4 Marks - Maths STD 12 Science Questions - Vidyadip

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Question 14 Marks

An open box with square base is to be made of a given quantity of card board of area c². Show that the maximum volume of the box is $\frac{\text{c}^2}{6\sqrt{3}}$ cubic units.

Answer

Suppose, the side of the square base of open box be x units and its height be y units.
Thus, the area of the metal used = x² + 4xy

Now, $\text{x}^2+4\text{xy}=\text{c}^2$
$\Rightarrow\ \text{y}=\frac{\text{c}^2-\text{x}^2}{4\text{x}}$
The volume of the box (V) is given by V = x²y
Substituting $\text{y}=\frac{\text{c}^2-\text{x}^2}{4\text{x}}$ in V = x², we get
$\text{V}=\text{x}^2\cdot\Big(\frac{\text{c}^2-\text{x}^2}{4\text{x}}\Big)$
$\Rightarrow\ \text{V}=\frac{1}{4}\text{x}(\text{c}^2-\text{x}^2)$
$\Rightarrow\ \text{V}=\frac{1}{4}(\text{c}^2-\text{x}-\text{x}^3)$
$\Rightarrow\ \frac{\text{dV}}{\text{dx}}=\frac{1}{4}(\text{c}^2-3\text{x}^2)$
Find the critical points by equating $\frac{\text{dV}}{\text{dx}}$ to zero.
$\therefore\ \text{c}^2-3\text{x}^2=0$
$\Rightarrow\ \text{x}^2=\frac{\text{c}^2}{3}$
$\Rightarrow\ \text{x}=\frac{\text{c}}{\sqrt{3}}$ [Neglecting negative value of x as side can’t be negative]
Consider, $\frac{\text{dV}}{\text{dx}}=\frac{1}{4}(\text{c}^2-3\text{x}^2)$
$\Rightarrow\ \frac{\text{d}^2\text{V}}{\text{dx}^2}=\frac{1}{4}(-6\text{x})=\frac{-3}{2}\text{x}$
$\Rightarrow\Big(\frac{\text{dv}^2}{\text{dx}^2}\Big)_{\text{x}=\frac{\text{C}}{\sqrt{3}}}=-\frac{3}{2}\Big(\frac{\text{G}}{\sqrt{3}}\Big)<0$
Thus, the volume (V) is maximum at $\text{x}=\frac{\text{c}}{\sqrt{3}}$
$\therefore$ Maximum volume of the box, $(\text{V})_{\text{x}=\frac{\text{c}}{\sqrt{3}}}=\frac{1}{4}\Big(\text{c}^2\cdot\frac{\text{c}}{\sqrt{3}}-\frac{\text{c}^3}{3\sqrt{3}}\Big)$
$=\frac{1}{4}\cdot\frac{(3\text{c}^3-\text{c}^3)}{3\sqrt{3}}=\frac{1}{4}\cdot\frac{2\text{c}^3}{3\sqrt{3}}$
$=\frac{\text{c}^2}{6\sqrt{3}}\text{ cu units}$

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Question 24 Marks

Find the points of local maxima, local minima and the points of inflection of the function f(x) = x⁵ - 5x⁴ + 5x³ - 1. Also find the corresponding local maximum and local minimum values.

Answer

Given that, f(x) = x⁵ - 5x⁴ + 5x³ -1
On differentiating w.r.t. x, we get
f'(x) = 5x⁴ - 20x³ + 15x²
For maxima or minima f'(x) = 0
⇒ 5x⁴ - 20x³ + 15x² = 0
⇒ 5x²(x² - 4x + 3) = 0
⇒ 5x²(x² - 3x - x + 3) = 0
⇒ 5x²[x(x - 3) -1(x - 3)] = 0
⇒ 5x²[(x - 1)(x - 3)] = 0
$\therefore$ x = 0, 1, 3
Sing scheme for $\frac{\text{dy}}{\text{dx}}=5\text{x}^2(\text{x}-1)(\text{x}-3)$
So, y has maximum value at x = 1 and minimum value at x = 3.
At x = 0, y has neither maximum nor minimum value, so x = 0 is the point of inflection
Maximum value of y = 1 - 5 + 5 - 1 = 0 (at x = 1)
and Minimum value of y = 3⁵ -5(3)⁴ + 5(3)³ -1 = 243 - 405 + 135 - 1 = -28 (at x = 3)

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Question 34 Marks

AB is a diameter of a circle and C is any point on the circle. Show that the area of $\triangle\text{ABC}$ is maximum, when it is isosceles.

Answer

We have, AB = 2r

And $\angle\text{ACB}=90^\circ$ [Since, angle in the semi-circle is always 90°]
Let AC = x and BC = y
$(2\text{r})^2 = \text{x}^2 + \text{y}^2$
$\Rightarrow\ \text{y}^2 = 4\text{r}^2 - \text{x}^2$
$\Rightarrow\ \text{y}=\sqrt{4\text{r}^2-\text{x}^2}\ \ \dots(\text{i})$
Now, area of $\triangle\text{ABC, A}=\frac{1}{2}\times\text{x}\times\text{y}$
$=\frac{1}{2}\times\text{x}\times(4\text{r}^2-\text{x}^2)^{\frac{1}{2}}$[using Eq. (i)]
Now, differentiating both sides w.r.t. x, we get
$\frac{\text{dA}}{\text{dx}}=\frac{1}{2}\Big[\text{x}\cdot\frac{1}{2}(4\text{r}^2-\text{x}^2)^{\frac{-1}{2}}\cdot(0-2\text{x})+(4\text{r}^2-\text{x}^2)^{\frac{-1}{2}}\cdot1\Big]$
$=\frac{1}2{}\Big[\frac{-2\text{x}^2}{2\sqrt{4\text{r}^2-\text{x}^2}}+(4\text{r}^2-\text{x}^2)^{\frac{1}{2}}\Big]$
$=\frac{1}{2}\Big[\frac{-\text{x}^2}{\sqrt{4\text{r}^2-\text{x}^2}}+\sqrt{4\text{r}^2-\text{x}^2}\Big]$
$=\frac{1}{2}\Big[\frac{-\text{x}^2+4\text{r}^2-\text{x}^2}{\sqrt{4\text{r}^2-\text{x}^2}}\Big]$ $=\frac{1}{2}\Big[\frac{-2\text{x}^2+4\text{r}^2}{\sqrt{4\text{r}^2-\text{x}^2}}\Big]$
$\frac{\text{dA}}{\text{dx}}=\Big[\frac{(-\text{x}^2+2\text{r}^2)}{\sqrt{4\text{r}^2-\text{x}^2}}\Big]$
Now, $\frac{\text{dA}}{\text{dx}}=0$
$-\text{x}^2+2\text{r}^2=0$
$\text{r}^2=\frac{1}{2}\text{x}^2$
$\text{r}=\frac{1}{\sqrt{2}}\text{x}^2$
$\text{x}=\text{r}\sqrt{2}$
Again, differentiating both sides w.r.t. x, we get
$\frac{\text{d}^2\text{A}}{\text{dx}^2}=\frac{\sqrt{4\text{r}^2-\text{x}^2}\cdot(2\text{x})+(2\text{r}^2-\text{x}^2)\cdot\frac{1}{2}(4\text{r}^2-\text{x}^2)^{\frac{-1}{2}}(-2\text{x})}{\big(\sqrt{4\text{r}^2-\text{x}^2}\big)^2}$
$=\frac{-2\text{x}\Big[\sqrt{4\text{r}^2-\text{x}^2}+(2\text{r}^2-\text{x}^2)\cdot\frac{1}{2\sqrt{4\text{r}^2-\text{x}^2}}\Big]}{\big(\sqrt{4\text{r}^2-\text{x}^2}\big)^2}$
$=\frac{-4\text{x}\cdot\big(\sqrt{4\text{r}^2-\text{x}^2}\big)^2+(2\text{r}^2-\text{x}^2)(-2\text{x})}{2\cdot(4\text{r}^2-\text{x}^2)^{\frac{3}{2}}}$
$=\frac{-4\text{x}(4\text{r}^2-\text{x}^2)^2+(2\text{r}^2-\text{x}^2)\cdot(-2\text{x})}{2\cdot(4\text{r}^2-\text{x}^2)^{\frac{3}{2}}}$
$=\frac{-16\text{xr}^2+4\text{x}^3+(2\text{r}^2-\text{x}^2)(-2\text{x})}{2\cdot(4\text{r}^2-\text{x}^2)^{\frac{3}{2}}}$
$\Big(\frac{\text{d}^2\text{A}}{\text{dx}^2}\Big)_{\text{x}=\text{r}\sqrt{2}}=\frac{-16\cdot\text{r}\sqrt{2}\cdot\text{r}^2+4\cdot\big(\text{r}\sqrt{2}\big)^3+\Big[2\text{r}^2-\big(\text{r}\sqrt{2}\big)^2\Big]\cdot\big(-2\cdot\text{r}\sqrt{2}\big)}{2\cdot(4\text{r}^2-2\text{r}^2)^{\frac{3}{2}}}$ $\big[\because\ \text{x}=\text{r}\sqrt{2}\big]$
$=\frac{-16\sqrt{2}\cdot\text{r}^3+8\sqrt{2}\text{r}^3}{2(2\text{r}^2)^{\frac{3}{2}}}=\frac{8\sqrt{2}\text{r}^2[\text{r}-2\text{r}]}{4\text{r}^3}$
$=\frac{-8\sqrt{2}\text{r}^3}{4\text{r}^3}=-2\sqrt{2}<0$
For $\text{x}=\text{r}\sqrt{2},$ the area of triangle is maximum.
For $\text{x}=\text{r}\sqrt{2},$ $\text{y}=\sqrt{4\text{r}^2\big(\text{r}\sqrt{2}\big)^2}=\sqrt{2\text{r}^2}=\text{r}\sqrt{2}$
Since, $\text{x}=\text{r}\sqrt{2}=\text{y}$
Hence, the triangle is isosceles.

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Question 44 Marks

A man, 2m tall, walks at the rate of $1\frac{2}{3}\text{m/s}$ towards a street light which is $5\frac{1}{3}\text{m}$ above the ground. At what rate is the tip of his shadow moving? at what rate is the length of the shadow changing when he is $3\frac{1}{3}\text{m}$ from the base of the light?

Answer

Let AB be the street light post and CD be the height of man i.e., CD = 2m.

Let BC = x m, CE = y m and $\frac{\text{dx}}{\text{dt}}=\frac{-5}{3}\text{m/s}$
From $\triangle\text{ABE}$ and $\triangle\text{DCE},$ we see that
$\triangle\text{ABE}\sim\triangle\text{DCE}$ [by AAA similarity]
$\therefore\ \frac{\text{AB}}{\text{DC}}=\frac{\text{BE}}{\text{CE}}$ $\Rightarrow\ \frac{\frac{16}{3}}{2}=\frac{\text{x}+\text{y}}{\text{y}}$
$\Rightarrow\ \frac{16}{6}=\frac{\text{x}+\text{y}}{\text{y}}$
$\Rightarrow\ 16\text{y}=6\text{x}+6\text{y}\Rightarrow10\text{y}=6\text{x}$
$\Rightarrow\ \text{y}=\frac{3}{5}\text{x}$
On differentiating both sides w.r.t. t, we get
$\frac{\text{dy}}{\text{dt}}=\frac{3}{5}\cdot\frac{\text{dx}}{\text{dt}}=\frac{3}{5}\cdot\Big(-1\frac{2}{3}\Big)$ [Since, the man is moving toward the light post]
$=\frac{3}{5}\cdot\Big(\frac{-5}{3}\Big)=-1\text{m/s}$
Let the distance of the tip of shadow at any instant from the foot of street light is z
$\Rightarrow\ \text{z}=\text{x}+\text{y}$
Now, differentiating both sides w.r.t. t, we get
$\frac{\text{dz}}{\text{dt}}=\frac{\text{dx}}{\text{dt}}+\frac{\text{dy}}{\text{dt}}=-\Big(\frac{5}{3}+1\Big)$
$=-\frac{8}{3}=-\Big(-2\frac{2}{3}\Big)\text{m/s}$
Hence, the tip of shadow is moving at the rate of $2\frac{2}{3}\text{m/s}$ towards the light source and length of the shadow is decreasing at the rate of 1m/ s.

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Question 54 Marks

If the sum of the lengths of the hypotenuse and a side of a right angled triangle is given, show that the area of the triangle is maximum when the angle between them is $\frac{\pi}{3}.$

Answer

Let ABC be a triangle with AC = h, AB = x and BC = y
Let h + x = k ...(1)
Also, $\angle\text{CAB}=\theta$
$\Rightarrow\ \cos\theta=\frac{\text{x}}{\text{h}}$
$\Rightarrow\ \text{x}=\text{h}\cos\theta$
Substituting $\text{x}=\text{h}\cos\theta$ in $\text{h}+\text{x}=\text{k},$ we get
$\text{h}+\text{h}\cos\theta=\text{k}$ [using (1)]
$\Rightarrow\ \text{h}=\frac{\text{k}}{(1+\cos\theta)}\ \ \dots(2)$
Now, area of $\triangle\text{ABC}=\frac{1}{2}(\text{AB}\cdot\text{BC})$
$\text{A}=\frac{1}{2}\cdot\text{x}\cdot\text{y}$
$=\frac{1}{2}\text{h}\cos\theta\cdot\text{h}\sin\theta$ $\Big[\because\ \text{x}=\text{h}\cos\theta\text{ and }\sin\theta=\frac{\text{y}}{\text{h}}\Big]$
$=\frac{1}{2}\text{h}^2\sin\theta\cos\theta$
$=\frac{2\text{h}^2}{4}\sin\theta\cdot\cos\theta$
$\therefore\ \text{A}=\frac{1}{4}\text{h}^2\sin2\theta\ \ \dots(3)$
Substituting $\text{h}=\frac{\text{k}}{1+\cos\theta}$ in $\text{A}=\frac{1}{4}\text{h}^2\sin2\theta,$ we get
$\text{A}=\frac{1}{4}\Big(\frac{\text{k}}{1+\cos\theta}\Big)^2\cdot\sin2\theta$
$\Rightarrow\ \text{A}=\frac{\text{k}^2}{4}\cdot\frac{\sin2\theta}{(1+\cos\theta)^2}\ \ \dots(4)$
$\Rightarrow\ \frac{\text{dA}}{\text{d}\theta}=\frac{\text{k}^2}{4}\bigg[\frac{(1+\cos\theta)^2\cdot\cos2\theta\cdot2-\sin2\theta\cdot2(1+\cos\theta)\cdot(0-\sin\theta)}{(1+\cos\theta)^4}\bigg]$
$\frac{\text{k}^2}{4}\Bigg\{\frac{2(1+\cos\theta)\big[(1+\cos\theta)\cdot\cos2\theta+\sin2\theta(\sin\theta)\big]}{(1+\cos\theta)^4}\Bigg\}$
$\frac{\text{k}^2}{4}\cdot\frac{2}{(1+\cos\theta)^3}\big[(1+\cos\theta)\cdot\cos2\theta+2\sin^2\theta\cdot\cos\theta\big]$
$=\frac{\text{k}^2}{2(1+\cos\theta)^3}\big[(1+\cos\theta)(1-2\sin^2\theta)+2\sin^2\theta\cdot\cos\theta\big]$
$=\frac{\text{k}^2}{2(1+\cos\theta)^3}\big[1+\cos\theta-2\sin^2\theta -2\sin^2\theta\cdot\cos\theta+2\sin^1\theta\cdot\cos\theta\big]$
$=\frac{\text{k}^2}{2(1+\cos\theta)^3}\big[(1+\cos\theta)-2\sin^2\theta\big]$
$ =\frac{\text{k}^2}{2(1+\cos\theta)^3}\big[1+\cos\theta-2+2\cos^2\theta\big]$
$=\frac{\text{k}^2}{2(1+\cos\theta)^3}(2\cos^2\theta+\cos\theta-1)\ \ \dots(5)$
For $\frac{\text{dA}}{\text{d}\theta}=0,$
$=\frac{\text{k}^2}{2(1+\cos\theta)^3}(2\cos^2\theta+\cos\theta-1)=0$
$\Rightarrow\ 2\cos^2\theta+\cos\theta-1=0$
$\Rightarrow\ 2\cos^2\theta+2\cos\theta-\cos\theta-1=0$
$\Rightarrow\ 2\cos^2\theta(\cos\theta+1)-1(\cos\theta+1)=0$
$\Rightarrow\ (2\cos\theta-1)(\cos\theta+1)=0$
$\Rightarrow\ \cos\theta=\frac{1}{2}\text{ or }\cos\theta=-1$
$\Rightarrow\ \theta=\frac{\pi}{3}$
$\theta=2\text{n}\pi\pm\pi$
Again, differentiating w.r.t $\theta$ in Eq. (v), we get
$\frac{\text{d}}{\text{d}\theta}\Big(\frac{\text{dA}}{\text{d}\theta}\Big)=\frac{\text{d}}{\text{d}\theta}\bigg[\frac{\text{k}^2}{2(1+\cos\theta)^3}(2\cos^2\theta+\cos\theta-1)\bigg]$
$\frac{\text{d}^2\text{A}}{\text{d}\theta^2}=\frac{\text{d}}{\text{d}\theta}\bigg[\frac{\text{k}^2(2\cos\theta-1)(1+\cos\theta)}{2(1+\cos\theta)^3}\bigg]=\frac{\text{d}}{\text{d}\theta}\bigg[\frac{\text{k}^2}{2}.\frac{(2\cos\theta-1)}{(1+\cos\theta)^2}\bigg]$
$=\frac{\text{k}^2}{2}\bigg[\frac{(1+\cos\theta)^2\cdot(-2\sin\theta)-2(1+\cos\theta)\cdot(-\sin\theta)(2\cos\theta-1)}{(1+\cos\theta)^4}\bigg]$
$=\frac{\text{k}^2}{2}\bigg[\frac{(1+\cos\theta)\cdot[1+\cos\theta](-2\sin\theta)+(2\sin\theta)(2\cos\theta-1)}{(1+\cos\theta)^4}\bigg]$
$=\frac{\text{k}^2}{2}\bigg[\frac{-2\sin\theta-2\sin\theta\cdot\cos\theta+4\sin\theta\cdot\cos\theta-2\sin\theta}{(1+\cos\theta)^3}\bigg]$
$=\frac{\text{k}^2}{2}\bigg[\frac{-4\sin\theta-\sin2\theta+2\sin2\theta}{(1+\cos\theta)^3}\bigg]=\frac{\text{k}^2}{2}\bigg[\frac{\sin2\theta-4\sin\theta}{(1+\cos\theta)^3}\bigg]$
$\bigg(\frac{\text{d}^2\text{A}}{\text{d}\theta^2}\bigg)_{\text{at}\theta=\frac{\pi}{3}}=\frac{\text{k}^2}{2}\Bigg[\frac{\sin\frac{2\pi}{3}-4\sin\frac{\pi}{3}}{\Big(1+\cos\theta\frac{\pi}{3}\Big)^3}\Bigg]=\frac{\text{k}^2}{2}\Bigg[\frac{\frac{\sqrt{3}}{2}-4\frac{\sqrt{3}}{2}}{\Big({1+\frac{1}{2}\Big)^3}}\Bigg]$
$=\frac{\text{k}^2}{2}\bigg[\frac{-3\sqrt{3}.8}{2.27}\bigg]=-\text{k}^2\Big(\frac{2\sqrt{3}}{9}\Big)<0$
Hence, area of the right angled triangle is maximum, when the angle between them is $\frac{\pi}{3}.$

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Question 64 Marks

Prove that the curves xy = 4 and x² + y² = 8 touch each other.

Answer

Given equation of curves are
xy = 4 ...(i)
and x² + y² = 8 ...(ii)
Let's first find the point of intersection of curves
From equation (i), $\text{y}=\frac{4}{\text{x}}$
Putting this is equation (ii), we get
$\text{x}^2+\frac{16}{\text{x}^2}=8$
$\Rightarrow\ \text{x}^4-8\text{x}^2+16=0$
$\Rightarrow\ (\text{x}^2-4)^2=0$
$\Rightarrow\ \text{x}^2=4$
$\Rightarrow\ \text{x}=\pm2$
$\Rightarrow\ \text{y}=\pm2$ (using (i))
$\therefore$ Points of intersection are (2, 2) and (-2, -2)
For curve (i), $\text{y}+\text{x}\Big(\frac{\text{dy}}{\text{dx}}\Big)_1=0$
$\Rightarrow\ \Big(\frac{\text{dy}}{\text{dx}}\Big)_1=-\frac{\text{y}}{\text{x}}$
At point (2, 2) and (-2, -2), $\Big(\frac{\text{dy}}{\text{dx}}\Big)_1=-1$
For curve (ii), $2\text{x}+2\text{y}\Big(\frac{\text{dy}}{\text{dx}}\Big)_2=0$
$\Rightarrow\ \Big(\frac{\text{dy}}{\text{dx}}\Big)_1=-\frac{\text{x}}{\text{y}}$
At point (2, 2) and (-2, -2), $\Big(\frac{\text{dy}}{\text{dx}}\Big)_2=-1$
Thus two curves touch each other at their points of intersection.

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Question 74 Marks

At what points on the curve x² + y² - 2x - 4y + 1 = 0, the tangents are parallel to the y-axis?

Answer

Given, equation of curve which is
x² + y² - 2x - 4y + 1 = 0 ...(i)
$\Rightarrow\ 2\text{x}+2\text{y}\frac{\text{dy}}{\text{dx}}-2-4\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}(2\text{y}-4)=2-2\text{x}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{2(1-\text{x})}{2(\text{y}-2)}$
Since, the tangents are parallel to the Y-axis i.e., $\tan\theta=\tan90^\circ=\frac{\text{dy}}{\text{dx}}.$
$\frac{1-\text{x}}{\text{y}-2}=\frac{1}{0}$
$\Rightarrow\ \text{y}-2=0$
$\Rightarrow\ \text{y}=2$
For y = 2 from Eq. (i), we get
$\text{x}^2+2^2-2\text{x}-4\times2+1=0$
$\Rightarrow\ \text{x}^2-2\text{x}-3=0$
$\Rightarrow\ \text{x}^2-3\text{x}+\text{x}-3=0$
$\Rightarrow\ \text{x}(\text{x}-3)+1(\text{x}-3)=0$
$\Rightarrow\ (\text{x}+1)(\text{x}-3)=0$
$\therefore\ \text{x}=-1,\text{ x}=3$
So, the required points are (-1, 2) and (3, 2).

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Question 84 Marks

Find the condition that the curves 2x = y² and 2xy = k intersect orthogonally.

Answer

We have, 2xy = k
$\Rightarrow\ \text{y}=\frac{\text{k}}{2\text{x}}$
Substituting $\text{y}=\frac{\text{k}}{2\text{x}}$ in 2x = y², we get
$2\text{x}=\Big(\frac{\text{k}}{2\text{x}}\Big)^2$
$\Rightarrow\ 8\text{x}^3=\text{k}^2$
$\Rightarrow\ \text{x}^3=\frac{1}{8}\text{k}^2$
$\Rightarrow\ \text{x}=\frac{1}{2}\text{k}^{\frac{2}{3}}$
Substituting $\text{x}=\frac{1}{2}\text{k}^{\frac{2}{3}}$ in $\text{y}=\frac{\text{k}}{2\text{x}},$ we get
$\text{y}=\frac{\text{k}}{2\text{k}}=\frac{\text{k}}{2\cdot\frac{1}{2}\text{k}^{\frac{2}{3}}}=\text{k}^{\frac{1}{3}}$
Thus, the point of intersection of two curves which is $\Big(\frac{1}{2}\text{k}^{\frac{2}{3}},\text{k}^{\frac{1}{3}}\Big)$
Consider, 2x = y² and 2xy = k
$\Rightarrow\ 2=2\text{y}\frac{\text{dy}}{\text{dx}}$ and $2\Big[\text{x}\cdot\frac{\text{dy}}{\text{dx}}+\text{y}\cdot1\Big]=0$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{1}{\text{y}}$ and $\Big(\frac{\text{dy}}{\text{dx}}\Big)=-\frac{\text{y}}{\text{x}}$
$\Rightarrow\ \Big(\frac{\text{dy}}{\text{dx}}\Big)_{\Big(\frac{1}{2}\text{k}^\frac{2}{3},\text{k}^\frac{1}{3}\Big)}=\frac{1}{\text{k}^{\frac{1}{3}}}$ [say m₁] and $\Big(\frac{\text{dy}}{\text{dx}}\Big)_{\Big(\frac{1}{2}\text{k}^\frac{2}{3},\text{k}^\frac{1}{3}\Big)}=\frac{-\text{k}^{\frac{1}{3}}}{\frac{1}{2}\text{k}^{\frac{2}{3}}}=-2\text{k}^{\frac{-1}{3}}$ [say m₂]
Since, the curves intersect orthogonally.
$\therefore\ \text{m}_1\cdot\text{m}_2=-1$
$\Rightarrow\ \frac{1}{\text{k}^{\frac{1}{3}}}\cdot\Big(-2\text{k}^{\frac{-1}{3}}\Big)=-1$
$\Rightarrow-2\text{k}^{\frac{-2}{3}=-1}$
$\Rightarrow\ \frac{2}{\text{k}^{\frac{2}{3}}}=1$
$\Rightarrow\ \text{k}^{\frac{2}{3}}=2$
$\therefore\ \text{k}^2=8$
which is the required condition.

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Question 94 Marks

Two men A and B start with velocities v at the same time from the junction of two roads inclined at 45° to each other. If they travel by different roads, find the rate at which they are being seperated.

Answer

Let two men start from the point C with velocity v each at the same time.
Also, $\angle\text{BCA}=45^\circ$
Since, A and B are moving with same velocity v, so they will cover same distance in same time.
Therefore, $\triangle\text{ABC}$ ABC is an isosceles triangle with AC = BC.
Now, draw $\text{CD}\bot\text{AB}$
Let at any instant t, the distance between them is AB

Let $\text{AC}=\text{BC}=\text{x}$ and $\text{AB}=\text{y}$
In $\triangle\text{ACD}$ and $\triangle\text{DCB},$
$\angle\text{CAD}=\angle\text{CBD}$ $[\because\text{AC}=\text{BC}]$
$\angle\text{CDA}=\angle\text{CDB}=90^\circ$
$\therefore\ \angle\text{ACD}=\angle\text{DCB}$
or $\angle\text{ACD}=\frac{1}{2}\times\angle\text{ACB}$
$\Rightarrow\ \angle\text{ACD}=\frac{1}{2}\times45^\circ$
$\Rightarrow\ \angle\text{ACD}=\frac{\pi}{8}$
$\therefore\ \sin\frac{\pi}{8}=\frac{\text{AD}}{\text{AC}}$
$\Rightarrow\ \sin\frac{\pi}{8}=\frac{\frac{\text{y}}{2}}{\text{x}}$ $\big[\because\text{AD}=\frac{\text{y}}{2}\big]$
$\Rightarrow\ \frac{\text{y}}{2}=\text{x}\sin\frac{\pi}{8}$
$\Rightarrow\ \text{y}=2\text{x}\sin\frac{\pi}{8}$
Now, differentiating both sides w.r.t. t, we get
$\frac{\text{dy}}{\text{dt}}=2.\sin\frac{\pi}{8}\cdot\frac{\text{dx}}{\text{dt}}$
$=2.\sin\frac{\pi}{8}.\text{v}$ $\Big[\because\text{v}=\frac{\text{dx}}{\text{dt}}\Big]$
$=2\text{v}.\frac{\sqrt{2-\sqrt{2}}}{2}$ $\Bigg[\because\ \sin\frac{\pi}{8}=\frac{\sqrt{2-\sqrt{2}}}{2}\Bigg]$
$=\sqrt{2-\sqrt{2}}\text{ v unit/s}$
which is the rate at which A and B are being separated.

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Question 104 Marks

Find the angle of intersection of the curves y = 4 - x² and y = x².

Answer

We have, y = 4 - x² ...(i)
and y = x² ...(ii)
Solving these equations, we get
$\text{x}^2=4-\text{x}^2$
$\Rightarrow\ \text{x}^2=2$
$\Rightarrow\ \text{x}=\pm\sqrt{2}$
$\therefore\ \text{y}=\text{x}^2=2$
So, the points of intersection are $\big(\sqrt{2},2\big)$ and $\big(-\sqrt{2},2\big).$
Now,
For first curve, $\text{m}_1=\Big(\frac{\text{dy}}{\text{dx}}\Big)_1=-2\text{x}$
For first curve, $\text{m}_2=\Big(\frac{\text{dy}}{\text{dx}}\Big)_2=2\text{x}$
At ponit $\big(\sqrt{2},2\big),\text{m}_1=-2\sqrt{2}$ and $\text{m}_2=2\sqrt{2}$
If angle between curves is $'\theta',$ then
$\therefore\ \tan\theta=\Big|\frac{\text{m}_1-\text{m}_2}{1+\text{m}_1\text{m}_2}\Big|=\Big|\frac{-2\sqrt{2}-2\sqrt{2}}{1-2\sqrt{2}\cdot2\sqrt{2}}\Big|=\Big|\frac{-4\sqrt{2}}{-7}\Big|$
$\therefore\ \theta=\tan^{-1}\frac{4\sqrt{2}}{7}$
For point $\big(-\sqrt{2},2\big),\text{m}_1=2\sqrt{2}$ and $\text{m}_2=-2\sqrt{2}$
$\therefore$ For this point also $\theta=\tan^{-1}\frac{4\sqrt{2}}7{}$

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Question 114 Marks

The volume of a cube increases at a constant rate. Prove that the increase in its surface area varies inversely as the length of the side.

Answer

Let the side of a cube be x unit.
$\therefore$ Volume of cube (V) = x³
$\Rightarrow\ \frac{\text{dV}}{\text{dt}}=3\text{x}^2\frac{\text{dx}}{\text{dt}}$
$\Rightarrow\ \text{k}=3\text{x}^2\frac{\text{dx}}{\text{dt}}$ (Since, the volume is increasing at constant rate)
$\Rightarrow\ \frac{\text{dx}}{\text{dt}}=\frac{\text{k}}{3\text{x}^2}$
Now, surface area of cube, S = 6x²
$\Rightarrow\ \frac{\text{dS}}{\text{dt}}=12\text{x}\cdot\frac{\text{dx}}{\text{dy}}$
$\Rightarrow\ \frac{\text{dS}}{\text{dt}}12\text{x}\cdot\frac{\text{k}}{3\text{x}^2}$ $\Big[\because\ \frac{\text{dx}}{\text{dt}}=\frac{\text{k}}{3\text{x}^2}\Big]$
$\Rightarrow\ \frac{\text{dS}}{\text{dt}}=\frac{12\text{k}}{3\text{x}}=4\Big(\frac{\text{k}}{\text{x}}\Big)$
$\Rightarrow\ \frac{\text{dS}}{\text{dt}}\propto\frac{1}{\text{x}}$
Thus, the surface area of the cube varies inversely as the length of the side.

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Question 124 Marks

If the sum of the surface areas of cube and a sphere is constant, what is the ratio of an edge of the cube to the diameter of the sphere, when the sum of their volumes is minimum?

Answer

Let length of one edge of cube be x units and radius of sphere be r units.
Now Surface area of cube $ = 6\text{x}^2$
And surface area of sphere $=4\pi\text{r}^2$
According to the question
$6\text{x}^2+4\pi\text{r}^2=\text{k},$ where k is constant
$\Rightarrow\ \text{x}=\Big[\frac{\text{k}-4\pi\text{r}^2}{6}\Big]^{\frac{1}{2}}$
Now, volume of cube $ =\text{x}^3$
and volume of sphere $=\frac{4}{3}\pi\text{r}^3$
The sum of volume of the cube and volume of the sphere is
$\text{S}=\text{x}^3+\frac{4}{3}\pi\text{r}^3$
$=\Big[\frac{\text{k}-4\pi\text{r}^2}{6}\Big]^{\frac{3}{2}}+\frac{4}{3}\pi\text{r}^3$
$\Rightarrow\ \frac{\text{dS}}{\text{dr}}=-2\pi\text{r}\Big[\frac{\text{k}-4\pi\text{r}^2}{6}\Big]^{\frac{1}{2}}+4\pi\text{r}^2\ \ \dots(\text{ii})$
$=-2\pi\text{r}\bigg[\Big\{\frac{\text{k}-4\pi\text{r}^2}{6}\Big\}^{\frac{1}{2}}-2\text{r}\bigg]$
Now, $\frac{\text{dS}}{\text{dr}}=0$
$\Rightarrow\ 2\text{r}=\Big(\frac{\text{k}-4\pi\text{r}^2}{6}\Big)^{\frac{1}{2}}\ (\text{as r}\neq0)$
$\Rightarrow\ 4\text{r}^2=\frac{\text{k}-4\pi\text{r}^2}{6}$
$\Rightarrow\ 24\text{r}^2=\text{k}-4\pi\text{r}^2$
$\Rightarrow\ \text{r}^2[24+4\pi]=\text{k}$
$\Rightarrow\ \text{r}=\sqrt{\frac{\text{k}}{24+4\pi}}=\frac{1}{2}\sqrt{\frac{\text{k}}{6+\pi}}$
Clearly, this is point of minima
For $\text{r}=\frac{1}{2}\sqrt{\frac{\text{k}}{6+\pi}},\text{x}=\Bigg[\frac{\text{k}-4\pi\cdot\frac{1}{4}\cdot\frac{\text{k}}{(6+\pi)}}{6}\Bigg]^{\frac{1}{2}}$
$=\bigg[\frac{(6+\pi)\text{k}-\pi\text{k}}{6(6+\pi)}\bigg]^{\frac{1}{2}}=\Big[\frac{\text{k}}{6+\pi}\Big]^{\frac{1}{2}}=2\text{r}$
Thus, the sum of their volume is minimum when x = 2r.
Hence, the ratio of an edge of cube to the diameter of the sphere is 1 : 1.

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Question 134 Marks

A kite is moving horizontally at a height of 151.5 meters. If the speed of kite is 10m/ s, how fast is the string being let out; when the kite is 250m away from the boy who is flying the kite? The height of boy is 1.5m.

Answer

Consider the figure,

We have, height (h) = 151.5m, speed of kite (v) = 10m/ s
Let CD be the height of kite and AB be the height of boy.
Let DB = x m = EA and AC = 250m
$\therefore\ \frac{\text{dx}}{\text{dt}}=10\text{m/s}$
From the figure, we have
EC = 151.5 - 1.5 = 150m and AE = x
Also, AC = 250m
In right angled $\triangle\text{CEA},$
AE² + EC² = AC²
⇒ x² + (150)² = y²
⇒ x² + (150)² = (250)²
⇒ x² = (250)² - (150)²
⇒ x² = (250 + 150)(250 - 150)
⇒ x² = 400 × 100
$\therefore$ x = 20 × 10 = 200
Consider, x² + (150)² = y²
$\Rightarrow\ 2\text{x}\cdot\frac{\text{dx}}{\text{dt}}+0=2\text{y}\frac{\text{dy}}{\text{dt}}$
$\Rightarrow\ 2\text{y}\frac{\text{dy}}{\text{dt}}=2\text{x}\frac{\text{dx}}{\text{dt}}$
$\Rightarrow\ \frac{\text{dy}}{\text{dt}}=\frac{\text{x}}{\text{y}}\cdot\frac{\text{dx}}{\text{dt}}$
$=\frac{200}{250}\cdot10=8\text{m/s}$ $\Big[\because\ \frac{\text{dx}}{\text{dt}}=10\text{m/s}\Big]$
So, the required rate at which the string is being let out is 8m/ s.

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Question 144 Marks

Find the co-ordinates of the point on the curve $\sqrt{\text{x}}+\sqrt{\text{y}}=4$ at which tangent is equally inclined to the axes.

Answer

We have, $\sqrt{\text{x}}+\sqrt{\text{y}}=4\ \ \dots(\text{i})$
$\Rightarrow\ \text{x}^{\frac{1}{2}}+\text{y}^{\frac{1}{2}}=4$
Difference,
$\frac{1}{2}\frac{1}{\text{x}{\frac{1}{2}}}+\frac{1}{2}\cdot\frac{1}{\text{y}{\frac{1}{2}}}=0$
$\frac{1}{2}\text{x}^{\frac{-1}{2}}+\frac{1}{2}\text{y}^{\frac{-1}{2}}\frac{\text{dy}}{\text{dx}}=0$
$\frac{1}{2}\text{y}^{\frac{-1}{2}}\frac{\text{dy}}{\text{dx}}=\frac{-1}{2}\text{x}^{\frac{-1}{2}}$
$$$\text{y}^{\frac{-1}{2}}\frac{\text{dy}}{\text{dx}}=2\times\frac{-1}{2}\text{x}^{\frac{-1}{2}}$
$\frac{\text{dy}}{\text{dx}}=-\frac{\text{x}^{\frac{-1}{2}}}{\text{y}^{\frac{-1}{2}}}$
$\frac{\text{dy}}{\text{dx}}=-\frac{\frac{1}{\text{x}^{\frac{1}{2}}}}{\frac{1}{\text{y}^{\frac{1}{2}}}}$
$\frac{\text{dy}}{\text{dx}}=-\frac{\text{y}^{\frac{1}{2}}}{\text{x}\frac{1}{2}}$
$\frac{\text{dy}}{\text{dx}}=-\sqrt{\frac{\text{y}}{\text{x}}}$
Since, tangent is equally inclined to the axes.
$\therefore\ \frac{\text{dy}}{\text{dx}}=\pm1$
$\Rightarrow-\sqrt{\frac{\text{y}}{\text{x}}}=\pm1$
$\Rightarrow\ \frac{\text{y}}{\text{x}}=1\Rightarrow\text{y}=\text{x}$
From Eq. (i), $\sqrt{\text{y}}+\sqrt{\text{y}}+=4$
$\Rightarrow\ 2\sqrt{\text{y}}=4$
$\Rightarrow4\text{y}=16$
$\therefore$ y = 4 and x = 4
when y = 4 and x = 4
So, the required coordinates are (4, 4).

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Question 154 Marks

If the straight line $\text{x}\cos\alpha+\text{y}\sin\alpha=\text{p}$ touches the curve $\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}=1,$ then prove that $\text{a}^2\cos^2\alpha+\text{b}^2\sin^2\alpha=\text{p}^2.$

Answer

We know that, if a line y = mx + c touches the ellipse $\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}=1,$ then
The required condition is c² = a² m² + b²
Here, given equation of the line is,
$\text{x}\cos\alpha+\text{y}\sin\alpha=\text{p}$
$\Rightarrow\ \text{y}=\frac{\text{p}-\text{x}\cos\alpha}{\sin\alpha}$
$=-\text{x}\cot\alpha+\frac{\text{p}}{\sin\alpha}$
$\Rightarrow\text{c}=\frac{\text{p}}{\sin\alpha}$
And $\text{m}=\cot\alpha$
$\therefore\Big(\frac{\text{P}}{\sin\alpha}\Big)^{2}=\text{a}^2(-\cot\alpha)^2+\text{b}^2$
$\Rightarrow\frac{\text{p}}{\sin^2\alpha}=\text{a}^2\frac{\cot^2\alpha}{\sin^2\alpha}+\text{b}$
$\Rightarrow\text{p}^2=\text{a}^2\cos^2\alpha+\text{b}^2\sin^2\alpha$
Hence proved.

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Question 164 Marks

Find the equation of the normal lines to the curve 3x² - y² = 8 which are parallel to the line x + 3y = 4.

Answer

We have, 3x² - y² = 8 ...(1)
$\Rightarrow\ 6\text{x}-2\text{y}\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{6\text{x}}{2\text{y}}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{3\text{x}}{\text{y}}=\text{m}_1$ (say)
And slope of normal $(\text{m}_2)=\frac{-1}{\text{m}_1}=\frac{-\text{y}}{3\text{x}}\ \ \dots(2)$
Consider, $\text{x}+3\text{y}=4$
$\Rightarrow\ \text{y}=\frac{4-\text{x}}{3}=\frac{-\text{x}}{3}+\frac{4}{3}$
$\therefore$ Slope of the line $(\text{m}_3)=\frac{-1}{3}$
Since, slope of normal to the curve should be equal to the slope of line x + 3y = 4, which is parallel to curve.
$\therefore\ \text{m}_2=\text{m}_3$
$\Rightarrow\ \frac{-\text{y}}{3\text{x}}=-\frac{1}{3}$
$\Rightarrow-3\text{y}=-3\text{x}$
$\Rightarrow\ \text{y}=\text{x}\ \ \dots(3)$
Substituting y = x in (1), we get
$3\text{x}^2-\text{x}^2=8$
$\Rightarrow\ \text{x}^2=4$
$\Rightarrow\ \text{x}=\pm2$
Substituting $\text{x}=\pm2$ in (3), we get
$\text{y}=\pm2$
Thus, the points at which normal to the curve is parallel to the line x + 3y = 4 are (2, 2) and (-2, -2).
Now, the equations of normal are given by
$\text{y}-2=\text{m}_2(\text{x}-2)$ and $\text{y}+2=\text{m}_2(\text{x}+2)$
$\Rightarrow\ \text{y}-2=\frac{-1}{3}(\text{x}-2)$ and $\text{y}+2=\frac{-1}{3}(\text{x}+2)$
$\Rightarrow\ 3\text{y}-6=-\text{x}+2$ and $3\text{y}+6=-\text{x}-2$
$\Rightarrow\ 3\text{y}+\text{x}=+8$ and $3\text{y}+\text{x}=-8$
So, the required equations are $3\text{y}+\text{x}=\pm8$

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Question 174 Marks

Prove that the curves y² = 4x and x² + y² - 6x + 1 = 0 touch each other at the point (1, 2).

Answer

We have, y² = 4x and x² + y² - 6x + 1 = 0
Since, both the curves touch each other at (1, 2) i.e., curves are passing through (1, 2).
$\therefore\ 2\text{y}\cdot\frac{\text{dy}}{\text{dx}}=4$
and $2\text{x}+2\text{y}\frac{\text{dy}}{\text{dx}}=6$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{4}{2\text{y}}$
and $\frac{\text{dy}}{\text{dx}}=\frac{6-2\text{x}}{2\text{y}}$
$\Rightarrow\ \Big(\frac{\text{dy}}{\text{dx}}\Big)_{(1,2)}=\frac{4}{4}=1$
and $\Big(\frac{\text{dy}}{\text{dx}}\Big)_{(1,2)}=\frac{6-2\cdot1}{2\cdot2}=\frac{4}{4}=1$
$\Rightarrow\ \text{m}_1=1$ and $\text{m}_2=1$
Thus, we see that slope of both the curves are equal to each other i.e., m₁ = m₂ = 1 at the point (1, 2).
Hence, both the curves touch each other.

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Question 184 Marks

Show that the line $\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}=1,$ touches the curve $\text{y}=\text{b}\cdot\text{e}^{\frac{-\text{x}}{\text{a}}}$ a e at the point where the curve intersects the axis of y.

Answer

Given equation of the curve, $\text{y}=\text{b}\cdot\text{e}^{\frac{-\text{x}}{\text{a}}}\ \ \dots(\text{i})$
It meets y-axis,
$\therefore\ \text{y}=\text{b e}^0=\text{b}$
$\therefore$ point where it meets y-axis is (0, b)
Differentiating (i), w.r.t. x, we get
$\frac{\text{dy}}{\text{dx}}=-\frac{\text{b}}{\text{a}}\text{e}^{\frac{-\text{x}}{\text{a}}}$
$\therefore\ \Big(\frac{\text{dy}}{\text{dx}}\Big)_{(0,\text{ b})}=-\frac{\text{b}}{\text{a}}\text{e}^{-0}=-\frac{\text{b}}{\text{a}}$
$\therefore$ Equation of tangent at point (0, b) is
$\text{y}-\text{b}=-\frac{\text{b}}{\text{a}}(\text{x}-0)$
$\therefore\ \text{ay}-\text{ab}=-\text{bx}$
or $\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}=1$
Hence, the line touches the curve at the point, where the curve intersects the axis of Y.

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Question 194 Marks

A swimming pool is to be drained for cleaning. If L represents the number of litres of water in the pool t seconds after the pool has been plugged off to drain and L = 200(10 - t)². How fast is the water running out at the end of 5 seconds? What is the average rate at which the water flows out during the first 5 seconds?

Answer

Let L represents the number of litres of water in the pool t seconds after the pool been plugged off to drain, then
$\text{L} = 200(10 - \text{t})^2$
$\therefore$ Rate at which the water is running out $=-\frac{\text{dL}}{\text{dt}}$
$=\frac{\text{dL}}{\text{dt}}=-200.2(10-\text{t}).(-1)$
$=400(10-\text{t})$
Rate at which the water is running out at the end of 5s
$=400(10-5)$
$=2000\text{L/s}$ Final rate
Since, initial rate $=-\Big(\frac{\text{dL}}{\text{dt}}\Big)_{\text{t}-0}=4000\text{L/s}$
$\therefore$ Aver agerate during 5s $=\frac{\text{Initial rate}+\text{Final rate}}{2}$
$=\frac{4000+2000}{2}$
$=3000\text{L/s}$

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Question 204 Marks

Find the dimensions of the rectangle of perimeter 36cm which will sweep out a volume as large as possible, when revolved about one of its sides. Also find the maximum volume.

Answer

Let breadth and length of the rectangle be x and y, respectively.

It is given that the perimeter of the rectangle = 36cm
⇒ 2x + 2y = 36
⇒ x + y = 18
⇒ y = 18 - x
Let the rectangle is being revolved about its length y.
Then, volume of resultant cylinder,
$\text{B}=\pi\text{x}^2\text{y}$
$\Rightarrow\ \text{V}=\pi\text{x}^2(18-\text{x})=\pi(18\text{x}^2-\text{x}^3)$
$\Rightarrow\ \frac{\text{dV}}{\text{dx}}=\pi(36\text{x}-3\text{x}^2)$
$\frac{\text{dv}}{\text{dx}}=0$
$\Rightarrow\ 36\text{x}=3\text{x}^2$
$\Rightarrow\ \text{x}=12\ (\text{ax x}\neq0)$
Also $\frac{\text{d}^2\text{V}}{\text{dx}^2}=\pi(36-6\text{x})$
$\Rightarrow\ \Big(\frac{\text{d}^2\text{V}}{\text{dx}^2}\Big)_{\text{x}=12}=\pi(36-72)=-36\pi<0$
$\therefore$ For x = 12, volume of the resultant cylinder is the maximum.
$\therefore$ y = 18 - 12 = 6
So, the dimensions of rectangle are 12cm and 6cm, respectively.
$\therefore$ Maximum volume of resultant cylinder,
$\text{V}=\pi\big[18\cdot(12)^2-(12)^3\big]$
$=\pi12^2(18-12)$
$=864\pi\text{cm}^3$

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4 Marks - Maths STD 12 Science Questions - Vidyadip