12π

126

cos 2x

None of these

(0, 2)

-\(\frac13\)

(1, 2)

77.66

0.09 x³ m³

(2 \(\sqrt2\),4)

R = {(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)}
It is seen that (a, a) ∈ R, for every a ∈{1, 2, 3, 4}.
∴ R is reflexive.
It is seen that (1, 2) ∈ R, but (2, 1) ∉ R.
∴ R is not symmetric.
Also, it is observed that (a, b), (b, c) ∈ R ⇒ (a, c) ∈ R for all a, b, c ∈ {1, 2, 3, 4}.
∴ R is transitive.
Hence, R is reflexive and transitive but not symmetric.
The correct answer is B.

\(\frac { 1 }{ 3 } \)

1

1 m/h

\(\frac67\)

1

x – y = 0

x + y = 3

\(\left( 4,\pm \frac { 8 }{ 3 } \right) \)

1368 Concern for children health and nutrient food for every child.

1368 Concern for children health and nutrient food for every child.

Let P be the perimeter and A be the area of the rectangle of length x and width y.
Given \(\frac{d x}{d t}=-5 \mathrm{~cm} / \mathrm{min}\) 
[negative (-) sign for decreasing rate]
\(x=8 \mathrm{~cm}, \frac{d y}{d t}=4 \mathrm{~cm} / \mathrm{min} \text { and } y=6 \mathrm{~cm}\) 
(ii) Perimeter of the rectangle.\(P=2(x+y)\) 
On differentiating both sides w.r.t. t, we get
\( \frac{d P}{d t} =2\left(\frac{d x}{d t}+\frac{d y}{d t}\right) \)
\(=2(-5+4)=-2 \mathrm{~cm} / \mathrm{min}\) 
So, perimeter decreases at the rate of 2 cm/min.
(ii) Area of the rectangle, A = xy
On differentiating both sides w.r.t. t, we get
\( \frac{d A}{d t} =x \frac{d y}{d t}+y \frac{d x}{d t} \)
\(=8 \times 4+6 \times(-5)=32-30=2 \mathrm{~cm}^{2} / \mathrm{min} \) 
Hence, area increases at the rate of 2 cm ² /min.

Let r be the radius, h be the height and V be the volume of the sand cone
Also given that, \(\frac{d V}{d t}=12 \mathrm{~cm}^{3} / \mathrm{s}, h=\frac{1}{6} r\) 
\(\Rightarrow r=6 h \text { and } h=4 \mathrm{~cm}\) 
Volume of sand cone,
\(V=\frac{1}{3} \pi r^{2} h\) 
\( \Rightarrow V=\frac{1}{3} \pi(6 h)^{2} h \)
\(\Rightarrow V=\frac{1}{3} \pi \times 36 h^{2} \times h=12 \pi h^{3} \) 
On differentiating both sides w.r.t. t, we get
\( \frac{d V}{d t}=12 \pi \times 3 h^{2} \frac{d h}{d t}=36 \pi h^{2} \frac{d h}{d t} \)
\(\Rightarrow 12=36 \pi(4)^{2} \frac{d h}{d t} \)
\(\Rightarrow \frac{d h}{d t}=\frac{12}{36 \pi \times 16}=\frac{1}{48 \pi} \mathrm{cm} / \mathrm{s} \) 
Hence, the height of the sand cone is increasing at the rate of \(\frac{1}{48 \pi} \mathrm{cm} / \mathrm{s}\). when the height is 4 cm,

Let △ABC be isosceles where BC is the base of fixed length  \( \underline{b} \)
Let the length of the two equal sides of  \( \triangle \mathrm{ABC} \text { be } a \text { . }\)
\(\text {Draw } \mathrm{AD} \perp \mathrm{BC}\)
\(\text {Now, in } \triangle A D C \) by applying the Pythagoras theorem, we have: 
\(\mathrm{AD}=\sqrt{a^{2}-\frac{b^{2}}{4}}\)
\(\therefore \text { Area of triangle }(A)=\frac{1}{2} b \sqrt{a^{2}-\frac{b^{2}}{4}}\)
 The rate of change of the area with respect to time (t) is given by 
\(\frac{d A}{d t}=\frac{1}{2} b \cdot \frac{2 a}{2 \sqrt{a^{2}-\frac{b^{2}}{4}}} \frac{d a}{d t}=\frac{a b}{\sqrt{4 a^{2}-b^{2}}} \frac{d a}{d t}\)
It is given that the two equal sides of the triangle are decreasing at the rate of 3 cm per second. 
\(\therefore \frac{d a}{d t}=-3 \mathrm{~cm} / \mathrm{s} \)
\(\therefore \frac{d A}{d t}=\frac{-3 a b}{\sqrt{4 a^{2}-b^{2}}} \)
Then, when a=b , we have: 
\(\frac{d A}{d t}=\frac{-3 b^{2}}{\sqrt{4 b^{2}-b^{2}}}=\frac{-3 b^{2}}{\sqrt{3 b^{2}}}=-\sqrt{3} b\)
Hence, if the two equal sides are equal to the base, then the area of the triangle is decreasing at the rate of  \( \sqrt{3} b \mathrm{~cm}^{2} / \mathrm{s}\) 
 

Let x be the length of a side, V be the volume, and s be the surface area of the cube.
Then, V = x³ and S = 6x² where x is a function of time t.
 It is given that  \( \frac{d V}{d t}=8 \mathrm{~cm}^{3} / \mathrm{s} \text { . }\)
Then, by using the chain rule, we have: 
\(\therefore 8=\frac{d V}{d t}=\frac{d}{d t}\left(x^{3}\right)=\frac{d}{d x}\left(x^{3}\right) \cdot \frac{d x}{d t}=3 x^{2} \cdot \frac{d x}{d t}\)
\(\Rightarrow \frac{d x}{d t}=\frac{8}{3 x^{2}} \ \text { (1) } \quad \text { [By chain rule] } \)
\(\text { Now, } \frac{d \mathrm{~S}}{d t}=\frac{d}{d t}\left(6 x^{2}\right)=\frac{d}{d x}\left(6 x^{2}\right) \cdot \frac{d x}{d t} \ 0 .\)
\(=12 x \cdot \frac{d x}{d t}=12 x .\left(\frac{8}{3 x^{2}}\right)=\frac{32}{x}\)
\(\text {Thus, when } x=12 \mathrm{~cm}, \frac{d S}{d t}=\frac{32}{12} \mathrm{~cm}^{2} / \mathrm{s}=\frac{8}{3} \mathrm{~cm}^{2} / \mathrm{s}\)
Hence, if the length of the edge of the cube is 12 cm, then the surface area is increasing at the rate of \(\frac{8}{3}\) cm²/s

The equation of the curve is given as: 
\(6 y=x^{3}+2\)
The rate of change of the position of the particle with respect to time (t) is given by, 
\(6 \frac{d y}{d t}=3 x^{2} \frac{d x}{d t}+0 \Rightarrow 2 \frac{d y}{d t}=x^{2} \frac{d x}{d t}\)
When the y -coordinate of the particle changes 8 times as fast as the 
\(x \text { -coordinate i.e., }\left(\frac{d y}{d t}=8 \frac{d x}{d t}\right), \text { we have: }\)
\(2\left(8 \frac{d x}{d t}\right)=x^{2} \frac{d x}{d t} \)
\(\Rightarrow 16 \frac{d x}{d t}=x^{2} \frac{d x}{d t} \)
\(\Rightarrow\left(x^{2}-16\right) \frac{d x}{d t}=0 \)
\(\Rightarrow x^{2}=16 \)
\(\Rightarrow x=\pm 4 \)
\(\text { When } x=4, y=\frac{4^{3}+2}{6}=\frac{66}{6}=11 \text { . }\)
\(\text { When } x=-4, y=\frac{(-4)^{3}+2}{6}=-\frac{62}{6}=-\frac{31}{3} \text { . }\)
Hence, the points required on the curve are (4,11) and 

We have
\(y=[x(x-2)]^{2}=\left[x^{2}-2 x\right]^{2} \)
\(\therefore \frac{d y}{d x}=y^{\prime}=2\left(x^{2}-2 x\right)(2 x-2)=4 x(x-2)(x-1) \)
\(\therefore \frac{d y}{d x}=0 \Rightarrow x=0, x=2, x=1 \)
\(\text {The points } x=0, x=1 \text { and } x=2 \) divide the real line into four disjoint intervals i.e.,
\((-\infty, 0),(0,1)(1,2), \text { and }(2, \infty) .\)
\(\text { In intervals }(-\infty, 0) \text { and }(1,2), \frac{d y}{d x}<0 \text { . }\)
∴y is strictly decreasing in intervals  \( (-\infty, 0) \text { and }(1,2) \text { . }\)
However, in intervals \( (0,1) \text { and }(2, \infty), \frac{d y}{d x}>0 .\)
\(\therefore y \text { is strictly increasing in intervals }(0,1) \text { and }(2, \infty) .\)
\(\therefore y \text { is strictly increasing for } 02 \text { . }\)
Increasing:\(\left( 0,1 \right) \cup \left( 2,\infty \right) \) ;Decreasing:\(\left( -\infty ,0 \right) \cup \left( 1,2 \right) ;\)points are(0,0),(1,1),(2,0).

We have
\(y=\frac{4 \sin \theta}{(2+\cos \theta)}-\theta \)
\(\therefore \frac{d y}{d x} =\frac{(2+\cos \theta)(4 \cos \theta)-4 \sin \theta(-\sin \theta)}{(2+\cos \theta)^{2}}-1 \)
\(=\frac{8 \cos \theta+4 \cos ^{2} \theta+4 \sin ^{2} \theta}{(2+\cos \theta)^{2}}-1 \)
\(=\frac{8 \cos \theta+4}{(2+\cos \theta)^{2}}-1 \)
\(\text { Now, } \frac{d y}{d x}=0 .\)
\(\Rightarrow \frac{8 \cos \theta+4}{(2+\cos \theta)^{2}}=1\)
\(\Rightarrow 8 \cos \theta+4=4+\cos ^{2} \theta+4 \cos \theta \)
\(\Rightarrow \cos ^{2} \theta-4 \cos \theta=0\)
\(\Rightarrow \cos \theta(\cos \theta-4)=0 \)
\(\Rightarrow \cos \theta=0 \text { or } \cos \theta=4\)
\(\text { Since } \cos \theta \neq 4, \cos \theta=0\)
\(\cos \theta=0 \Rightarrow \theta=\frac{\pi}{2}\)
\(\text { Now, }\frac{d y}{d x}=\frac{8 \cos \theta+4-\left(4+\cos ^{2} \theta+4 \cos \theta\right)}{(2+\cos \theta)^{2}}=\frac{4 \cos \theta-\cos ^{2} \theta}{(2+\cos \theta)^{2}}=\frac{\cos \theta(4-\cos \theta)}{(2+\cos \theta)^{2}}\)
\(\text { In interval }\left(0, \frac{\pi}{2}\right), \text { we have } \cos \theta>0 . \text { Also, } 4>\cos \theta \Rightarrow 4-\cos \theta>0 \text { . }\)
\(\therefore \cos \theta(4-\cos \theta)>0 \text { and also }(2+\cos \theta)^{2}>0 \)
\(\Rightarrow \frac{\cos \theta(4-\cos \theta)}{(2+\cos \theta)^{2}}>0 \)
\(\Rightarrow \frac{d y}{d x}>0 \)
 Therefore, y is strictly increasing in interval \( \left(0, \frac{\pi}{2}\right) \text { . }\)
Also, the given function is continuous at  \( x=0 \text { and } x=\frac{\pi}{2} \text { . }\)
Hence, y is increasing in interval  \( \left[0, \frac{\pi}{2}\right] \text { . }\)

We have,
\( \frac{d y}{d x}=\frac{1}{1+x} \frac{d}{d x}(1+x)-\left[\frac{(2+x) \frac{d}{d x}(2 x)-2 x \frac{d}{d x}(2+x)}{(2+x)^2}\right] \\ =\frac{1}{1+x}-\left[\frac{(2+x) 2-2 x}{(2+x)^2}\right] \\ =\frac{1}{1+x}-\frac{(4+2 x-2)}{(2+x)^2} \\ =\frac{1}{1+x}-\frac{4}{(2+x)^2} \)
this implies
\( \frac{d y}{d x}=\frac{(2+x)^2-4(1+x)}{(1+x)(2+x)^2} \\ =\frac{x^2}{(1+x)(2+x)^2} \)
Domain of the given function is given to be x > -1
x + 1 > 0
Also (2+x)² > 0 and x² \(\geq 0\)
From equation (1), \(\frac{d y}{d x} \geq 0\) for all x in domain  x>-1 and f is an increasing function.

Note that on x-axis, y = 0. So the equation of the curve, when y = 0, gives x = 7. Thus, the curve cuts the x-axis at (7, 0). Now differentiating the equation of the curve with respect to x, we obtain
\(\frac{d y}{d x} =\frac{1-y(2 x-5)}{(x-2)(x-3)} \)
\(\left.\frac{d y}{d x}\right]_{(7,0)} =\frac{1-0}{(5)(4)}=\frac{1}{20}\)
Therefore, the slope of the tangent at (7, 0) is \(\frac{1}{20}\) Hence, the equation of the tangent at (7, 0) is
\(y-0=\frac{1}{20}(x-7) \quad \text { or } \quad 20 y-x+7=0\)

\(\text {We have } x=a \cos \theta+a \theta \sin \theta \)
\(\therefore \frac{d x}{d \theta}=-a \sin \theta+a \sin \theta+a \theta \cos \theta=a \theta \cos \theta \)
\(y=a \sin \theta-a \theta \cos \theta \)
\(\therefore \frac{d y}{d \theta}=a \cos \theta-a \cos \theta+a \theta \sin \theta=a \theta \sin \theta \)
\(\therefore \frac{d y}{d x}=\frac{d y}{d \theta} \cdot \frac{d \theta}{d x}=\frac{a \theta \sin \theta}{a \theta \cos \theta}=\tan \theta \)
Slope of the normal at any point \( \theta \text { is }-\frac{1}{\tan \theta} \text { . }\)
The equation of the normal at a given point (x, y) is given by,
\(y-a \sin \theta+a \theta \cos \theta=\frac{-1}{\tan \theta}(x-a \cos \theta-a \theta \sin \theta) \)
\(\Rightarrow y \sin \theta-a \sin ^{2} \theta+a \theta \sin \theta \cos \theta=-x \cos \theta+a \cos ^{2} \theta+a \theta \sin \theta \cos \theta \)
\(\Rightarrow x \cos \theta+y \sin \theta-a\left(\sin ^{2} \theta+\cos ^{2} \theta\right)=0 \)
\(\Rightarrow x \cos \theta+y \sin \theta-a=0 \)
Now, the perpendicular distance of the normal from the origin is
\(\frac{|-a|}{\sqrt{\cos ^{2} \theta+\sin ^{2} \theta}}=\frac{|-a|}{\sqrt{1}}=|-a|, \text { which is independent of } \theta\)
 Hence, the perpendicular distance of the normal from the origin is constant.

The equation of the given curve is \( y=\frac{1}{x-1}, x \neq 1 \text { . }\)
The slope of the tangents to the given curve at anv point (x, b is given by )
\(\frac{d y}{d x}=\frac{-1}{(x-1)^{2}}\)
If the slope of the tangent is −1, then we have
\(\frac{-1}{(x-1)^{2}}=-1 \)
\(\Rightarrow(x-1)^{2}=1 \)
\(\Rightarrow x-1=\pm 1 \)
\(\Rightarrow x=2,0 \)
When x = 0, y = −1 and  when x = 2, y = 1 .
Thus, there are two tangents to the given curve having slope -1. These are passing through the points (0, -1) and (2, 1) 
The equation of the tangent through (0,-1) is given by,
\(y-(-1)=-1(x-0) \)
\(\Rightarrow y+1=-x \)
\(\Rightarrow y+x+1=0 \)
∴ The equation of the tangent through (2,1) is given by,
\(y-1=-1(x-2) \)
\(\Rightarrow y-1=-x+2 \)
\(\Rightarrow y+x-3=0 \)
Hence, the equations of the required lines are y+x+1=0 and y+x−3=0 .

Let r be the radius of the sphere and Δr be the error in measuring the radius.
Then, r = 9 m and Δr = 0.03 m
Now, the surface area of the sphere (S) is given by,
\(S=4 \pi r^{2} \)
\(\therefore \frac{d S}{d r}=8 \pi r \)
\(\therefore d S=\left(\frac{d S}{d r}\right) \Delta r \)
\(=(8 \pi r) \Delta r \)
\(=8 \pi(9)(0.03) \mathrm{m}^{2} \)
\(=2.16 \pi \mathrm{m}^{2} \)
Hence, the approximate error in calculating the surface area is \( 2.16 \pi \mathrm{m}^{2} \text { . }\)

Let r and h be the radius and the height (altitude) of the cone respectively.
Then, the volume (V) of the cone is given as:
V = 13πr²h
⇒ h = 3Vπr²V
=13πr²h
⇒h = 3Vπr²
The surface area (S) of the cone is given by
S = πrl (where l is the slant height)
\(\therefore \frac{d S}{d r} =\frac{r \cdot \frac{6 \pi^{2} r^{5}}{2 \sqrt{\pi^{2} r^{6}+9 V^{2}}}-\sqrt{\pi^{2} r^{6}+9 V^{2}}}{r^{2}} \)
\(=\frac{3 \pi^{2} r^{6}-\pi^{2} r^{6}-9 V^{2}}{r^{2} \sqrt{\pi^{2} r^{6}+9 V^{2}}} \)
\(=\frac{2 \pi^{2} r^{6}-9 V^{2}}{r^{2} \sqrt{\pi^{2} r^{6}+9 V^{2}}} \)
\(=\frac{2 \pi^{2} r^{6}-9 V^{2}}{r^{2} \sqrt{\pi^{2} r^{6}+9 V^{2}}} \)
\(\text { Now }, \frac{d S}{d r}=0 \Rightarrow 2 \pi^{2} r^{6}=9 V^{2} \Rightarrow r^{6}=\frac{9 V^{2}}{2 \pi^{2}}\)
Thus, it can be easily verified that when \(r^{6}=\frac{9 V^{2}}{2 \pi^{2}}, \frac{d^{2} S}{d r^{2}}>0\)
∴ By second derivative test, the surface area of the cone is the least when \(r^{6}=\frac{9 V^{2}}{2 \pi^{2}}\)
\(\text { When } r^{6}=\frac{9 V^{2}}{2 \pi^{2}}, h=\frac{3 V}{\pi r^{2}}=\frac{3}{\pi r^{2}}\left(\frac{2 \pi^{2} r^{6}}{9}\right)^{\frac{1}{2}}=\frac{3}{\pi r^{2}} \cdot \frac{\sqrt{2} \pi r^{3}}{3}=\sqrt{2} r \text { . }\)
Hence, for a given volume, the right circular cone of the least curved surface has an altitude equal to \(\sqrt{2}\) times the radius of the base.

Let a piece of length l be cut from the given wire to make a square.
Then, the other piece of wire to be made into a circle is of length (28 − l) m.
Now, side of square =.\(\frac{l}{4}\)
Let r be the radius of the circle. Then, \(2 \pi r=28-l \Rightarrow r=\frac{1}{2 \pi}(28-l)\)
The combined areas of the square and the circle (A) is given by,
\(A=(\text { side of the square })^{2}+\pi r^{2} \)
\(=\frac{l^{2}}{16}+\pi\left[\frac{1}{2 \pi}(28-l)\right]^{2} \)
\(=\frac{l^{2}}{16}+\frac{1}{4 \pi}(28-l)^{2} \)
\(\therefore \frac{d A}{d l}=\frac{2 l}{16}+\frac{2}{4 \pi}(28-l)(-1)=\frac{l}{8}-\frac{1}{2 \pi}(28-l) \)
\(\frac{d^{2} A}{d l^{2}}=\frac{1}{8}+\frac{1}{2 \pi}>0 \)
\(\text { Now }, \frac{d A}{d l}=0 \Rightarrow \frac{l}{8}-\frac{1}{2 \pi}(28-l)=0 \)
\(\Rightarrow \frac{\pi l-4(28-l)}{8 \pi}=0 \)
\(\Rightarrow(\pi+4) l-112=0 \)
\(\Rightarrow l=\frac{112}{\pi+4} \)
Thus, when \(l=\frac{112}{\pi+4}, \frac{d^{2} \mathrm{~A}}{d l^{2}}>0 .\)
By second derivative test, the area (A) is the minimum when.\(l=\frac{112}{\pi+4}\)
Hence, the combined area is the minimum when the length of the wire in making the square is \(\frac{112}{\pi+4}\)m while the length of the wire in making the circle is
28 − 112π + 4 = 28π + 4m

Let r and h be the radius and height of the cylinder respectively.
Then, the surface area (S) of the cylinder is given by
\(S= 2 \pi r^{2}+2 \pi r h \)
\(\Rightarrow h =\frac{S-2 \pi r^{2}}{2 \pi r} \)
\(=\frac{S}{2 \pi}\left(\frac{1}{r}\right)-r \)
Let V be the volume of the cylinder. Then,
\(V=\pi r^{2} h=\pi r^{2}\left[\frac{S}{2 \pi}\left(\frac{1}{r}\right)-r\right]=\frac{S r}{2}-\pi r^{3} \)
\(\text { Then, } \frac{d V}{d r}=\frac{S}{2}-3 \pi r^{2}, \frac{d^{2} V}{d r^{2}}=-6 \pi r \)
\(\text { Now, } \frac{d V}{d r}=0 \Rightarrow \frac{S}{2}=3 \pi r^{2} \Rightarrow r^{2}=\frac{S}{6 \pi} \)
\(\text { When } r^{2}=\frac{S}{6 \pi}, \text { then } \frac{d^{2} V}{d r^{2}}=-6 \pi\left(\sqrt{\frac{S}{6 \pi}}\right)<0 \)
∴ By second derivative test, the volume is the maximum when \(r^{2}=\frac{S}{6 \pi}\)
\(\text { Now, when } r^{2}=\frac{S}{6 \pi} \text { , then } h=\frac{6 \pi r^{2}}{2 \pi}\left(\frac{1}{r}\right)-r=3 r-r=2 r \text { . }\)
Hence, the volume is the maximum when the height is twice the radius i.e., when the height is equal to the diameter

Let r and h be the radius and height of the cone respectively inscribed in a sphere of radius R.
Let V be the volume of the cone.
Then, \(V=\frac{1}{3} \pi r^{2} h\)
Height of the cone is given by
\(h=R+A B=R+\sqrt{R^{2}-r^{2}} \quad[\mathrm{ABC} \text { is a right triangle }] \)
\(\therefore V=\frac{1}{3} \pi r^{2}\left(R+\sqrt{R^{2}-r^{2}}\right) \)
\(=\frac{1}{3} \pi r^{2} R+\frac{1}{3} \pi r^{2} \sqrt{R^{2}-r^{2}} \)
\(\therefore \frac{d V}{d r}=\frac{2}{3} \pi r R+\frac{2}{3} \pi r \sqrt{R^{2}-r^{2}}+\frac{1}{3} \pi r^{2} \cdot \frac{(-2 r)}{2 \sqrt{R^{2}-r^{2}}} \)
\(=\frac{2}{3} \pi r R+\frac{2}{3} \pi r \sqrt{R^{2}-r^{2}}-\frac{1}{3} \pi \frac{r^{3}}{\sqrt{R^{2}-r^{2}}} \)
\(=\frac{2}{3} \pi r R+\frac{2 \pi r\left(R^{2}-r^{2}\right)-\pi r^{3}}{3 \sqrt{R^{2}-r^{2}}} \)
\(=\frac{2}{3} \pi r R+\frac{2 \pi r R^{2}-3 \pi r^{3}}{3 \sqrt{R^{2}-r^{2}}} \)
\(\frac{d^{2} V}{d r^{2}}=\frac{2 \pi R}{3}+\frac{3 \sqrt{R^{2}-r^{2}}\left(2 \pi R^{2}-9 \pi r^{2}\right)-\left(2 \pi r R^{2}-3 \pi r^{3}\right) \cdot \frac{(-2 r)}{6 \sqrt{R^{2}-r^{2}}}}{9\left(R^{2}-r^{2}\right)} \)
\(=\frac{2}{3} \pi R+\frac{9\left(R^{2}-r^{2}\right)\left(2 \pi R^{2}-9 \pi r^{2}\right)+2 \pi r^{2} R^{2}+3 \pi r^{4}}{27\left(R^{2}-r^{2}\right)^{\frac{3}{2}}} \)
\(\text { Now, } \frac{d V}{d r}=0 \Rightarrow \frac{2}{\pi} r R=\frac{3 \pi r^{3}-2 \pi r R^{2}}{3 \sqrt{R^{2}-r^{2}}} \)
\(\Rightarrow 2 R=\frac{3 r^{2}-2 R^{2}}{\sqrt{R^{2}-r^{2}}} \Rightarrow 2 R \sqrt{R^{2}-r^{2}}=3 r^{2}-2 R^{2} \)
\(\Rightarrow 4 R^{2}\left(R^{2}-r^{2}\right)=\left(3 r^{2}-2 R^{2}\right)^{2} \)
\(\Rightarrow 4 R^{4}-4 R^{2} r^{2}=9 r^{4}+4 R^{4}-12 r^{2} R^{2} \)
\(\Rightarrow 9 r^{4}=8 R^{2} r^{2} \)
\(\Rightarrow r^{2}=\frac{8}{9} R^{2} \)
\(\text { When } r^{2}=\frac{8}{9} R^{2}, \text { then } \frac{d^{2} V}{d r^{2}}<0 \)
∴ By second derivative test, the volume of the cone is the maximum when \(r^{2}=\frac{8}{9} R^{2}\)
\(\text { When } r^{2}=\frac{8}{9} R^{2}, h=R+\sqrt{R^{2}-\frac{8}{9} R^{2}}=R+\sqrt{\frac{1}{9} R^{2}}=R+\frac{R}{3}=\frac{4}{3} R \text { . }\)
\(\text { Therefore, }=\frac{1}{3} \pi\left(\frac{8}{9} R^{2}\right)\left(\frac{4}{3} R\right)\)
\(=\frac{8}{27}\left(\frac{4}{3} \pi R^{3}\right)\)
\(=\frac{8}{27} \times \)(Volume of the sphere)
 Hence, the volume of the largest cone that can be inscribed in the sphere is \( \frac{8}{27} \)

Let radius of cone = r, Height of cone = h ,
Slant height of cone = I, Semi-vertical angle = \(\alpha\) 
Surface area = 5 and volume of cone
On squaring both sides, we get
\(V^{2}=\frac{1}{9} \pi^{2} r^{4} h^{2}=\frac{1}{9} \pi^{2} r^{4}\left(l^{2}-r^{2}\right)\) ...(ii)
\(\left[\because \text { in } \Delta A O B \text { , using Pythagoras theorem, } l^{2}=r^{2}+h^{2}\right]\) 
Given, surface area of cone \(S=\pi r l+\pi r^{2}\) 
\(\Rightarrow \pi r l=S-\pi r^{2} \)
\(\Rightarrow l=\frac{S-\pi r^{2}}{\pi r} \) 
On putting the value of 1 in Eq. (ii), we get
\(V^{2}=\frac{1}{9} \pi^{2} r^{4}\left[\left(\frac{S-\pi r^{2}}{\pi r}\right)^{2}-r^{2}\right]\) 
\(\Rightarrow V^{2}=\frac{1}{9} \pi^{2} r^{4}\left[\frac{S^{2}-2 \pi S r^{2}+\pi^{2} r^{4}}{\pi^{2} r^{2}}-r^{2}\right] \)
\(\Rightarrow V^{2}=\frac{1}{9} \pi^{2} r^{4}\left[\frac{S^{2}-2 \pi S r^{2} \neq \pi^{2} r^{4}-\pi^{2} r^{4}}{\pi^{2} r^{2}}\right. \)
\(\Rightarrow V^{2}=\frac{1}{9} r^{2}\left[S^{2}-2 \pi S r^{2}\right] \)
\(\Rightarrow V^{2}=\frac{1}{9}\left[S^{2} r^{2}-2 \pi S r^{4}\right]\) 
Let  \(V^{2}=f(r)\) then f(r) is maximum or minimum accordingly as volume (V) is maximum or minimum 
Now, \(f(r)=\frac{1}{9}\left(S^{2} r^{2}-2 \pi S r^{4}\right)\) 
For maximum or minimum value put \(f^{\prime}(r)=0\) 
\(\Rightarrow \frac{1}{9}\left(2 S^{2} r-8 \pi S r^{3}\right)=0 \)
\(\Rightarrow 2 \operatorname{Sr}\left(S-4 \pi r^{2}\right)=0 \)
\(\Rightarrow 2 S r=0 \text { and } S-4 \pi r^{2}=0 \Rightarrow S \neq 0 \)
\(\Rightarrow r=0 \text { and } S-4 \pi r^{2}=0 \)
\(\Rightarrow r=0 \quad \text { and } S=4 \pi r^{2} \) 
Since, r = 0 if; not possible
\(\therefore \ S=4 \pi r^{2} \Rightarrow r^{2}=\frac{S}{4 \pi} \Rightarrow r=\sqrt{\frac{S}{4 \pi}}\) 
Again, differentiating both sides of Eq( v) w.r.t. r, we get
\(f^{\prime \prime}(r)=\frac{1}{9}\left(2 S^{2}-24 \pi S r^{2}\right) \)
\(\text { At } r=\sqrt{\frac{S}{4 \pi}},\left[f^{\prime \prime}(r)\right]_{r=\sqrt{\frac{s}{4 \pi}}}=\frac{1}{9}\left[2 S^{2}-24 \pi S \cdot \frac{S}{4 \pi}\right] \)
\(=\frac{1}{9}\left[2 S^{2}-6 S^{2}\right]=\frac{-4}{9} S^{2}<0 \) 
\(\therefore\) Volume of cone is maximum, when \(n=\sqrt{\frac{S}{4 \pi}}\) 
\(S=4 \pi r^{2}\) 
On putting the value of S in Eq. (ill), we get
\(l=\frac{4 \pi r^{2}-\pi r^{2}}{\pi r}=\frac{3 \pi r^{2}}{\pi r}=3 r\) 
In \(\Delta O B A, \sin \alpha=\frac{r}{I}\) 
\(\Rightarrow \sin \alpha=\frac{r}{3 r}=\frac{1}{3} \Rightarrow \alpha=\sin ^{-1}\left(\frac{1}{3}\right)\) 
Hence, the semi-vertical angle of cone is \(\sin ^{-1}\left(\frac{1}{3}\right)\)

Let we have a sphere of radius R and a cylinder is inscribed in it. EB = r be the radius of the cylinder and BC = h be the height of cylinder.
In right angled \(\Delta A B C\) we have
\((A C)^{2}=(A B)^{2}+(B C)^{2}\) [using Pythagoras theorem]
\(\Rightarrow(2 R)^{2}=(2 r)^{2}+h^{2} \Rightarrow 4 R^{2}=4 r^{2}+h^{2} \)
\(\Rightarrow \cdot r^{2}=\frac{1}{4}\left(4 R^{2}-h^{2}\right) \) 
volume of cylinder \(V=\pi r^{-2} h\) 
\(\Rightarrow V=\pi\left[\frac{1}{4}\left(4 R^{2}-h^{2}\right)\right] h\) 
\(\Rightarrow V=\frac{\pi}{4}\left(4 R^{2} h-h^{3}\right)\)   ...(ii)
On differentiating V w.r.t. II, we get
\(\frac{d V}{d h}=\frac{\pi}{4}\left(4 R^{2}-3 h^{2}\right)\) 
For maxima or minima put \(\frac{d V}{d h}=0\) 
\(\Rightarrow \frac{\pi}{4}\left(4 R^{2}-3 h^{2}\right)=0 \Rightarrow 4 R^{2}=3 h^{2}\) 
\(\Rightarrow h=\frac{2 R}{\sqrt{3}} \ [\because \text { height cannot be negative }] \) 
On differentiating both sides of Eq. (iii) w.r.t. h, we get
\( \frac{d^{2} V}{d h^{2}} =-\frac{3}{2} \pi h \)
\(\text { At } h=\frac{2 R}{\sqrt{3}}, \frac{d^{2} V}{d h^{2}} =-\frac{3}{2} \pi \times \frac{2 R}{\sqrt{3}}=-\sqrt{3} \pi R<0[\because R>0] \) 
\(\therefore\) Volume of cylinder is maximum at \(h=\frac{2 R}{\sqrt{3}}\) 
From Eq. (i), we get \(r^{2}=\frac{1}{4}\left(4 R^{2}-\frac{4 R^{2}}{3}\right)=\frac{2 R^{2}}{3}\) 
\(\therefore\) Volume of largest cylinder inscribed in a sphere,
\( V =\pi\left(\frac{2 R^{2}}{3}\right) \times \frac{2 R}{\sqrt{3}} \)
\(=\frac{1}{\sqrt{3}} \times\left(\frac{4}{3} \pi R^{3}\right) \)
\(=\frac{1}{\sqrt{3}} \times \text { Volume of sphere } \) 
Hence, height of the cylinder of maximum volume is \(\frac{2 R}{\sqrt{3}}\) 
and volume of largest cylinder inscribed in sphere is \(\frac{1}{\sqrt{3}}\)  times volume of sphere. 

Let the length and breadth of the tank be x and y m, respectively
Then, volume = 75m³
\(\Rightarrow 3 x y=75 \quad[\because \text { depth of tank }=3 \mathrm{~m}]\)
\(\Rightarrow y=\frac{25}{x} \) 
Let C be the cost of the tank.
Then, \(C^{\prime}=100 x y+50(3 \times 2 x+3 \times 2 y)\) 
= 100xy + 300x + 300y
\( =100 x \times \frac{25}{x}+300 x+300 \times \frac{25}{x} \quad\left[\because y=\frac{25}{x}\right] \)
\(\Rightarrow C =2500+300 x+\frac{7500}{ } \) 
On differentiating twice w.r.t. x, we get
\( \frac{d C}{d x}=300-\frac{7500}{x^{2}} \)
\(\text { and } \ \frac{d^{2} C}{d x^{2}}=\frac{15000}{x^{3}}>0 \) 
For minimum value put \(\frac{d C}{d x}=0\) 
\( \Rightarrow 300-\frac{7500}{x^{2}}=0 \Rightarrow x^{2}=25\)
\(\Rightarrow x=5 \quad[\because \text { length cannot be negative }]\)
\(\text { At } x=5, \frac{d^{2} C}{d x^{2}}=\frac{15000}{5^{3}}=120>0 \) 
So, e is minimum.
When x = 5, then e = 2500 + 1500 + 1500 = 5500
Hence, the cost of least expensive tank is Rs. 5500.

Let S(x) be the selling price of x items and let C(x) be the cost price of x items. Then, we have

\(\mathrm{S}(x)=\left(5-\frac{x}{100}\right) x=5 x-\frac{x^{2}}{100} \)
\(\mathrm{C}(x)=\frac{x}{5}+500 \)
Thus, the profit function P(x) is given by
\(\mathrm{P}(x)=\mathrm{S}(x)-\mathrm{C}(x)=5 x-\frac{x^{2}}{100}-\frac{x}{5}-500 \)
\(i.e., \mathrm{P}(x)=\frac{24}{5} x-\frac{x^{2}}{100}-500 \)
\(or \ \mathrm{P}^{\prime}(x)=\frac{24}{5}-\frac{x}{50} \)
\(\text { Now } \mathrm{P}^{\prime}(x)=0 \text { gives } x=240 \text { . Also } \mathrm{P}^{\prime \prime}(x)=\frac{-1}{50} \text { . So } \mathrm{P}^{\prime \prime}(240)=\frac{-1}{50}<0\)
Thus, x = 240 is a point of maxima. Hence, the manufacturer can earn maximum profit, if he sells 240 items.

Let length of rectangle = 2x m and breadth = y m

Given, perimeter of the window = 10 m
\(\because\) Perimeter of rectangle + Perimeter of semi-circle = 10m
\( \Rightarrow 2 y+2 x+\frac{1}{2}(2 \pi x)=10 \)
\(\Rightarrow 2 y=10-x(\pi+2)\) 
Let A be area of the window, then
A = Area of semi-circle + Area of rectangle
\(=\frac{1}{2} \pi x^{2}+2 x y\) 
\(\Rightarrow A=\frac{1}{2}\left(\pi x^{2}\right)+x[10-x(\pi+2)]\) [using Eq. (i)]
\(=\frac{1}{2}\left(\pi x^{2}\right)+10 x-x^{2} \pi-2 x^{2} \)
\(=10 x-\frac{\pi x^{2}}{2}-2 x^{2} \) 
On differentiating twice w.r.t. r, we get
\(\frac{d A}{d x}=10-\pi x-4 x\) ...(ii)
and \(\frac{d^{2} A}{d x^{2}}=-\pi-4\) ...(iii)
For maximum or minima,put \(\frac{d A}{d x}=0\) 
\( \Rightarrow 10-\pi x-4 x=0 \Rightarrow 10=(4+\pi) x \)
\( \Rightarrow x=\frac{10}{4+\pi} \) 
on putting \(x=\frac{10}{4+\pi}\) in Eq.(iii) we get
\(\frac{d^{2} A}{d x^{2}}=\text { Negative }\) 
Thus, A has local maxima, when \(x=\frac{10}{4+\pi}\) 
\(\therefore\) Radius of semi-circle \(x=\frac{10}{4+\pi}\) 
·and one side of rectangle \(2 x=\frac{2 \times 10}{4+\pi}=\frac{20}{4+\pi}\) 
Now, from Eq. (i), we get
\(y=\frac{1}{2}[10-x(\pi+2)]\) 
\(=\frac{1}{2}\left[10-\left(\frac{10}{\pi+4}\right)(\pi+2)\right] \quad[\text { from } \mathrm{Eq}(iv)]\) 
\(=\frac{10 \pi+40-10 \pi-20}{2(\pi+4)} \)
\(=\frac{20}{2(\pi+4)}=\frac{10}{\pi+4} \) 
So, other side of rectangle \(y=\frac{10}{\pi+4}\) 
Light is maximum, when area is maximum.
Hence, dimensions of the window are length \(=\frac{20}{\pi+4} \mathrm{~m}\) and breadth \(=\frac{10}{\pi+4} \mathrm{~m}\)

Let the side of the square to be cut off be x cm. Then, the length and the breadth of the box will be (18 − 2x) cm each and the height of the box is x cm.
Therefore, the volume V(x) of the box is given by,
V(x) = x(18 − 2x)²
\(\therefore V^{\prime}(x)=(18-2 x)^{2}-4 x(18-2 x) \)
\(=(18-2 x)[18-2 x-4 x] \)
\(=(18-2 x)(18-6 x) \)
\(=6 \times 2(9-x)(3-x) \)
\(=12(9-x)(3-x) \)
\(\text { And, } V^{\prime \prime}(x) =12[-(9-x)-(3-x)] \)
\(=-12(9-x+3-x) \)
\(=-12(12-2 x) \)
\(=-24(6-x) \)
\(\text { Now, } V^{\prime}(x)=0 \Rightarrow x=9 \text { or } x=3\)
If x = 9, then the length and the breadth will become 0.
x ≠ 9.
x = 3. Now, \(V^{\prime \prime}(3)=-24(6-3)=-72<0\)
By second derivative test, x = 3 is the point of maxima of V.
Hence, if we remove a square of side 3 cm from each corner of the square tin and make a box from the remaining sheet, then the volume of the box obtained is the largest possible.

The slant height is given by
\(1=\sqrt{h^2+r^2}\)
Now volume is
\( \mathrm{V}=\frac{\pi}{3}\left(\mathrm{r}^2\right) \mathrm{h}\\ =\frac{\pi}{3}\left(1^2-h^2\right) h\\ \frac{\mathrm{dV}}{\mathrm{dh}}=\frac{\pi}{3}\left[1^2-\mathrm{h}^2-\mathrm{h}(2 \mathrm{~h})\right] \)
Now for critical points
\( \frac{\mathrm{dV}}{\mathrm{dh}}=0 \)
\(Or 1^2-h^2-2 h^2=0 \)
\(1^2=3 h^2 1=\sqrt{3} h\\ \frac{\mathrm{h}}{1}=\frac{1}{\sqrt{3}}=\cos \theta \)
Hence
\( \tan \theta=\frac{\sqrt{1-\cos ^2 \theta}}{\cos \theta} \)
\(=\frac{\frac{\sqrt{2}}{\sqrt{3}}}{\frac{1}{\sqrt{3}}} \)
\(=\sqrt{2} \)
\(Therefore\ \tan \theta=\sqrt{2} \)
\(Or\ \theta=\tan ^{-1}(\sqrt{2}) \)

Let one of the numbers be x. Then the other number is (15 – x). Let S(x) denote the sum of the squares of these numbers. Then
S(x) = x² + (15 – x)² = 2x² – 30x + 225
\(\left\{\begin{array}{l} \mathrm{S}^{\prime}(x)=4 x-30 \\ \mathrm{~S}^{\prime \prime}(x)=4 \end{array}\right.\)
\(\text { Now } S^{\prime}(x)=0 \text { gives } x=\frac{15}{2} \text { . Also } S^{\prime}\left(\frac{15}{2}\right)=4>0\)
 Therefore, by second derivative test \(x=\frac{15}{2}\)  is the point of local minima of S. Hence the sum of squares of numbers is minimum when the numbers are \(\frac{15}{2} \text { and } 15-\frac{15}{2}=\frac{15}{2}\)

Let (h, k) be any point on the parabola y = x². Let D be the required distance between (h, k) and (0, c). Then
\(\mathrm{D}=\sqrt{(h-0)^{2}+(k-c)^{2}}=\sqrt{h^{2}+(k-c)^{2}}\)
Since (h, k) lies on the parabola y = x², we have k = h². So (1) gives
\(D \equiv D(k)=\sqrt{k+(k-c)^{2}} \)
\(D^{\prime}(k)=\frac{1+2(k-c)}{2 \sqrt{k+(k-c)^{2}}} \)
\(D^{\prime}(k)=0 \text { gives } k=\frac{2 c-1}{2} \)
Observe that when \(k<\frac{2 c-1}{2}, \text { then } 2(k-c)+1<0 \text { , i.e., } \mathrm{D}^{\prime}(k)<0 \text { . Also when }\)
\(k>\frac{2 c-1}{2}, \text { then } \mathrm{D}^{\prime}(k)>0 . \) So, by first derivative test, D(k) is minimum at \(k=\frac{2 c-1}{2}\)
Hence, the required shortest distance is given by
\(\mathrm{D}\left(\frac{2 c-1}{2}\right)=\sqrt{\frac{2 c-1}{2}+\left(\frac{2 c-1}{2}-c\right)^{2}}=\frac{\sqrt{4 c-1}}{2}\)

Let R be a point on AB such that AR = x m.
Then RB = (20 – x) m (as AB = 20 m).
we have RP² = AR² + AP²
and RQ² = RB² + BQ²
Therefore RP² + RQ² = AR² + AP² + RB² + BQ²
= x² + (16)² + (20 – x)² + (22)²
= 2x² – 40x + 1140
Let S \(\equiv\) S(x) = RP² + RQ² = 2x² – 40x + 1140.
Therefore S'(x) = 4x – 40.
Now S'(x) = 0 gives x = 10. Also S''(x) = 4 > 0, for all x and so S''(10) > 0.
Therefore, by second derivative test, x = 10 is the point of local minima of S. Thus, the distance of R from A on AB is AR = x = 10 m.

The given function is f(x) = sin x + cos x.
\(\therefore f^{\prime}(x)=\cos x-\sin x\)
\(\text { Now, }f^{\prime}(x)=0 \Rightarrow \sin x=\cos x \Rightarrow \tan x=1 \Rightarrow x=\frac{\pi}{4}\)
Then, we evaluate the value of f at critical point and at the end points\(x=\frac{\pi}{4}\) of the interval [0, π].
\(f\left(\frac{\pi}{4}\right)=\sin \frac{\pi}{4}+\cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2} \)
\(f(0)=\sin 0+\cos 0=0+1=1 \)
\(f(\pi)=\sin \pi+\cos \pi=0-1=-1 \)
Hence, we can conclude that the absolute maximum value of f on [0, π] is \(\sqrt{2}\) occurring at \(x=\frac{\pi}{4}\)and the absolute minimum value of f on [0, π] is −1 occurring at x = π.

The given function is \(f(x)=4 x-\frac{1}{2} x^{2}\)
\(\therefore f^{\prime}(x)=4-\frac{1}{2}(2 x)=4-x\)
\(\text { Now, }f^{\prime}(x)=0 \Rightarrow x=4\)
Then, we evaluate the value of f at critical point x = 4 and at the end points of the interval \(\left[-2, \frac{9}{2}\right]\)
\(f(4)=16-\frac{1}{2}(16)=16-8=8 \)
\(f(-2)=-8-\frac{1}{2}(4)=-8-2=-10 \)
\(f\left(\frac{9}{2}\right)=4\left(\frac{9}{2}\right)-\frac{1}{2}\left(\frac{9}{2}\right)^{2}=18-\frac{81}{8}=18-10.125=7.875 \)
Hence, we can conclude that the absolute maximum value of f on \(\left[-2, \frac{9}{2}\right]\) is 8 occurring at x = 4 and the absolute minimum value of f on \(\left[-2, \frac{9}{2}\right]\) is −10 occurring at x = −2

Let x metre be the length of a side of the removed squares. Then, the height of the box is x, length is 8 – 2x and breadth is 3 – 2x. If V(x) is the volume of the box, then
V(x) = x(3 – 2x) (8 – 2x)
= 4x³ – 22x² + 24x
\(\therefore \left\{\begin{array}{l} \mathrm{V}^{\prime}(x)=12 x^{2}-44 x+24=4(x-3)(3 x-2) \\ \mathrm{V}^{\prime \prime}(x)=24 x-44 \end{array}\right.\)
\(\text {Now } \ \mathrm{V}^{\prime}(x)=0 \text {gives } x=3, \frac{2}{3} . \text { But } x \neq 3 \text { (Why?) }\)
\(\text {Thus, we have } x=\frac{2}{3}. \text {Now } \mathrm{V}^{\prime \prime}\left(\frac{2}{3}\right)=24\left(\frac{2}{3}\right)-44=-28<0 \text { . }\)
Therefore, \(x=\frac{2}{3}\)is the point of maxima, i.e., if we remove a square of side \(\frac{2}{3}\) metre from each corner of the sheet and make a box from the remaining sheet, then the volume of the box such obtained will be the largest and it is given by \(\frac{2}{3}\)
\(V\left(\frac{2}{3}\right) =4\left(\frac{2}{3}\right)^{3}-22\left(\frac{2}{3}\right)^{2}+24\left(\frac{2}{3}\right) \)
\(=\frac{200}{27} \mathrm{~m}^{3} \)

Slope of the tangent \(y=x-11\) is 1.
\(y=x^3-11x+5\)
\(\Rightarrow \frac{dy}{dx}=3x^2-11\)
If the points is (\(x_1,y_1\)), then
\(3x_1^2-11=1\)
\(\Rightarrow x_1=\pm2\)
When \(x_1=2,y_1=8-22+5=-9\)
when \(x_1=-2,y_1=-8+22+5=19\)
Since (-2, 19) does not lie on the tangent y = x-11
∴ Required points is (2, -9)

Let   \(y=\sqrt x \)
\(\frac{dy}{dx}=\frac{1}{2\sqrt x}\)
\(\therefore y+ \Delta y=\sqrt{x+\Delta x}\)
\(\Rightarrow y+\frac{dy}{dx}\cdot \Delta x=\sqrt{x+\Delta x}\)
\(\Rightarrow \sqrt x+\frac{1}{2\sqrt x}\cdot \Delta x=\sqrt{x+\Delta x}\)
\(\sqrt{49}+\frac{1}{2\sqrt{49}\cdot(0.5)}=\sqrt{49.5}\)
\(\Rightarrow\sqrt{49.5}=7+\frac{1}{28}=7.0357\)

\(f(x) =\frac{3}{10}x^4-\frac{4}{5}x^3+\frac{36}{5}\)
\(=\frac{6}{5}(x-1)(x+2)(x-3)\)

\(f'(x)>0\Rightarrow x=1,-2,3\)
ஃ Intervals are (-∞,- 2), (- 2,1), (1,3) and (3,∞)
\(\because\ \ \ \ f'(x)>0\ for (-2,1)\cup(3,\infty)\)
f(x) is strictly increasing in (-2,1) and (3, ∞)
\(\because\ \ \ \ f'(x)<0\ for (-\infty,-2)\ and\ (1,3)\)
f(x) is strictly decreasing in (- ∞,- 2) and (1, 3)

\(\frac { { x }^{ 2 } }{ { a }^{ 2 } } -\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1\)
\(\Rightarrow \quad \frac { 2x }{ { a }^{ 2 } } -\frac { 2y }{ { b }^{ 2 } } \frac { dy }{ dx } =0\)
\(\Rightarrow \quad \frac { dy }{ dx } =\frac { { b }^{ 2 }x }{ { a }^{ 2 }y } \)
Slope of tangent at \(\left( \sqrt { 2 } a,\quad b \right) =\frac { \sqrt { 2 } b }{ a } \)
Slope of normal at \(\left( \sqrt { 2 } a,b \right) =-\frac { a }{ \sqrt { 2 } b } \)
Equation of tangent is
\(y-b=\frac { \sqrt { 2 } b }{ a } (x-\sqrt { 2 } a\)
i.e., \(\sqrt { 2 } bx-ay=ab\)
and equation of normal is
\(y-b=-\frac { a }{ \sqrt { 2 } b } (x-\sqrt { 2 } a)\)
i.e., \(\sqrt { 2 } bx-ay=ab\)
and equation of normal is
\(y-b=-\frac { a }{ \sqrt { 2 } b } (x-\sqrt { 2 } a)\)
\(i.e.,\quad ax+\sqrt { 2 } by=\sqrt { 2 } ({ a }^{ 2 }+{ b }^{ 2 })\)

Let y = f(x) = x^3/2, x = 4,
x + \(\Delta \)x = 3.968
\(\therefore \ \Delta x=-0.032\)
\(\Delta y={ \left[ \frac { dy }{ dx } \right] }_{ x=4 }\Delta x\)
\(\Rightarrow \Delta y={ \left[ \frac { 3 }{ 2 } { x }^{ 1/2 } \right] }_{ x=4 }\Delta x\)
\(\Rightarrow \Delta y=\frac { 3 }{ 2 } \times 2(-0.032)=-0.096\)
(3.968)^3/2 = f(x+\(\Delta \)x)
= f(x) + \(\Delta \)y = 8-0.096
= 7.904

Given \(f(x)=\frac{4 \sin x-2 x-x \cos x}{2+\cos x}\) 
\( =\frac{4 \sin x-x(2+\cos x)}{2+\cos x} \)
\(=\frac{4 \sin x}{2+\cos x}-\frac{x(2+\cos x)}{2+\cos x} \)
\(=\frac{4 \sin x}{2+\cos x}-x \) 
On differentiating both sides w.r.t. x, we get
\( f^{\prime}(x) =4\left\{\frac{(2+\cos x) \cos x-\sin x(0-\sin x)}{(2+\cos x)^{2}}\right\}-1 \)
\(=4\left\{\frac{2 \cos x+\cos ^{2} x+\sin ^{2} x}{(2+\cos x)^{2}}\right\}-1 \)
\(\left[\because \cos ^{2} x+\sin ^{2} x=1\right] \)
\(=\frac{8 \cos x+4}{(2+\cos x)^{2}}-1 \)
\(=\frac{8 \cos x+4-(2+\cos x)^{2}}{(2+\cos x)^{2}} \)
\(=\frac{8 \cos x+4-4-\cos ^{2} x-4 \cos x}{(2+\cos x)^{2}} \) 
\(=\frac{4 \cos x-\cos ^{2} x}{(2+\cos x)^{2}} \)
\( =\frac{\cos x(4-\cos x)}{(2+\cos x)^{2}} \) 
We know that \(-1 \leq \cos x \leq 1\) 
\(\Rightarrow 4-\cos x>0 \text { and }(2+\cos x)^{2}>0\) 
(a) For f(x) to be increasing,
\(f^{\prime}(x) \geq 0, \text { when } \cos x \geq 0\) 
[\(\because\)cos x is positive in I and IV quadrants]
Thus, f(x) is increasing in the interval
\(\left(0, \frac{\pi}{2}\right] \cup\left[\frac{3 \pi}{2}, 2 \pi\right)\) 
(b) For f(x) to be decreasing
\(f^{\prime}(x) \leq 0, \text { when } \cos x \leq 0\) 
\([\because \cos x \text { is negative in II and III quadrants] }\) 
Hence, f(x) is decreasing in the interval \(\left[\frac{\pi}{2}, \frac{3 \pi}{2}\right]\).

\(f(x)=x^{3}+\frac{1}{x^{3}} \)
\(\therefore f^{\prime}(x)=3 x^{2}-\frac{3}{x^{4}}=\frac{3 x^{6}-3}{x^{4}} \)
\(\text {Then, } f^{\prime}(x)=0 \Rightarrow 3 x^{6}-3=0 \Rightarrow x^{6}=1 \Rightarrow x=\pm 1 \)
\(\text {Now, the points } x=1 \text { and } x=-1 \)divide the real line into three disjoint intervals i.e.
\((-\infty,-1),(-1,1), \text { and }(1, \infty)\)
\(\text {In intervals }(-\infty,-1) \text { and }(1, \infty) \text { i.e., when } x<-1 \text { and } x>1, f^{\prime}(x)>0 \text { . }\)
\(\text {Thus, when } x<-1 \text { and } x>1, f \text { is increasing. }\)

Given the equation of curve is
\(y=x^{3}+2 x+6\)   ..(i)
and equation of line is
x+14y+4 = 0
On differentiating both sides of (i) W.r.t. x, we get
\(\frac{d y}{d x}=3 x^{2}+2\) 
\(\therefore \text { Slope of normal }=\frac{-1}{\left(\frac{d y}{d x}\right)}=\frac{-1}{3 x^{2}+2}\) 
Also, slope of the line \(x+14 y+4=0 \text { is }-\frac{1}{14}\) 
\(\left[\because \text { slope of the line } A x+B y+C=0 \text { is }\left(-\frac{A}{B}\right)\right]\) 
Since, the normal to the curve is parallel to the line.
\(\therefore\) Slope of line = Slope of normal
\(\begin{array}{ll} \Rightarrow & \frac{-1}{14}=\frac{-1}{3 x^{2}+2} \Rightarrow 3 x^{2}+2=14 \\ \Rightarrow & 3 x^{2}=12 \Rightarrow x^{2}=4 \Rightarrow x=\pm 2 \end{array}\) 
From Eq. (i), at x = 2, we get
y = (2)³ +2(2)+6 = 8+4+6 =18
and at x = - 2, we get 
\(y=(-2)^{3}+2(-2)+6=-8-4+6=-6\) 
\(\therefore\) Normal passes through the points (2, 18)and 
Also, slope of normal \(\frac{-1}{14}\) 
Hence, equation of normal at point (2, 18) is
\( y-y_{1}=\text { Slope of normal } \times\left(x-x_{1}\right) \)
\(\Rightarrow y-18=\frac{-1}{14}(x-2) \)
\(\Rightarrow 14 y-252 =-x+2 \Rightarrow x+14 y=254 \) 
and equation of normal at point (- 2, - 6) is
\( y+6 =-\frac{1}{14}(x+2) \)
\(\Rightarrow14 y+84 =-x-2 \)
\( \Rightarrow x+14 y =-86 \) 
Hence, the two equations of normal are x + 14y = 254 and x+14y = -86

We have,\((255)^{1 / 4}\) 
It can be written as \((255)^{1 / 4}=(256-1)^{1 / 4}\) 
Let \(x=256 \text { and } \Delta x=-1, \text { such that, } y=f(x)=(x)^{1 / 4}\) 
\( \text { Now, } \Delta y=f(x+\Delta x)-f(x) \)
\(\therefore \Delta y=(x+\Delta x)^{1 / 4}-(x)^{1 / 4} \)
\(\Rightarrow \Delta y=(256-1)^{1 / 4}-(256)^{1 / 4} \)
\(\Rightarrow \Delta y=(255)^{1 / 4}-4 \)
\(\Rightarrow (255)^{1 / 4}=4+\Delta y \) 
Now,
\( d y =\left(\frac{d y}{d x}\right) \Delta x=\frac{1}{4} x^{-3 / 4} \Delta x \)
\(=\frac{1}{4 x^{3 / 4}} \Delta x \quad\left[\because y=x^{1 / 4} \Rightarrow \frac{d y}{d x}=\frac{1}{4} x^{-3 / 4}\right] \)
\(=\frac{1}{4(256)^{3 / 4}} \times(-1)=\frac{-1}{4\left(4^{4}\right)^{3 / 4}} \)
\(=\frac{-1}{4(4)^{3}}=\frac{-1}{256}=-0.0039\) 
Then, from Eq. (i), we get
\( (255)^{1 / 4} \approx 4+(-0.0039) \quad[\because d y \approx \Delta y] \)
\(=4-0.0039=3.9961 \) 
Hence, the approximate value of \((255)^{1 / 4} \text { is } 3.9961\)

\(\text { Consider } v=(x) \frac{1}{10} \text { . Let } x=1 \text { and } \Delta x=-0.001\)
\(\text { Then, }\Delta y=(x+\Delta x)^{\frac{1}{10}}-(x)^{\frac{1}{10}}=(0.999)^{\frac{1}{10}}-1\)
\(\Rightarrow(0.999)^{\frac{1}{10}}=1+\Delta y\)
\(\text { Now, } d y \text { is approximately equal to } \Delta y \text { and is given by, }\)
\(d y=\left(\frac{d y}{d x}\right) \Delta x =\frac{1}{10(x)^{\frac{9}{10}}}(\Delta x) \quad\left[\text { as } y=(x)^{\frac{1}{10}}\right] \)
\(=\frac{1}{10}(-0.001)=-0.0001\)
\(\text { Hence, the approximate value of }(0.999)^{\frac{1}{10}} \text { is } 1+(-0.0001)=0.9999 .\)

\(\text { Consider } y=x^{\frac{1}{4}} \text { . Let } x=81 \text { and } \Delta x=0.5\)
\(\text { Then, }\Delta y=(x+\Delta x)^{\frac{1}{4}}-(x)^{\frac{1}{4}}=(81.5)^{\frac{1}{4}}-(81)^{\frac{1}{4}}=(81.5)^{1 /}-3\)
\(\Delta y=(x+\Delta x)^{\frac{1}{4}}-(x)^{\frac{1}{4}}=(81.5)^{\frac{1}{4}}-(81)^{\frac{1}{4}}=(81.5)^{4}-3 \)
\(\Rightarrow(81.5)^{\frac{1}{4}}=3+\Delta y \)
\(\text { Now, } d y \text { is approximately equal to } \Delta y \text { and is given by, }\)
\(d y=\left(\frac{d y}{d x}\right) \Delta x =\frac{1}{4(x)^{\frac{3}{4}}}(\Delta x) \quad\left[\text { as } y=x^{\frac{1}{4}}\right] \)
\(=\frac{1}{4(3)^{3}}(0.5)=\frac{0.5}{108}=0.0046 \)
\(\text { Hence, the approximate value of }(81.5)^{\frac{1}{4}} \text { is } 3+0.0046=3.0046\)

The given function is \( f(x)=(x-2)^{4}(x+1)^{3} \text { . }\)
\(\therefore f^{\prime}(x) =4(x-2)^{3}(x+1)^{3}+3(x+1)^{2}(x-2)^{4} \)
\(=(x-2)^{3}(x+1)^{2}[4(x+1)+3(x-2)] \)
\(=(x-2)^{3}(x+1)^{2}(7 x-2) \)
\(\text { Now, } f^{\prime}(x)=0 \Rightarrow x=-1 \text { and } x=\frac{2}{7} \text { or } x=2\)
Now, for values of x close to \( \frac{2}{7} \)and to the left of \(\frac{2}{7}, f^{\prime}(x)>0\). Also, for values of x close to \(\frac{2}{7}\)and to
\(\text { the right of } \frac{2}{7}, f^{\prime}(x)<0\)
\(\text {Thus, } x=\frac{2}{7} \)is the point of local maxima
Now, for values of x close to 2 and to the left of 2,f′(x)<0. Also, for values of x close to 2 and to the right of 2,f′(x)>0

 
 Now, as the value of x varies through −1,f′(x) does not changes its sign. 
 Thus, x=−1 is the point of inflexion.

Let f(x) = x⁴ − 62x² + ax + 9
\(\therefore f^{\prime}(x)=4 x^{3}-124 x+a\)
It is given that function f attains its maximum value on the interval [0, 2] at x = 1.
\(\therefore f^{\prime}(1)=0 \)
\(\Rightarrow 4-124+a=0 \)
\(\Rightarrow a=120 \)
Hence, the value of a is 120

\(f(x)=|x+2|-1\)
\(\text {We know that }|x+2| \geq 0 \text { for every } x \in \mathbf{R} \text { . }\)
\(\text {Therefore, } f(x)=|x+2|-1 \geq-1 \text { for every } x \in \mathbf{R} \text { . }\)
The minimum value of f is attained when \( |x+2|=0 \text { . }\)
\(|x+2|=0 \)
\(\Rightarrow x=-2 \)
\(\therefore \text { Minimum value of } f=f(-2)==|-2+2|-1=-1\)
Minimum value = -1, no maximum value

\(g(x)=-|x+1|+3\)
\(\text {We know that }-|x+1| \leq 0 \text { for every } x \in \mathbf{R} \text { . }\)
\(\text {Therefore, } g(x)=-|x+1|+3 \leq 3 \text { for every } x \in \mathbf{R} \text { . }\)
The maximum value of g is attained when \( |x+1|=0 \text { . }\)
\(|x+1|=0 \)
\(\Rightarrow x=-1 \)
\(\therefore \text { Maximum value of } g=g(-1)=-|-1+1|+3=3\)
Maximum value = 3, no minimum value

\(h(x)=x+1, x \in(-1,1)\)
\(\text { Here, if a point } x_{0} \text { is closest to }-1 \text { , then we find } \frac{x_{0}}{2}+1
\(\text {Also, if } x_{1} \text { is closest to } 1 \text { , then } x_{1}+1<\frac{x_{1}+1}{2}+1 \text { for all } x_{1} \in(-1,1) \text { . }\)
Hence, function h(x) has neither maximum nor minimum value in (-1,1) \(\)

Let one number be x. Then, the other number is (16 − x).
Let the sum of the cubes of these numbers be denoted by S(x). Then, Now,
\(S(x)=x^{3}+(16-x)^{3} \)
\(\therefore S^{\prime}(x)=3 x^{2}-3(16-x)^{2}, S^{\prime \prime}(x)=6 x+6(16-x) \)
\(\text { Now, } S^{\prime}(x)=0 \Rightarrow 3 x^{2}-3(16-x)^{2}=0 \)
\(\Rightarrow x^{2}-(16-x)^{2}=0 \)
\(\Rightarrow x^{2}-256-x^{2}+32 x=0 \)
\(\Rightarrow x=\frac{256}{32}=8 \)
\(\text { Now, } S^{\prime \prime}(8)=6(8)+6(16-8)=48+48=96>0\)
∴ By second derivative test, x = 8 is the point of local minima of S.
Hence, the sum of the cubes of the numbers is the minimum when the numbers are 8 and 16 − 8 = 8.

Let the equation of an ellipse be \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) then any point on the ellipse is \(P(a \cos \theta, b \sin \theta)\) 

From P, draw PM \(\perp\) OX and produce it to meet the ellipse at Q, then APQ is an isosceles triangle, let S be its area then 
\( S =2 \times \frac{1}{2} \times A M \times M P \)
\( =(O A-O M) \times M P \)
\(=(a-a \cos \theta) \cdot b \sin \theta \)
\(\Rightarrow S =a b(\sin \theta-\sin \theta \cos \theta)=a b\left(\sin \theta-\frac{1}{2} \sin 2 \theta\right) \) 
On differentiating twice w.r.t. θ, we get
\( \frac{d S}{d \theta} =a b(\cos \theta-\cos 2 \theta) \)
\(\text { and } \frac{d^{2} S}{d \theta^{2}}=a b(-\sin \theta+2 \sin 2 \theta) \) 
For maxima or minima, put \(d S / d \theta=0\) 
\(\Rightarrow \cos \theta=\cos 2 \theta \Rightarrow 2 \theta=2 \pi-\theta\) 
\(\Rightarrow 3 \theta=2 \pi \Rightarrow \theta=\frac{2 \pi}{3}\) 
At \(\theta=\frac{2 \pi}{3}, \frac{d^{2} S}{d \theta^{2}}=a b\left[-\sin \frac{2 \pi}{3}+2 \sin \left(2 \times \frac{2 \pi}{3}\right)\right]\) 
\( =a b\left[-\sin \left(\pi-\frac{\pi}{3}\right)+2 \sin \left(\pi+\frac{\pi}{3}\right)\right] \)
\(= a b\left(-\sin \frac{\pi}{3}-2 \sin \frac{\pi}{3}\right)\left[\begin{array}{c} \because \sin \left(\pi-\frac{\pi}{3}\right)=\sin \frac{\pi}{3} \\ \text { and } \sin \left(\pi+\frac{\pi}{3}\right)=\frac{-\sin \pi}{3} \end{array}\right] \)
\(=a b\left(-\frac{\sqrt{3}}{2}-\frac{2 \sqrt{3}}{2}\right)=a b\left(\frac{-3 \sqrt{3}}{2}\right)=\frac{-3 \sqrt{3} a b}{2}<0 \) 
\(\therefore\) S is maximum, when \(\theta=\frac{2 \pi}{3}\) and maximum value of
\( S =a b\left(\sin \frac{2 \pi}{3}-\frac{1}{2} \cdot 2 \sin \frac{2 \pi}{3} \cos \frac{2 \pi}{3}\right) \)
\(=a b\left[\sin \left(\pi-\frac{\pi}{3}\right)-\sin \left(\pi-\frac{\pi}{3}\right) \cos \left(\pi-\frac{\pi}{3}\right)\right] \)
\(=a b\left[\sin \frac{\pi}{3}-\sin \frac{\pi}{3}\left(-\cos \frac{\pi}{3}\right)\right] \) 
\(=a b\left[\sin \left(\pi-\frac{\pi}{3}\right)-\sin \left(\pi-\frac{\pi}{3}\right) \cos \left(\pi-\frac{\pi}{3}\right)\right. \)
\(=a b\left[\sin \frac{\pi}{3}-\sin \frac{\pi}{3}\left(-\cos \frac{\pi}{3}\right)\right] \)
\(=a b\left(\sin \frac{\pi}{3}+\sin \frac{\pi}{3} \cdot \cos \frac{\pi}{3}\right) \)
\(=a b\left(\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2} \times \frac{1}{2}\right) \) 
\(=a b\left(\frac{2 \sqrt{3}+\sqrt{3}}{4}\right) \)
\(=\frac{3 \sqrt{3}}{4} a b \text { sq units } \) 
Hence, the maximum area of isosceles triangle is \(\frac{3 \sqrt{3}}{4} a b \text { sq units. }\)

Let P be a point on the hypotenuse AC of right angled
\(\Delta A B C \text { . Such that } P L \perp A B=a \text { and } P M \perp B C=b\) 
Let \(\angle A P L=\angle A C B=\theta(\text { say })\) 
Then,\(A P=a \sec \theta, P C=b \operatorname{cosec} \theta\) 
 
Let I be the length of the hypotenuse, then
\(l=A P+P C \)
\(\Rightarrow l=a \sec \theta+b \operatorname{cosec} \theta, 0<\theta<\frac{\pi}{2} \) 
On differentiating both sides w.r.t. θ, we get
\(\frac{d l}{d \theta}=a \sec \theta \tan \theta-b \operatorname{cosec} \theta \cot \theta\) 
For maxima or minima,put \(\frac{d l}{d \theta}=0\) 
\(\Rightarrow a \sec \theta \tan \theta=b \operatorname{cosec} \theta \cot \theta \)
\(\Rightarrow\frac{a \sin \theta}{\cos ^{2} \theta}=\frac{b \cos \theta}{\sin ^{2} \theta} \Rightarrow \tan \theta=\left(\frac{b}{a}\right)^{1 /} \) 
Again, differentiating both sides of Eq. (i) w.r.t. θ, we get
\( \frac{d^{2} l}{d \theta^{2}}=a\left(\sec \theta \times \sec ^{2} \theta+\tan \theta \times \sec \theta \tan \theta\right) -b\left[\operatorname{cosec} \theta\left(-\operatorname{cosec}^{2} \theta\right)+\cot \theta(-\operatorname{cosec} \theta \cot \theta)\right] \)
\(=a \sec \theta\left(\sec ^{2} \theta+\tan ^{2} \theta\right) +b \operatorname{cosec} \theta\left(\operatorname{cosec}^{2} \theta+\cot ^{2} \theta\right) \) 
For \(0<\theta<\frac{\pi}{2}\) ,all trigonometric ratios are positive

Also \(2 a>0 \text { and } b>0\) 
\(\therefore \ \frac{d^{2} l}{d \theta^{2}} \text { is positive. }\) 
and least value of
\(l=a \sec \theta+b \operatorname{cosec} \theta\) 
\(=a \frac{\sqrt{a^{2 / 3}+b^{2 / 3}}}{a^{1 / 3}}+b \frac{\sqrt{a^{2 / 3}+b^{2 / 3}}}{b^{1 / 3}} \)
\(=\sqrt{a^{2 / 3}+b^{2 / 3}}\left(a^{2 / 3}+b^{2 / 3}\right)=\left(a^{2 / 3}+b^{2 / 3}\right)^{3 / 2} \)
\({\left[\because\right. \text { in } \Delta E F G, \tan \theta=\frac{b^{1 / 3}}{a^{1 / 3}}, \sec \theta=\frac{\sqrt{a^{2 / 3}+b^{2 /}}}{a^{1 / 3}}} \text { and } \left.\operatorname{cosec} \theta=\frac{\sqrt{a^{2 / 3}+b^{2 / 3}}}{b^{1 / 3}}\right] \) 
Hence proved.