Maths NCERT Exemplar Solutions Class 12th Chapter Four

$f\left(x\right)=a{x}^2+bx+c$  

g (x) = px + q

$f\left(g\left(x\right)\right)=a{\left(px+q\right)}^2+b\left(p\right)\left(+q\right)+c$                

$\Rightarrow 8x^2-2x=a{\left(px+q\right)}^2+b\left(px+q\right)+c$                

Compare 8 = ap² …………… (i)

-2 = a (2pq) + bp

0 = aq² + bq + c

$9\left(f\left(x\right)\right)=p\left(a{x}^2+bx+c\right)+q$                

? 4x² + 6x + 1 = apx² + bpx + cp + q

? Andhra Pradesh = 4 ……………. (ii)

6 = bp

1 = cp + q

From (i) & (ii), p = 2, q = -1

? b = 3, c = 1, a = 2

f (x) = 2x² + 3x + 1

f (2) = 8 + 6 + 1 = 15

g (x) = 2x – 1

g (2) = 3

$N=M^2+M^4+.....+M^{98}$

$=\left(-{\alpha }^{2}I\right)+{\left(-{\alpha }^{2}I\right)}^2+....+{\left(-{\alpha }^{2}I\right)}^{49}$

$=I\left(-{\alpha }^2+{\alpha }^4-{\alpha }^6+....-{\alpha }^{98}\right)$

N = $-I\left({\alpha }^2-{\alpha }^4+{\alpha }^6.......+{\alpha }^{98}\right)$

$=-I\frac{{\alpha }^2\left(1-{\left(-{\alpha }^2\right)}^{49}\right)}{1-\left(-{\alpha }^2\right)}$

N = $\frac{-I{\alpha }^2\left(1+{\alpha }^{98}\right)}{1+{\alpha }^2}$

Now $\left(I-m^2\right)N=-2I$

$\left(I+{\alpha }^{2}I\right)\frac{\left(-I{\alpha }^2\cdot (1+{\alpha }^{98}\right)}{1+{\alpha }^2}=-2I$

? α¹⁰⁰ + α² = 2

? α = $\pm 1$

$2si{n}^2\theta =cos2\theta =1-2si{n}^2\theta$

$\Rightarrow 4si{n}^2\theta =1\Rightarrow sin\theta =\pm \frac{1}{2}$

$\frac{\pi }{6},\frac{5\pi }{6},\frac{7\pi }{6},\frac{11\pi }{6}$

$Sum\frac{7\pi }{6}+\frac{11\pi }{6}=3\pi \Rightarrow k=3$

Let equation of normal to x² = y at Q (t, t²) is x + 2ty = t + 2t³

It passes through the point (1, -1) so, 2t³ + 3t – 1 = 0

Let f(t) = 2t³ + 3t – 1 f   $\left(\frac{1}{4}\right)f\left(\frac{1}{3}\right)<0\Rightarrow t\in \left(\frac{1}{4},\frac{1}{3}\right)$

Let P(1 – sin q, -1 + cos q) $\therefore$  slope of normal = slope of CP $-\frac{1}{2t}=\frac{cos\theta }{-sin\theta }\Rightarrow 2t$ Þ = tan q according to question $x=1-sin\theta =1-\frac{2t}{\sqrt[]{1+4t^2}}=g\left(t\right)\therefore g\left(t\right)=1-\frac{2t}{\sqrt[]{1+4t^2}}$ ,  

Þ g'(t) < 0 g(t) is decreasing function in  $t\in \left(\frac{1}{4},\frac{1}{3}\right)\Rightarrow g\left(t\right)\in \left(0.440,0.485\right)\in \left(\frac{1}{4},\frac{1}{2}\right)$

Let

P : Ramesh listens to music

Q : Ramesh is out of his village

R : It is Sunday

S : It is Saturday

Then the statement “Ramesh listens to music only if he is in his village and it is Sunday or Saturday can be expressed as

$P\equiv$ Ramesh listens to music

$\sim q\equiv$ He is in village.

$r\vee s\equiv$ Saturday or Sunday

$P\Rightarrow \left(\left(\sim q\right)\wedge \left(r\vee s\right)\right)$

$D=\left|\begin{array}{ccc}1& -2& 3\\ 2& 1& 1\\ 1& -7& a\end{array}\right|=0\Rightarrow a=8$

also,  $D_1=\left|\begin{array}{ccc}9& -2& 3\\ b& 1& 1\\ 24& -7& 8\end{array}\right|=0\Rightarrow b=3$

hence,  $a-b=8-3=5$

Let TV (r) denotes truth value of a statement $r$ .

$\text{Now, if}TV\left(p\right)=TV\left(q\right)=T$

$\Rightarrow \mathrm{T}\mathrm{V}\left({\mathrm{S}}_1\right)=\mathrm{F}$

Also, if $TV\left(p\right)=T$ and $TV\left(q\right)=F$

$\Rightarrow \mathrm{T}\mathrm{V}\left({\mathrm{S}}_2\right)=\mathrm{T}$

$\bar{x}=10$

$\Rightarrow \stackrel{?}{\mathrm{x}}=\frac{63+\mathrm{a}+\mathrm{b}}{8}=10$

$\Rightarrow a+b=17$

Since, variance is independent of origin.

So, we subtract 10 from each observation.

So,  ${\sigma }^2=13.5=\frac{79+(a-10{)}^2+(b-10{)}^2}{8}$

$\Rightarrow {\mathrm{a}}^2+{\mathrm{b}}^2-20(\mathrm{a}+\mathrm{b})=-171$

$\Rightarrow a^2+b^2=169$

From (1) and (2) ; $a=12$ and $b=5$