Hint

\(
\begin{aligned}
&\quad(\mathbf{a}, \mathbf{c}, \mathbf{d}) \mathrm{z}=\frac{1}{a+i b t}=x+i y \\
&\Rightarrow x+i y=\frac{a-i b t}{a^2+b^2 t^2} \\
&\Rightarrow x=\frac{a}{a^2+b^2 t^2}, y=\frac{-b t}{a^2+b^2 t^2} \\
&\Rightarrow x^2+y^2=\frac{1}{a^2+b^2 t^2}=\frac{x}{a}
\end{aligned}
\)
\(
\Rightarrow x^2+y^2-\frac{x}{a}=0
\)
\(\therefore\) Locus of \(z\) is a circle with centre \(\left(\frac{1}{2 a}, 0\right)\) and radius \(\frac{1}{2|a|}\) irrespective of ‘ a’ +ve or -ve
Also for \(b=0, a \neq 0\), we get, \(y=0\)
\(\therefore\) locus is \(\mathrm{x}\)-axis
and for \(\mathrm{a}=0, \mathrm{~b} \neq 0\) we get \(\mathrm{x}=0\)
\(\therefore\) locus is y-axis.
Hence, a, c, and d are the correct options.

Hint

(c, d) We have \(w=\frac{\sqrt{3}+i}{2}=\cos \frac{\pi}{6}+i \sin \frac{\pi}{6}\)
\(\Rightarrow w=\cos \frac{n \pi}{6}+i \sin \frac{n \pi}{6}\)
\(\therefore \mathrm{P}\) contains all those points which lie on unit circle and have arguments \(\frac{\pi}{6}, \frac{2 \pi}{6}, \frac{3 \pi}{6}\) and so on.
Since, \(\mathrm{z}_1 \in \mathrm{P} \cap \mathrm{H}_1\) and \(\mathrm{z}_2 \in \mathrm{P} \cap \mathrm{H}_2\), therefore \(\mathrm{z}_1\) and \(\mathrm{z}_2\) can have possible positions as shown in the figure.
\(
\therefore \angle \mathrm{z}_1 \mathrm{Oz}_2 \text { can be } \frac{2 \pi}{3} \text { or } \frac{5 \pi}{6} \text {. }
\)

Hint

(d) We have \(\left(1+\omega+\omega^2\right)^7=\left(-\omega^2-\omega^2\right)^7\)
\(
=(-2)^7\left(\omega^2\right)^7=-128 \omega^{14}=-128 \omega^2
\)

Hint

\((\mathrm{A}) \rightarrow(\mathrm{q}, \mathrm{r})\)
\(|z-i| z||=|z+i| z \mid\)
\(\Rightarrow z\) is equidistant from two points \((0,|z|)\) and
\((0,-|z|)\), which lie on imaginary axis.
\(\therefore z\) must lie on real axis \(\Rightarrow \operatorname{Im}(z)=0\). Also \(\left|I_m(z)\right| \leq 1\)
(B) \(\rightarrow\) p
\(z=\omega-\frac{1}{\omega}=2(\cos \theta+i \sin \theta)-\frac{1}{2}(\cos \theta-i \sin \theta)\)
\(\Rightarrow x+i y=\frac{3}{2} \cos \theta+i \frac{5}{2} \sin \theta\)
Here, \(|z|=\sqrt{\frac{9+25}{4}}=\sqrt{\frac{34}{4}} \leq 3\) and \(\left|R_e(z)\right| \leq 2\)
Also \(x=\frac{3}{2} \cos \theta, y=\frac{5}{2} \sin \theta \Rightarrow \frac{4 x^2}{9}+\frac{4 y^2}{25}=1\)
Which is an ellipse with \(e=\sqrt{1-\frac{9}{25}}=\frac{4}{5}\)
\((\mathrm{D}) \rightarrow(\mathrm{q}, \mathrm{r}, \mathrm{s}, \mathrm{t})\)
Let \(\omega=\cos \theta+i \sin \theta\) then \(z=2 \cos \theta \Rightarrow \operatorname{Im} z=0\) Also \(|z| \leq 3\) and \(|\operatorname{Im}(z)| \leq 1,\left|\mathrm{R}_e(z)\right| \leq 2\)

Hint

Given : \(z \neq 0 \quad\) Let \(z=a+i b\)
\(\operatorname{Re}(\mathrm{z})=0 \Rightarrow z=i b \Rightarrow z^2=-b^2\)
\(\therefore \operatorname{Im}(z^2)=0\)
\(\therefore(\mathrm{A})\) corresponds to (q)
\(\operatorname{Arg} z=\frac{\pi}{4} \Rightarrow a=\mathrm{b} \Rightarrow z=a+i a\) \(\Rightarrow z^2=a^2-a^2+2 i a^2 \Rightarrow z^2=2 i a^2 \Rightarrow \operatorname{Re}(z^2)=0\)
\(\therefore(B)\) corresponds to \((\mathrm{p})\).

Hint

The given circle is \(|\mathrm{z}-1|=\sqrt{2}\), where \(\mathrm{z}_0=1\) is the centre and \(\sqrt{2}\) is radius of circle. \(z_1\) is one of the vertex of the square inscribed in the given circle.
Clearly \(z_2\) can be obtained by rotating \(z_1\) by an angle \(90^{\circ}\) in anticlockwise direction, about centre \(z_0\)
Thus, \(z_2-z_0=\left(z_1-z_0\right) e^{i \pi / 2}\)
or \(z_2-1=(2+i \sqrt{3}-1) i \Rightarrow z_2=i-\sqrt{3}+1\)
\(
z_2=(1-\sqrt{3})+i
\)
Again rotating \(z_2\) by \(90^{\circ}\) about \(z_0\), we get
\(
\begin{aligned}
&z_3-z_0=\left(z_2-z_0\right) i \\
&\Rightarrow z_3-1=[(1-\sqrt{3})+i-1] i=-\sqrt{3} i-1 \Rightarrow z_3=-i \sqrt{3}
\end{aligned}
\)
And similarly \(1=(-i \sqrt{3}-1) i=\sqrt{3}-i\)
\(
\Rightarrow \quad z_4=(\sqrt{3}+1)-i
\)
Hence, the remaining vertices are
\(
(1-\sqrt{3})+i,-i \sqrt{3},(\sqrt{3}+1)-i
\)

Hint

\(
\text { Given : }\left|\frac{z-\alpha}{z-\beta}\right|=k \Rightarrow|z-\alpha|=k|z-\beta|
\)
Let pt. A represents complex number \(\alpha\) and \(B\) that of \(\beta\), and \(P\) represents \(z\). then \(|z-\alpha|=k|z-\beta|\)
\(\Rightarrow z\) is the complex number whose distance from \(A\) is \(k\) times its distance from \(B\).
i.e. \(P A=k P B\)
\(\Rightarrow P\) divides \(A B\) in the ratio \(k: 1\) internally or externally (at \(P^{\prime}\) ).
Then \(P=\left(\frac{k \beta+\alpha}{k+1}\right)\) and \(P^{\prime}=\left(\frac{k \beta-\alpha}{k-1}\right)\)
Now through \(P P^{\prime}\) a number of circles can pass, but with given data we can find radius and centre of that circle for which \(P P^{\prime}\) is diameter.
\(
\begin{aligned}
&\therefore \text { Centre }=\text { mid. point of } P P^{\prime}=\left(\frac{k \beta+\alpha}{k+1}+\frac{k \beta-\alpha}{k-1}\right) \\
&= \\
&=\frac{k^2 \beta+k \alpha-k \beta-\alpha+k^2 \beta-k \alpha+k \beta-\alpha}{2\left(k^2-1\right)} \\
&=\frac{k^2 \beta-\alpha}{k^2-1}=\frac{\alpha-k^2 \beta}{1-k^2}
\end{aligned}
\)
Also radius \(=\frac{1}{2}\left|P P^{\prime}\right|\)
\(
\begin{aligned}
&=\frac{1}{2}\left|\frac{k \beta+\alpha}{k+1}-\frac{k \beta-\alpha}{k-1}\right| \\
&=\frac{1}{2}\left|\frac{k^2 \beta+k \alpha-k \beta-\alpha-k^2 \beta+k \alpha-k \beta+\alpha}{k^2-1}\right|=\frac{k|\alpha-\beta|}{\left|1-k^2\right|}
\end{aligned}
\)

Hint

Let us consider, \(\sum_{r=1}^n a_r z^r=1\) where \(\left|a_r\right|<2\)
\(
\begin{aligned}
&\Rightarrow a_1 z+a_2 z^2+a_3 z^3+\ldots+a_n z^n=1 \\
&\Rightarrow\left|a_1 z+a_2 z^2+a_3 z^3+\ldots+a_n z^n\right|=1
\end{aligned}
\)
But we know that \(\left|z_1+z_2\right| \leq\left|z_1\right|+\left|z_2\right|\)
\(\therefore \quad\) Using its generalised form, we get
\(
\begin{gathered}
\left|a_1 z+a_2 z^2+a_3 z^3+\ldots+a_n z^n\right| \\
\quad \leq\left|a_1 z\right|+\left|a_2 z^2\right|+\ldots+\left|a_n z^n\right| \\
\Rightarrow 1 \leq\left|a_1\right||z|+\left|a_2\right|\left|z^2\right|+\left|a_3\right||z|+\ldots+\left|a_n\right|\left|z^n\right| \\
\quad\left[\operatorname{using~eq}{ }^{\mathrm{n}}(\mathrm{i})\right]
\end{gathered}
\)
But given that \(\left|a_r\right|<2 \forall r=1, \ldots, n\)

\(
\begin{aligned}
\therefore & 1<2\left[|z|+|z|^2+|z|^3+\ldots+|z|^n\right] \\
\Rightarrow & \quad 1<2\left[\frac{|z|\left(1-|z|^n\right)}{1-|z|}\right] \Rightarrow 2\left[\frac{|z|-|z|^{n+1}}{1-|z|}\right]>1 \\
\Rightarrow & 2\left[|z|-|z|^{n+1}\right]>1-|z| \quad(\because 1-|z|>0 \text { as }|z|<1 / 3) \\
\Rightarrow & {\left[|z|-|z|^{n+1}\right]>\frac{1}{2}-\frac{1}{2}|z| } \\
\Rightarrow & \frac{3}{2}|z|>\frac{1}{2}+|z|^{n+1} \\
\Rightarrow &|z|>\frac{1}{3}+\frac{2}{3}|z|^{n+1} \Rightarrow|z|>\frac{1}{3}
\end{aligned}
\)
which is a contradiction as given that \(|z|<\frac{1}{3}\)
\(
\therefore \quad \text { There exists no such complex number. }
\)

Hint

The given equation can be written as
\(
\left(z^p-1\right)\left(z^q-1\right)=0
\)
\(
\therefore \quad z=(1)^{1 / p} \quad \text { or } \quad(1)^{1 / q}
\)
where \(\mathrm{p}\) and \(\mathrm{q}\) are distinct prime numbers.
Hence both the equations will have distinct roots and as \(z \neq 1\), both will not be simultaneously zero for any value of \(z\) given by equations in (i)
Also \(1+\alpha+\alpha^2+\ldots+\alpha^{p-1}=\frac{1-\alpha^p}{1-\alpha}=0(\alpha \neq 1)\)
or \(1+\alpha+\alpha^2+\ldots+\alpha^p=\frac{1-\alpha^q}{1-\alpha}=0(\alpha \neq 1)\)
Because of (i) either \(\alpha^p=1\) and if \(\alpha^q=1\) but not both simultaneously as \(p\) and \(q\) are distinct primes.

Hint

\(
\begin{aligned}
&|z|^2 \omega-|\omega|^2 z=z-\omega \dots(i)\\
&z \bar{z} \omega-\omega \bar{\omega} z=z-\omega \\
&\Rightarrow \quad z \omega(\bar{z}-\bar{\omega})=z-\omega .
\end{aligned}
\)
Taking modulus, \(|z \cdot \omega||\bar{z}-\bar{\omega}|=z-\omega\)
\(
\begin{aligned}
&|z \omega||z-\omega|=z-\omega . \\
&\Rightarrow \quad|z-\omega|(|z \omega|-1)=0
\end{aligned}
\)
If \(|z-\omega|=0\) then \(z-\omega=0 \quad \therefore z=\omega\).
If \(|z \omega|-1=0\) then \(z \omega=1 \quad \therefore|z|=\frac{1}{|\omega|}=r\)
(say)
Let \(z=r e^{i \theta}, \omega=\frac{1}{r} e^{i \phi}\)
From (i) \(r^2\left(\frac{1}{r} e^{i \phi}\right)-\frac{1}{r^2}\left(r e^{i \theta}\right)=r e^{i \theta}-\frac{1}{r} e^{i \phi}\)
\(
\begin{aligned}
&\therefore \quad\left(r+\frac{1}{r}\right) e^{i \phi}=\left(r+\frac{1}{r}\right) e^{i \theta} \\
&e^{i \phi}=e^{i \theta} \Rightarrow \theta=\phi \\
&\therefore \quad z \bar{\omega}=\left(r e^{i \theta}\right)\left(\frac{1}{r} e^{-i \theta}\right)=1 \\
&\therefore \quad z=\omega \text { or } z \bar{\omega}=1
\end{aligned}
\)

Hint

True

\(z^2+p z+q=0\)
\(z_1+z_2=-p, z_1 z_2=\mathrm{q}\)
By rotation through \(\alpha\) in anticlockwise direction,
\(
\begin{aligned}
&z_2=z_1 e^{i \alpha} \\
&\frac{z_2}{z_1}=\frac{e^{i \alpha}}{1}=\frac{\cos \alpha+i \sin \alpha}{1}
\end{aligned}
\)
Add 1 in both sides to get \(z_1+z_2=-\mathrm{p}\)
\(
\begin{aligned}
&\therefore \frac{z_1+z_2}{z_1}=\frac{1+\cos \alpha+i \sin \alpha}{1}=2 \cos \frac{\alpha}{2}\left[\cos \frac{\alpha}{2}+i \sin \frac{\alpha}{2}\right] \\
&\Rightarrow \frac{\left(z_2+z_1\right)}{z_1}=2 \cos \frac{\alpha}{2} e^{i \alpha / 2}
\end{aligned}
\)
On squaring, \(\left(z_2+z_1\right)^2=4 \cos ^2(\alpha / 2) z_1^2 \cdot e^{i \alpha}\)
\(
=4 \cos ^2 \frac{\alpha}{2} z_1^2 \cdot \frac{z_2}{z_1}=4 \cos ^2 \frac{\alpha}{2} z_1 z_2
\)
\(
\Rightarrow \quad p^2=4 q \cos ^2 \frac{\alpha}{2}
\)

Hint

Let \(z=x+i y\) then \(\bar{z}=i z^2\)
\(
\begin{aligned}
&\Rightarrow x-i y=i\left(x^2-y^2+2 i x y\right) \\
&\Rightarrow x-i y=i\left(x^2-y^2\right)-2 x y \\
&\Rightarrow x(1+2 y)=0 ; x^2-y^2+y=0 \\
&\Rightarrow x=0 \text { or } y=-\frac{1}{2} \Rightarrow x=0, y=0,1
\end{aligned}
\)
or \(y=-\frac{1}{2} \quad x=\pm \frac{\sqrt{3}}{2}\)
For non zero complex number \(z\),
\(
\begin{aligned}
&x=0, y=1 ; x=\frac{\sqrt{3}}{2}, y=-\frac{1}{2} ; x=\frac{-\sqrt{3}}{2}, y=-\frac{1}{2} \\
&\therefore \quad z=i, \frac{\sqrt{3}}{2}-\frac{i}{2},-\frac{\sqrt{3}}{2}-\frac{i}{2}
\end{aligned}
\)

Hint

Let \(z=r_1\left(\cos \theta_1+i \sin \theta_1\right)\) and \(w=r_2\left(\cos \theta_2+i \sin \theta_2\right)\)
We have, \(|z|=r_1,|w|=r_2, \arg (z)=\theta_1\) and \(\arg (w)=\theta_2\)
Given, \(|z| \leq 1,|w|<1\)
\(\Rightarrow \quad r_1 \leq 1\) and \(r_2 \leq 1\)
Now, \(z-w=\left(r_1 \cos \theta_1-r_2 \cos \theta_2\right)+i\left(r_1 \sin \theta_1-r_2 \sin \theta_2\right)\)
\(\Rightarrow|z-w|^2=\left(r_1 \cos \theta_1-r_2 \cos \theta_2\right)^2+\left(r_1 \sin \theta_1-r_2 \sin \theta_2\right)^2\)
\(=3 \cos ^2 \theta_1+r_2^2 \cos ^2 \theta_2-2 r_1 r_2 \cos \theta_1 \cos \theta_2\)
\(+7 \sin ^2 \theta_1+r_2^2 \sin ^2 \theta_2-2 r_1 r_2 \sin \theta_1 \sin \theta_2\)
\(=r_1\left(\cos ^2 \theta_1+\sin ^2 \theta_1\right)+r_2^2\left(\cos ^2 \theta_2+\sin ^2 \theta_2\right)\)
\(-2 r_1 r_2\left(\cos \theta_1 \cos \theta_2+\sin \theta_1 \sin \theta_2\right)\)
\(=r_1+r_2^2-2 r_1 r_2 \cos \left(\theta_1-\theta_2\right)\)
\(=\left(r_1-r_2\right)^2+2 r_1 r_2\left[1-\cos \left(\theta_1-\theta_2\right)\right]\)
\(=\left(r_1-r_2\right)^2+4 r_1 r_2 \sin ^2\left(\frac{\theta_1-\theta_2}{2}\right)\)
\(
\leq\left|r_1-r_2\right|^2+4\left|\sin \left(\frac{\theta_1-\theta_2}{2}\right)\right|^2 \quad\left[ r_1, r_2 \leq 1\right]
\)
and \(|\sin \theta| \leq|\theta|, \forall \theta \in \mathrm{R}\)
Therefore, \(|z-w|^2 \leq\left|r_1-r_2\right|^2+4\left|\frac{\theta_1-\theta_2}{2}\right|^2\)
\(
\leq\left|r_1-r_2\right|^2+\left|\theta_1-\theta_2\right|^2
\)
\(
\Rightarrow \quad|z-w|^2 \leq(|z|-|w|)^2+(\arg z-\arg w)^2
\)

Hint

Dividing throughout by \(i\) and knowing that \(\frac{1}{i}=-i\), we get
\(
\begin{aligned}
&z^3-i z^2+i z+1=0 \\
&\Rightarrow \quad z^2(z-i)+i(z-i)=0 \quad \text { as } \quad 1=-i^2 \\
&\Rightarrow \quad(z-i)\left(z^2+i\right)=0 \quad \therefore z=i \quad \text { or } \quad z^2=-i \\
&\therefore \quad|z|=|i|=1 \text { or }\left|z^2\right|=|z|^2=|-i|=1 \Rightarrow|z|=1
\end{aligned}
\)
Hence, in either case \(|z|=1\)

Hint

\(1, a_1, a_2, \ldots, a_{n-1}\) are the \(n\) roots of unity. Therefore they are roots of eq. \(x^n-1=0\)
Therefore by factor theorem,
\(x^n-1=(x-1)\left(x-a_1\right)\left(x-a_2\right) \ldots .\left(x-a_{n-1}\right) \dots(i)\)
\(\Rightarrow \frac{x^n-1}{x-1}=\left(x-a_1\right)\left(x-a_2\right) \ldots .\left(x-a_{n-1}\right) \dots(ii)\) On differentiating both sides of eq. (i), we get
\(
n x^{n-1}=\left(x-a_1\right)\left(x-a_2\right) \ldots\left(x-a_{n-1}\right)+(x-1)\left(x-a_2\right)
\)
\(\ldots .\left(x-a_{n-1}\right)+\ldots .+(x-1)\left(x-a_1\right) \ldots\left(x-a_{n-2}\right)\)
For \(x=1\), we get \(n=\left(1-a_1\right)\left(1-a_2\right) \ldots\left(1-a_{n-1}\right)\)
[Since the terms except first, contain \((x-1)\) and hence become zero for \(x\) \(=1]\)

Hint

We know that if \(z_1, z_2, z_3\) are vertices of an equilateral triangle, then
\(
\frac{z_1-z_2}{z_3-z_2}=\frac{z_3-z_1}{z_2-z_1}
\)
Here \(z_3=0\),
\(
\therefore \frac{z_1-z_2}{-z_2}=\frac{-z_1}{z_2-z_1}
\)
\(
\Rightarrow-z_1^2-z_2^2+2 z_1 z_2=z_1 z_2, \quad \therefore z_1^2+z_2^2-z_1 z_2=0 \text {. }
\)

Hint

Let us consider the equilateral triangle with each side of length \(2 a\) and having two of its vertices \(A(-a, 0)\) and \(B(a, 0)\) on \(x\)-axis, then third vertex \(C\) will clearly lie on y-axissuch that \(O C=2 a \sin 60^{\circ}=a \sqrt{3}, \therefore\) \(C=(0, a \sqrt{3)}\).
Now if \(A, B\) and \(C\) are represented by complex number \(z_1, z_2, z_3\) then \(z_1=-a ; z_2=a ; z_3=a \sqrt{3} i\)
Since in an equilateral triangle, centriod and circumcentre coincide,
\(
\begin{aligned}
&\therefore \text { Circumcentre, } z_0=\frac{z_1+z_2+z_3}{3} \\
&\Rightarrow z_0=\frac{-a+a+a \sqrt{3} i}{3}=\frac{i a}{\sqrt{3}}
\end{aligned}
\)
Now, \(z_1^2+z_2^2+z_3^2=a^2+a^2-3 a^2=-a^2\)
and \(3 z_0^2=(i a)^2=-a^2\)
\(\therefore\) Clearly \(3 z_0^2=z_1^2+z_2^2+z_3^2\)

Hint

Since, \(\beta\) and \(\gamma\) are the complex cube roots of unity therefore, we can suppose \(\beta=\omega\) and \(\gamma=\omega^2\) so that \(\omega+\omega^2+1=0\) and \(\omega^3=1\).
Then \(x y z=(a+b)\left(a \omega^2+b \omega\right)\left(a \omega+b \omega^2\right)\)
\(=(a+b)\left(a^2 \omega^3+a b \omega^4+a b \omega^2+b^2 \omega^3\right)\)
\(=(a+b)\left(a^2+a b \omega+a b \omega^2+b^2\right)\left(\mathrm{using} \omega^3=1\right)\)
\(=(a+b)\left(a^2+a b\left(\omega+\omega^2\right)+b^2\right)\)
\(=(a+b)\left(a^2-a b+b^2\right)=a^3+b^3\)

Hint

(a) \(\alpha+\beta=64, \alpha \beta=256\)
\(
\frac{\alpha^{3 / 8}}{\beta^{5 / 8}}+\frac{\beta^{3 / 8}}{\alpha^{5 / 8}}=\frac{\alpha+\beta}{(\alpha \beta)^{5 / 8}}=\frac{64}{\left(2^8\right)^{5 / 8}}=\frac{64}{32}=2
\)

Hint

(b) Let \(\alpha\) and \(\beta\) be the roots of the given quadratic equation, \(2 x^2+2 x-1=0\)
Then, \(\alpha+\beta=-\frac{1}{2} \Rightarrow-1=2 \alpha+2 \beta\) and \(4 \alpha^2+2 \alpha-1=0 \quad[\because \alpha\) is root of eq. (i) \(]\)
\(
\Rightarrow 4 \alpha^2+2 \alpha+2 \alpha+2 \beta=0 \Rightarrow \beta=-2 \alpha(\alpha+1)
\)

Hint

(b) Let \(|x|=y\) then
\(
\begin{aligned}
&9 y^2-18 y+5=0 \\
&\Rightarrow 9 y^2-15 y-3 y+5=0 \\
&\Rightarrow(3 y-1)(3 y-5)=0 \\
&\Rightarrow y=\frac{1}{3} \text { or } \frac{5}{3} \Rightarrow|x|=\frac{1}{3} \text { or } \frac{5}{3}
\end{aligned}
\)
Roots are \(\pm \frac{1}{3}\) and \(\pm \frac{5}{3}\)
\(
\therefore \text { Product }=\frac{25}{81}
\)

Hint

\(
\text { (d) Let } \alpha \text { and } \beta \text { be the roots of the quadratic equation }
\)
\(
\begin{aligned}
&7 x^2-3 x-2=0\\
&\therefore \alpha+\beta=\frac{3}{7}, \alpha \beta=\frac{-2}{7}\\
&\text { Now, } \frac{\alpha}{1-\alpha^2}+\frac{\beta}{1-\beta^2}\\
&=\frac{\alpha-\alpha \beta(\alpha+\beta)+\beta}{1-\left(\alpha^2+\beta^2\right)+(\alpha \beta)^2}\\
&=\frac{(\alpha+\beta)-\alpha \beta(\alpha+\beta)}{1-(\alpha+\beta)^2+2 \alpha \beta+(\alpha \beta)^2}\\
&=\frac{\frac{3}{7}+\frac{2}{7} \times \frac{3}{7}}{1-\frac{9}{49}+2 \times \frac{-2}{7}+\frac{4}{49}}=\frac{27}{16}
\end{aligned}
\)

Hint

(d) \(u=\frac{2(x+i y)+i}{(x+i y)-k i}=\frac{2 x+i(2 y+1)}{x+i(y-k)}\)
Real part of \(u=\operatorname{Re}(u)=\frac{2 x^2+(y-K)(2 y+1)}{x^2+(y-K)^2}\)
Imaginary part of \(u\)
\(
\begin{aligned}
&=\operatorname{Im}(u)=\frac{-2 x(y-K)+x(2 y+1)}{x^2+(y-K)^2} \\
&\because \operatorname{Re}(u)+\operatorname{Im}(u)=1 \\
&\Rightarrow 2 x^2+2 y^2-2 K y+y-K-2 x y+2 K x+2 x y+x \\
&=x^2+y^2+K^2-2 K y
\end{aligned}
\)
Since, the curve intersect at \(y\)-axis
\(
\begin{aligned}
&\therefore x=0 \\
&\Rightarrow y^2+y-K(K+1)=0
\end{aligned}
\)
Let \(y_1\) and \(y_2\) are roots of equations if \(x=0\)
\(
\begin{aligned}
&\because y_1+y_2=-1 \\
&y_1 y_2=-\left(K^2+K\right)
\end{aligned}
\)
\(
\begin{aligned}
&\therefore\left(y_1-y_2\right)^2=\left(1+4 K^2+4 K\right) \\
&\text { Given } P Q=5 \Rightarrow\left|y_1-y_2\right|=5 \\
&\quad \Rightarrow 4 K^2+4 K-24=0 \Rightarrow K=2 \text { or }-3 \\
&\text { as } K>0, \therefore K=2
\end{aligned}
\)

Hint

(b) Since \(\alpha\) is common root of \(x^2-x+2 \lambda=0\) and \(3 x^2-10 x+27 \lambda=0\) \(\therefore 3 \alpha^2-10 \alpha+27 \lambda=0\) \(3 \alpha^2-3 \alpha+6 \lambda=0\)
\(\therefore\) On subtract, we get \(\alpha=3 \lambda\)
Now, \(\alpha \beta=2 \lambda \Rightarrow 3 \lambda \cdot \beta=2 \lambda \Rightarrow \beta=\frac{2}{3}\)
\(
\begin{aligned}
&\Rightarrow \alpha+\beta=1 \Rightarrow 3 \lambda+\frac{2}{3}=1 \Rightarrow \lambda=\frac{1}{9} \text { and } \\
&\alpha \gamma=9 \lambda \Rightarrow 3 \lambda \cdot \gamma=9 \lambda \Rightarrow \gamma=3 \\
&\therefore \frac{\beta \gamma}{\lambda}=18
\end{aligned}
\)

Hint

(d) \(\alpha \cdot \beta=2\) and \(\alpha+\beta=-p\) also \(\frac{1}{\alpha}+\frac{1}{\beta}=-q\)
\(
\Rightarrow p=2 q
\)
\(
\text { Now } \begin{aligned}
&\left(\alpha-\frac{1}{\alpha}\right)\left(\beta-\frac{1}{\beta}\right)\left(\alpha+\frac{1}{\beta}\right)\left(\beta+\frac{1}{\alpha}\right) \\
&=\left[\alpha \beta+\frac{1}{\alpha \beta}-\frac{\alpha}{\beta}-\frac{\beta}{\alpha}\right]\left[\alpha \beta+\frac{1}{\alpha \beta}+\right] \\
=& \frac{9}{2}\left[\frac{5}{2}-\frac{\alpha^2+\beta^2}{2}\right]=\frac{9}{4}\left[5-\left(p^2-4\right)\right]
\end{aligned}
\)
\(
=\frac{9}{4}\left(9-p^2\right) \quad\left[\because \alpha^2+\beta^2=(\alpha+\beta)^2-2 \alpha \beta\right]
\)

Hint

(c) The given quadratic equation is
\(
\left(\lambda^2+1\right) x^2-4 \lambda x+2=0
\)
\(\because\) One root is in the interval \((0,1)\)
\(\therefore f(0) f(1) \leq 0\)
\(\Rightarrow 2\left(\lambda^2+1-4 \lambda+2\right) \leq 0\)
\(\Rightarrow 2\left(\lambda^2-4 \lambda+3\right) \leq 0\)
\((\lambda-1)(\lambda-3) \leq 0 \Rightarrow \lambda \in[1,3]\)
But at \(\lambda=1\), both roots are 1 so \(\lambda \neq 1\)
\(\therefore \lambda \in(1,3]\)

Hint

(c) Since, \(\alpha\) and \(\beta\) are the roots of the equaton \(5 x^2+6 x-2=0\)
Then, \(5 \alpha^2+6 \alpha-2=0,5 \beta^2+6 \beta-2=0\)
\(
\begin{aligned}
&5 \alpha^2+6 \alpha=2 \\
&\begin{aligned}
5 S_6 &+6 S_5=5\left(\alpha^6+\beta^6\right)+6\left(\alpha^5+\beta^5\right) \\
&=\left(5 \alpha^4+6 \alpha^5\right)+\left(5 \beta^6+6 \beta^5\right) \\
&=\alpha^4\left(5 \alpha^2+6 \alpha\right)+\beta^4\left(5 \beta^2+6 \beta\right) \\
&=2\left(\alpha^4+\beta^4\right)=2 S_4
\end{aligned}
\end{aligned}
\)

Hint

(a) Let \(e^x=t \in(0, \infty)\)
Given equation
\(
\begin{aligned}
&t^4+t^3-4 t^2+t+1=0 \\
&\Rightarrow t^2+t-4+\frac{1}{t}+\frac{1}{t^2}=0 \\
&\Rightarrow \quad\left(t^2+\frac{1}{t^2}\right)+\left(t+\frac{1}{t}\right)-4=0
\end{aligned}
\)
Let \(t+\frac{1}{t}=y\)
\(\begin{aligned}\left(y^2-2\right)+y-4 &=0 \quad \Rightarrow \quad \\ y^2+y-6=0 \end{aligned} \quad \Rightarrow \quad \begin{array}{r}y^2+y-6=0 \\ y=-3,2\end{array}\)
\(\Rightarrow y=2 \quad \Rightarrow t+\frac{1}{t}=2\)
\(\Rightarrow e^x+e^{-x}=2\)
\(x=0\), is the only solution of the equation
Hence, there is only one solution to the given equation.

Hint

(a) Let \(z=\alpha \pm i \beta\) be the complex roots of the equation
So, sum of roots \(=2 \alpha=-b\) and
Product of roots \(=\alpha^2+\beta^2=45\)
\(
(\alpha+1)^2+\beta^2=40
\)
Given, \(|z+1|=2 \sqrt{10}\)
\(
\begin{array}{lll}
\Rightarrow & (\alpha+1)^2-\alpha^2=-5 & {\left[Q \beta^2=45-\alpha^2\right]} \\
\Rightarrow & 2 \alpha+1=-5 \quad \Rightarrow & 2 \alpha=-6
\end{array}
\)
Hence, \(b=6\) and \(b^2-b=30\)

Hint

\(
\begin{aligned}
&(d) \alpha^5=5 \alpha+3 \\
&\beta^5=5 \beta+3 \\
&p_5=5(\alpha+\beta)+6=5(1)+6 \\
&\quad\left[\mathrm{Q} \text { from } x^2-x-1=0, \alpha+\beta=\frac{-b}{a}=1\right] \\
&p_5=11 \text { and } p_5=\alpha^2+\beta^2=\alpha+1+\beta+1 \\
&p_2=3 \text { and } p_3=\alpha^3+\beta^3=2 \alpha+1+2 \beta+1 \\
&=2(1)+2=4 \\
&p_2 \times p_3=12 \text { and } p_5=11 \Rightarrow p_5 \neq p_2 \times p_3
\end{aligned}
\)

Hint

(b) \((k+1) \tan ^2 x-\sqrt{2} \lambda \tan x+(k-1)=0\)
\(
\begin{gathered}
\tan \alpha+\tan \beta=\frac{\sqrt{2} \lambda}{k+1} \\ 
\tan \alpha \cdot \tan \beta=\frac{k-1}{k+1} \\ 
\therefore \quad \tan (\alpha+\beta)=\frac{\frac{\sqrt{2} \lambda}{k+1}}{1-\frac{k-1}{k+1}}=\frac{\lambda}{\sqrt{2}} \\
\text { [Product of root } = \frac{k-1}{k+1}] \\
\tan ^2(\alpha+\beta)=\frac{\lambda^2}{2}=50 \\
\lambda=10 .
\end{gathered}
\)

Hint

(d) Consider the quadratic polynomials in the form of the equation
\(
\begin{aligned}
&x^2+20 x-2020=0 \dots(i)\\
&x^2-20 x+2020=0 \dots(ii)
\end{aligned}
\)
Since, \(a\) and \(b\) are roots of the equation(i), then
\(
a+b=-20, a b=-2020
\)
\(c\) and \(d\) are the roots of the equation (ii), then \(c+d=20, c d=2020\)
Now,
\(
\begin{aligned}
& \quad a c(a-c)+a d(a-d)+b c(b-c)+b d(b-d) \\
=& a^2 c-a c^2+a^2 d-a d^2+b^2 c-b c^2+b^2 d-b d^2 \\
=& a^2(c+d)+b^2(c+d)-c^2(a+b)-d^2(a+b) \\
=&(c+d)\left(a^2+b^2\right)-(a+b)\left(c^2+d^2\right) \\
=&(c+d)\left((a+b)^2-2 a b\right)-(a+b)\left((c+d)^2-2 c d\right) \\
=& 20\left[(20)^2+4040\right]+20\left[(20)^2-4040\right] \\
=& 20 \times 800=16000
\end{aligned}
\)
Note: We know that, for a quadratic polynomial of the form \(\mathrm{ax}^2+\mathrm{bx}+\mathrm{c}\), Sum of roots \(=\frac{-\text { Coefficient of } x}{\text { Coefficient of } x^2}=\frac{-b}{a}\),
And product of roots \(=\frac{\text { Constant term }}{\text { Coefficient of } \mathrm{x}^2}=\frac{\mathrm{c}}{\mathrm{a}}\)

Hint

(b) Given equation is, \(x^2+x \sin \mathrm{\theta}-2 \sin \mathrm{\theta}=0\)
\(a+b=-\sin \theta\) and \(a b=-2 \sin \theta\)
\(
\begin{aligned}
&\frac{\left(\alpha^{12}+\beta^{12}\right) \alpha^{12} \beta^{12}}{\left(\alpha^{12}+\beta^{12}\right)(\alpha-\beta)^{24}}=\frac{(\alpha \beta)^{12}}{(\alpha-\beta)^{24}} \\
&|\alpha-\beta|=\sqrt{(\alpha+\beta)^2-4 \alpha \beta}=\sqrt{\sin ^2 \theta+8 \sin \theta} \\
& \frac{(\alpha \beta)^{12}}{(\alpha-\beta)^{24}}=\frac{(2 \sin \theta)^{12}}{\sin ^{12} \theta(\sin \theta+8)^{12}}=\frac{2^{12}}{(\sin \theta+8)^{12}}
\end{aligned}
\)

Hint

According to question,
\(
5+\left|2^x-1\right|=2^x\left(2^x-2\right)
\)
Case I
\(
\begin{aligned}
&2^x \geq 1 \\
&\Rightarrow 5+2^x-1=2^x\left(2^x-2\right)
\end{aligned}
\)
Let \(2^x=\mathrm{t}\)
\(\Rightarrow 5+\mathrm{t}-1=\mathrm{t}(\mathrm{t}-2)\)
\(=t^2-3 t-4=0\)
\(\Rightarrow \mathrm{t}=4\) or \(-1\) (invalid) \(\left(\because 2^{\mathrm{x}} \neq-1\right)\)
\(\Rightarrow 2^x=4\)
\(\Rightarrow x=2\)
\(\therefore\) only one solution
Case II
\(2^x<1\)
\(\Rightarrow 5+1-2^x=2^x\left(2^x-2\right)\)
Let \(2^x=t\)
\(\Rightarrow 5+1-t=t(t-2)\)
\(\Rightarrow \mathrm{t}^2-\mathrm{t}-6=0\)
\(\Rightarrow(\mathrm{t}-3)(\mathrm{t}-2)=0\)
\(\Rightarrow \mathrm{t}=3,-2\)
\(2^x=3\) (invalid as \(2^x<1\) ), or \(2^x=-2\) (invalid)
Therefore, number of real roots is one.

Hint

(d) Since \(2-\sqrt{3}\) is a root of the quadratic equation \(\mathrm{x}^2+\mathrm{px}+\mathrm{q}=0\)
\(\therefore 2+\sqrt{3}\) is the other root
\(\Rightarrow x^2+p x+q=[x-(2-\sqrt{3})[x-(2+\sqrt{3})]\)
\(
\begin{aligned}
&=x^2-(2+\sqrt{3}) x-(2-\sqrt{3}) x+\left(2^2-(\sqrt{3})^2\right) \\
&=x^2-4 x+1
\end{aligned}
\)
Now, by comparing \(p=-4, q=1\)
\(
p^2-4 q-12=16-4-12=0
\)

Hint

(d) Let \(\sqrt{x}=a\)
given equation will become:
\(
\begin{aligned}
&|a-2|+a(a-4)+2=0 \\
&\quad \Rightarrow|a-2|+a^2-4 a+4-2=0 \\
&\quad \Rightarrow|a-2|+(a-2)^2-2=0
\end{aligned}
\)
Let \(|a-2|=y(\) Clearly \(y \geq 0)\)
\(
\begin{aligned}
&\Rightarrow \quad y+y^2-2=0 \\
&\Rightarrow \quad y=1 \text { or }-2 \text { (rejected) } \\
&\Rightarrow \quad|a-2|=1 \Rightarrow a=1,3
\end{aligned}
\)
When \(\sqrt{x}=1 \Rightarrow x=1\)
When \(\sqrt{x}=3 \Rightarrow x=9\)
Hence, the required sum of solutions of the equation \(=10\)

Hint

(c) The given quadratic equation is \(x^2-2 x+2=0\)
Then, the roots of the this equation are \(\frac{2 \pm \sqrt{-4}}{2}=1 \pm i\)
Now, \(\frac{\alpha}{\beta}=\frac{1-i}{1+i}=\frac{(1-i)^2}{1-i^2}=i\)
or \(\frac{\alpha}{\beta}=\frac{1-i}{1+i}=\frac{(1-i)^2}{1-i^2}=-i \quad\) So, \(\frac{\alpha}{\beta}=\pm i\)
Now, \(\left(\frac{\alpha}{\beta}\right)^n=1 \Rightarrow(\pm \mathrm{i})^n=1\)
\(\Rightarrow n\) must be a multiple of 4.
Hence, the required least value of \(n=4\).

Hint

\(
\text { (b) Let roots of the quadratic equation are } \alpha, \beta \text {. }
\)

Given, \(\lambda=\frac{\alpha}{\beta}\) and \(\lambda+\frac{1}{\lambda}=1 \Rightarrow \frac{\alpha}{\beta}+\frac{\beta}{\alpha}=1\) \(\frac{(\alpha+\beta)^2-2 \alpha \beta}{\alpha \beta}=1 \ldots\) (i)
The quadratic equation is, \(3 m^2 x^2+m(m-4) x+2=0\) \( \mathrm{a}+\mathrm{b}=\frac{m(4-m)}{3 m^2}=\frac{4-m}{3 m}\) and \(\alpha \beta=\frac{2}{3 m^2}\)
Put these values in eq (1),
\(
\begin{aligned}
&\frac{\left(\frac{4-m}{3 m}\right)^2}{\frac{2}{3 m^2}}=3 \\
&\Rightarrow(m-4)^2=18 \Rightarrow m=4 \pm \sqrt{18}
\end{aligned}
\)
Therefore, least value is \(4-\sqrt{18}=4-3 \sqrt{2}\)

Hint

(d) Consider the quadratic equation
\(
(c-5) x^2-2 c x+(c-4)=0
\)
Now, \(f(0) . f(3)>0\) and \(f(0) . f(2)<0\)
\(
\begin{aligned}
&\Rightarrow \quad(c-4)(4 c-49)>0 \text { and }(c-4)(c-24)<0 \\
&\Rightarrow \quad c \in(-\infty, 4) \cup\left(\frac{49}{4}, \infty\right) \text { and } c \in(4,24) \\
&\Rightarrow \quad c \in\left(\frac{49}{4}, 24\right)
\end{aligned}
\)
Integral values in the interval \(\left(\frac{49}{4}, 24\right)\) are \(13,14, \ldots, 23\).
\(
\therefore \quad S=\{13,14, \ldots, 23\}
\)

Hint

(a) Consider the equation
\(
\begin{aligned}
&x^2+2 x+2=0 \\
&x=\frac{-2 \pm \sqrt{4-8}}{2}=-1 \pm i
\end{aligned}
\)
Let \(\alpha=-1+i, \beta=-1-i\)
\(
\begin{aligned}
&\alpha^{15}+\beta^{15}=(-1+i)^{15}+(-1-i)^{15} \\
&=\left(\sqrt{2} e^{i \frac{3 \pi}{4}}\right)^{15}+\left(\sqrt{2} e^{-i \frac{3 \pi}{4}}\right)^{15} \\
&=(\sqrt{2})^{15}\left[e^{\frac{i 45 \pi}{4}}+e^{\frac{-i 45 \pi}{4}}\right] \\
&=(\sqrt{2})^{15} \cdot 2 \cos \frac{45 \pi}{4}=(\sqrt{2})^{15} \cdot 2 \cos \frac{3 \pi}{4} \\
&=\frac{-2}{\sqrt{2}}(\sqrt{2})^{15} \\
&=-2(\sqrt{2})^{14}=-256
\end{aligned}
\)

Hint

\(
6 x^2-11 x+ \alpha=0
\)
If roots are rational numbers, then discriminant should be a perfect square.
\(
D=121-24 \alpha
\)
So, If \(\alpha=1, D=97\), which is not a perfect square.
\(\alpha=2, D=73\), which is not a perfect square.
\(\alpha=3, D=49\), it is a perfect square.
\(\alpha=4, D=25\), it is a perfect square.
\(\alpha=5, D=1\), it is a perfect square.
So, number of possible values of \(\alpha\) is 3.

Hint

(b) Here, \(9 x^2+27 x+20=0\)
\(
\begin{aligned}
&\therefore x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a} \\
&\Rightarrow x=\frac{-27 \pm \sqrt{27^2-4 \times 9 \times 20}}{2 \times 9} \\
&\Rightarrow x=-\frac{4}{3},-\frac{5}{3}
\end{aligned}
\)
Given, \(\cos A=-\frac{3}{5}\)
\(\therefore \sec A=\frac{1}{\cos A}=-\frac{5}{3}\)
Here, A is an obtuse angle.
\(
\therefore \tan A=-\sqrt{\sec ^2 A-1}=-\frac{4}{3} .
\)

Hint

(b) As \(\tan A\) and \(\tan B\) are the roots of \(3 x^2-10 x-25=0\), So, \(\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A \tan B}=\frac{\frac{10}{3}}{1+\frac{25}{3}}=\frac{10 / 3}{28 / 3}=\frac{5}{14}\)
Now, \(\cos 2(A+B)=-1+2 \cos ^2(A+B)\)
\(
\begin{aligned}
=& \frac{1-\tan ^2(A+B)}{1+\tan ^2(A+B)} \Rightarrow \cos ^2(A+B)=\frac{196}{221} \\
& \therefore 3 \sin ^2(A+B)-10 \sin (A+B) \cos (A+B)-25 \cos ^2(A+B) \\
=& \cos ^2(A+B)\left[3 \tan ^2(A+B)-10 \tan (A+B)-25\right] \\
=& \frac{75-700-4900}{196} \times \frac{196}{221}=-\frac{5525}{196} \times \frac{196}{221}=-25
\end{aligned}
\)

Hint

(b) \(\alpha, \beta\) are roots of \(x^2-x+1=0\)
\(\therefore \quad \alpha=-\omega\) and \(\beta=-\omega^2\)
where \(\omega\) is cube root of unity
\(
\begin{gathered}
\therefore \quad \alpha^{101}+\beta^{107}=(-\omega)^{101}+(-\omega)^{107} \\
=-\left[\omega^2+\omega\right]=-[-1]=1
\end{gathered}
\)

Hint

\(
\begin{aligned}
&\text { (a) We have, } \sum_{r=1}^n(x+r-1)(x+r)=10 n\\
&\sum_{r=1}^n\left(x^2+x r+(r-1) x+r^2-r\right)=10 n\\
&\Rightarrow \quad \sum_{\mathrm{r}=1}^{\mathrm{n}}\left(\mathrm{x}^2+(2 \mathrm{r}-1) \mathrm{x}+\mathrm{r}(\mathrm{r}-1)\right)=10 \mathrm{n}\\
&\Rightarrow \mathrm{nx}^2+\{1+3+5+\ldots .+(2 \mathrm{n}-1)\} \mathrm{x}\\
&+\{1.2+2.3+\ldots .+(n-1) n\}=10 n
\end{aligned}
\)
\(
\begin{aligned}
&\Rightarrow \quad \mathrm{nx}^2+\mathrm{n}^2 \mathrm{x}+\frac{(\mathrm{n}-1) \mathrm{n}(\mathrm{n}+1)}{3}=10 \mathrm{n} \\
&\Rightarrow \quad \mathrm{x}^2+\mathrm{nx}+\frac{\mathrm{n}^2-31}{3}=0
\end{aligned}
\)
Let \(\alpha\) and \(\alpha+1\) be its two solutions
\(\Rightarrow \quad \alpha+(\alpha+1)=-\mathrm{n} \quad\) ( \(\because\) it has two consequtive integral solutions) \(\Rightarrow \quad \alpha=\frac{-\mathrm{n}-1}{2} \dots(i)\) Also \(\alpha(\alpha+1)=\frac{\mathrm{n}^2-31}{3} \quad\)…(ii)
Putting value of (i) in (ii), we get
\(
\begin{aligned}
&-\left(\frac{\mathrm{n}+1}{2}\right)\left(\frac{1-\mathrm{n}}{2}\right)=\frac{\mathrm{n}^2-31}{3} \\
\Rightarrow \quad \mathrm{n}^2 &=121 \Rightarrow \mathrm{n}=11
\end{aligned}
\)

Hint

(c) \((x-1)\left(x^2+5 x-50\right)=0\)
\(\Rightarrow \quad(\mathrm{x}-1)(\mathrm{x}+10)(\mathrm{x}-5)=0\)
\(\Rightarrow \quad x=1,5,-10\)
Sum \(=-4\)

Hint

(c) \(\left(x^2-5 x+5\right)^{x^2+4 x-60}=1\)
Case I
\(x^2-5 x+5=1\) and \(x^2+4 x-60\) can be any real number \(\Rightarrow \mathrm{x}=1,4\)
Case II
\(x^2-5 x+5=-1\) and \(x^2+4 x-60\) has to be an even number \(\Rightarrow \mathrm{x}=2,3\)
where 3 is rejected because for \(x=3\),
\(x^2+4 x-60\) is odd.
Case III
\(x^2-5 x+5\) can be any real number and
\(x^2+4 x-60=0\)
\(\Rightarrow \mathrm{x}=-10,6\)
\(\Rightarrow\) Sum of all values of \(x\)
\(
=-10+6+2+1+4=3
\)

Hint

(a) \(\sqrt{2 x+1}-\sqrt{2 x-1}=1 \dots(i)\)
\(\Rightarrow \quad 2 x+1+2 x-1-2 \sqrt{4 x^2-1}=1\)
\(\Rightarrow \quad 4 x-1=2 \sqrt{4 x^2-1}\)
\(\Rightarrow \quad 16 x^2-8 x+1=16 x^2-4\)
\(\Rightarrow \quad 8 x=5\)
\(\Rightarrow \quad x=\frac{5}{8}\) which satisfies equation (i)
So, \(\sqrt{4 x^2-1}=\frac{3}{4}\)

Hint

(a) \(\alpha, \beta=\frac{6 \pm \sqrt{36+8}}{2}=3 \pm \sqrt{11}\) \(\alpha=3+\sqrt{11}, \beta=3-\sqrt{11}\)
\(
\begin{aligned}
& \therefore \mathrm{a}_{\mathrm{n}}=(3+\sqrt{11})^{\mathrm{n}}-(3-\sqrt{11})^{\mathrm{n}} \frac{\mathrm{a}_{10}-2 \mathrm{a}_8}{2 \mathrm{a}_9} \\
=& \frac{(3+\sqrt{11})^{10}-(3-\sqrt{11})^{10}-2(3+\sqrt{11})^8+2(3-\sqrt{11})^8}{2\left[(3+\sqrt{11})^9-(3-\sqrt{11})^9\right]} \\
=& \frac{(3+\sqrt{11})^8\left[(3+\sqrt{11})^2-2\right]+(3-\sqrt{11})^8\left[2-(3-\sqrt{11})^2\right]}{2\left[(3+\sqrt{11})^9-(3-\sqrt{11})^9\right]} \\
=& \frac{(3+\sqrt{11})^8(9+11+6 \sqrt{11}-2)+(3-\sqrt{11})^8(2-9-11+6 \sqrt{11})}{2\left[(3+\sqrt{11})^9(3-\sqrt{11})^9\right]} \\
=& \frac{6(3+\sqrt{11})^9-6(3-\sqrt{11})^9}{2\left[(3+\sqrt{11})^9-(3-\sqrt{11})^9\right]}=\frac{6}{2}=3
\end{aligned}
\)

Hint

(c) \(x^2-2 x \sec \theta+1=0 \Rightarrow x=\sec \theta \pm \tan \theta\) and \(\mathrm{x}^2+2 \mathrm{x} \tan \theta-1=0 \Rightarrow \mathrm{x}=-\tan \theta \pm \sec \theta\)
\(
\begin{aligned}
&\because-\frac{\pi}{6}<\theta<-\frac{\pi}{12} \\
&\Rightarrow \sec \frac{\pi}{6}>\sec \theta>\sec \frac{\pi}{12}
\end{aligned}
\)
and \(-\tan \frac{\pi}{6}<\tan \theta<-\tan \frac{\pi}{12}\)
Also \(\tan \frac{\pi}{12}<-\tan \theta<\tan \frac{\pi}{6}\)
Since, \(\alpha_1, \beta_1\) are roots of \(\mathrm{x}^2-2 \mathrm{x} \sec \theta+1=0\) and \(\alpha_1>\beta_1\)
\(\therefore \alpha_1=\sec \theta-\tan \theta\) and \(\beta_1=\sec \theta+\tan \theta\)
Since, \(\alpha_2, \beta_2\) are roots of \(\mathrm{x}^2+2 \mathrm{x} \tan \theta-1=0\) and \(\alpha_2>\beta_2\)
\(
\begin{aligned}
&\therefore \alpha_2=-\tan \theta+\sec \theta, \beta_2=-\tan \theta-\sec \theta \\
&\therefore \alpha_1+\beta_2=\sec \theta-\tan \theta-\tan \theta-\sec \theta=-2 \tan \theta
\end{aligned}
\)

Hint

(c) Consider \(-3(x-[x])^2+2[x-[x])+a^2=0\)
\(\Rightarrow \quad 3\{x\}^2-2\{x\}-a^2=0 \quad(x-[x]=\{x\})\)
\(
\begin{aligned}
&\Rightarrow \quad 3\left(\{x\}^2-\frac{2}{3}\{x\}\right)=a^2, a \neq 0 \\
&\Rightarrow \quad a^2=3\{x\}\left(\{x\}-\frac{2}{3}\right)
\end{aligned}
\)
Now, \(\{x\} \in(0,1)\) and \(\frac{-2}{3} \leq a^2<1\)
(by graph)
Since, \(x\) is not an integer
\(
\therefore \quad a \in(-1,1)-\{0\}
\)
\(
\Rightarrow a \in(-1,0) \cup(0,1)
\)

Hint

(c) \(x^2+|2 x-3|-4=0\)
\(|2 x-3|=\left\{\begin{array}{ccc}(2 x-3) & \text { if } & \mathrm{x}>\frac{3}{2} \\ -(2 x-3) & \text { if } & \mathrm{x}<\frac{3}{2}\end{array}\right.\)
for \(x>\frac{3}{2}, x^2+2 x-3-4=0\)
\(
x^2+2 x-7=0
\)
\(
x=\frac{-2 \pm \sqrt{4+28}}{2}=\frac{-2 \pm 4 \sqrt{2}}{2}=-1 \pm 2 \sqrt{2}
\)
Here \(x=2 \sqrt{2}-1 \quad\left\{2 \sqrt{2}-1<\frac{3}{2}\right\}\)
for \(x<\frac{3}{2}\)
\(
\begin{aligned}
&x^2-2 x+3-4=0 \\
&\Rightarrow x^2-2 x-1=0 \\
&\Rightarrow x=\frac{2 \pm \sqrt{4+4}}{2}=\frac{2 \pm 2 \sqrt{2}}{2}=1 \pm \sqrt{2}
\end{aligned}
\)
Here \(x=1-\sqrt{2} \quad\left\{(1-\sqrt{2})<\frac{3}{2}\right\}\)
Sum of roots: \((2 \sqrt{2}-1)+(1-\sqrt{2})=\sqrt{2}\)

Hint

(d) Quadratic equation with real coefficients and purely imaginary roots can be considered as \(p(x)=x^2+a=0\) where \(a>0\) and \(a \in R\)
\(
\begin{aligned}
&\text { The } p[p(x)]=0 \Rightarrow\left(x^2+a\right)^2+a=0 \\
&\quad \Rightarrow x^4+2 a x^2+\left(a^2+a\right)=0 \\
&\Rightarrow x^2=\frac{-2 a \pm \sqrt{4 a^2-4 a^2-4 a}}{2} \\
&\Rightarrow x^2=-a \pm \sqrt{a} i
\end{aligned}
\)
\(\Rightarrow x=\sqrt{-a \pm \sqrt{a} i}=\alpha \pm i \beta\), where \(\alpha, \beta \neq 0\)
\(\therefore p[p(x)]=0\), has complex roots which are neither purely real nor purely imaginary.

Hint

(b) Given \(\alpha^3+\beta^3=-p\) and \(\alpha \beta=q\)
Let \(\frac{\alpha^2}{\beta}\) and \(\frac{\beta^2}{\alpha}\) be the root of required quadratic equation.
So, \(\frac{\alpha^2}{\beta}+\frac{\beta^2}{\alpha}=\frac{\alpha^3+\beta^3}{\alpha \beta}=\frac{-p}{q}\)
and \(\frac{\alpha^2}{\beta} \times \frac{\beta^2}{\alpha}=\alpha \beta=q\)
Hence, required quadratic equation is
\(
\begin{aligned}
&x^2-\left(\frac{-p}{q}\right) x+q=0 \\
&\Rightarrow x^2+\frac{p}{q} x+q=0 \Rightarrow q x^2+p x+q^2=0
\end{aligned}
\)

Hint

(c) \(\because \alpha, \beta\) are the roots of \(x^2-6 x-2=0\)
\(
\begin{aligned}
&\therefore \alpha^2-6 \alpha-2=0 \\
&\quad \Rightarrow \alpha^{10}-6 \alpha^9-2 \alpha^8=0 \\
&\quad \Rightarrow \alpha^{10}-2 \alpha^8=6 \alpha^9 \quad \ldots \text { (i) } \\
&\text { Similarly } \beta^{10}-2 \beta^8=6 \beta^9 \quad \ldots \text { (ii) } \\
&\text { On subtracting (ii) from (i), } \\
&\alpha^{10}-\beta^{10}-2\left(\alpha^8-\beta^8\right)=6\left(\alpha^9-\beta^9\right) \\
&\quad \Rightarrow a_{10}-2 a_8=6 a_9 \Rightarrow \frac{a_{10}-2 a_8}{2 a_9}=3
\end{aligned}
\)

Hint

\(
(2 x)^{\ln 2}=(3 y)^{\ln 3}
\)
Taking \(\ln\) both sides,
\(
\begin{aligned}
&\Rightarrow(\ln 2)(\ln 2 x)=(\ln 3)(\ln 3 y) \\
&\Rightarrow(\ln 2)(\ln 2+\ln x)=(\ln 3)(\ln 3+\ln y) \rightarrow(1)
\end{aligned}
\)
Now, we will take the second equation,
\(
3^{\ln x}=2^{\ln y}
\)
Taking \(\ln\) both sides,
\(
\begin{aligned}
&\Rightarrow(\ln x)(\ln 3)=(\ln y)(\ln 2) \\
&\Rightarrow(\ln y)=\frac{(\ln x)(\ln 3)}{\ln 2}
\end{aligned}
\)
Putting value of \(\ln y\) in (1),
\(
\begin{aligned}
&(\ln 2)(\ln 2+\ln x)=(\ln 3)\left(\ln 3+\frac{(\ln x)(\ln 3)}{\ln 2}\right) \\
&\Rightarrow(\ln x)\left(\ln 2-\left(\frac{(\ln 3)^2}{\ln 2}\right)\right)=(\ln 3)^2-(\ln 2)^2 \\
&\Rightarrow \frac{\ln x}{\ln 2}\left((\ln 2)^2-(\ln 3)^2\right)=(\ln 3)^2-(\ln 2)^2 \\
&\Rightarrow \frac{\ln x}{\ln 2}=-1 \\
&\Rightarrow(\ln x)=\ln (2)^{-1} \\
&\Rightarrow x=2^{-1} \Rightarrow x=\frac{1}{2}, \text { which is the required value of } x_0 \text {. }
\end{aligned}
\)

Hint

(b) 

\(
\begin{aligned}
& \alpha^3+\beta^3=q \\
& \Rightarrow(\alpha+\beta)^3-3 \alpha \beta(\alpha+\beta)=q \\
& \Rightarrow-p^3+3 p \alpha \beta=q \\
& \Rightarrow \alpha \beta=\frac{q+p^3}{3 p} \\
& x^2-\left(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}\right) x+\frac{\alpha}{\beta} \cdot \frac{\beta}{\alpha}=0 \\
& \Rightarrow x^2-\frac{p^2-2\left(\frac{p^3+q}{3 p}\right)}{\frac{p^3+q}{3 p}} x+1=0 \\
& \Rightarrow x^2-\left(\frac{(\alpha+\beta)^2-2 \alpha \beta}{\alpha \beta}\right) x+1=0 \\
& \Rightarrow\left(p^3+q\right) x^2-\left(3 p^3-2 p^3-2 q\right) x+\left(p^3+q\right)=0 \\
& \Rightarrow\left(p^3+q\right) x^2-\left(p^3-2 q\right) x+\left(p^3+q\right)=0
\end{aligned}
\)

Hint

(d) Since \(\alpha\) and \(\beta\) are the roots of \(x^2-p x+r=0\) \(\therefore \quad \alpha+\beta=p \dots(i)\)
and \(\alpha \beta=r \dots(ii)\)
Also \(\frac{\alpha}{2}\) and \(2 \beta\) are the roots of \(x^2-q x+r=0\)
\(
\therefore \quad \frac{\alpha}{2}+2 \beta=q \Rightarrow \alpha+4 \beta=2 q \dots(iii)
\)
Solving (i) and (iii) for \(\alpha\) and \(\beta\), we get
\(
\beta=\frac{1}{3}(2 q-p) \text { and } \alpha=\frac{2}{3}(2 q-q)
\)
On substituting the values of \(\alpha\) and \(\beta\), in equation (ii), we get \(\frac{2}{9}(2 p-q)(2 q-p)=r\).

Hint

(a) \(\because a, b, c\) are sides of a triangle and \(a \neq b \neq c\) \(\therefore \quad|a-b|<|c| \Rightarrow a^2+b^2-2 a b<c^2 \dots(i)\)
Similarly,
\(
b^2+c^2-2 b c<a^2 \quad \ldots \text {. (ii) } ; c^2+a^2-2 c a<b^2 \dots(iii)
\)
On adding, (i), (ii) and (iii) we get
\(
\begin{aligned}
&a^2+b^2+c^2<2(a b+b c+c a) \\
&\Rightarrow \quad \frac{a^2+b^2+c^2}{a b+b c+c a}<2 \dots(iv) \\
&\because \quad \text { Roots of the given equation are real } \\
&\therefore \quad(a+b+c)^2-3 \lambda(a b+b c+c a) \geq 0 \\
&\Rightarrow \quad \frac{a^2+b^2+c^2}{a b+b c+c a} \geq 3 \lambda-2 \dots(v)
\end{aligned}
\)
From (iv) and (v), \(3 \lambda-2<2 \Rightarrow \lambda<\frac{4}{3}\)

Hint

(a) \(x^2+p x+q=0\)
Let roots be \(\alpha\) and \(\alpha^2\), then
\(
\alpha+\alpha^2=-p, \alpha \alpha^2=q \Rightarrow \alpha=q^{1 / 3}
\)
\(
\therefore \quad(q)^{1 / 3}+\left(q^{1 / 3}\right)^2=-p
\)
On taking cube on both sides, we get
\(
\begin{aligned}
&q+q^2+3 q\left(q^{1 / 3}+q^{2 / 3}\right)=-p^3 \\
&\Rightarrow q+q^2-3 p q=-p^3 \Rightarrow p^3+q^2-q(3 p-1)=0
\end{aligned}
\)

Hint

(c) Let \(\alpha, \alpha^2\) be the roots of \(3 x^2+p x+3\).
\(\therefore \quad \alpha+\alpha^2=-p / 3\) and \(\alpha^3=1\)
\(\Rightarrow(\alpha-1)\left(\alpha^2+\alpha+1\right)=0 \Rightarrow \alpha=1\) or \(\alpha^2+\alpha=-1\)
If \(\alpha=1\), then \(p=-6\), which is not possible as \(p>0\)
If \(\alpha^2+\alpha=-1 \Rightarrow-p / 3=-1 \Rightarrow p=3\).

Hint

(d) Given: \((x-a)(x-b)-1=0, b>a\).
or \(x^2-(a+b) x+(a b-1)=0\)
Let \(f(x)=x^2-(a+b) x+(a b-1)\)
\(
\begin{aligned}
D=(a+b)^2-4(a b-1) \\
=(a-b)^2+1>0
\end{aligned}
\)
Since coeff. of \(x^2\) i.e. \(1>0, \therefore f(x)\) represents upward parabola, intersecting \(x\)-axis at two points corresponding to two real roots, \(D\) being \(+\mathrm{ve}\).
\(\operatorname{Also} f(a)=f(b)=-1\)
\(\Rightarrow\) curve is below \(x\)-axis at \(a\) and \(b\)
\(\therefore \quad a\) and \(b\) both lie between the roots.
Therefore, the graph of the given equation is as shown.

It is clear from graph, that one root of the equation lies in \((-\infty, a)\) and other in \((b, \infty)\).

Hint

(b) Given : \(\mathrm{c}<0<\mathrm{b}\) and \(\alpha+\beta=-\mathrm{b} \dots(i)\)
\(\alpha \beta=\mathrm{c} \quad\)….(ii)
From (ii), c \(<0 \Rightarrow \alpha \beta<0 \Rightarrow\) Either \(\alpha\) is -ve or \(\beta\) is – ve and second quantity is positive.
From (i), \(b>0 \Rightarrow-\mathrm{b}<0 \Rightarrow \alpha+\beta<0\)
\(\Rightarrow\) the sum is negative
\(\Rightarrow\) (Modules of negative quantity) \(>\) (Modulus of positive quantity)
But given \(\alpha<\beta\). Therefore, it is clear that \(\alpha\) is negative and \(\beta\) is positive and modulus of \(\alpha\) is greater than modulus of \(\beta\)
\(
\Rightarrow \alpha<0<\beta<|\alpha|
\)

Hint

(a) If both roots of a quadratic equation \(a x^2+b x+c=0\) are less than \(\mathrm{k}\), then \(f(k)>0, \mathrm{D} \geq 0, \alpha+\beta<2 \mathrm{k}\)
\(
\begin{aligned}
f(x) &=x^2-2 a x+a^2+a-3=0 \\
f(3)>0, & \alpha+\beta<6, D \geq 0 . \\
& \Rightarrow a^2-5 a+6>0, a<3,-4 a+12 \geq 0 \\
& \Rightarrow a<2 \text { or } a>3, a<3, a<3 \Rightarrow a<2
\end{aligned}
\)

Hint

(c) For the equation \(p x^2+q x+1=0\) to have real roots \(D \geq 0 \Rightarrow q^2 \geq 4 p\)
If \(p=1\) then \(q^2 \geq 4 \quad \Rightarrow q=2,3,4\)
If \(p=2\) then \(q^2 \geq 8 \quad \Rightarrow q=3,4\)
If \(p=3\) then \(q^2 \geq 12 \Rightarrow q=4\)
If \(p=4\) then \(q^2 \geq 16 \Rightarrow q=4\)
\(\therefore \quad\) Number of required equations \(=7\)

Hint

(c) \(\alpha, \beta\) are roots of the equation \((x-a)(x-b)=c, c \neq 0\)
\(\therefore \quad(x-a)(x-b)-c=(x-\alpha)(x-\beta)\)
\(\Rightarrow(x-\alpha)(x-\beta)+c=(x-\mathbf{a})(x-b)\)
\(\Rightarrow\) Roots of \((x-\alpha)(x-\beta)+c=0\) are \(a\) and \(b\)

Hint

(d) If \(f(\alpha)\) and \(f(\beta)\) are of opposite signs then there must lie a value \(\gamma\) between \(\alpha\) and \(\beta\) such that \(f(\gamma)=0\).
\(a, b, c\) are real numbers and \(a \neq 0\).
Since \(\alpha\) is a root of \(a^2 x^2+b x+c=0\)
\(
\therefore a^2 \alpha^2+b \alpha+c=0 \dots(i)
\)
Also \(\beta\) is a root of \(a^2 x^2-b x-c=0\) \(\therefore a^2 \beta^2-b \beta-c=0 \quad\)… (ii)
Now, let \(f(x)=a^2 x^2+2 b x+2 c\)
Then \(f(\alpha)=a^2 \alpha^2+2 b \alpha+2 c=a^2 \alpha^2+2(b \alpha+\mathrm{c})\) \(=a^2 \alpha^2+2\left(-a^2 \alpha^2\right)\)
[using eq. (i) \(]\) \(=-a^2 \alpha^2\).
\(
\begin{aligned}
&\text { and } f(\beta)=a^2 \beta^2+2 b \beta+2 c=a^2 \beta^2+2(b \beta+c) \\
&=a^2 \beta^2+2\left(a^2 \beta^2\right) \\
&=3 a^2 \beta^2>0 \text {. } \\
&\text { Since } f(\alpha) \text { and } f(\beta) \text { are of opposite signs and } \gamma \text { is a root of equation } f(x)= \\
&\quad 0 \\
&\quad \therefore \gamma \text { must lie between } \alpha \text { and } \beta \\
&\quad \Rightarrow \alpha<\gamma<\beta \text {. }
\end{aligned}
\)

Hint

(a) Given : \(x-\frac{2}{x-1}=1-\frac{2}{x-1}\)
Clearly \(x \neq 1\) for the given equation to be defined. If \(x \neq 1\), we can cancel the common term \(\frac{-2}{x-1}\) on both sides to get \(x=1\), but it is not possible. So given equation has no roots.

Hint

(c) Since, \(\left(x^2+p x+1\right)\) is a factor of \(a x^3+b x+c\), hence we can assume that zeros of \(x^2+p x+1\) are \(\alpha, \beta\) and that of \(a x^3+b x+c\) be \(\alpha, \beta, \gamma\)
\(
\begin{gathered}
\therefore \quad \alpha+\beta=-p \dots(i)\\
\alpha \beta=1 \quad \ldots .(\text { ii) } \\
\text { and } \alpha+\beta+\gamma=0 \dots(iii) \\
\alpha \beta+\beta \gamma+\gamma \alpha=\frac{b}{a} \dots(iv) \\
\alpha \beta \gamma=\frac{-c}{a} \dots(v)
\end{gathered}
\)
On solving (ii) and (v), we get \(\gamma=-c / a\).
On solving (i) and (iii), we get \(\gamma=p\) \(\therefore \quad p=\gamma=-c / a\)
Using equations (i), (ii) and (iv), we get
\(
\begin{aligned}
&1+\gamma(-p)=\frac{b}{a} \\
&\Rightarrow 1+\left(-\frac{c}{a}\right)\left(\frac{c}{a}\right)=\frac{b}{a} \quad( \gamma=p=-c / a) \\
&\Rightarrow 1-\frac{c^2}{a^2}=\frac{b}{a} \Rightarrow a^2-c^2=a b
\end{aligned}
\)

Hint

(b) Given :

\(
\begin{aligned}
&(x-b)(x-c)+(x-a)(x-c)+(x-a)(x-b)=0 \\
& \Rightarrow 3 x^2-2(a+b+c) x+(a b+b c+c a)=0 \\
D=& 4(a+b+c)^2-12(a b+b c+c a) \\
&=4\left[a^2+b^2+c^2-a b-b c-c a\right] \\
&=2\left[(a-b)^2+(b-c)^2+(c-a)^2\right] \geq 0 \quad \forall a, b, c \\
& \therefore \quad \text { Roots of given equation are always real. }
\end{aligned}
\)

Hint

\((l-m) x^2-5(l+m) x-2(l-m)=0\) is the given equation.
Comparing given equation with standard form of quadratic equation \(a x^2+b x+c=0\)
we get \(a=l-m, b=-5(l+m), c=-2(l-m)\)
The nature of the roots of a quadratic equation depends on the value of where \(D\) is given as
\(
D=b^2-4 a c
\)
Substituting the values of \(a, b, c\) for the given equation we get
\(
\begin{aligned}
&\Rightarrow D=5^2(l+m)^2-4(l-m)(-2)(l-m) \\
&\Rightarrow D=25(l+m)^2+8(l-m)^2
\end{aligned}
\)
For all real values of \(l, m(l+m)^2>0,(l-m)^2>0\) as \(l \neq m\) \(\Rightarrow D>0\)
When \(D>0\) for a quadratic equation, the equation has real and unequal roots.
Hence, the given quadratic equation \((l-m) x^2-5(l+m) x-2(l-m)=0\) has real and unequal roots,

Hint

(8) Since, \(2 x^2+(a-10) x+\frac{33}{2}=2 a\) has real roots, \(\therefore \quad D \geq 0\)
\(
\Rightarrow \quad(a-10)^2-4(2)\left(\frac{33}{2}-2 a\right) \geq 0
\)
\(
\begin{aligned}
&\Rightarrow \quad(a-10)^2-4(33-4 a) \geq 0 \\
&\Rightarrow \quad a^2-4 a-32 \geq 0 \\
&\Rightarrow \quad(a-8)(a+4) \geq 0
\end{aligned}
\)
\(\Rightarrow \quad a \leq-4 \cup a \geq 8\)
\(\Rightarrow \quad a \in(-\infty,-4] \cup[8, \infty)\)

Hint

(2) The given equation is
\(
x^2-8 k x+16\left(k^2-k+1\right)=0
\)
\(\because \quad\) Both the roots are real and distinct.
\(
\begin{aligned}
&\therefore \quad D>0 \Rightarrow(8 k)^2-4 \times 16\left(k^2-k+1\right)>0\\
&\Rightarrow \mathrm{k}>1 \dots(i) \\
&\because \quad \text { Both the roots are greater than or equal to } 4\\
&\therefore \quad \alpha+\beta>8 \text { and } f(4) \geq 0\\
&\Rightarrow k>1 \dots(ii) \\
&\text { and } 16-32 k+16\left(k^2-k+1\right) \geq 0\\
&\Rightarrow k^2-3 k+2 \geq 0 \Rightarrow(k-1)(k-2) \geq 0\\
&\Rightarrow k \in(-\infty, 1] \cup[2, \infty) \dots(iii) \\
&\text { Combining (i), (ii) and (iii), we get } k \geq 2\\
&\therefore \quad \text { Smallest value of } k=2 .
\end{aligned}
\)

Hint

The given equation : \(x^2-3 k x+2 e^{2 \ln k}-1=0\)
\(
\Rightarrow x^2-3 k x+\left(2 k^2-1\right)=0
\)
Now, product of roots \(=2 k^2-1\)
\(
\therefore 2 \mathrm{k}^2-1=7 \Rightarrow \mathrm{k}^2=4 \quad \Rightarrow \mathrm{k}=2,-2
\)
For real roots, \(D \geq 0\)
\(
\Rightarrow 9 k^2-4\left(2 k^2-1\right) \geq 0 \quad \Rightarrow k^2+4 \geq 0
\)
which is true for all \(k\). Thus \(k=2,-2\)
But for \(k=-2, \ln k\) is not defined
We reject \(k=-2\), we get \(k=2\).

Hint

Since, \(p\) and \(q\) are real and one root is \(2+\mathrm{i} \sqrt{3}\), therefore other root should be \(2-i \sqrt{3}\)
\(\therefore \quad p=-(\) sum of roots \()=-4, q=\) product of roots \(=4+3=7\)

Hint

(True) \(f(x)=(x-a)(x-c)+2(x-b)(x-d)\).
\(f(\mathrm{a})=+v e ; f(b)=-v e ; f(c)=-v e ; f(d)=+v e\)
\(\therefore \quad\) There exists two real and distinct roots one in the interval \((a, b)\) and other in \((c, d)\).

Hint

(False) \(2 x^2+3 x+1=0 \Rightarrow x=-1,-1 / 2\); both are rationals \(\therefore \quad\) Statement is false.

Hint

(b) \(\alpha^2=\alpha+1\)
\(
\beta^2=\beta+1
\)
\(a_n=p \alpha^n+q \beta^n\)
\(
\begin{aligned}
=& p\left(\alpha^{n-1}+\alpha^{n-2}\right)+q\left(\beta^{n-1}+\beta^{n-2}\right) \\
=a_{n-1} &+a_{n-2} \\
\therefore \quad & a_{12}=a_{11}+a_{10}
\end{aligned}
\)

Hint

\(
\begin{aligned}
{ (d) } \alpha=\frac{1+\sqrt{5}}{2}, \beta=\frac{1-\sqrt{5}}{2} \\
 a_4=a_3+a_2 \\
= 2 a_2+a_1 \\
\quad=3 a_1+2 a_0 \\
28= p(3 \alpha+2)+q(3 \beta+2) \\
28=(p+q)\left(\frac{3}{2}+2\right)+(p-q)\left(\frac{3 \sqrt{5}}{2}\right) \\
\therefore \quad p-q=0 \text { and }(p+q) \times \frac{7}{2}=28 \\
\Rightarrow  p+q=8 \Rightarrow p=q=4 \\
\therefore p+2 q=12
\end{aligned}
\)

Hint

(b) As \(a, b, c, p, q, \in R\) and the two given equations have exactly one common root
\(\Rightarrow\) Either both equations have real roots
or both eqations have imaginary roots
or \(p^2-q \leq 0\) and \(b^2-a c \leq 0\)
\(
\Rightarrow\left(p^2-q\right)\left(b^2-a c\right) \geq 0
\)
\(\therefore\) Statement 1 is true.
Also we have \(\alpha \beta=q\) and \(\frac{\alpha}{\beta}=\frac{c}{a}\)
\(
\therefore \frac{\alpha \beta}{\alpha / \beta}=\frac{q}{c} \times a \Rightarrow \beta^2=\frac{q a}{c}
\)
As \(\hat{\mathrm{a}} \neq 1\) or \(-1 \Rightarrow \hat{\mathrm{a}}^2 \neq 1 \Rightarrow \frac{q a}{c} \neq 1\) or \(c \neq q a\)
Again, as exactly one root á is common, and \(\hat{a} \neq 1\)
\(
\therefore \text { á }+\hat{\mathrm{a}} \neq \mathrm{á}+\frac{1}{\hat{\mathrm{a}}} \Rightarrow \frac{-2 b}{a} \neq-2 p \quad \Rightarrow b \neq a p
\)
\(\therefore\) Statement 2 is correct.
But Statement 2 is not a correct explanation of Statement 1.

Hint

Roots of \(x^2-10 c x-11 d=0\) are \(a\) and \(b\) \(\Rightarrow a+b=10 c\) and \(a b=-11 d\)
Similarly \(c\) and \(d\) are the roots of \(x^2-10 a x-11 b=0 \Rightarrow c+d=10 a\) and \(c d\)
\(
\begin{aligned}
&=-11 b \\
&\Rightarrow a+b+c+d=10(a+c) \text { and } a b c d=121 b d \\
&\Rightarrow b+d=9(a+c) \text { and } a c=121
\end{aligned}
\)
Also we have \(a^2-10 a c-11 d=0\) and \(c^2-10 a c-11 b=0\)
\(
\begin{aligned}
&\Rightarrow a^2+c^2-20 a c-11(b+d)=0 \\
&\Rightarrow(a+c)^2-22 \times 121-99(a+c)=0 \\
&\Rightarrow a+c=121 \text { or }-22
\end{aligned}
\)
For \(a+c=-22\), we get \(a=c\)
\(\therefore \quad\) Rejecting this value we have \(a+c=121\)
\(
\therefore \quad a+b+c+d=10(a+c)=1210
\)

Hint

Given :
\(
x^2+(a-b) x+(1-a-b)=0, a, b \in R
\)
For this equation to have unequal real roots for all value of \(b\)
\(
\text { if } \quad \begin{aligned}
D &>0 \\
& \Rightarrow(a-b)^2-4(1-a-b)>0 \\
& \Rightarrow a^2+b^2-2 a b-4+4 a+4 b>0 \\
& \Rightarrow b^2+b(4-2 a)+a^2+4 a-4>0
\end{aligned}
\)
Which is a quadratic expression in \(b\), and it will be true for all \(b \in R\), if discriminant of above equation is less than zero.
\(
\text { i.e., } \begin{aligned}
&(4-2 a)^2-4\left(a^2+4 a-4\right)<0 \\
& \Rightarrow \quad(2-a)^2-\left(a^2+4 a-4\right)<0 \\
& \Rightarrow \quad 4-4 a+a^2-a^2-4 a+a<0 \\
& \Rightarrow \quad-8 a+8<0 \therefore a>1
\end{aligned}
\)

Hint

We know \((\alpha-\beta)^2=[(\alpha+\delta)-(\beta+\delta)]^2\)
\(
\Rightarrow \quad(\alpha+\beta)^2-4 \alpha \beta=(\alpha+\delta+\beta+\delta)^2-4(\alpha+\delta)(\beta+\delta)
\)
\(
\Rightarrow \frac{b^2}{a^2}-\frac{4 c}{a}=\frac{B^2}{A^2}-\frac{4 C}{A} \Rightarrow \frac{4 a c-b^2}{a^2}=\frac{4 A C-B^2}{A^2}
\)
\(\left[\operatorname{Here} \alpha+\beta=-\frac{b}{a}, \alpha \beta=\frac{c}{a}\right.\),
\(
\left.(\alpha+\delta)(\beta+\delta)=-\frac{B}{A} \operatorname{and}(\alpha+\delta)(\beta+\delta)=\frac{C}{A}\right]
\)

Hint

Given : For \(a, b, c \in R, a x^2+b x+c=0\) has two real roots \(\alpha\) and \(\beta\), where \(\alpha<-1\) and \(\beta>1\). There may be two cases depending upon the value of a, as shown below.
In each of cases (i) and (ii) \(a f(-1)<0\) and \(a f(1)<0\)

\(\Rightarrow a(a-b+c)<0\) and \(a(a+b+c)<0\)
Dividing by \(a^2(>0)\), we get
\(
1-\frac{b}{a}+\frac{c}{a}<0 \dots(i)
\)
and \(1+\frac{b}{a}+\frac{c}{a}<0 \dots(ii)\)
On combining (i) and (ii) we get
\(
1+\left|\frac{b}{a}\right|+\frac{c}{a}<0 \quad \text { or } \quad 1+\frac{c}{a}+\left|\frac{b}{a}\right|<0
\)

Hint

Given \(
\left|x^2+4 x+3\right|+2 x+5=0
\)
Here two cases are possible.
Case I: \(x^2+4 x+3 \geq 0 \Rightarrow(x+1)(x+3) \geq 0\)
\(\Rightarrow x \in(-\infty,-3] \cup[-1, \infty) \dots(i)\)
Then the given equation becomes,
\(
\begin{aligned}
&\Rightarrow x^2+6 x+8=0 \\
&\Rightarrow(x+4)(x+2)=0, \quad \therefore x=-4,-2
\end{aligned}
\)
But \(x=-2\) does not satisfy (i) and hence rejected.
\(\therefore \quad\) Solution is \(x=-4\)
Case II : \(x^2+4 x+3<0\)
\(\Rightarrow \quad(x+1)(x+3)<0\)
\(\Rightarrow \quad x \in(-3,-1) \dots(ii)\)
Then the given equation becomes,
\(
-\left(x^2+4 x+3\right)+2 x+5=0
\)
\(
\begin{aligned}
&\Rightarrow-x^2-2 x+2=0 \Rightarrow x^2+2 x-2=0 \\
&\Rightarrow x=\frac{-2 \pm \sqrt{4+8}}{2} \quad \therefore x=-1+\sqrt{3},-1-\sqrt{3}
\end{aligned}
\)
But \(x=-1+\sqrt{3}\) does not satisfy (ii) and hence rejected.
\(\therefore \quad\) Solution is \(x=-1-\sqrt{3}\)
On combining solutions in the two cases, we get the solutions : \(x=-4,-1-\) \(\sqrt{3}\).

Hint

Given:
\(x^2-2 a|x-a|-3 a^2=0 \dots(i)\)
Here two cases are possible.
Case I : \(x-a>0\), then \(|x-a|=x-a\)
Hence, Eq. (i) becomes
\(
\begin{aligned}
&x^2-2 a(x-a)-3 a^2=0 \\
&\Rightarrow \quad x^2-2 a x-a^2=0 \Rightarrow \quad x=\frac{2 a \pm \sqrt{4 a^2+4 a^2}}{2} \\
&\therefore \quad x=a \pm a \sqrt{2}
\end{aligned}
\)
Case II : \(x-a<0\), then \(|x-a|=-(x-a)\)
Hence, Eq. (i) becomes
\(
\begin{aligned}
&x^2+2 a(x-a)-3 a^2=0 \\
&\Rightarrow \quad x^2+2 a x-5 a^2=0 \Rightarrow \quad x=\frac{-2 a \pm \sqrt{4 a^2+20 a^2}}{2} \\
&\therefore \quad x=\frac{-2 a \pm 2 a \sqrt{6}}{2} \Rightarrow \quad x=-a \pm a \sqrt{6}
\end{aligned}
\)
Hence, the solution set is \(\{a \pm a \sqrt{2},-a \pm a \sqrt{6}\}\)

Hint

Given, \((5+2 \sqrt{6})^{x^2-3}+(5-2 \sqrt{6})^{x^2-3}=10 \dots(i)\)
Put \(y=(5+2 \sqrt{6})^{x^2-3} \Rightarrow(5-2 \sqrt{6})^{x^2-3}=\frac{1}{y}\)
From Eq. (i), \(y+\frac{1}{y}=10\)
\(
\begin{array}{ll}
& \Rightarrow y^2-10 y+1=0 \Rightarrow y=5 \pm 2 \sqrt{6} \\
& \Rightarrow(5+2 \sqrt{6})^{x^2-3}=5+2 \sqrt{6} \\
\text { or } & (5+2 \sqrt{6})^{x^2-3}=5-2 \sqrt{6} \\
& \Rightarrow x^2-3=1 \quad \text { or } x^2-3=-1 \\
& \Rightarrow x=\pm 2 \text { or } x=\pm \sqrt{2} \\
& \Rightarrow x=\pm 2, \pm \sqrt{2}
\end{array}
\)

Hint

Given \(a>0\), so we have to consider two cases :
\(a \neq 1\) and \(a=1\).
Also it is clear that \(x>0\)
and \(x \neq 1, a x \neq 1, a^2 x \neq 1\).
Case I : If \(a>0, \neq 1\)
then given equation can be simplified as
\(
\frac{2}{\log _a x}+\frac{1}{1+\log _a x}+\frac{3}{2+\log _a x}=0
\)
Putting \(\log _a x=y\), we get
\(
2(1+y)(2+y)+y(2+y)+3 y(1+y)=0
\)
\(
\begin{aligned}
&\Rightarrow \quad 6 y^2+11 y+4=0 \Rightarrow y=-4 / 3 \text { and }-1 / 2 \\
&\Rightarrow \log _a x=-4 / 3 \text { and } \log _a x=-1 / 2 \\
&\Rightarrow x=a^{-4 / 3} \text { and } x=a^{-1 / 2}
\end{aligned}
\)

Hint

\(\sqrt{x+1}=1+\sqrt{x-1}\)
Squaring both sides, we get
\(
\begin{aligned}
&x+1=1+x-1+2 \sqrt{x-1} \Rightarrow 1=2 \sqrt{x-1} \\
&\Rightarrow 1=4(x-1) \\
&\Rightarrow \quad x=5 / 4
\end{aligned}
\)

Hint

(a) \(a x^2-2 b x+5=0\),
If \(\alpha\) and \(\alpha\) are roots of equations, then sum of roots
\(
2 \alpha=\frac{2 b}{a} \quad \Rightarrow \quad \alpha=\frac{b}{a}
\)
and product of roots \(=\alpha^2=\frac{5}{a} \Rightarrow \frac{b^2}{a^2}=\frac{5}{a}\)
\(\quad \Rightarrow \quad b^2=5 a \quad(a \neq 0)\)
For \(x^2-2 b x-10=0\)
\(\alpha+\beta=2 b\)
and \(\alpha \beta=-10\)
\(\alpha=\frac{b}{a}\) is also root of \(x^2-2 b x-10=0\)
\(\Rightarrow \quad b^2-2 a b^2-10 a^2=0\)
By eqn. (i) \(\Rightarrow 5 a-10 a^2-10 a^2=0\)
\(
\Rightarrow 20 a^2=5 a \Rightarrow a=\frac{1}{4} \text { and } b^2=\frac{5}{4}
\)
\(\alpha^2=20\) and \(\beta^2=5\)
Now, \(\alpha^2+\beta^2=5+20=25\)

Hint

The triangle is possible if,
\(
\begin{aligned}
&5+5 r>5 r^2 \cdots(1) \\
&5 r+5 r^2>5 \cdots(2) \\
&5 r^2+5>5 r \cdots(3)
\end{aligned}
\)
From eqn (1)
\(
\begin{aligned}
&5 r^2-5 r-5<0 \\
&\Rightarrow r^2-r-1<0 \\
&\Rightarrow\left[r-\left(\frac{1-\sqrt{5}}{2}\right)\right]\left[r-\left(\frac{1+\sqrt{5}}{2}\right)\right]<0 \\
&\Rightarrow 0<r<\frac{1+\sqrt{5}}{2} \quad \cdots(4) \\
&\because r>0
\end{aligned}
\)
From eqn (2)
\(
5 r^2+5 r-5>0
\)
\(
\begin{aligned}
&r^2+r-1>0 \\
&\left(r-\left(\frac{-1-\sqrt{5}}{2}\right)\right)\left(r-\left(\frac{-1+\sqrt{5}}{2}\right)\right)>0 \\
&r>\frac{-1+\sqrt{5}}{2} \quad \cdots(5) 
\end{aligned}
\)
\(
\because r>0
\)
From eqn (3)
\(
\begin{aligned}
&5 r^2-5 r+5>0 \\
&\Rightarrow r^2-r+1>0 \\
&\Rightarrow r \in R^{+} \cdots(6) \\
&(\because D=1-4=-3<0)
\end{aligned}
\)
Therefore, from (4), (5) and (6), we get \(\left(\frac{-1+\sqrt{5}}{2}\right)<r<\left(\frac{1+\sqrt{5}}{2}\right)\) \(r \in(0.618,1.618)\)
Hence, \(r \neq \frac{7}{4}\)

Hint

(b) \(|z-(3-2 i)| \leq 4\) represents a circle whose centre is \((3,-2)\) and radius \(=4\).
\(|z|=|z-0|\) represents the distance of point ‘ \(z\) ‘ from origin \((0,0)\)

Suppose \(R S\) is the normal of the circle passing through origin ‘ \(O\) ‘ and \(G\) is its center \((3,-2)\).
Here, \(O R\) is the least distance and \(O S\) is the greatest distance
\(O R=R G-O G\) and \(O S=O G+G S \dots(i)\)
As, \(R G=G S=4\)
\(
O G=\sqrt{3^2+(-2)^2}=\sqrt{9+4}=\sqrt{13}
\)
From (i), \(O R=4-\sqrt{13}\) and \(O S=4+\sqrt{13}\)
So, required difference \(=(4+\sqrt{13})-(4-\sqrt{13})\)
\(
=\sqrt{13}+\sqrt{13}=2 \sqrt{13}
\)

Hint

\(x^2+b x-1=0 \dots(i)\) common root
\(x^2+x+b=0 \dots(ii)\)
By solving eqn(i) & (ii) we get, \({x=\frac{b+1}{b-1}}\)
Put \(x=\frac{b+1}{b-1}\) in equation
\(
\begin{aligned}
&\left(\frac{b+1}{b-1}\right)^2+\left(\frac{b+1}{b-1}\right)+b=0 \\
&(b+1)^2+(b+1)(b-1)+b(b-1)^2=0 \\
&b^2+1+2 b+b^2-1+b\left(b^2-2 b+1\right)=0 \\
&2 b^2+2 b+b^3-2 b^2+b=0 \\
&b^3+3 b=0 \\
&b\left(b^2+3\right)=0 \\
&b^2=-3 \\
&b=\pm \sqrt{3} i \\
&|b|=\sqrt{3}
\end{aligned}
\)

Hint

(d) We have
\(
f(x)=x^2+2 b x+2 c^2
\)
and \(g(x)=-x^2-2 c x+b^2,(x \in R)\)
\(
\Rightarrow f(x)=(x+b)^2+2 c^2-b^2
\)
and \(g(x)=-(x+c)^2+b^2+c^2\)
Now, \(f_{\min }=2 c^2-b^2\) and \(g_{\max }=b^2+c^2\)
\(
\text { Given : } \begin{aligned}
& \min f(x)>\max g(x) \\
& \Rightarrow 2 c^2-b^2>b^2+c^2 \\
& \Rightarrow c^2>2 b^2 \Rightarrow|c|>|b| \sqrt{2} \\
& \Rightarrow \frac{|c|}{|b|}>\sqrt{2} \Rightarrow\left|\frac{c}{b}\right|>\sqrt{2} \Rightarrow\left|\frac{c}{b}\right| \in(\sqrt{2}, \infty) .
\end{aligned}
\)

Hint

(a) Given equations are
\(
\begin{aligned}
&x^2+2 x+3=0 \quad \text {… (i) } \\
&a x^2+b x+c=0 \quad \text {… (ii) }
\end{aligned}
\)
Roots of equation (i) are imaginary roots in order pair.
According to the question (ii) will also have both roots same as (i). Thus
\(
\frac{a}{1}=\frac{b}{2}=\frac{c}{3}=\lambda \text { (say) }
\)
\(
\Rightarrow \quad a=\lambda, b=2 \lambda, c=3 \lambda
\)
Hence, the required ratio is \(1: 2: 3\)

Hint

(b) Let \(\alpha\) be the common root of given equations, then \(a^2+b \alpha-1=0 \quad\)…(i)
and \(\alpha^2+\alpha+b=0\)…(ii)
On subtracting (ii) from (i), we get
\(
\begin{gathered}
(b-1) \alpha-(b+1)=0 \\
\Rightarrow \alpha=\frac{b+1}{b-1}
\end{gathered}
\)
Substituting this value of \(\alpha\) in equation (i), we get
\(
\begin{aligned}
&\left(\frac{b+1}{b-1}\right)^2+b\left(\frac{b+1}{b-1}\right)-1=0 \Rightarrow b^3+3 b=0 \\
\Rightarrow \quad & b=0, i \sqrt{3},-i \sqrt{3}
\end{aligned}
\)

Hint

(b) \(f(x)=a x^2+b x+c\) has same sign as that of a if \(D<0\).
Since \(x^2+2 a x+10-3 a>0 \forall x\)
\(
\begin{aligned}
&\therefore \quad D<0 \Rightarrow 4 a^2-4(10-3 a)<0 \Rightarrow a^2+3 a-10<0 \\
&\Rightarrow \quad(a+5)(a-2)<0 \Rightarrow a \in(-5,2)
\end{aligned}
\)

Hint

\(
\begin{aligned}
&x^2+a x+b=0 \dots(1) \\
&x^2+b x+a=0 \dots(2)
\end{aligned}
\)
then a will satisfy for both the equations
\(
\begin{aligned}
&a^2+a a+b=a^2+b a+a=a(a-b)=a-b \\
&a=1 \\
&a=1 \text { is the common root from equation } 1 \\
&(1)^2+a(1)+b=0 \\
&1+a+b=0 \\
&a+b=-1
\end{aligned}
\)

Hint

(True) \(P(x) \cdot Q(x)=\left(a x^2+b x+c\right)\left(-a x^2+b x+c\right)\)
\(\Rightarrow \quad D_1=b^2-4 a c\) and \(D_2=b^2+4 a c\)
clearly, \(\mathrm{D}_1+\mathrm{D}_2=2 \mathrm{~b}^2 \geq 0\)
\(\therefore\) At least one of \(\mathrm{D}_1\) and \(\mathrm{D}_2\) is positive. Hence, at least two real roots. 

Hint

(a, d) Given, \(x_1\) and \(x_2\) are roots of \(\alpha x^2-x+\alpha=0\). \(\therefore \quad x_1+x_2=\frac{1}{\alpha}\) and \(x_1 x_2=1\)
Also, \(\left|x_1-x_2\right|<1\)
\(\Rightarrow\left|x_1-x_2\right|^2<1 \Rightarrow\left(x_1-x_2\right)^2<1\)
or \(\left(x_1+x_2\right)^2-4 x_1 x_2<1\)
\(\Rightarrow \quad \frac{1}{\alpha^2}-4<1\) or \(\frac{1}{\alpha^2}<5\)
or \(5 \alpha^2-1>0\) or \((\sqrt{5} \alpha-1)(\sqrt{5} \alpha+1)>0\)
\(\therefore \quad \alpha \in\left(-\infty,-\frac{1}{\sqrt{5}}\right) \cup\left(\frac{1}{\sqrt{5}}, \infty\right) \dots(i)\)
Also, \(D>0\)
\(\Rightarrow 1-4 \alpha^2>0\) or \(\alpha \in\left(-\frac{1}{2}, \frac{1}{2}\right) \dots(ii)\)
From Eqs. (i) and (ii), we get
\(
\alpha \in\left(-\frac{1}{2}, \frac{-1}{\sqrt{5}}\right) \cup\left(\frac{1}{\sqrt{5}}, \frac{1}{2}\right)
\)

Hint

(b) Given : \(a, b, c, d, p\) are real and distinct numbers such that
\(
\begin{aligned}
&\left(a^2+b^2+c^2\right) p^2-2(a b+b c+c d) p+\left(b^2+c^2+d^2\right) \leq 0 \\
&\Rightarrow \quad\left(a^2 p^2+b^2 p^2+c^2 p^2\right)-(2 a b p+2 b c p+2 c d p)+\left(b^2+c^2+d^2\right) \leq 0 \\
&\Rightarrow \quad\left(a^2 p^2-2 a b p+b^2\right)+\left(b^2 p^2-2 b c p+c^2\right)+\left(c^2 p^2-2 c d p+d^2\right) \leq 0
\end{aligned}
\)
\(
\Rightarrow(a p-b)^2+(b p-c)^2+(c p-d)^2 \leq 0
\)
Since, LHS is the sum of perfect squares, therefore LHS can never be -ve.
\(
\therefore(a p-b)^2+(b p-c)^2+(c p-d)^2=0
\)
Which is possible only when each term is zero individually
i.e. \(a p-b=0 ; b p-c=0 ; c p-d=0\)
\(
\begin{aligned}
&\Rightarrow \quad \frac{b}{a}=p ; \frac{c}{b}=p ; \frac{d}{c}=p \Rightarrow \frac{b}{a}=\frac{c}{b}=\frac{d}{c}=p \\
&\therefore \quad a, b, c, d \text { are in G.P. }
\end{aligned}
\)

Hint

(c, d) Let \(y=\frac{(x-a)(x-b)}{(x-c)}\)
\(
\begin{aligned}
&\Rightarrow \quad(x-c) y=x^2-(a+b) x+a b \\
&\Rightarrow \quad x^2-(a+b+y) x+a b+c y=0
\end{aligned}
\)
Here, \(D=(a+b+y)^2-4(a b+c y)\)
\(
=y^2+2 y(a+b-2 c)+(a-b)^2
\)
Since \(x\) is real and y assumes all real values.
\(\therefore \quad D \geq 0\) for all real values of \(y\)
\(
\Rightarrow \quad y^2+2 y(a+b-2 c)+(a-b)^2 \geq 0
\)
As we know that the sign of a quadratic polynomial is same as that of coefficient of \(y^2\) if its descriminant \(<0\)
\(
\begin{aligned}
&\therefore \quad 4(a+b-2 c)^2-4(a-b)^2<0 \\
&\Rightarrow \quad 4(a+b-2 c+a-b)(a+b-2 c-a+b)<0 \\
&\Rightarrow \quad 16(a-c)(b-c)<0 \\
&\Rightarrow \quad 16(c-a)(c-b)<0 \dots(i)
\end{aligned}
\)
If \(a<b\) then from inequation (i), we get \(\mathrm{c} \in(\mathrm{a}, \mathrm{b})\)
\(
\Rightarrow a<c<b
\)
If \(a>b\) then from inequation (i), we get \(c \in(b, a)\)
\(
\Rightarrow a>c>b
\)
Thus, both (c) and (d) are the correct answer.

Hint

Given : \(a x^2+b x+c=0 \dots(i)\)
and \(\quad a^3 x^2+a b c x+c^2=0 \dots(ii)\)
\(
\therefore \quad \alpha+\beta=-\frac{b}{a}, \alpha \cdot \beta=\frac{c}{a}
\)
Divide the equation (ii) by \(a^3\), we get
\(
\begin{aligned}
& x^2+\frac{b}{a} \cdot \frac{c}{a} \cdot x+\left(\frac{c}{a}\right)^3=0 \\
\Rightarrow & x^2-(\alpha+\beta) \cdot(\alpha \beta) x+(\alpha \beta)^3=0 \\
\Rightarrow & x^2-\alpha^2 \beta x-\alpha \beta^2 x+(\alpha \beta)^3=0 \\
\Rightarrow & x\left(x-\alpha^2 \beta\right)-\alpha \beta^2\left(x-\alpha^2 \beta\right)=0 \\
\Rightarrow &\left(x-\alpha^2 \beta\right)\left(x-\alpha \beta^2\right)=0 \\
\Rightarrow & x=\alpha^2 \beta, \alpha \beta^2
\end{aligned}
\)

Hint

\(
\begin{aligned}
&x^2-3 x+2>0 \text { and } x^2-3 x-4 \leq 0 \\
&x^2-3 x+2>0 \\
&\Rightarrow(x-1)(x-2)>0 \\
&\Rightarrow x \in(-\infty, 1) \cup(2, \infty) \cdots(1) \\
&x^2-3 x-4 \leq 0 \\
&\Rightarrow(x+1)(x-4) \leq 0 \\
&\Rightarrow x \in[-1,4] \cdots(2) \\
&\text { From }(1) \text { and }(2) \\
&x \in[-1,1) \cup(2,4]
\end{aligned}
\)

Hint

\(\alpha, \beta\) are the roots of \(x^2+p x+q=0\) \(\alpha+\beta=-p, \quad \alpha \beta=q\)
\(\gamma, \delta\) are the roots of \(x^2+r x+s=0\) \(\gamma+\delta=-r, \gamma \delta=s\)
Now, \((\alpha-\gamma)(\alpha-\delta)(\beta-\gamma)(\beta-\delta)\)
\(
\begin{aligned}
=& {\left[\alpha^2-(\gamma+\delta) \alpha+\gamma \delta\right]\left[\beta^2-(\gamma+\delta) \beta+\gamma \delta\right] } \\
=\quad & {\left[\alpha^2+r \alpha+s\right]\left[\beta^2+r \beta+s\right] } \\
& \because \quad \alpha, \beta \text { are roots of } x^2+p x+q=0 \\
& \therefore \quad \alpha^2+p \alpha+q=0 \text { and } \beta^2+p \beta+q=0 \\
& \Rightarrow \quad \alpha^2=-p \alpha-q \text { and } \beta^2=-p \beta-q
\end{aligned}
\)
\(
\begin{aligned}
& \quad \therefore(\alpha-\gamma)(\alpha-\delta)(\beta-\gamma)(\beta-\delta)=\left[\alpha^2+r \alpha+s\right]\left[\beta^2+r \beta+s\right] \\
=& {[(r-p) \alpha+(s-q)][(r-p) \beta+(s-q)] } \\
=\quad &(r-p)^2 \alpha \beta+(r-p)(s-q)(\alpha+\beta)+(s-q)^2 \\
=& q(r-p)^2-p(r-p)(s-q)+(s-q)^2
\end{aligned}
\)
Now if the equations \(x^2+p x+q=0\) and \(x^2+r x+s=0\) have a common root say \(\alpha\), then \(\alpha^2+p \alpha+q=0\) and \(\alpha^2+r \alpha+s=0\)
\(
\begin{aligned}
&\Rightarrow \frac{\alpha^2}{p s-q r}=\frac{\alpha}{q-s}=\frac{1}{r-p} \\
&\Rightarrow \alpha^2=\frac{p s-q r}{r-p} \text { and } \alpha=\frac{q-s}{r-p}
\end{aligned}
\)
\(\Rightarrow(q-s)^2=(r-p)(p s-q r)\), which is the required condition.

Hint

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

Hint

\(
\begin{aligned}
& px ^2+ qx – r =0<_\beta^\alpha \\
& p=A, q=A R, r=A R^2 \\
& A x^2+A R x-A R^2=0 \\
& x^2+R x-R^2=0<_\beta^\alpha
\end{aligned}
\)
\(
\begin{aligned}
& \because \frac{1}{\alpha}+\frac{1}{\beta}=\frac{3}{4} \\
& \therefore \frac{\alpha+\beta}{\alpha \beta}=\frac{3}{4} \Rightarrow \frac{- R }{- R ^2}=\frac{3}{4} \Rightarrow R =\frac{4}{3} \\
& (\alpha-\beta)^2=(\alpha+\beta)^2-4 \alpha \beta= R ^2-4\left(- R ^2\right)=5\left(\frac{16}{9}\right) \\
& =80 / 9
\end{aligned}
\)

Hint

\(
\begin{aligned}
& x ^2- x -1=0 \\
& S _{ n }=2023 \alpha^{ n }+2024 \beta^{ n } \\
& S _{ n -1}+ S _{ n -2}=2023 \alpha^{ n -1}+2024 \beta^{ n -1}+2023 \alpha^{ n -2}+2024 \beta^{ n -2} \\
& =2023 \alpha^{ n -2}[1+\alpha]+2024 \beta^{ n -2}[1+\beta] \\
& =2023 \alpha^{ n -2}\left[\alpha^2\right]+2024 \beta^{ n -2}\left[\beta^2\right] \\
& =2023 \alpha^{ n }+2024 \beta^{ n } \\
& S _{ n -1}+ S _{ n -2}= S _{ n } \\
& \text { Put } n =12 \\
& S _{11}+ S _{10}= S _{12}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& x^2-2^y=2023 \\
& \Rightarrow x=45, y=1 \\
& \sum_{(x, y)=C}(x+y)=46 .
\end{aligned}
\)
Explanation:

First, let’s consider the equation \(x^2-2^y=2023\) where \(x\) and \(y\) are natural numbers. Our goal is to find all the pairs \((x, y)\) that satisfy this equation and then sum the values of \(x+y\) for each pair in set \(C\).
Since 2023 is an odd number, and \(x^2\), the square of any natural number, is even when \(x\) is even and odd when \(x\) is odd, we can determine that for the left-hand side of the equation to be odd (thus equal to 2023), \(x\) must be odd since the right-hand side of the equation \(\left(2^y\right)\) is always even as it represents a power of two.
Also, 2023 can be factored into prime factors to further analyze the possible solutions:
\(
2023=7 \times 17 \times 17
\)
Thus, allowing us to rewrite the equation as:
\(
x^2-2^y=7 \times 17^2
\)
The next step is to check for potential values of \(x\) that would fit the equation, keeping in mind that \(x\) must be odd. We can try to express \(x^2\) as \(7 \times 17^2\) plus a power of 2 , recognizing that we are looking for the decomposition of the form:
\(
x^2=7 \times 17^2+2^y
\)
By examining the powers of 2 and keeping in mind that they grow very quickly, we can reason that \(y\) cannot be very large because \(x^2\) must not exceed 2023 by a large margin.
Let’s start by trying the lowest values for \(y\) since that would make \(2^y\) small and \(x\) has a better chance of being a natural number:
For \(y=1\) :
\(
x^2=2023+2^1=2023+2=2025
\)
Surprisingly, we find a perfect square since \(45^2=2025\). Therefore, \((x, y)=(45,1)\) is one solution.
For \(y=2\) or higher:
\(2^y\) becomes at least 4 and increases exponentially, so \(x^2\) must be at least 2027 or higher in such cases. There’s no natural number between 45 and 46 , and \(46^2\) far exceeds the target (2116), making it impossible for \(x^2\) to be less than 2116 for any larger \(y\).
Hence, it appears there is only one possible solution: \((x, y)=(45,1)\).
Therefore, the sum \(\sum_{(x, y) \in C}(x+y)\) for this set will consist of only this one pair:
\(
\sum_{(x, y) \in C}(x+y)=45+1=46
\)

Hint

\(
\begin{aligned}
& x^2-70 x+\lambda=0 \\
& \alpha+\beta=70 \\
& \alpha \beta=\lambda \\
& \therefore \alpha(70-\alpha)=\lambda
\end{aligned}
\)
Since, 2 and 3 does not divide \(\lambda\)
\(
\therefore \alpha=5, \beta=65, \lambda=325
\)
By putting value of \(\alpha, \beta, \lambda\) we get the required value 60 .

Hint

\(
x=0 \text { and } x^2+3|x|+5|x-1|+6|x-2|=0
\)
Here all terms are + ve except at \(x=0\)
So there is no value of \(x\)
Satisfies this equation
Only solution \(x =0\)
No of solution 1 .

Hint

\(
\begin{aligned}
& f(x)=(a+b-2 c) x^2+(b+c-2 a) x+(c+a-2 b) \\
& f(x)=a+b-2 c+b+c-2 a+c+a-2 b=0 \\
& f(1)=0 \\
& \therefore \alpha \cdot 1=\frac{c+a-2 b}{a+b-2 c} \\
& \alpha=\frac{c+a-2 b}{a+b-2 c} \\
& \text { If }-1<\alpha<0 \\
& -1<\frac{c+a-2 b}{a+b-2 c}<0
\end{aligned}
\)
\(
b+c<2 a \text { and } b>\frac{a+c}{2}
\)
therefore, \(b\) cannot be G.M. between \(a\) and \(c\).
If, \(0<\alpha<1\)
\(
0<\frac{c+a-2 b}{a+b-2 c}<1
\)
\(b > c\) and \(b <\frac{ a + c }{2}\)
Therefore, \(b\) may be the G.M. between \(a\) and \(c\).

Hint

\(
\begin{aligned}
& \frac{a x^2+2(a+1) x+9 a+4}{x^2-8 x+32}<0 \\
& D=64-4 \times 32<0 \\
& \& a=1>0 \\
& \therefore x^2-8 x+32>0 \forall x \in R \\
& a x^2+2(a+1) x+9 a+4<0 \forall x \in R
\end{aligned}
\)

Only possible when
\(
a <0 \& D <0
\)
but we need positive integral value of \(a\).
So, No solution

Hint

\(
\begin{aligned}
& \text { Take } e^{\sin x}=t(t>0) \\
& \Rightarrow t-\frac{2}{t}=2 \\
& \Rightarrow \frac{t^2-2}{t}=2 \\
& \Rightarrow t^2-2 t-2=0 \\
& \Rightarrow t^2-2 t+1=3 \\
& \Rightarrow(t-1)^2=3 \\
& \Rightarrow t=1 \pm \sqrt{3} \\
& \Rightarrow t=1 \pm 1.73 \\
& \Rightarrow t=2.73 \text { or }-0.73 \text { (rejected as } t>0 \text { ) } \\
& \Rightarrow e^{\sin x}=2.73 \\
& \Rightarrow \log _e e^{\sin x}=\log _e 2.73 \\
& \Rightarrow \sin x=\log _e 2.73>1
\end{aligned}
\)
So no solution.

Hint

\(
\begin{aligned}
& \left(a^2+b^2\right) x^2-2 b(a+c) x+b^2+c^2=0 \\
& \Rightarrow a^2 x^2-2 a b x+b^2+b^2 x^2-2 b c x+c^2=0 \\
& \Rightarrow(a x-b)^2+(b x-c)^2=0 \\
& \Rightarrow a x-b=0, \quad b x-c=0 \\
& \Rightarrow a+b>c \quad b+c>a \quad c+a>b \\
& a+a x>b x \quad|a x+b x>a| a x^2+a>a x \\
& a+a x>a x^2 \left\lvert\, \begin{array}{l}
a x+a x^2>a \\
x^2-x+1>0
\end{array}\right. \\
&
\end{aligned}
\)
\(
x^2-x-1<0\left|x^2+x-1>0\right| \text { always true }
\)
\(
\begin{aligned}
& \frac{1-\sqrt{5}}{2}<x<\frac{1+\sqrt{5}}{2} \\
& x<\frac{-1-\sqrt{5}}{2}, \text { or } x>\frac{-1+\sqrt{5}}{2} \\
& \Rightarrow \frac{\sqrt{5}-1}{2}< x <\frac{\sqrt{5}+1}{2} \\
& \Rightarrow \alpha=\frac{\sqrt{5}-1}{2}, \beta=\frac{\sqrt{5}+1}{2} \\
& 2\left(\alpha^2+\beta^2\right)=12\left(\frac{(\sqrt{5}-1)^2+(\sqrt{5}+1)^2}{4}\right)=36
\end{aligned}
\)

Hint

\(
\Delta=0=
\)

\(
\left[\begin{array}{ccc}
\alpha^2 & \alpha & 1 \\
1 & 1 & 1 \\
a & b & c
\end{array}\right]
\)

\(\alpha^2( c – b )-\alpha( c – a )+( b – a )=0\)
It is singular when \(\alpha=1\)
\(
\begin{aligned}
& \frac{(a-c)^2}{(b-a)(c-b)}+\frac{(b-a)^2}{(a-c)(c-b)}+\frac{(c-b)^2}{(a-c)(b-a)} \\
& \frac{(a-b)^3+(b-c)^3+(c-a)^3}{(a-b)(b-c)(c-a)} \\
& =3 \frac{(a-b)(b-c)(c-a)}{(a-b)(b-c)(c-a)}=3
\end{aligned}
\)

Hint

\(
\begin{aligned}
& |x|^2-2|x|+|\lambda-3|=0 \\
& |x|^2-2|x|+|\lambda-3|-1=0 \\
& (|x|-1)^2+|\lambda-3|=1
\end{aligned}
\)
At \(\lambda=3, x=0\) and 2 ,
at \(\lambda=4\) or 2 , then
\(
x =1 \text { or }-1
\)
So maximum value of \(x+\lambda=5\)

Hint

\(
\begin{aligned}
& 3\left(x^2+\frac{1}{x^2}\right)-2\left(x+\frac{1}{x}\right)+5=0 \\
& 3\left[\left(x+\frac{1}{x}\right)^2-2\right]-2\left(x+\frac{1}{x}\right)+5=0 \\
& \text { Let } x+\frac{1}{x}=t \\
& 3 t^2-2 t-1=0 \\
& 3 t^2-3 t+t-1=0 \\
& 3 t(t-1)+1(t-1)=0 \\
& (t-1)(3 t+1)=0 \\
& t=1,-\frac{1}{3} \\
& x+\frac{1}{x}=1,-\frac{1}{3} \Rightarrow \text { No solution. }
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \log _2\left(9^{2 \alpha-4}+13\right)-\log _2\left(\frac{5}{2} \cdot 3^{2 \alpha-4}+1\right)=2 \\
& \Rightarrow \frac{9^{2 \alpha-4}+13}{\frac{5}{2} 3^{2 \alpha-4}+1}=4
\end{aligned}
\)
\(
\Rightarrow \alpha=2 \quad \text { or } \quad 3
\)
\(
\sum_{\alpha \in S} \alpha=5 \text { and } \sum_{\alpha \in S}(\alpha+1)^2=25
\)
\(
\begin{aligned}
& \Rightarrow x^2-50 x+25 \beta=0 \text { has real roots } \\
& \Rightarrow \beta \leq 25 \\
& \Rightarrow \beta_{\max }=25
\end{aligned}
\)

Hint

\(x^2+60^{\frac{1}{4}} x+a=0 \)
\(
\alpha+\beta=-60^{\frac{1}{4}} \quad \& \quad \alpha \beta= a
\)
Given \(\alpha^4+\beta^4=-30\)
\(
\begin{aligned}
& \Rightarrow\left(\alpha^2+\beta^2\right)^2-2 \alpha^2 \beta^2=-30 \\
& \Rightarrow\left\{(\alpha+\beta)^2-2 \alpha \beta\right\}^2-2 a^2=-30
\end{aligned}
\)
\(
\Rightarrow\left\{60^{\frac{1}{2}}-2 a \right\}^2-2 a ^2=-30
\)
\(
\begin{aligned}
& \Rightarrow 60+4 a ^2-4 a \times 60^{\frac{1}{2}}-2 a ^2=-30 \\
& \Rightarrow 2 a ^2-4.60^{\frac{1}{2}} a +90=0 \\
& \text { Product }=\frac{90}{2}=45
\end{aligned}
\)

Hint

\(
14 x^2-31 x+3 \lambda=0
\)
\(\alpha+\beta=\frac{31}{14} \dots(1)\)
and \(\alpha \beta=\frac{3 \lambda}{14} \dots(2)\)
\(
35 x^2-53 x+4 \lambda=0
\)
\(\alpha+\gamma=\frac{53}{35} \dots(3)\)
and \(\alpha \gamma=\frac{4 \lambda}{35} \dots(4)\)
\(
\begin{aligned}
& \frac{(2)}{(4)} \Rightarrow \frac{\beta}{\gamma}=\frac{3 \times 35}{4 \times 14}=\frac{15}{8} \Rightarrow \beta=\frac{15}{8} \gamma \\
& \text { (1) }-(3) \Rightarrow \beta-\gamma=\frac{31}{14}-\frac{53}{35}=\frac{155-106}{70}=\frac{7}{10} \\
& \frac{15}{8} \gamma-\gamma=\frac{7}{10} \Rightarrow \gamma=\frac{4}{5} \\
& \Rightarrow \beta=\frac{15}{8} \times \frac{4}{5}=\frac{3}{2}
\end{aligned}
\)
\(
\begin{aligned}
& \Rightarrow \alpha=\frac{31}{14}-\beta=\frac{31}{14}-\frac{3}{2}=\frac{5}{7} \\
& \Rightarrow \lambda=\frac{14}{3} \alpha \beta=\frac{14}{3} \times \frac{5}{7} \times \frac{3}{2}=5 \\
& \text { so, sum of roots } \frac{3 \alpha}{\beta}+\frac{4 \alpha}{\gamma}=\left(\frac{3 \alpha \gamma+4 \alpha \beta}{\beta \gamma}\right) \\
& =\frac{\left(3 \times \frac{4 \lambda}{35}+4 \times \frac{3 \lambda}{14}\right)}{\beta \gamma}=\frac{12 \lambda(14+35)}{14 \times 35 \beta \gamma}
\end{aligned}
\)
\(
=\frac{49 \times 12 \times 5}{490 \times \frac{3}{2} \times \frac{4}{5}}=5
\)
Product of roots
\(
=\frac{3 \alpha}{\beta} \times \frac{4 \alpha}{\gamma}=\frac{12 \alpha^2}{\beta \gamma}=\frac{12 \times \frac{25}{49}}{\frac{3}{2} \times \frac{4}{5}}=\frac{250}{49}
\)
So, required equation is \(x^2-5 x+\frac{250}{49}=0\)
\(
\Rightarrow 49 x^2-245 x+250=0
\)