Maths NCERT Exemplar Solutions Class 12th Chapter Two

Case 1: 1 $\le \frac{1}{4x^2-1}\le 1$

4x2 – 1 $\ge 1or4x^2-1\le -1$

$x^2\ge \frac{2}{4}or{x}^2\le 0$

So x $\in (-\infty ,-\frac{1}{\sqrt[]{2}}]\cup [\frac{1}{\sqrt[]{2}},\infty )\cup \left\{0\right\}$

 $Let\frac{si{n}^{-1}x}{\alpha }=\frac{co{s}^{-1}x}{\beta }=k\Rightarrow si{n}^{-1}x+co{s}^{-1}x=k\left(\alpha +\beta \right)\Rightarrow \alpha +\beta =\frac{\pi }{2k}$

Now $\frac{2\pi \alpha }{\alpha +\beta }=\frac{2\pi \alpha }{\frac{\pi }{2k}}\Rightarrow 4k\alpha =4si{n}^{-1}x.Heresin\left(\frac{2\pi \alpha }{\alpha +\beta }\right)=sin\left(4si{n}^{-1}x\right)$

$\Rightarrow sin4\theta =4x\sqrt[]{1-x^2}\left(1-2x^2\right)$

 $f\left(0\right)+3+\lambda +4=14$

$\therefore f\left(0\right)=7-\lambda =c$

$f\left(1\right)=a+b+c=3$ ……. (i)

$f\left(3\right)=9a+3b+c=4$ …… (ii)

$f\left(-2\right)=4a-2b+c=\lambda$ …. (iii)

(ii) – (iii)

$a+b=\frac{4-\lambda }{5},$ put in equation (i)

$6\lambda =24\lambda =4$

 $[x{]}^2+2[x+2]-7=0$

$\begin{array}{rr}& \Rightarrow [x{]}^2+2[x]+4-7=0\\ & \Rightarrow \left[x\right]=1,-3\\ & \Rightarrow x\in \left[\mathrm{1,2}\right)\cup [-3,-2)\end{array}$

R = { (a, b): b = pq, where p, q $\ge 3$ are prime}

$\Rightarrow 60*11=660$

p, q $\in$  {3, 5, 7, 11, 13, 17, 19, 23, 29, 3, 37, 41, 43, 47, 53, 59} total 16

p, q $\in$  {3, 5, 7, 11, 13, 17, 19}

{3, 5, 7, 11, 13, 17, 19}

7 + 3 + 1 = 11

${\int }_0^3?g\left(x\right)-f\left(x\right)={\int }_0^3?\left|\right|x-2|-2|dx-{\int }_0^3?|x-2|dx$

$=\left(\frac{1}{2}*2*2+1+\frac{1}{2}*1*1\right)-\left(\frac{1}{2}*2*2+\frac{1}{2}*1*1\right)$

$=\left(2+1+\frac{1}{2}\right)-\left(2+\frac{1}{2}\right)=1$

 $f\left(x\right)=\left[1+x\right]+\frac{{\alpha }^{2\left|x\right|}+\left\{x\right\}+\left[x\right]-1}{2\left[x\right]+\left\{x\right\}}$

$li{m}_{x\to 0}-f\left(x\right)=\alpha -\frac{4}{3}$

$\Rightarrow li{m}_{h\to 0}1-1+\frac{{\alpha }^{-h-1}-1-1}{-h-1}=\alpha -\frac{4}{3}$

$\therefore \frac{{\alpha }^{-1}-2}{-1}\alpha -\frac{4}{3}$

32 - 10 + 3 = 0

$\therefore \alpha =3or1/3$

$?\alpha$ in integer, hence = 3