Class 12th

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.

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

Differentiate with respect to x, we have:

$\begin{array}{l}\frac{2dy}{dx}+2x=0\\ \Rightarrow \frac{dy}{dx}=-x\\ \therefore {\frac{dy}{dx}]}_{(1,1)}=-1\end{array}$

The slope of the normal to the given curve at point (1,1) is

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

Hence, the equation of the normal to the given curve at (1,1) is given as:

$\begin{array}{l}\Rightarrow y-1=1(x-1)\\ \Rightarrow y-1=x-1\\ \Rightarrow x-y=0\end{array}$

Therefore, option (B) is correct.

The equation of the tangent to the given curve is $y=mx+1.$

Now, substituting $y=mx+1.$ in $y^2=4x,$ we get:

$\begin{array}{l}\Rightarrow {\left(mx+1\right)}^2=4x\\ \Rightarrow m^2{x}^2+1+2mx-4x=0\\ \Rightarrow m^2{x}^2+x(2m-4)+1=0..........(i)\end{array}$

Since a tangent touches the curve at one point, the roots of equation (i) must be equal.

Therefore, we have:

Discriminant = 0

$\begin{array}{l}{(2m-4)}^2-4{(m)}^2(1)=0\\ \Rightarrow 4m^2+16-16m-4m^2=0\\ \Rightarrow 16-16m=0\\ \Rightarrow m=1\end{array}$

Hence, the required value of m is 1.

Therefore, option (A) is correct.

Let x be the depth of the wheat inside the cylindrical tank is with radius = 10 cm

Then, Volume V of the cylindrical tank is

V = π (10)²x = 100πpx m³

As $\frac{d}{dt}=314\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$h$}\right.$

$\Rightarrow 100\pi \frac{dx}{dt}=314$

$\frac{dx}{dt}=\frac{314}{100\pi }=\frac{314}{100\pi 3.14}=\frac{314}{314}=1.\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$h$}\right.$ i e, rate of increasing of depth

option (A) is correct

Kindly go through the solution

Kindly go through the solution

We have,

f (x) defined on [a, b]

And f (x) > 0 ∀ x ∈ [a, b].

Let x₁, x₂ ∈ [a, b] and x₂>x₁

In the internal x₁, x₂], f (x) will also be continuous and differentiable.

Hence by mean value theorem, there exist c [x₁, x₂] such that

$f(c)=\frac{f\left(x_2\right)-f\left(x_1\right)}{x_2-x_{1.}}$

f (x) > 0 ∀ x ∈ [a, b].

Then, f (c) > 0.

$\Rightarrow \frac{f\left(x_2\right)-f\left(x_1\right)}{x_2-x_1}>0$

i.e., f (x₂) -f (x₁) > 0

f (x₂) >f (x₁).

Hence, the function f (x) is always increasing on [a, b]

This is a short answer type question as classified in NCERT Exemplar

"A sheet of cellulose acetate laid over a suitable support" is the material used to make a semipermeable membrane for reverse osmosis.

Let x be the radius of the sphere ad x be radius of the right circular cone.

Let height of cone = y

Then, in ΔOBA,

(y-r)² + x² = r²

y² + r²- 2ry + x² = r²

x²=2ry - y²

So, the volume V of the cone is

$V=\frac{1}{3}\pi ·x^2·y$ $\left\{=\frac{1}{3}\pi {(radius)}^2*hight\right\}$

$=\frac{1}{3}*\left(2xy-y^2\right)y$

$=\frac{1}{3}x\left(2x{y}^2-y^3\right).$

So, $\frac{dv}{dy}=\frac{\pi }{3}\left[4xy-3y^2\right]$

And $\frac{d^2}{d{y}^2}=\frac{x}{3}[4x-6y].$

At $\frac{dV}{dy}=0.$

$\Rightarrow \frac{\pi }{3}\left[4xy-3y^2\right]=0$

4x -y- 3y² = 0

$\Rightarrow y=\frac{4x}{3}$ asy> 0.

At y = $\frac{4\pi }{3},\frac{d^{2}v}{d{y}^2}=\frac{\pi }{3}[4x-8x]$ = $-\frac{4\pi r}{3}<0.$

Ø V is maximum when y = $\frac{4\pi }{3}$