Maths NCERT Exemplar Solutions Class 11th Chapter Eight

Let a, b, c be direction ratios of plane containing lines

$\frac{x}{2}=\frac{y}{3}=\frac{z}{5}$

and

$\frac{x}{3}=\frac{y}{7}=\frac{z}{8}$

Equation of plane P is : 1 (x – 3) 1 (y + 4) + 2 (z – 7) = 0

$\Rightarrow x-y+2z-21=0$

Distance from point (2, 5, 11) is

$d=\frac{\left|2+5+22-2\right|}{\sqrt[]{6}}$

$\therefore d^2=\frac{32}{3}$

Radius of circle S touching x-axis and centre  $\left(\alpha ,\beta \right)$ $$ is |. According to given conditions

${\alpha }^2+{\left(\beta -1\right)}^2={\left(\left|\beta \right|+1\right)}^2$

${\alpha }^2=4\beta as\beta >0$

$$ $\therefore$  Required locus is L : x2 = 4y

The area of shaded region = $2{\displaystyle {\int }_0^{4}2\sqrt[]{y}dy}$ ${\displaystyle {}_{}^{}\text{exception 25:}}$

= $4.{\left[\frac{\frac{y^{\frac{3}{2}}}{3}}{2}\right]}_0^4$ ${\frac{\frac{{}^{\frac{}{}}}{}}{}}_{}^{}$

= $\frac{64}{3}$ $\frac{}{}$ square units.

Let inclination of required line is,

So the coordinates of point B can be assumed as

$\left(4-\frac{\sqrt[]{29}}{3}cos\theta ,3-\frac{\sqrt[]{29}}{3}sin\theta \right)$

which satisfies x – y – 2 = 0

$sin\theta -cos\theta =\frac{3}{\sqrt[]{29}}$

By squaring

$sin2\theta =\frac{20}{29}=\frac{2tan\theta }{1+ta{n}^2\theta }$

Point B : $\left(\frac{10}{3},\frac{4}{3}\right)$ $\frac{}{}\frac{}{}$

which also satisfies x + 2y = 6

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

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

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

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

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

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

SSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS

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

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

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

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

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

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

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

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

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

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

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

This is a True or False Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l}Since,thenumberoftermsinthegivenexpressionis{\left(a+b\right)}^{n}is1morethanni.e.,n+1\\ Hence,thegivenstatementis\text{'}False\text{'}.\end{array}$

This is a True or False Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l}Givenexpressionis{\left(x-\frac{1}{x^2}\right)}^{2n}\\ T_{r+1}={}_{2n}{C}_r{\left(x\right)}^{2n-r}{\left(-\frac{1}{x^2}\right)}^r={}_{2n}{C}_r{\left(x\right)}^{2n-r}{\left(-1\right)}^r.\frac{1}{x^{2r}}\\ ={}_{2n}{C}_r{\left(x\right)}^{2n-r-2r}{\left(-1\right)}^r={}_{2n}{C}_r{\left(x\right)}^{2n-3r}{\left(-1\right)}^r\\ Forthetermindependentofx,2n-3r=0\\ \therefore r=\frac{2n}{3}whichisnotanintegerandtheexpressionisnotpossibletobetrue.\\ Hence,thegivenstatementis\text{'}False\text{'}.\end{array}$