Ncert Solutions Maths class 12th

$a=\left|\frac{4.2-3\left(-1\right)-21}{5}\right|$

$=\left|\frac{8+3-21}{5}\right|=2$

$\therefore L=4a=8$

$\overrightarrow{a}=\alpha \widehat{i}+2\widehat{j}-\widehat{k}\&\overrightarrow{b}=-2\widehat{i}+\alpha \widehat{j}+\widehat{k}$

$\left|\overrightarrow{a}*\overrightarrow{b}\right|=\sqrt[]{15\left(a^2+4\right)}$

$2{\left|\overrightarrow{a}\right|}^2+\left(\overrightarrow{a}.\overrightarrow{b}\right){\left|\overrightarrow{b}\right|}^2=?$

$\overrightarrow{a}.\overrightarrow{b}=\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|cos\theta$

$cos\theta =\frac{-1}{\left(a^2+5\right)}$ $\therefore \left|sin\theta \right|=\frac{\sqrt[]{{\left(a^2+5\right)}^2-1}}{\left(a^2+5\right)}$

$\left|\overrightarrow{a}*\overrightarrow{b}\right|=\left|\stackrel{?}{a}\right|*\left|\overrightarrow{b}\right|*\left|sin\theta \right|$

$=\left(a^5+5\right)*\frac{\sqrt[]{{\left(a^2+5\right)}^2-1}}{\left(a^2+5\right)}=\sqrt[]{15\left(a^2+4\right)}$

${\left(a^2+5\right)}^2-1=15\left(a^2+4\right)$

$\Rightarrow \left(\alpha =\pm 3\right)$

$2{\left|\overrightarrow{a}\right|}^2+\left(\overrightarrow{a}.\overrightarrow{b}\right){\left|\overrightarrow{b}\right|}^2$

= $2\left(a^2+5\right)-\left(a^2+5\right)$

$\left(a^2+5\right)=14$

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

$e=\sqrt[]{1+\frac{b^2}{a^2}}$ $e\text{'}=\sqrt[]{1+\frac{a^2}{b^2}}$

$\mathcal{l}=\frac{2b^2}{a}$ $\mathcal{l}\text{'}=\frac{2a^2}{b}$

$\left(1+\frac{b^2}{a^2}\right)=\frac{11}{14}*\frac{2b^2}{a}$ $\left(1+\frac{a^2}{b^2}\right)=\frac{11}{8}*\frac{2a^2}{b}$

$7*\left\{{\left(\frac{7b}{4}\right)}^2+b^2\right\}=11*b^2*\frac{7b}{4}\Rightarrow a*77=65$

65b² = 44b³

65 = b * 44

$\therefore 77a+44b=65*2=130$

$\frac{dy}{dx}+2ytanx=2sinx$  

  $I.F.=e^{2{\displaystyle \int tanxdx}}=se{c}^{2}x$             

$\therefore$  Solution y sec²x =   $2{\displaystyle \int sinxse{c}^{2}xdx=2{\displaystyle \int secxtanxdx}}$

y sec² x = 2sec x + c

y = 2 cos x + c cos²x passes $\left(\frac{\pi }{4},0\right)=B$     C =   $-2\sqrt[]{2}$

y = 2 cos x  $2\sqrt[]{2}co{s}^{2}x$ $\therefore {\displaystyle \underset{0}{\overset{\pi /2}{\int }}ydx}=2-\frac{\pi }{\sqrt[]{2}}$                              

  $f\left(x\right)+f\left(x+k\right)=n\forall x\in R,\&k>0............\left(i\right)$

Replace x by x + k.          

$f\left(x+k\right)+f\left(x+2k\right)=n...............\left(ii\right)$

From (i) & (ii), f(x + 2k) = f(x).

$\therefore f\left(x\right)$  is periodic with period = 2k.

$I_1={\displaystyle \underset{0}{\overset{4nk}{\int }}f\left(x\right)}dx=2x{\displaystyle \underset{0}{\overset{2k}{\int }}f\left(x\right)dx}...................\left(ii\right)$

$I_2={\displaystyle \underset{-k}{\overset{3k}{\int }}f\left(x\right)}dx$ put x = t + k

$={\displaystyle \underset{-2k}{\overset{2k}{\int }}t\left(t+k\right)dt}=2{\displaystyle \underset{0}{\overset{2k}{\int }}f\left(t+k\right)dt}$

$=2n{\displaystyle \underset{0}{\overset{2k}{\int }}ndx}=4n^{2}k.$

  $f\left(\frac{r}{4}\right)=\sqrt[]{2},f\left(\frac{r}{2}\right)=0\&f\text{'}\left(\frac{r}{2}\right)=1$

g(x)   ${\displaystyle \underset{}{\overset{}{\int }}\left(f\text{'}\left(t\right)sect+tantsectf\left(t\right)\right)dt}$

$={\left[f\left(t\right)sect\right]}_x^{\pi /4}$

=   $\sqrt[]{2}*\sqrt[]{2}-f\left(x\right)secx$

$=\frac{2cosx-f\left(x\right)}{cosx}$

=   $\frac{2sinx+1}{sinx}=3$

$f\left(x\right)=\{\begin{array}{cc}\hfill \left[x\right],x<0\hfill & \hfill \left|1-x\right|,x\ge 0\hfill \end{array}$ $g\left(x\right)=\{\begin{array}{cc}\hfill c^x-x,x<0\hfill & \hfill {\left(x-1\right)}^2-1,x\ge 0\hfill \end{array}$

fog

$fog(\begin{array}{}\end{array}\left[e^x-x\right],\left(e^x-x<0\right)\left(x<0\right)\left[{\left(x-1\right)}^2-1\right]{\left(x-1\right)}^2-1<0x\ge 0\left|1-e^x+x\right|,e^x-x\ge 0x<0\left|1-{\left(x+1\right)}^2+1\right|,{\left(x-1\right)}^2-1\ge 0x\ge 0,x\in \left(2,\infty \right)($

 Not possible as of inequalities give $?.$  

e^x – x < 0, x < 0                 (x 1)² – 1 < 0

Not possible                     (x – 1 + 1)(x – 1 – 1) < 0

(x)(x – 2) < 0

x   $\in \left(0,2\right)$

continuous  $\forall$ x < 0

$fog\{\begin{array}{ccc}\hfill \left|1-e^x+x\right|,x<0\hfill & \hfill \left|1-{\left(x-1\right)}^2+1\right|,x=0\hfill & \hfill \left|1-{\left(x-1\right)}^2+1\right|,x\ge 2\hfill \end{array}$ Discontinuous at 0

continuous   $\forall x$

  $\therefore$  fog is discontinuous at 0

$f\left(x\right)=a{x}^2+bx+c$  

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

f (x) is an even function

$f\left(-\frac{1}{4}\right)=f\left(-\frac{1}{2}\right)=f\left(\frac{1}{2}\right)=f\left(\frac{1}{4}\right)=0$

So, f (x) has at least four roots in (-2, 2)

$g\left(\frac{-3}{4}\right)=g\left(\frac{3}{4}\right)=0$

So, g (x) has at least two roots in (2, 2)

now number of roots of f (x) $\cdot g"\left(x\right)=f\text{'}\left(x\right)\cdot g\text{'}\left(x\right)=0$

It is same as number of roots of $\frac{d}{dx}\left(f\left(x\right)\cdot g\text{'}\left(x\right)\right)=0$ will have atleast 4 roots in (2, 2)