Ncert Solutions Maths class 12th

$\begin{array}{l}=a.{\displaystyle \int x^2.}dx+b{\displaystyle \int x}.dx+c{\displaystyle \int 1.dx}\\ =\frac{a·x^3}{3}+b\frac{x^2}{2}+cx+d.\end{array}$

$\begin{array}{l}\int x^2-\frac{x^2}{x^2}dx\\ ={\displaystyle \int x^2}–1dx={\displaystyle \int x^2}dx–{\displaystyle \int 1dx}\end{array}$

$=\frac{x^3}{3}-x+C$

= $\begin{array}{l} {\displaystyle \int 4e^{3x}dx}+{\displaystyle \int 1dx}\\ \frac{4·e^{3x}}{3}+x+C\end{array}$

$\frac{d}{dx}\left(-\frac{1}{2}cos2x-\frac{4}{3}{e}^{3x}\right)=sin2x-4e^{3x}$

Therefore, an anti-derivative of $\left(sin2x-4e^{3x}\right)is-\frac{1}{2}cos2x-\frac{4}{3}{e}^{3x}$

$\frac{dx}{}{(ax+b)}^3=3a{(ax+b)}^2$

${\left(ax+b\right)}^2=\frac{1}{3a}.\frac{a}{dx}{(ax+b)}^3$

${\left(ax+b\right)}^2=\frac{d}{dx}\left(\frac{1}{3a}{(ax+b)}^3\right)$

Therefore, an anti-derivative of ${(ax+b)}^{3}is\frac{1}{3a}(ax+b){}^3$

$\begin{array}{l}\frac{d}{dx}\left(e^{2}x\right)=2e^{2x}\\ e^{2x}=\frac{1}{2}-\frac{d}{dx}\left(e^{2x}\right)\\ e^{2x}=\frac{d}{dx}\left(\frac{1}{2}{e}^{2x}\right)\end{array}$

Therefore, an anti-derivative of $e^{2x}is\frac{1}{2}{e}^{2x}$

$\begin{array}{l}\frac{dx}{}sin3x=3cos3x \\ cos3x=\frac{1}{3}dx\left(sin3x\right) \\ cos3x= \frac{d}{dx}\left(\frac{1}{3}sin3x\right)\end{array}$

Therefore, an anti-derivative of $cos3xis\frac{1}{3}sin3x$

$\begin{array}{l}\frac{d}{dx}cos2x=–2sin2x \\ sin2x=–\frac{1}{2}\frac{d}{dx}\left(cos2x\right) \\ sin2x =\frac{d}{dx}\left(\frac{1}{2}cos2x\right)\end{array}$

Therefore, an anti-derivative of $sin2xis-\frac{1}{2}cos2x$

The equation of the given curve is $9y^2=x^3$

Differentiate with respect to x, we have:

$\begin{array}{l}9\left(2y\right)\frac{dy}{dx}=3x^2\\ \Rightarrow \frac{dy}{dx}=\frac{x^2}{6y}\end{array}$

The slope of the normal to the given curve at point $\left(x_1,y_1\right)$ is

$\frac{-1}{{\frac{dy}{dx}]}_{\left(x_1,y_1\right)}}=-\frac{6y_1}{{x_1}^2}$

$\therefore$ The equation of the normal to the curve at $\left(x_1,y_1\right)$ is

$\begin{array}{l}y-y_1=-\frac{6y_1}{{x_1}^2}(x-x_1)\\ \Rightarrow {x_1}^{2}y-{x_1}^2{y}_1=-6x{y}_1+6x_1{y}_1\\ \Rightarrow 6x{y}_1+{x_1}^{2}y=6x_1{y}_1+{x_1}^2{y}_1\\ \Rightarrow \frac{6x{y}_1}{6x_1{y}_1+{x_1}^2{y}_1}+\frac{{x_1}^{2}y}{6x_1{y}_1+{x_1}^2{y}_1}=1\\ \Rightarrow \frac{x}{\frac{x_1(6+x_1)}{6}}+\frac{y}{\frac{y_1(6+x_1)}{x_1}}\end{array}$

It is given that the normal makes intercepts with the axes.

Therefore, we have:

$\begin{array}{l}\therefore \frac{x_1(6+x_1)}{6}=\frac{y_1(6+x_1)}{x_1}\\ \Rightarrow \frac{x_1}{6}=\frac{y_1}{x_1}\\ \Rightarrow {x_1}^2=6y_1..........(i)\end{array}$

Also, the point $\left(x_1,y_1\right)$ lies on the curve, so we have

$9{y_1}^2={x_1}^3..........(ii)$

From (i) and (ii), we have:

$9\left(\frac{{x_1}^2}{6}\right)={x_1}^3\Rightarrow \frac{{x_1}^4}{4}={x_1}^3\Rightarrow x_1=4$

From (ii), we have:

$\begin{array}{l}9{y_1}^2={(4)}^3=64\\ \Rightarrow {y_1}^2=\frac{64}{9}\\ \Rightarrow y_1=\pm \frac{8}{3}\end{array}$

Hence, the required points are $\left(4,\pm \frac{8}{3}\right)$ .

Therefore, option (A) is correct.

The equation of the given curve is $x^2=4y$

Differentiating with respect to x, we have:

$\begin{array}{l}2x=4.\frac{dy}{dx}\\ \Rightarrow \frac{dy}{dx}=\frac{x}{2}\end{array}$

The slope of the normal to the given curve at point (h,k) is given by,

$\frac{-1}{{\frac{dy}{dx}]}_{(h,k)}}=-\frac{2}{h}$

$\therefore$ Equation of the normal at point (h,k) is given as:

$y-k=\frac{-2}{h}(x-h)$

Now, it is given that the normal passes through the point (1,2).

Therefore, we have:

$2-k=\frac{-2}{h}(1-h)or,k=2+\frac{2}{h}(1-h)..........(i)$

Since (h,k) lies on the curve $x^2=4y$ ,we have $h^2=4k$

$\Rightarrow k=\frac{h^2}{4}$

From equation (i), we have:

$\begin{array}{l}\frac{h^2}{4}=2+\frac{2}{h}(1-h)\\ \Rightarrow \frac{h^3}{4}=2h+2-2h=2\end{array}$

$\begin{array}{l}\Rightarrow h^3=8\\ \Rightarrow h=2\\ \therefore k=\frac{h^2}{4}\Rightarrow k=1\end{array}$

Hence, the equation of the normal is given as:

$\begin{array}{l}\Rightarrow y-1=\frac{-2}{2}(x-2)\\ \Rightarrow y-1=-(x-2)\\ \Rightarrow x+y=3\end{array}$

Therefore, option (A) is correct.