Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

= A + B (say)

(1 + x)¹⁵ =

differentiating 15 (1 + x)¹⁴ =   $\sum r{.}^{15}{C}_r{x}^{r-1}$

put  = x = -1

$\Rightarrow 0={-}^{15}{C}_1+2{.}^{15}{C}_2-.......=A$      

  $B{=}^{14}{C}_{13}=2^{13}$            

$\therefore A+B=2^{13}-14$           

Let direction ratio of the normal to the required plane are l, m, n

$\begin{array}{l}\therefore 3l+m-2n=0\\ \therefore 2l-5m-n=0\\ \therefore \frac{l}{-11}=\frac{m}{-1}=\frac{n}{-17}\end{array}$             

Equation of required plane

11 (x – 1) + 1 (y – 2) + 17 (z + 3) = 0

$\Rightarrow 11x+y+17z+38=0$           

$e^{\left(co{s}^{2}x+co{s}^{4}x+co{s}^{6}x+.....\infty \right)}lo{g}_{e}2$  

$=e^{\frac{co{s}^{2}x}{1-co{s}^{2}x}lo{g}_{e}2}=e^{co{t}^{2}xlo{g}_{e}2}=2^{co{t}^{2}x}$          

t² – 9t + 8 = 0

(t – 8) (t – 1) = 0

$t=2^{co{t}^{2}x}=8=2^3$

$\Rightarrow co{t}^{2}x=3=co{t}^2\frac{\pi }{6}$

$\frac{2sinx}{sinx+\sqrt[]{3}cosx}=\frac{2*\frac{1}{2}}{\frac{1}{2}+\sqrt[]{3}*\frac{\sqrt[]{3}}{2}}=\frac{1}{2}$            

Any point on line $\frac{x-3}{1}=\frac{y-4}{2}=\frac{z-5}{2}=r$ is P (r + 3, 2r + 4, 2r + 5) lies on x + y + z = 17,

5r + 12 = 17

r = 1

$P\left(4,6,7\right)A\left(1,1,9\right)distAP=\sqrt[]{9+25+4}=\sqrt[]{38}$       

x² + y² = 36         ……(i)

y² = 9x      .(ii)

Solving (i) & (ii)

x² + 9x – 36 = 0

(x + 12) (x – 3) = 0

x = 3

Let $A={\displaystyle \underset{0}{\overset{3}{\int }}\left(\sqrt[]{36-x^2}-3\sqrt[]{x}\right)dx}$  

$={\left[\frac{x\sqrt[]{36-x^2}}{2}+18si{n}^{-1}\frac{x}{6}\right]}_0^3-3.{\frac{x^{3/2}}{3/2}]}_0^3$   

  $=3\pi -\frac{3\sqrt[]{3}}{2}$           

$\therefore$ Required area = $=\frac{1}{2}\pi {\left(6\right)}^2+2\left(3\pi -\frac{3\sqrt[]{3}}{2}\right)=\left(24\pi -3\sqrt[]{3}\right)sq.unit$

$f\left(x\right)=\left[x-1\right]cos\left(\frac{2x-}{2}\right)\pi$

$Ifx=k,k\in I$

then f(x) = 0 as $cos\left(\frac{2k-1}{2}\right)\pi =0,\forall k\in I$  

LHL = $\underset{h\to 0}{Lt}\left[k-h-1\right]cos\left(\frac{2k-2h-1}{2}\right)\pi =\underset{h\to 0}{Lt}\left(k-2\right)cos\left(\frac{2k-1}{2}\right)\pi \to 0$  

$RHL=\underset{h\to 0}{Lt}\left[k+h-1\right]cos\left(\frac{2k+2h-1}{2}\right)\pi =\underset{h\to 0}{Lt}\left(k-1\right)cos\left(\frac{2k-1}{2}\right)\pi \to 0$           

  $\therefore f\left(x\right)$  is continuous $\forall x\in R$  

  $f\left(x\right)=2x-1g:R-\left\{1\right\}\to R$       

  $g\left(x\right)=\frac{x-1/2}{x-1}$         

  $f\left\{g\left(x\right)\right\}=2\left(\frac{x-1/2}{x-1}\right)-1=\frac{2x-1-x+1}{x-1}=\frac{x}{x-1},x\ne 1$          

$y=\frac{x}{x-1}\Rightarrow x=\frac{y}{y-1}\therefore y\ne 1$           

One – one but not onto

BC = CN = x = 75

$cot\theta =\frac{3h}{75}$            

$tan\theta =\frac{h}{75}$          

$\frac{3h^2}{75*75}=1\Rightarrow h=\frac{75}{\sqrt[]{3}}=25\sqrt[]{3}$

$\underset{x\to 0}{lim}\frac{{\displaystyle \underset{0}{\overset{x^2}{\int }}\left(sin\sqrt[]{t}\right)dt}}{x^3}$ $\left(\frac{0}{0}\right)$   aplying L' Hospital rule

$=\underset{x\to 0}{lim}\frac{sin\left(\sqrt[]{x^2}\right).2x}{3x^2}$

$=\frac{2}{3},forx>0$

p + q = 2

p⁴ + q⁴ = 272

  ${\left(p^2+q^2\right)}^2-2p^2{q}^2=272$          

${\left[{\left(p+q\right)}^2-2pq\right]}^2-2p^2{q}^2=272$           

Let pq = t -> (4 – 2t)² – 2t² = 272

2t² – 16 t – 256 = 0

->t = pq = 16

Required equation x² – 2x + 16 = 0