Maths NCERT Exemplar Solutions Class 11th Chapter Eight

  $\frac{2}{-4}=\frac{-1}{2}$

y – 4 = 2 (x – 3)

y = 2x – 2

x² + (2x – 2)² = 25

  $5x^2-8x-21=0$

$z\left(\frac{-7}{5},\frac{-24}{5}\right)$              

x = 2t $A\left(2t,\frac{t^2}{3}\right)$ $\frac{{}^{}}{}$

$\frac{{}^{}}{}$ $y=\frac{t^2}{3}$  S (0, 3)

${\frac{}{}}^{}$  $3y={\left(\frac{x}{2}\right)}^2$ $B\left(0,\lambda \right)$ $$

$3k=\frac{t^2}{3}+3+\lambda =\frac{2t^2}{3}+3-\frac{12t^2}{9-t^2}$

$\underset{t\to 1}{lim}3k=\frac{2}{3}+3-\frac{12}{8}=3-\frac{5}{6}=\frac{13}{6}$

$y^{2}dx+\left(x^2-xy+y^2\right)dy=0$

$\frac{dx}{dy}+\frac{x^2-xy+y^2}{y^2}=0\Rightarrow \frac{dx}{dy}+{\left(\frac{x}{y}\right)}^2-\left(\frac{x}{y}\right)+1=0$

Put x = vy $\Rightarrow v+y\frac{dv}{dy}+v^2-v+1=0$

$\Rightarrow \frac{ydv}{dy}+v^2+1=0$

$\Rightarrow \frac{\pi }{6}+\mathcal{l}n\left|y\right|=\frac{\pi }{4}$

$\mathcal{l}n\left|y\right|=\frac{\pi }{12}$

From  $\Delta APQ$ $$

$$ $\frac{x+15}{y}=tan75°$  …. (i)

From $\Delta RQA$ $$

$\frac{}{}$  $\frac{15}{y}=tan60°$ …… (ii)

From (i) and (ii)

$=5\left(2\sqrt[]{3}+3\right)m$

Number of solution of the equation |cos x| = sin x for $x\in \left[-4\pi ,4\pi \right]$ $$ will be equal to 4 times the number of solutions of the same equation for $x\in \left[0,2\pi \right]$ $$

Graphs of y = |cos x| and y = sin x are as shown below.

Hence, two solutions of given equation in [0, 2]

$$ $\Rightarrow$  total of 8 solutions in [4, 4]

For $x^2+\alpha x=\beta >0\forall x\in R$  to hold, we should have   ${\alpha }^2-4\beta <0$

If $\alpha =1,\beta$  can be 1, 2, 3, 4, 5, 6 i.e., 6 choices

If a  = 4, b can be 5 or 6 i.e., 2 choices

hence total favourable outcomes

= 6 + 5 + 4 + 2 + 0 + 0 = 17

Required probability =  $\frac{17}{36}$

If n is number of trials, p is probability of success and q is probability of unsuccess then,

Mean = np and variance = npq.

Here np + npq = 24         …. (i)

np. npq = 128                   …. (ii)

and q = 1 – p                    …… (iii)

From eq. (i), (ii) and (iii) :

$=\left(32+\frac{32*31}{2}\right).\frac{1}{2^{32}}$      

$=\frac{33}{2^{28}}$

  $?\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=0$ …….(i)

then

$\overrightarrow{a}+\overrightarrow{c}=-\overrightarrow{b}$

then

$\left(\overrightarrow{a}+\overrightarrow{c}\right)*\overrightarrow{b}=-\overrightarrow{b}*\overline{b}$

$\stackrel{}{}\stackrel{}{}\stackrel{}{}\stackrel{}{}\stackrel{}{}$  $\therefore \overrightarrow{a}*\overrightarrow{b}+\overrightarrow{c}*\overrightarrow{b}=\overrightarrow{0}$ …….(ii)

Now (S1) : $\stackrel{}{}\stackrel{}{}\stackrel{}{}\stackrel{}{}\stackrel{}{}\text{exception 25:}$

$\left|\overrightarrow{a}*\overrightarrow{b}+\overrightarrow{c}*\overrightarrow{b}\right|-\left|\overrightarrow{c}\right|=6\left(2\sqrt[]{2}-1\right)$

$\therefore cos\left(\angle ACB\right)=\sqrt[]{\frac{2}{3}}$

$\therefore \angle ACB=co{s}^{-1}\sqrt[]{\frac{2}{3}}$

$$ $\therefore S\left(2\right)$  is correct