Ncert Solutions Maths class 12th

Let f(x) = $\frac{x^2-9}{x-5}$  

  $f\text{'}\left(x\right)=\frac{2x\left(x-5\right)-\left(x^2-9\right)}{{\left(x-5\right)}^2}$

$=\frac{x^2-10x+9}{{\left(x-5\right)}^2}=\frac{\left(x-1\right)\left(x-9\right)}{{\left(x-5\right)}^2}$

$\alpha =f\left(1\right)=2,\beta =\left\{f\left(0\right),f\left(2\right)\right\}=\frac{5}{3}$

$\therefore {\displaystyle \underset{-1}{\overset{3}{\int }}max}\left\{\frac{x^2-9}{x-5},x\right\}dx+{\left[\frac{x^2}{2}\right]}_{9/5}^3$           

a₁ = 18, a₂ = 16

a₁ + a₂ = 34

  $f\left(\theta \right)=sin\theta +{\displaystyle \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}\left(sin\theta +tcos\theta \right)f\left(t\right)dt}$

$=sin\theta +sin\theta {\displaystyle \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}f\left(t\right)dt+cos\theta {\displaystyle \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}tf\left(t\right)dt}}$

  $=\left(1+{\displaystyle \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}f\left(t\right)dt}\right)sin\theta +\left({\displaystyle \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}tf\left(t\right)dt}\right)cos\theta$              

f(q) = a sin q + b cos q

  $\therefore a=1+{\displaystyle \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}f\left(t\right)dt=1+{\displaystyle \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}\left(asint+bcost\right)dt}}$

$\therefore a=2b+1..........(i)$

$b={\displaystyle \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}tf\left(t\right)dt}={\displaystyle \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}\left(asint+bcost\right)dt}$

 Solving (i) & (ii) we get a =   $-\frac{1}{3},b=-\frac{2}{3}$

    $\therefore \left|{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}f\left(\theta \right)d\theta }\right|=1$           

$y=2\left|x^2-\frac{3}{2}x-\frac{7}{2}\right|$

$=2\left|{\left(x-\frac{3}{4}\right)}^2-\frac{65}{16}\right|$

  $\frac{x-7}{3}=\frac{y-1}{-1}=\frac{z+2}{1}=r_1$

$A\left(3r_1+7,1-r_1,-2+r_1\right)$

and   $\frac{x}{2}=\frac{y-7}{3}=\frac{z}{1}=r_2$  

  $B\left(2r_2,7+3r_2,r_2\right)$

$A/q,\frac{3r_1-2r_2+7}{1}=\frac{3r_2+r_1+6}{4}=\frac{r_1-r_2-2}{2}$             

  $\Rightarrow r_1=-5,r_2=-3$

$\therefore A\left(-8,6,-7\right)andB\left(-6,-2,-3\right)$

AB² = 84

  $f\left(x\right)=\{\begin{array}{l}\left|2x^2-3x-7\right|,x\le -1\\ \left[4x^2-1\right],-1<x<1\\ \left|x+1\right|+\left|x-2\right|,x\ge 1\end{array}$              

f(-1) = 1

f(1) = 3

Hence f(x) will be discontinuous at x = 1 and also 4x² – 1 = 0 , 1 , 2

$\Rightarrow x=\pm \frac{1}{2},\pm \frac{1}{\sqrt[]{2}},\pm \frac{\sqrt[]{3}}{2}$          

f(b) = 2f(a) + 3f(c) + f(d)

Value of f(c)       Value of f(a)      Number of functions

$\left(p\Delta q\right)\Rightarrow \left(p\Delta \sim q\right)\vee \left(\sim p\Delta q\right)$

Case I

When  $\Delta$  is same as $\vee .$  

Then  $\left(p\Delta \sim q\right)\vee \left(-p\Delta q\right)$ becomes

$\left(p\vee \sim q\right)\vee \left(\sim p\vee q\right)$ which is always true, so x becomes tautology.

Case II

When  $\Delta$  is same as $\vee$  

Then  $\left(p\wedge q\right)\Rightarrow \left(p\wedge \sim q\right)\vee \left(\sim p\wedge q\right)$  becomes $p\wedge q$ is T, then $\left(p\wedge \sim q\right)\vee \left(\sim p\wedge q\right)$ is false, so x cannot be tautology.

Case III

When  $\Delta$  is same as $\Rightarrow$  

Then&

$f\left(x+y\right)=2f\left(x\right)f\left(y\right),x,y\in N$

$f\left(1\right)=2$

${\displaystyle \sum _{k=1}^{10}f\left(\alpha +k\right)={\displaystyle \sum _{k=1}^{10}2f\left(\alpha \right)f\left(k\right)}}$

$=2f\left(\alpha \right){\displaystyle \sum _{k=1}^{10}f\left(k\right)=2.2^{2\alpha -1}.\frac{2}{3}\left(4^{10}-1\right)}$

$=\frac{2^{2\alpha +1}}{3}.\left(4^{10}-1\right)$

$\Rightarrow 2\alpha +1=9$

=4

  $l={\displaystyle \underset{0}{\overset{\pi }{\int }}\frac{e^{cosx}sinxdx}{\left(1+co{s}^{2}x\right)\left(e^{cosx}+e^{-cosx}\right)}}$ ${\displaystyle \underset{}{\overset{}{}}\frac{{}^{}}{{}^{}{}^{}{}^{}}}$

$2l={\displaystyle \underset{0}{\overset{\pi }{\int }}\frac{sindx}{1+co{s}^{2}x}=2{\displaystyle \underset{0}{\overset{\pi /2}{\int }}\frac{sinxdx}{1+co{s}^{2}x}}}$

$l={\displaystyle \underset{1}{\overset{0}{\int }}\frac{-dt}{1+t^2}={\displaystyle \underset{0}{\overset{1}{\int }}\frac{dt}{1+t^2}=\frac{\pi }{4}}}$

n = 33, p = success, q = failure

3P (x = 0) = P (x = 1)

${\Rightarrow }^{33}{C}_0{p}^0{q}^{33}{=}^{33}{C}_{1}p{q}^{32}$            

$\Rightarrow p=\frac{1}{12},q=\frac{11}{12}\Rightarrow \frac{q}{p}=11$          

………. (i)

Subtracting, (ii) – (i), we get 1320

$\left|\widehat{a}\right|=1,\left|\widehat{b}\right|=1$

${\left|\widehat{b}\right|}^2={\left|\overrightarrow{c}+2\left(\overrightarrow{c}\times \widehat{a}\right)\right|}^2$

$\Rightarrow 1={\left|\overrightarrow{c}\right|}^2+4{\left|\overrightarrow{c}\right|}^2{\left(\frac{\sqrt[]{3}-1}{2\sqrt[]{2}}\right)}^2$

${\left|\overrightarrow{c}\right|}^2\left[1+4\times \frac{{\left(\sqrt[]{3}-1\right)}^2}{8}\right]={\left|\overrightarrow{c}\right|}^2\left(3-\sqrt[]{3}\right)$

$\therefore {\left|\overrightarrow{c}\right|}^2=\frac{1}{3-\sqrt[]{3}}=\frac{3+\sqrt[]{3}}{6}$

$\Rightarrow {\left|6\overrightarrow{c}\right|}^2=6\left(3+\sqrt[]{3}\right)$