Ncert Solutions Maths class 12th

Option (A)

              Option (B)

              Option (C)

              Option (D)

${\left[e^x\right]}^2+\left[e^x\right]-2=0$              

Let $\left[e^x\right]=t$   

->t² + t – 2 = 0 Þ (t + 2) (t – 1) = 0

t = -2 $\left[e^{-x}\right]$ -> = -2 not possible

$\therefore t=\left[e^x\right]=1\Rightarrow 1\le e^x<2$

$\Rightarrow 0\le x<ln2$              

${\displaystyle \underset{}{\overset{}{\int }}\frac{si{n}^{2}xdx}{e^{\left\{\frac{x}{\pi }\right\}}}=l}$

$f\left(x\right)=\frac{si{n}^{2}x}{e^{\left\{\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$\pi $}\right.\right\}}}$ is periodic with period =

$l=100{\displaystyle \underset{0}{\overset{\pi }{\int }}\frac{si{n}^{2}x}{e^{x/\pi }}dx.}$

$=50{\displaystyle \underset{0}{\overset{\pi }{\int }}\frac{1-cos2x}{e^{x/\pi }}}dx=50{\displaystyle \underset{0}{\overset{\pi }{\int }}{e}^{-x/\pi }dx-50}{\displaystyle \underset{0}{\overset{\pi }{\int }}{e}^{-x/\pi }}cos2xdx$

$=50\pi \left(1-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$e$}\right.\right)\left[1-\frac{1}{1+4{\pi }^2}\right]=\frac{50\left(1-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$e$}\right.\right)4{\pi }^3}{1+4{\pi }^2}$

$\therefore \alpha =200\left(1-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$e$}\right.\right)$

$\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-1}{3}.......\left(i\right)$

$\frac{x-2}{1}=\frac{y-\lambda }{2}=\frac{z-3}{4}......\left(ii\right)$

$\overrightarrow{OA}=\left(i+2j+k\right)\overrightarrow{OB}=\left(2\widehat{i}+\lambda \widehat{j}+3\widehat{k}\right)$

$\overrightarrow{a}=2i+j+3k\&\overrightarrow{b}=i+2j+4k$

$\therefore$ shortest distance = $\left|\frac{\left(\overrightarrow{OB}-\overrightarrow{OA}\right).\left(\overrightarrow{a}\times \overrightarrow{b}\right)}{\overrightarrow{a}\times \overrightarrow{b}}\right|=\frac{1}{\sqrt[]{38}}$

$\Rightarrow \left|14-5\lambda \right|=1$           

$\Rightarrow \lambda =3\&\frac{13}{5}$          

Integer value of $\lambda$ = 3

f (t) = t³ – 6t² + 9t + 3

$\Rightarrow f\text{'}\left(t\right)=3t^2-12t+9=3\left(t-1\right)\left(t-3\right)$

$\therefore f\text{'}\left(t\right)=0\Rightarrow t=1,3$

$\Rightarrow f\left(1\right)=1,f\left(3\right)=-3$

$g\left(x\right)=\{\begin{array}{l}f\left(x\right),0\le x<1\\ 1,1\le x\le 3g\left(x\right)iscontinuous\\ 4-x,3<x\le 4\end{array}$

Hence g (x) is not differentiable at only x = 3.

$\frac{1}{\alpha \left(\alpha +1\right)\left(\alpha +2\right)..........\left(\alpha +20\right)}=\frac{A_0}{\alpha }+\frac{A_1}{\alpha +1}+\frac{A_2}{\alpha +2}+......+\frac{A_{20}}{\alpha +20}$

Solving by partial fraction, we get

$A_{13}=\frac{-1}{14!7!},A_{14}=\frac{-1}{14!6!}and{A}_{15}=\frac{-1}{14!5!}$

$\therefore \frac{A_{14}+A_{15}}{A_{13}}=\frac{3}{10}\Rightarrow {\left(\frac{A_{14}+A_{15}}{A_{13}}\right)}^2={\left(\frac{3}{10}\right)}^2\Rightarrow 100{\left(\frac{A_{14}+A_{15}}{A_{13}}\right)}^2=9$

$?\left|{\overrightarrow{v}}_1\right|=\left|{\overrightarrow{v}}_2\right|\Rightarrow p^2-p-2=0\Rightarrow p-1,2$ we take only p = 2 (p > 0)

$\therefore cos\theta =\frac{\overrightarrow{v_2}.\overrightarrow{v_2}}{\left|\overrightarrow{v_1}\right|\left|\overrightarrow{v_2}\right|}=\frac{4\sqrt[]{3}+3}{13}\Rightarrow tan\theta =\frac{6\sqrt[]{3}-2}{4\sqrt[]{3}+3}=\frac{\alpha \sqrt[]{3}-2}{4\sqrt[]{3}+3}\therefore \alpha =6$

Let equation of normal PQ is

$y=-tx+3t+\frac{3}{2}{t}^3$ and it is passing through

$P\left(3,\frac{3}{2}\right)\therefore t=1$

$\therefore Q\left(\frac{3}{2},3\right)\Rightarrow a=\frac{3}{2}andb=3$     

2 (a + b) = 9