
Sum of roots \(=\frac{-b}{a}\)
Product of roots taken two at a time \(=\frac{c}{a}\)
Product of roots \(=\frac{-d}{a}\)
Given cubic equation is :
\(
x^3+b x+c=0
\)
\(\because \alpha, \beta, \gamma\) are the roots of above equation.
And \(\beta \gamma=1=-\alpha\)
So, product of roots \(=-c\)
\(
\begin{aligned}
& \Rightarrow \alpha \beta \gamma=-c \\
& \Rightarrow(-1)(1)=-c \\
& \Rightarrow c=1
\end{aligned}
\)
Since, \(\alpha=-1\) is the root. So,
\(
\begin{aligned}
& \Rightarrow-1-b+c=0 \\
& \Rightarrow c-b=1 \\
& \Rightarrow 1-b=1 \Rightarrow b=0
\end{aligned}
\)
The given equation becomes \(x^3+1=0\)
So, roots are \(-1,-\omega,-\omega^2\)
\(
\begin{aligned}
& \therefore b^3+2 c^3-3 \alpha^3-6 \beta^3-8 \gamma^3 \\
& =0+2-3(-1)^3-6(-\omega)^3-8\left(-\omega^2\right)^3 \\
& =2+3+6 \omega^3+8 \omega^6 \\
& =5+6+8=19
\end{aligned}
\)

Hint

We have,
\(
\begin{aligned}
& A=\{x \in R:[x+3]+[x+4] \leq 3\} \\
& \text { Here, }[x+3]+[x+4] \leq 3 \\
& \Rightarrow[x]+3+[x]+4 \leq 3 \\
& (\because[x+n]=[x]+n, n \in I) \\
& \Rightarrow 2[x]+4 \leq 0 \Rightarrow[x] \leq-2 \\
& \Rightarrow x \in(-\infty,-1)
\end{aligned}
\)
\(
A \equiv(-\infty,-1) \dots(i)
\)
Also,
\(
B=\left\{x \in R: 3^x\left(\sum_{r=1}^{\infty} \frac{3}{10^x}\right)^{x-3}<3^{-3 x}\right\}
\)
Here,
\(
\begin{aligned}
& 3^x\left(\frac{3}{10}+\frac{3}{10^2}+\frac{3}{10^3}+\ldots\right)^{x-3}<3^{-3 x} \\
\Rightarrow & 3^x\left(\frac{\frac{3}{10}}{1-\frac{1}{10}}\right)^{x-3}<3^{-3 x} \\
\Rightarrow & 3^x\left(\frac{1}{3}\right)^{x-3}<3^{-3 x} \\
\Rightarrow & 3^{x-x+3}<3^{-3 x} \Rightarrow 3^3<3^{-3 x}
\end{aligned}
\)
\(
\begin{aligned}
& \Rightarrow \quad-3 x>3 \Rightarrow x<-1 \\
& \Rightarrow \quad x \in(-\infty,-1) \\
& \Rightarrow \quad B \equiv(-\infty,-1) \ldots \ldots . . .(i i)
\end{aligned}
\)
From equations (i) and (ii), we get
\(
A=B
\)

Hint

We have, \(\left|x^2-8 x+15\right|-2 x+7=0\)
\(
\Rightarrow|(x-3)(x-5)|-2 x+7=0
\)
Now, when \(x \leq 3\) or \(x \geq 5\), then
\(
\begin{gathered}
x^2-8 x+15-2 x+7=0 \\
\Rightarrow x^2-10 x+22=0 \\
\Rightarrow x^2-10 x+25-3=0 \\
\Rightarrow(x-5)^2-3=0 \\
\Rightarrow(x-5)= \pm \sqrt{3} \\
\Rightarrow x=5+\sqrt{3}, 5-\sqrt{3}
\end{gathered}
\)
Since, \(x \leq 3\) or \(x \geq 5\)
\(
\therefore x=5+\sqrt{3}
\)
When, \(3<x<5\), then
\(
\begin{aligned}
& -\left(x^2-8 x+15\right)-2 x+7=0 \\
& \Rightarrow-x^2+8 x-15-2 x+7=0 \\
& \Rightarrow-x^2+6 x-8=0 \\
& \Rightarrow x^2-6 x+8=0 \\
& \Rightarrow(x-4)(x-2)=0 \Rightarrow x=2,4
\end{aligned}
\)
Since, \(3<x<5\)
\(
\therefore x=4
\)
\(\therefore\) Sum of roots \(=(5+\sqrt{3})+4=9+\sqrt{3}\)

Hint

\(
2 x^2-8 x+k=0
\)
\(
\begin{aligned}
& f(1)>0 \Rightarrow k>6 \\
& f(2)<0 \Rightarrow k<8 \\
& f(3)>0 \Rightarrow k>6 \\
& k \in(6,8)
\end{aligned}
\)
Only 1 integral value of \(k\) is 7 .

Hint

Let \((\sqrt{3}+\sqrt{2})^{x^2-4}=t\)
\(
\begin{aligned}
& t+\frac{1}{t}=10 \\
\Rightarrow & t^2-10 t+1=0 \\
\Rightarrow & t=\frac{10 \pm \sqrt{100-4}}{2}=5 \pm 2 \sqrt{6}
\end{aligned}
\)

Case-I:
\(
\begin{gathered}
t=5+2 \sqrt{6}=(\sqrt{3}+\sqrt{2})^2 \\
\Rightarrow(\sqrt{3}+\sqrt{2})^{x^2-4}=(\sqrt{3}+\sqrt{2})^2 \\
\Rightarrow x^2-4=2 \Rightarrow x^2=6 \Rightarrow x= \pm \sqrt{6}
\end{gathered}
\)

Case-II :
\(
\begin{aligned}
& t=5-2 \sqrt{6}=(\sqrt{3}-\sqrt{2})^2 \\
& (\sqrt{3}+\sqrt{2})^{x^2-4}=(\sqrt{3}-\sqrt{2})^2 \\
& \Rightarrow\left((\sqrt{3}-\sqrt{2})^{-1}\right)^{x^2-4}=(\sqrt{3}-\sqrt{2})^2 \\
& \Rightarrow 4-x^2=2 \\
& \Rightarrow x^2=2 \\
& \Rightarrow x= \pm \sqrt{2}
\end{aligned}
\)

Hint

\(
e^{4 x}+8 e^{3 x}+13 e^{2 x}-8 e^x+1=0
\)
Let \(e ^{ x }= t\)
Now, \(t ^4+8 t ^3+13 t ^2-8 t +1=0\)
Dividing equation by \(t ^2\)
\(
\begin{aligned}
& t^2+8 t+13-\frac{8}{t}+\frac{1}{t^2}=0 \\
& t^2+\frac{1}{t^2}+8\left(t-\frac{1}{t}\right)+13=0 \\
& \left(t-\frac{1}{t}\right)^2+2+8\left(t-\frac{1}{t}\right)+13=0
\end{aligned}
\)
Let \(t -\frac{1}{ t }= z\)
\(
\begin{aligned}
& z^2+8 z+15=0 \\
& (z+3)(z+5)=0 \\
& z=-3 \text { or } z=-5
\end{aligned}
\)
So, \(t-\frac{1}{t}=-3\) or \(t-\frac{1}{t}=-5\)
\(t^2+3 t-1=0\) or \(t^2+5 t-1=0\)
\(t =\frac{-3 \pm \sqrt{13}}{2}\) or \(t =\frac{-5 \pm \sqrt{29}}{2}\)
as \(t=e^x\) so \(t\) must be positive,
\(t=\frac{\sqrt{13}-3}{2}\) or \(\frac{\sqrt{29}-5}{2}\)
So, \(x=\ln \left(\frac{\sqrt{13}-3}{2}\right)\) or \(x=\ln \left(\frac{\sqrt{29}-5}{2}\right)\)
Hence two solutions and both are negative.

Hint

\(
\begin{aligned}
& x^2-4 x+[x]+3=x[x] \\
& \Rightarrow x^2-4 x+[x]+3-x[x]=0 \\
& \Rightarrow(x-1)(x-3)-[x](x-1)=0 \\
& \Rightarrow(x-1)(x-[x]-3)=0 \\
& \therefore x=1
\end{aligned}
\)
Or
\(
\begin{aligned}
& x-[x]-3=0 \\
& \Rightarrow\{x\}-3=0[\text { As }\{x\}=x-[x]] \\
& \Rightarrow\{x\}=3
\end{aligned}
\)
But we know, \(0<\{x\}<1\)
\(
\therefore\{x\} \neq 3
\)
\(\therefore x\) has only one solution in \((-\infty, \infty)\) which is \(x=1\).

Hint

\(
\begin{aligned}
& \frac{1}{20}\left(\frac{1}{20-a}-\frac{1}{40-a}+\frac{1}{40-a}-\frac{1}{60-a}+\ldots .+\frac{1}{180-a}-\frac{1}{200-a}\right)=\frac{1}{256} \\
& \Rightarrow \frac{1}{20}\left(\frac{1}{20-a}-\frac{1}{200-a}\right)=\frac{1}{256} \\
& \Rightarrow \frac{1}{20}\left(\frac{180}{(20-a)(200-a)}\right)=\frac{1}{256} \\
& \Rightarrow(20-a)(200-a)=9.256
\end{aligned}
\)

OR \(a^2-220 a+1696=0\)
\(
\Rightarrow a=212,8
\)

Hint

\(
\begin{aligned}
& \left|x^2\right|-7|x|+9 \leq 0 \\
& \Rightarrow|x| \in\left[\frac{7-\sqrt{13}}{2}, \frac{7+\sqrt{13}}{2}\right]
\end{aligned}
\)
As \(x \in Z\)
So, \(x\) can be \(\pm 2, \pm 3, \pm 4, \pm 5\)
Out of these values of \(x\),
\(
x=3,-4,-5
\)
satisfy \(S\) as well
\(
n(S \cap T)=3
\)

Hint

\(
\begin{aligned}
& \alpha+\beta=\sqrt{2}, \alpha \beta=\sqrt{6} \\
& \frac{1}{\alpha^2}+1+\frac{1}{\beta^2}+1=2+\frac{\alpha^2+\beta^2}{6} \\
& =2+\frac{2-2 \sqrt{6}}{6}=-a \\
& \left(\frac{1}{\alpha^2}+1\right)\left(\frac{1}{\beta^2}+1\right)=1+\frac{1}{\alpha^2}+\frac{1}{\beta^2}+\frac{1}{\alpha^2 \beta^2} \\
& =\frac{7}{6}+\frac{2-2 \sqrt{6}}{6}=b \\
& \Rightarrow a+b=\frac{-5}{6}
\end{aligned}
\)
So, equation is \(x^2+\frac{17 x}{6}+\frac{7}{6}=0\)
Or \(6 x^2+17 x+7=0\)
Both roots of equation are – ve and distinct

Hint

\(
\begin{aligned}
& 3^{\sqrt{\log _3 5}}-5^{\sqrt{\log _5 3}}=3^{\sqrt{\log _3 5}}-\left(3^{\log _3 5}\right)^{\sqrt{\log _5 3}} \\
& 3^{\left(\log _3 5\right)^{\frac{1}{3}}}-5^{\left(\log _5 3\right)^{\frac{2}{3}}}=5^{\left(\log _5 3\right)^{\frac{2}{3}}}-5^{\left(\log _5 3\right)^{\frac{2}{3}}}=0
\end{aligned}
\)
Note: In the given equation ‘ \(x\) ‘ is missing.

So \(x^2-5 x+3(-1)=0 <_\beta^\alpha\)
\(
\begin{aligned}
& \alpha+\beta+\frac{1}{\alpha}+\frac{1}{\beta}=(\alpha+\beta)+\frac{\alpha+\beta}{\alpha \beta} \\
& =5-\frac{5}{3}=\frac{10}{3} \\
& \left(\alpha+\frac{1}{\beta}\right)\left(\beta+\frac{1}{\alpha}\right)=2+\alpha \beta+\frac{1}{\alpha \beta}=2-3-\frac{1}{3}=\frac{-4}{3}
\end{aligned}
\)

Hint

\(
x^2+(3-a) x+1=2 a <_\beta^\alpha
\)
\(
\begin{aligned}
& \alpha+\beta=a-3, \alpha \beta=1-2 a \\
& \Rightarrow \alpha^2+\beta^2=(a-3)^2-2(1-2 a) \\
& =a^2-6 a+9-2+4 a \\
& =a^2-2 a+7 \\
& =(a-1)^2+6
\end{aligned}
\)
So, \(\alpha^2+\beta^2 \geq 6\)

Hint

When, \(x^5=1\)
then \(x^5-1=0\)
\(
\Rightarrow(x-1)\left(x^4+x^3+x^2+x+1\right)=0
\)
Given, \(x^4+x^3+x^2+x+1=0\) has roots \(\alpha, \beta, \gamma\) and 8 .
\(\therefore\) Roots of \(x^5-1=0\) are \(1, \alpha, \beta, \gamma\) and 8 .
We know, Sum of \(p^{\text {th }}\) power of \(n^{\text {th }}\) roots of unity \(=0\). (If \(p\) is not multiple of \(n\) ) or \(n\) (If \(p\) is multiple of \(n\) )
\(\therefore\) Here, Sum of \(p ^{\text {th }}\) power of \(n ^{\text {th }}\) roots of unity
Here, \(p=2021\), which is not multiple of 5 .
\(
\begin{aligned}
& \therefore 1^{2021}+\alpha^{2021}+\beta^{2021}+\gamma^{2021}+8^{2021}=0 \\
& \Rightarrow \alpha^{2021}+\beta^{2021}+\gamma^{2021}+8^{2021}=-1
\end{aligned}
\)

Hint

Given,
\(
\frac{(x+2)\left(x^2+3 x+5\right)}{-2+3 x-x^2} \geq 0
\)
\(x^2+3 x+5\) is a quadratic equation
\(
\begin{aligned}
& a=1>0 \text { and } D=(-3)^2-4 \cdot 1 \cdot 5=-11<0 \\
& \therefore x^2+3 x+5>0 \text { (always) }
\end{aligned}
\)
So, we can ignore this quadratic term
\(
\begin{aligned}
& \frac{(x+2)}{-2+3 x-x^2} \geq 0 \\
\Rightarrow & \frac{x+2}{-\left(x^2-3 x+2\right)} \geq 0 \\
\Rightarrow & \frac{x+2}{x^2-3 x+2} \leq 0 \\
\Rightarrow & \frac{x+2}{x^2-2 x-x+2} \leq 0 \\
\Rightarrow & \frac{x+2}{(x-1)(x-2)} \leq 0
\end{aligned}
\)
\(
\begin{aligned}
& \therefore x \in(-\alpha,-2] \cup(1,2) \\
& \therefore S_1=(-\alpha,-2] \cup(1,2)
\end{aligned}
\)
Now,
\(
\begin{aligned}
& 3^{2 x}-3^{x+1}-3^{x+2}+27 \leq 0 \\
& \Rightarrow\left(3^x\right)^2-3 \cdot 3^x-3^2 \cdot 3^x+27 \leq 0
\end{aligned}
\)
Let \(3^x=t\)
\(
\begin{aligned}
& \Rightarrow t^2-3 \cdot t-3^2 \cdot t+27 \leq 0 \\
& \Rightarrow t(t-3)-9(t-3) \leq 0 \\
& \Rightarrow(t-3)(t-9) \leq 0
\end{aligned}
\)
\(
\begin{aligned}
& \therefore 3 \leq t \leq 9 \\
& \Rightarrow 3^1 \leq 3^x \leq 3^2 \\
& \Rightarrow 1 \leq x \leq 2 \\
& \therefore x \in[1,2] \\
& \therefore S_2=[1,2] \\
& \therefore S_1 \cup S_2=(-\alpha, 2] \cup(1,2) \cup[1,2] \\
& =(-\alpha, 2] \cup[1,2]
\end{aligned}
\)

Hint

Given quadratic equation,
\(
3 x^2+(\alpha-6) x+(\alpha+3)=0
\)
Let, \(a\) and \(b\) are the roots of the equation,
\(
\therefore a+b=-\frac{\alpha-6}{3}
\)
and \(a b=\frac{\alpha+3}{3}\)
For real roots,
\(
\begin{aligned}
& D \geq 0 \\
& \Rightarrow(\alpha-6)^2-4 \cdot 3 \cdot(\alpha+9) \geq 0 \\
& \Rightarrow \alpha^2-12 \alpha+36-12 \alpha-36 \geq 0 \\
& \Rightarrow \alpha^2-24 \alpha \geq 0 \\
& \Rightarrow \alpha(\alpha-24) \geq 0
\end{aligned}
\)
\(
\therefore \alpha>24 \text { or } \alpha<0
\)
\(\therefore\) Real roots of the equation possible for \(\alpha>24\) or \(\alpha<0\).
Now, sum of square of roots
\(
\begin{aligned}
& =a^2+b^2 \\
& =(a+b)^2-2 a b \\
& =\frac{(\alpha-6)^2}{9}-2 \cdot \frac{(\alpha+3)}{3} \\
& =\frac{\alpha^2-12 \alpha+36-6 \alpha-18}{9} \\
& =\frac{\alpha^2-18 \alpha+18}{9}=f(x)
\end{aligned}
\)
\(\therefore\) Sum of square of roots are minimum when \(a^2+b^2=\) minimum.
\(
\therefore f(\alpha)_{\min }=\frac{\alpha^2-18 \alpha+18}{9}
\)
Value of quadratic equation \(\alpha^2-18 \alpha+18\) is minimum at \(\alpha=-\frac{b}{2 a}=-\frac{(-18)}{2.1}=9\)
But for real roots \(\alpha\) should be less than 0 or greater than 24 .
So, there is no value of \(\alpha\) in the range \(\alpha>24 \cup \alpha<0\) where sum of squares of two real roots is minimum.
\(\therefore S\) is an empty set.

Hint

\(
\begin{aligned}
& f(x)=x^4-4 x+1=0 \\
& f^{\prime}(x)=4 x^3-4 \\
& =4(x-1)\left(x^2+1+x\right)
\end{aligned}
\)
\(
f^{\prime}(x)=0 \Rightarrow x=1
\)
\(x=1\) is point of minima.
\(
\begin{aligned}
& f(1)=-2 \\
& f(0)=1
\end{aligned}
\)

Hint

\(
A=(-3,1) \text { and } B=(-\infty,-1] \cup[3, \infty)
\)
So, \(A-B=(-1,1)\)
\(
B-A=(-\infty,-3] \cup[3, \infty)=R-(-3,3)
\)
\(A \cap B=(-3,-1]\)
and \(A \cup B=(-\infty, 1) \cup[3, \infty)=R-[1,3)\)

Hint

Given equation \(x^7-7 x-2=0\)
Let \(f(x)=x^7-7 x-2\)
\(
f^{\prime}(x)=7 x^6-7=7\left(x^6-1\right)
\)
and \(f^{\prime}(x)=0 \Rightarrow x=+1\)
and \(f(-1)=-1+7-2=5>0\)
\(
f(1)=1-7-2=-8<0
\)
So, roughly sketch of \(f(x)\) will be
So, number of real roots of \(f(x)=0\) and 3

\(
f(x)=2 \text { has } 3 \text { real distinct solution. }
\)

Hint

\(
\begin{aligned}
& x+1-2 \log _2\left(3+2^x\right)+2 \log _4\left(10-2^{-x}\right)=0 \\
& \log _2\left(2^{x+1}\right)-\log _2\left(3+2^x\right)^2+\log _2\left(10-2^{-x}\right)=0 \\
& \log _2\left(\frac{2^{x+1} \cdot\left(10-2^{-x}\right)}{\left(3+2^x\right)^2}\right)=0 \\
& \frac{2\left(10.2^{-x}-1\right)}{\left(3+2^x\right)^2}=1 \\
& \Rightarrow 20.2^x-2=9+2^{2 x}+6.2^x \\
& \therefore\left(2^x\right)^2-14\left(2^x\right)+11=0
\end{aligned}
\)
Roots are \(2^{x_1} \& 2^{x_2}\)
\(
\begin{aligned}
& \therefore 2^{x_1} \cdot 2^{x_2}=11 \\
& x_1+x_2=\log _2(11)
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \alpha=\max _{x \in R}\left\{8^{2 \sin 3 x} \cdot 4^{4 \cos 3 x}\right\} \\
& =\max \left\{2^{6 \sin 3 x} \cdot 2^{8 \cos 3 x}\right\} \\
& =\max \left\{2^{6 \sin 3 x+8 \cos 3 x}\right\}
\end{aligned}
\)
and \(\beta=\min \left\{8^{2 \sin 3 x} \cdot 4^{4 \cos 3 x}\right\}=\min \left\{2^{6 \sin 3 x+8 \cos 3 x}\right\}\)
Now range of \(6 \sin 3 x+8 \cos 3 x\)
\(
\begin{aligned}
& =\left[-\sqrt{6^2+8^2},+\sqrt{6^2+8^2}\right]=[-10,10] \\
& \alpha=2^{10} \& \beta=2^{-10}
\end{aligned}
\)
So, \(\alpha^{1 / 5}=2^2=4\)
\(
\Rightarrow \beta^{1 / 5}=2^{-2}=1 / 4
\)
quadratic \(8 x^2+b x+c=0\)
\(
\begin{aligned}
& -\frac{b}{8}=\frac{17}{4} \Rightarrow b=-34 \\
& \frac{c}{8}=1 \Rightarrow c=8 \\
& \therefore c-b=8+34=42
\end{aligned}
\)

Hint

\(
\begin{aligned}
& x^2+(20)^{1 / 4} x+(5)^{1 / 2}=0 \\
& \Rightarrow x^2+\sqrt{5}=-(20)^{1 / 4} x
\end{aligned}
\)
Squaring both sides, we get
\(
\begin{aligned}
& \left(x^2+\sqrt{5}\right)^2=\sqrt{20} x^2 \\
& \Rightarrow x^4=-5 \Rightarrow x^8=25 \\
& \Rightarrow \alpha^8+\beta^8=50
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \sum_{n=8}^{100}\left[\frac{(-1)^n n}{2}\right] \\
& =[4]+[-4.5]+[5]+[-5.5]+[6]+\ldots . .+[-49.5]+[50] \\
& =4-5+5-6+6 \ldots . .-50+50 \\
& =4
\end{aligned}
\)

Hint

\(
\begin{aligned}
& |x|^2-|x|-12=0 \\
& \Rightarrow(|x|+3)(|x|-4)=0 \\
& \Rightarrow|x|=4 \\
& \Rightarrow x= \pm 4
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \text { Let } e^{6 x}-e^{4 x}-2 e^{3 x}-12 e^{2 x}+e^x+1=0 \\
& \Rightarrow e^{6 x}-2 e^{3 x}+1-e^x\left[e^{3 x}+12 e^x-1\right]=0 \\
& \Rightarrow\left(e^{3 x}-1\right)^2-e^x\left(e^{3 x}-1\right)-12 e^{2 x}=0 \\
& \Rightarrow\left(e^{3 x}-1\right)^2-4 e^x\left(e^{3 x}-1\right)+3 e^x\left(e^{3 x}-1\right)-12 e^{2 x}=0 \\
& \Rightarrow\left(e^{3 x}-1\right)\left[e^{3 x}-1-4 e^x\right]+3 e^x\left[e^{3 x}-1-4 e^x\right]=0 \\
& \Rightarrow\left(e^{3 x}-1+3 e^x\right)\left(e^{3 x}-1-4 e^x\right)=0 \\
& \text { Either } e^{3 x}=1+4 e^x
\end{aligned}
\)

From the above graphs, only two roots are possible.

Hint

\(
\begin{aligned}
& {\left[e^x\right]^2+\left[e^x+1\right]-3=0} \\
& \Rightarrow\left[e^x\right]^2+\left[e^x\right]+1-3=0 \\
& \text { Let }\left[e^x\right]=t \\
& \Rightarrow t^2+t-2=0 \\
& \Rightarrow t=-2,1 \\
& {\left[e^x\right]=-2 \text { (Not possible) }} \\
& \text { or }\left[e^x\right]=1 \therefore 1 \leq e^x<2 \\
& \Rightarrow \ln (1) \leq x<\ln (2) \\
& \Rightarrow 0 \leq x<\ln (2) \\
& \Rightarrow x \in[0, \ln 2)
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \text { As, }\left(\alpha^2+\sqrt{3}\right)=-(3)^{1 / 4} . \alpha \\
& \Rightarrow\left(\alpha^4+2 \sqrt{3} \alpha^2+3\right)=\sqrt{3} \alpha^2 \text { (On squaring) } \\
& \therefore\left(\alpha^4+3\right)=(-) \sqrt{3} \alpha^2 \\
& \Rightarrow \alpha^8+6 \alpha^4+9=3 \alpha^4 \text { (Again squaring) } \\
& \therefore \alpha^8+3 \alpha^4+9=0 \\
& \Rightarrow \alpha^8=-9-3 \alpha^4
\end{aligned}
\)
(Multiply by \(\alpha^4\) )
So, \(\alpha^{12}=-9 \alpha^4-3 \alpha^8\)
\(
\begin{aligned}
& \therefore \alpha^{12}=-9 \alpha^4-3\left(-9-3 \alpha^4\right) \\
& \Rightarrow \alpha^{12}=-9 \alpha^4+27+9 \alpha^4
\end{aligned}
\)
Hence, \(\alpha^{12}=(27)^2\)
\(
\begin{aligned}
& \Rightarrow\left(\alpha^{12}\right)^8=(27)^8 \\
& \Rightarrow \alpha^{96}=(3)^{24}
\end{aligned}
\)
\(
\therefore \alpha^{96}\left(\alpha^{12}-1\right)+\beta^{96}\left(\beta^{12}-1\right)=(3)^{24} \times 52
\)

Hint

\(
\begin{aligned}
& x^2-\sqrt{2} x-\sqrt{3}=0 \\
& P_n=\alpha^n-\beta^n
\end{aligned}
\)
\(\alpha\) and \(\beta\) are the roots of the equation
Using Newton’s theorem
\(
\begin{aligned}
& P_{n+2}-\sqrt{2} P_{n+1}-\sqrt{3} P_n=0 \\
& \text { Put } n=10 \\
& P_{12}-\sqrt{2} P_{11}-\sqrt{3} P_{10}=0 \\
& P_{12}=\sqrt{2} P_{11}+\sqrt{3} P_{10}
\end{aligned}
\)
Put \(n=9\)
\(
\begin{aligned}
& P_{11}-\sqrt{2} P_{10}-\sqrt{3} P_9=0 \\
& P_{11}=\sqrt{2} P_{10}+\sqrt{3} P_9 \\
& (11 \sqrt{3}-10 \sqrt{2}) P_{10}+(11 \sqrt{2}+10) P_{11}-11 P_{12}
\end{aligned}
\)
Put the value of \(P_{12} \& P_{12}\) in above equation.
\(
\begin{aligned}
& =(11 \sqrt{3}-10 \sqrt{2}) P_{10}+(11 \sqrt{2}+10)\left(\sqrt{2} P_{10}+\sqrt{3} P_9\right)-11\left(\sqrt{2} P_{11}+\sqrt{3} P_{10}\right. \\
& =11 \sqrt{3} P_{10}-10 \sqrt{2} P_{10}+22 P_{10}+10 \sqrt{2} P_{10}+11 \sqrt{6} P_9+10 \sqrt{3} P_9-11 \sqrt{2} P_{11} \\
& =22 P_{10}+11 \sqrt{6} P_9+10 \sqrt{3} P_9-11 \sqrt{2}\left(\sqrt{2} P_{10}+\sqrt{3} P_9\right) \\
& =22 P_{10}+11 \sqrt{6} P_9+10 \sqrt{3} P_9-22 P_{10}-11 \sqrt{6} P_9 \\
& =10 \sqrt{3} P_9
\end{aligned}
\)

Hint

\(
\begin{aligned}
& x^2+2 \sqrt{2 x}-1=0 \\
& \alpha+\beta=-2 \sqrt{2} \text { and } \alpha \beta=-1 \\
& \alpha^2+\beta^2=(\alpha+\beta)^2-2 \alpha \beta \\
& =8+2=10 \\
& \alpha^4+\beta^4=\left(\alpha^2+\beta^2\right)^2-2(\alpha \beta)^2 \\
& =100-2=98 \\
& \alpha^6+\beta^6=\left(\alpha^2+\beta^2\right)^3-3 \alpha^2 \beta^2\left(\alpha^2+\beta^2\right) \\
& =1000-3(10) \\
& =970 \\
& \therefore \quad \frac{1}{10}\left(\alpha^6+\beta^6\right)=97
\end{aligned}
\)
Equation whose roots are \(\alpha^4+\beta^4\) and \(\frac{1}{10}\left(\alpha^6+\beta^6\right)\) is
\(
\begin{aligned}
& x^2-(98+97) x+98 \times 97=0 \\
& x^2-195 x+9506=0
\end{aligned}
\)

Hint

First, let’s start by substituting \(y=(8)^x\) in the given equation. By substituting, the equation \(8^{2 x}-16 \cdot 8^x+48=0\) will be transformed into
\(
y^2-16 y+48=0
\)
Now, we have a quadratic equation in \(y\). To find the roots of this quadratic equation, we can use the quadratic formula:
\(
y=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}
\)
In this case, \(a=1, b=-16\), and \(c=48\). Substituting these values into the formula, we get:
\(
\begin{aligned}
& y=\frac{16 \pm \sqrt{256-192}}{2} \\
& y=\frac{16 \pm \sqrt{64}}{2} \\
& y=\frac{16 \pm 8}{2}
\end{aligned}
\)
Solving for the two possible values of \(y\), we have:
\(
\begin{aligned}
& y=\frac{16+8}{2}=12 \\
& y=\frac{16-8}{2}=4
\end{aligned}
\)
Now, recall that we substituted \(y=(8)^x\). So, we need to solve for \(x\) when \(y=12\) and \(y=4\).
\(
8^x=12
\)
\(
\begin{aligned}
& x=\log _8(12) \\
& 8^x=4 \\
& x=\log _8(4)
\end{aligned}
\)
Therefore, the solutions for \(x\) are \(\log _8(12)\) and \(\log _8(4)\). The sum of these solutions is:
\(
\log _8(12)+\log _8(4)
\)
Using the logarithmic property that \(\log _b(m)+\log _b(n)=\log _b(m \cdot n)\), we get:
\(
\begin{aligned}
& \log _8(12)+\log _8(4)=\log _8(12 \cdot 4) \\
& =\log _8(48)
\end{aligned}
\)
Now, we note that:
\(
48=8 \cdot 6
\)
Thus,
\(
\log _8(48)=\log _8(8 \cdot 6)=\log _8(8)+\log _8(6)=1+\log _8(6)
\)
Therefore, the sum of all the solutions of the equation is: \(1+\log _8(6)\)

Hint

\(
\begin{aligned}
& x^2-\left(t^2-5 t+6\right) x+1=0 \\
& \therefore a_{2025}-\left(t^2-5 t+6\right) a_{2024}+a_{2023}=0 \\
& \Rightarrow \frac{a_{2025}+a_{2023}}{a_{2024}}=t^2-5 t+6 \\
& =\left(t+\frac{5}{2}\right)^2+\left(\frac{-1}{4}\right) \\
& \text { Minimum value }=\frac{-1}{4}
\end{aligned}
\)

Hint

Equation is \(a x^2+b x+c=0\)
\(D >0\) [for roots to be real & distinct]
\(
\Rightarrow b^2-4 a c>0
\)
For \(b<2\) no value of \(a \& c\) are possible
For \(b=3 \Rightarrow a c<\frac{9}{4}\)
\(
(a, c) \in\{(1,1),(1,2),(2,1)\} \Rightarrow 3 \text { cases }
\)
For \(b=4 \Rightarrow a c<4\)
\(
(a, c) \in\{(1,1),(1,2),(2,1),(3,1),(1,3)\} \Rightarrow 5 \text { cases }
\)
For \(b=5 \Rightarrow a c<\frac{25}{4}\)
\(
\begin{aligned}
& (a, c) \in\{(1,1),(1,2),(2,1),(3,1),(1,3),(2,2), \\
& (4,1),(1,4),(3,2),(2,3),(5,1),(1,5),(1,6), \\
& (6,1)\}=14 \text { cases }
\end{aligned}
\)
For \(b=6 \Rightarrow a c<9\)
\(
\begin{aligned}
& (a, c) \in\{(1,1),(1,2),(2,1),(3,1),(1,3),(2,2) \\
& (4,1),(1,4),(3,2),(2,3),(5,1),(1,5),(1,6), \\
& (6,1),(2,4),(4,2)\}=16 \text { cases }
\end{aligned}
\)
Total cases \(=3+5+14+16=38\) cases
\(\Rightarrow\) Probability, \(p=\frac{38}{216}\)
\(
\Rightarrow 216 p=38
\)

Hint

\(
\begin{aligned}
& (x-2)(x-6)=0 \\
& \Rightarrow x^2-8 x+12=0 \\
& \Rightarrow \frac{x^2}{12}-\frac{8 x}{12}+1=0 \\
& \therefore a=\frac{1}{12}, b=\frac{-2}{3} \\
& \frac{1}{2 a+b}=\frac{1}{\frac{1}{6}-\frac{2}{3}} \Rightarrow \frac{6}{-3}=-2 \\
& \frac{1}{6 a+b}=\frac{1}{\frac{1}{2}-\frac{2}{3}} \Rightarrow \frac{6}{-1}=-6 \\
& \therefore(x+2)(x+6)=0 \\
& \Rightarrow x^2+8 x+12=0
\end{aligned}
\)

Hint

\(
\begin{aligned}
& x^2-x-1=0 \\
& S_n=2023 \alpha^n+2024 \beta^n \\
& S_{n-1}+S_{n-2}=2023 \alpha^{n-1}+2024 \beta^{n-1}+2023 \alpha^{n-2}+2024 \beta^{n-2} \\
& =2023 \alpha^{n-2}[1+\alpha]+2024 \beta^{n-2}[1+\beta] \\
& =2023 \alpha^{n-2}\left[\alpha^2\right]+2024 \beta^{n-2}\left[\beta^2\right] \\
& =2023 \alpha^n+2024 \beta^n \\
& S_{n-1}+S_{n-2}=S_n \\
& P_{u t} n=12 \\
& S_{11}+S_{10}=S_{12}
\end{aligned}
\)

Hint

\(
|x+1||x+3|-4|x+2|+5=0
\)
(I) If \(x<-3\)
\(
\begin{aligned}
& x^2+4 x+3+4 x+8+5=0 \\
& x^2+8 x+16=0 \Rightarrow x=-4 \Rightarrow \text { one solution }
\end{aligned}
\)
\(
\begin{aligned}
& \text { (II) If }-3 \leq x<-2 \\
& -x^2-4 x-3+4 x+8+5=0 \\
& x^2-10=0 \Rightarrow x= \pm \sqrt{10} \Rightarrow \text { do not satisfy }-3 \leq \\
& x<-2
\end{aligned}
\)
\(
\begin{aligned}
& \text { (III) If }-2 \leq x<-1 \\
& -x^2-4 x-3-4 x-8+5=0 \\
& x^2-8 x+6=0 \\
& (x+4)^2=10 \Rightarrow x=-4 \pm \sqrt{10} \Rightarrow \text { do not satisfy } \\
& -2 \leq x<-1
\end{aligned}
\)
(IV) If \(x \geq-1\)
\(
\begin{aligned}
& x^2+4 x+3-4 x-8+5=0 \\
& x^2=0 \\
& x=0 \text { (One solution) }
\end{aligned}
\)
\(x=0\) (One solution)
\(\Rightarrow\) The number of distinct real roots are two

Hint

\(
\begin{aligned}
& 4 x^4+8 x^3-17 x^2-12 x+9=0 \\
& (x+1)\left(4 x^3+4 x^2-21 x+9\right)=0 \\
& (x+1)(x+3)\left(4 x^2-8 x+3\right)=0 \\
& (x+1)(x+3)\left(4 x^2-6 x-2 x+3\right)=0 \\
& (x+1)(x+3)(2 x(2 x-3)-1(2 x-3))=0
\end{aligned}
\)
\(
\frac{(x+1)}{x=-1}
\)
\(
\begin{gathered}
\underset{\downarrow}{(x+1)} \\
x=-3
\end{gathered}
\)
\(
\frac{(2 x-1)}{x=\frac{1}{2}}
\)
\(
\begin{aligned}
& (2 x-3) \\
& x=\frac{3}{2}=0
\end{aligned}
\)
\(
\begin{aligned}
& (4+1)(4+9)\left(4+\frac{1}{4}\right)\left(4+\frac{9}{4}\right)=\frac{125}{16} m \\
& 5 \times 13 \times\left(\frac{17}{4}\right) \times\left(\frac{25}{4}\right)=\frac{125}{16} m \\
& \frac{125}{16} \times[13 \times 17]=\frac{125}{16} m \\
& m=13 \times 17 \\
& m=221
\end{aligned}
\)

Hint

\(
x|x+5|+2|x+7|-2=0
\)
\(
\begin{aligned}
& \text { (i) } d x \geq-5 \Rightarrow x(x+5)+2(x+7)-2=0 \\
& x^2+7 x+12=0 \Rightarrow x=-3,-4
\end{aligned}
\)
(ii)
\(
\begin{aligned}
& x \in(-7,-5) \\
& x(-x-5)+2(x+7)-2=0 \\
& -x^2-3 x+12=0 \\
& \Rightarrow x^2+3 x-12=0 \\
& \Rightarrow x=\frac{-3-\sqrt{57}}{2} \text { satisfy }
\end{aligned}
\)
(iii)
\(
\begin{aligned}
& x \leq-7 \\
& \Rightarrow x(-x-5)+2(-x-7)-2=0 \\
& -x^2-7 x-16=0 \Rightarrow x^2+7 x+16=0
\end{aligned}
\)
No solution

Hint

\(
|x| \quad|x+2|-5|x+1|-1=0
\)
\(
\begin{aligned}
& \text { (I) if } x<-2 \\
& x^2+2 x+5 x+5-1=0 \\
& x^2+7 x+4=0 \Rightarrow \text { one root satisfying } x<-2
\end{aligned}
\)
(I) if \(x<-2\)
\(
\begin{aligned}
& \text { (II) if }-2 \leq x<-1 \\
& -x^2-2 x+5 x+5-1=0 \\
& x^2-3 x-4=0 \Rightarrow \text { not root satisfying }-2 \leq x<-1
\end{aligned}
\)
\(
\begin{aligned}
& \text { (III) if }-1 \leq x<0 \\
& -x^2-2 x-5 x-5-1=0 \\
& x^2+7 x+6=0 \\
& x=-1 \text { is only root satisfying }-1 \leq x<0
\end{aligned}
\)
(IV) if \(x \geq 0\)
\(
\begin{aligned}
& x^2+2 x-5 x-5-1=0 \\
& x^2-3 x-6=0
\end{aligned}
\)
one root satisfying \(x \geq 0\)
\(\Rightarrow\) The number of distinct real roots are three.

Hint

The given equation is \(x\left(x^2+3|x|+5|x-1|+6|x-2|\right)=0\), which can be solved by analyzing it in parts. It can be broken down into: \(x=0\) and
\(
x^2+3|x|+5|x-1|+6|x-2|=0 \text {. }
\)
For \(x=0\), it’s clear that it is a solution to the equation since it makes the entire expression equal to zero.
Case (I)
\(
\begin{aligned}
& x<0 \\
& x^2-3 x-5(x-1)-6(x-2)=0 \\
& x^2-14 x+17=0
\end{aligned}
\)
\(\because\) All roots are positive \(\Rightarrow\) no solution
Case (II)
\(
\begin{aligned}
& 0<x<1 \\
& x^2+3 x-5(x-1)-6(x-2)=0 \\
& x^2-8 x+17=0 \\
& \because D<0 \Rightarrow \text { no solution }
\end{aligned}
\)
\(
\begin{aligned}
& \text { Case (III) } \\
& 1<x<2 \\
& x^2+3 x+5(x-1)-6(x-2)=0 \\
& x^2+2 x+7=0 \\
& \Rightarrow \text { no solution } \\
& \text { Case (IV) } \\
& x>2 \\
& x^2+3 x+5(x-1)+6(x-2)=0 \\
& x^2+14 x-19=0 \\
& \text { All roots less than } 2 \\
& \Rightarrow \text { no solution }
\end{aligned}
\)
Here \(x=0\) is only solution.

Hint

\(
\begin{aligned}
& \because(a-b)^2+(b-c)^2+(c-a)^2>0 \\
& \Rightarrow 2\left(a^2+b^2+c^2-a b-b c-c a\right)>0 \\
& \Rightarrow 2>2(a b+b c+c a) \Rightarrow a b+b c+c a<1
\end{aligned}
\)