class 12 maths chapter 6 Miscellaneous ncert solution

 Miscellaneous(Exercise 6)

Question 1: Using differentials; find the approximate value of each of the following.

(a) \left(\frac{17}{81}\right)^{\frac{1}{4}}

(b) (33)^{-\frac{1}{5}}

Answer : (a) Let \mathrm{y}=x^{\frac{1}{4}}

Let x=\frac{16}{81} and \Delta x=\frac{1}{81}

Then, \Delta y=(x+\Delta x)^{\frac{1}{4}}-x^{\frac{1}{4}}

=\left(\frac{17}{81}\right)^{\frac{1}{4}}-\left(\frac{16}{81}\right)^{\frac{1}{4}}

=\left(\frac{17}{81}\right)^{\frac{1}{4}}-\frac{2}{3}

\therefore\left(\frac{17}{81}\right)^{\frac{1}{4}}=\frac{2}{3}+\Delta y

Now, dy is approximately equal to \Delta y and given by,

d y=\left(\frac{d y}{d x}\right) \Delta x=\frac{1}{4(x)^{\frac{1}{4}}}(\Delta y)

=\frac{1}{4\left(\frac{16}{81}\right)^{\frac{1}{4}}}\left(\frac{1}{81}\right)

=\frac{27}{4 \times 8} \times \frac{1}{81}=\frac{1}{32 \times 3}

=\frac{1}{96}=0.010

Hence, the approximate value of \left(\frac{17}{81}\right)^{\frac{1}{4}} is \frac{2}{3}+0.010=0.677

(b) Let \mathrm{y}=x^{-\frac{1}{5}}

Let x=32 and \Delta x=1

Then, \Delta y=(x+\Delta x)^{-\frac{1}{5}}-x^{-\frac{1}{5}}

=(33)^{-\frac{1}{5}}-(32)^{-\frac{1}{5}}=(33)^{-\frac{1}{5}}-\frac{1}{2}

\therefore(33)^{-\frac{1}{5}}=\frac{1}{2}+\Delta y

Now, dy is approximately equal to \Delta y and is given by,

d y=\left(\frac{d y}{d x}\right) \Delta x=\frac{-1}{5(x)^{\frac{6}{5}}}(\Delta x) \quad\left(\text { as } y=x^{-\frac{1}{5}}\right)

=-\frac{1}{5(2)^6}(1)=-\frac{1}{320}=-0.003

Hence, the approximate value of (33)^{-\frac{1}{5}} is \frac{1}{2}+(-0.003)=0.5-0.003=0.497

Question 2: Show that the function given by f(x)=\frac{\log x}{x} has maximum at \mathrm{x}=\mathrm{e}.

Answer : The given function is f(x)=\frac{\log x}{x}

f^{\prime}(x)=\frac{x\left(\frac{1}{x}\right)-\log x}{x^2}=\frac{1-\log x}{x^2}

Now,f^{\prime}(x)=0

1-\log x=0  \Rightarrow \log x=1

\log x=\log e \Rightarrow  x=e^{\log x}

Now, f''(e)=\frac{x^2\left(-\frac{1}{x}\right)-(1-\log x)(2 x)}{x^4}

=\frac{x-2 x(1-\log x)}{x^4}=\frac{-3+2 \log x}{x^3}

Now, f^{\prime \prime}(e)=\frac{-3+2 \log e}{e^3}

=\frac{-3+2}{e^3}=\frac{-1}{e^3}<0

Therefore, by second derivative test, f is the maximum at x=e.

Question 3: The two equal sides of an isosceles triangle with fixed base b are decreasing at the rate of 3 \mathrm{~cm} per second .how fast is the area decreasing when the two equal sides are equal to the base?

Answer : Let \triangle A B C be isosceles where \mathrm{BC} is the base of fixed length b. Let the length of the two equal sides of \triangle A B C be ‘a’, draw \mathrm{AD} \perp B C

Now, in \triangle A D C, by applying the Pythagoras theorem,

\mathrm{AD}=\sqrt{a^2-\frac{b^2}{4}}

Area of triangle (\mathrm{A})=\frac{1}{2} b \sqrt{a^2-\frac{b^2}{4}}

The rate of change of the area with respect to time (\mathrm{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 \mathrm{~cm} per second.

\frac{d a}{d t}=-3 \mathrm{~cm} / \mathrm{sec}

\Rightarrow \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 a b}{\sqrt{4 a^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} / \mathrm{sec}^2.

Question 4: Find the equation of the normal to curve y^2=4 x at the point (1,2).

Answer : The equation of the given curve is y^2=4 x,

Differentiating with respect to x, we have

2 y \frac{d y}{d x}=4

\frac{d y}{d x}=\frac{4}{2 y}=\frac{2}{y}

\left.\therefore \frac{d y}{d x}\right]_{(1,2)}=\frac{2}{2}=1

Now, the slope of the normal at point (1,2) is \frac{-1}{\left.\frac{d y}{d x}\right]_{(1,2)}}=\frac{-1}{1}=-1

Equation of the normal at

(1,2) is y-2=-1(x-1) \Rightarrow y-2=-x+1 i.e. x+y-3=0.

Question 5: Show that the normal at any point \theta to the curve
x=a \cos \theta+a \theta \sin \theta, \quad y=a \sin \theta-a \theta \cos \theta is at a constant distance from the origin.

Answer : We have x=a \cos \theta+a \theta \sin \theta

\Rightarrow \frac{d x}{d \theta}=a(- \sin \theta+ \sin \theta+ \theta \cos \theta)=a \theta \cos \theta

Since, y=a \sin \theta-a \theta \cos \theta

\Rightarrow \frac{d y}{d \theta}=a (\cos \theta- \cos \theta+\theta \sin \theta)=a \theta \sin \theta

Now, \frac{d y}{d x}=\frac{d y/d\theta}{dx/d \theta}

=\frac{a \theta \sin \theta}{a \theta \cos \theta}=\tan \theta

Slope of the normal= -\frac{1}{\tan \theta}

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 constant.

\frac{|-a|}{\sqrt{\sin ^2 \theta+\cos ^2 \theta}}=\frac{|-a|}{\sqrt{1}}=|-a|, which is independent of \theta.

Hence, the perpendicular distance of the normal from the origin is constant.

Question 6: Find the intervals in which the function f given by f(x)=\frac{4 \sin x-2 x-x \cos x}{2+\cos x} is (i) Increasing (ii) Decreasing.

Answer : f(x)=\frac{4 \sin x-2 x-x \cos x}{2+\cos x}

\Rightarrow f^{\prime}(x)=\frac{(2+\cos x)(4 \cos x-2-\cos x+x \sin x)-(4 \sin x-2 x-x \cos x)(-\sin x)}{(2+\cos x)^2}

\Rightarrow f^{\prime}(x)=\frac{(2+\cos x)(3 \cos x-2+x \sin x)+\sin x(4 \sin x-2 x-x \cos x)}{(2+\cos x)^2}

\Rightarrow f^{\prime}(x)=\frac{6 \cos x-4+2 x \sin x+3 \cos ^2 x-2 \cos x+x \sin x \cos x+4 \sin ^2 x-2 x \sin x-x \sin x \cos x}{(2+\cos x)^2}

\Rightarrow f^{\prime}(x)=\frac{4 \cos x-4+3 \cos ^2 x+4 \sin ^2 x}{(2+\cos x)^2}

\Rightarrow f^{\prime}(x)=\frac{4 \cos x-4+3 \cos ^2 x+4-4 \cos ^2 x}{(2+\cos x)^2}

\Rightarrow f^{\prime}(x)=\frac{4 \cos x-\cos ^2 x}{(2+\cos x)^2}

=\frac{\cos x(4-\cos x)}{(2+\cos x)^2}

Now f^{\prime}(x)=0 \Rightarrow \cos x=0 or \cos x=4

But \cos x \neq 4

\cos x=0 \text { i.e. } x=\frac{\pi}{2}, \frac{3 \pi}{2}

Now, x=\frac{\pi}{2} and x=\frac{3 \pi}{2} divides (0,2 \pi) into three disjoint intervals, i.e.

\left(0, \frac{\pi}{2}\right),\left(\frac{\pi}{2}, \frac{3 \pi}{2}\right) \text { and }\left(\frac{3 \pi}{2}, 2 \pi\right)

In intervals \left(0, \frac{\pi}{2}\right) and \left(\frac{\pi}{2}, \frac{3 \pi}{2}\right), f^{\prime}(x)>0.

Thus, f(x) is increasing for 0<x<\frac{\pi}{2} and \frac{3 \pi}{2}<x<2 \pi.

In the interval \left(\frac{\pi}{2}, \frac{3 \pi}{2}\right), f^{\prime}(x)<0.

Thus, f(x) is decreasing for \frac{\pi}{2}<x<\frac{3 \pi}{2}.

Question 7: Find the intervals in which the function f given by f(x)=x^3+\frac{1}{x^3}, x \neq 0 is (i) increasing (ii) decreasing.

Answer : f(x)=x^3+\frac{1}{x^3}

\Rightarrow f^{\prime}(x)=3 x^2-\frac{3}{x^4}=\frac{3 x^6-3}{x^4}

Then, f^{\prime}(x)=0 i.e. 3 x^6-3=0

x^6=1,=x=\pm 1

Now, the points x=1 and x=-1 divide the real line into three disjoint intervals i.e., (-\infty,-1),(-1,1) and (1, \infty)

In intervals (-\infty,-1) and (1, \infty) i.e., when x<-1 and x>1, f^{\prime}(x)>0

Thus, when x<-1 and x>1, f is increasing.

In interval (-1,1) i.e., when -1<\mathrm{x}<1, \mathrm{f}^{\prime}(\mathrm{x})<0

Thus, when -1<x<1, f is decreasing.

Question 8: Find the maximum area of an isosceles triangle inscribed in the ellipse \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 with its vertex at one end of the major axis.

Answer : The given ellipse is \frac{x^2}{a^2}+\frac{y^2}{b^2}=1.

Let the major axis be along the x-axis.

class 12 maths chapter 6 Miscellaneous ncert solution

Let A B C be the triangle inscribed in the ellipse where
vertex A is at (a, 0).

Let the point B(x,y)

y=\pm \frac{b}{a} \sqrt{a^2-x^2}

BC = 2y and AM = a - x

Area of \triangle ABC=\frac{1}{2}\times 2y(a-x)

A = y(a-x)=\frac{b}{a} \sqrt{a^2-x^2}(a-x)--(i)

Squaring both side

A^2 = S(Let) = \frac{b^2}{a^2}(a^2-x^2)(a-x)^2

Differntiate with respect to x

\frac{dS}{dx}= \frac{b^2}{a^2}[(a^2-x^2)2(a-x)(-1)+(a-x)^2(-2x)]

\Rightarrow \frac{dS}{dx} = \frac{b^2}{a^2}(a-x)^2[-2(a+x)-2x]

\Rightarrow \frac{dS}{dx}= \frac{b^2}{a^2}(a-x)^2[-2a-4x]--(ii)

For max and minima \frac{dS}{dx}=0

\frac{b^2}{a^2}(a-x)^2[-2a-4x]=0

\Rightarrow a-x=0 or -2a -4x =0

\Rightarrow x = a or x = -\frac{a}{2}

Since, x = a is not possible

Again differentiate of eq (ii)

\frac{d^2S}{dx^2}=\frac{b^2}{a^2}[(a-x)^2(-4)+(-2a-4x)2(a-x)(-1)]

\Rightarrow \frac{d^S}{dx^2}=\frac{b^2}{a^2}(a-x)[-4(a-x)-2(-2a-4x)]

At x = -\frac{a}{2}

\frac{d^2S}{dx^2}=\frac{b^2}{a^2}[-4(a-\frac{a}{2})-2(-2a-4(-\frac{a}{2}))]

= \frac{b^2}{a^2}[-4(\frac{a}{2})+0]= \frac{b^2}{a^2}(-2a)<0

Hence area of triangle is max when x = -\frac{a}{2}

Putting in eq (i)

Max area = \frac{b}{a} \sqrt{a^2-(-\frac{a}{2})^2}[a-(-\frac{a}{2})]

= \frac{b}{a}\sqrt{\frac{3a^2}{4}}[\frac{3a}{2}]

= \frac{3\sqrt{3}ab}{4}

Question 9: A tank with rectangular base and rectangular sides, open at the top is to be constructed so that its depth is 2 \mathrm{~m} and volume is 8 \mathrm{~m}^3. If building of tank costs ₹70 per square meters for the base and ₹45 per square metre for sides. What is the cost of least expensive tank?

Answer : Let \mathrm{l}, \mathrm{b}, and \mathrm{h} represent the length, breadth, and height of the tank respectively.

Then, we have height (h)=2 \mathrm{~m}

Volume of the \operatorname{tank}=8 \mathrm{~m}^3

Volume of the tank =l \times b \times h

\therefore 8=l \times b \times 2

l b=4 \quad \Rightarrow b=\frac{4}{l}

Now, area of the base =l b=4

\Rightarrow b = \frac{4}{l}

Total cost (C) = 70lb + 2(l+b)h\times 45

C = 70l\times \frac{4}{l}+90(l+\frac{4}{l})\times 2

\Rightarrow C = 280 + 180(l+\frac{4}{l})--(i)

Differentiate with respect to ‘l’

\frac{dC}{dl }= 0 +180(1-\frac{4}{l^2})

For max and minima \frac{dC}{dl}=0

180(1-\frac{4}{l^2})=0

1 = \frac{4}{l^2}

\Rightarrow l^2 = 4

\Rightarrow l = 2

Again differentiate with respect to l

\frac{d^2C}{dl^2}= 180(0-4(\frac{-2}{l^3}))

\Rightarrow 180(\frac{8}{l^3})

At l = 2

\frac{d^2C}{dl^2}=180(\frac{8}{(2)^3})=180>0

Therefore cost of tank is min when l = 2

Putting in (i)

minimum cost (C) = 280+180(2+\frac{4}{2})

= 280+180(4) = 280+720

= ₹ 1000

Question 10: The sum of the perimeter of a circle and square is \mathrm{k}, where \mathrm{k} is some constant. Prove that the sum of their areas is least when the side of square is double the radius of the circle.

Answer : Let r be the radius of the circle and a be the side of the square. Then, we have:

2 \pi r+4 a=k( where k is constant )

\Rightarrow a=\frac{k-2 \pi r}{4}

The sum of the areas of the circle and the square (A) is given by,

A=\pi r^2+a^2=\pi r^2+\frac{(k-2 \pi r)^2}{16}

\frac{d A}{d r}=2 \pi r+\frac{2(k-2 \pi r)(-2 \pi)}{16}=2 \pi r-\frac{\pi(k-2 \pi r)}{4}

Now \frac{d A}{d r}=0.

\Rightarrow 2 \pi r=\frac{\pi(k-2 \pi r)}{4}

\Rightarrow 8 r=k-2 \pi r

\Rightarrow(8+2 \pi) r=k

\Rightarrow r=\frac{k}{8+2 \pi}=\frac{k}{2(4+\pi)}

Now, \frac{d^2 A}{d r^2}=2 \pi+\frac{\pi^2}{2}>0.

\therefore When r=\frac{k}{2(4+\pi)},

\frac{d^2 A}{d r^2}>0.

The sum of the area is least when r=\frac{k}{2(4+\pi)}

a=\frac{k-2 \pi\left[\frac{k}{2(4+\pi)}\right]}{4}

=\frac{k(4+\pi)-\pi k}{4(4+\pi)}=\frac{4 k}{4(4+\pi)}=2 r

Hence, it has been proved that the sum of their areas is least when the side of the square is double the radius of the circle.

Question 11: A window is in the form of rectangle surmounted by a semicircular opening. The total perimeter of the window is 10 \mathrm{~m}. Find the dimensions of the window to admit maximum light through the whole opening

Answer: Let x and y be the length and breadth of the rectangular window.

class 12 maths chapter 6 Miscellaneous ncert solution

Radius of the semicircular opening =\frac{x}{2}

It is given that the perimeter of the window is 10 \mathrm{~m}.

\Rightarrow x+2 y+\frac{\pi x}{2}=10 \Rightarrow x\left(1+\frac{\pi}{2}\right)+2 y=10

\Rightarrow 2 y=10-x\left(1+\frac{\pi}{2}\right)

\Rightarrow y=5-x\left(\frac{1}{2}+\frac{\pi}{4}\right)

Area of the window (A) is given by,

A=x y+\frac{\pi}{2}\left(\frac{x}{2}\right)^2

\Rightarrow A=x\left[5-x\left(\frac{1}{2}+\frac{\pi}{4}\right)\right]+\frac{\pi x^2}{8}

=5 x-x^2\left(\frac{1}{2}+\frac{\pi}{4}\right)+\frac{\pi x^2}{8}

\Rightarrow \frac{d A}{d x}=5-2 x\left(\frac{1}{2}+\frac{\pi}{4}\right)+\frac{\pi x}{4}

=5-x\left(1/2+\frac{\pi}{2}\right)+\frac{\pi}{4} x

\Rightarrow \frac{d^2 A}{d x^2}=-\left(\frac{1}{2}+\frac{\pi}{2}\right)+\frac{\pi}{4}=-1/2-\frac{\pi}{4}

Now, \frac{d A}{d x}=0

\Rightarrow 5-x\left(1+\frac{\pi}{2}\right)+\frac{\pi}{4} x=0

\Rightarrow x\left(1+\frac{\pi}{4}\right)=5

\Rightarrow x=\frac{5}{\left(1+\frac{\pi}{4}\right)}=\frac{20}{\pi+4}

Thus, when x=\frac{20}{\pi+4} then \frac{d^2 A}{d x^2}<0.

Therefore, by second derivative test, the area is the maximum when length x=\frac{20}{\pi+4} m.

Now, y=5-\frac{20}{\pi+4}\left(\frac{2+\pi}{4}\right)

=5-\frac{5(2+\pi)}{\pi+4}=\frac{10}{\pi+4} m

Hence, the required dimensions of the window to admit maximum light is given by

Length x=\frac{20}{\pi+4} m and breath =\frac{10}{\pi+4} m.

Question 12: A point on the hypotenuse of a triangle is at distance a and b from the sides of the triangle. Show that the minimum length of the hypotenuse is \left(a^{\frac{2}{3}}+b^{\frac{2}{3}}\right)^{\frac{3}{2}}

Answer: Let A B C be right-angled at B. Let A B=x and B C=y.

class 12 maths chapter 6 Miscellaneous ncert solution

Let \mathrm{P} be a point on the hypotenuse of the triangle such that \mathrm{P} is at a distance of a and b from the sides \mathrm{AB} and \mathrm{BC} respectively. Let \angle c=\theta

We have, \mathrm{AC}=\sqrt{x^2+y^2}

Now, P C=b \operatorname{cosec} \theta and A P=a \sec \theta

\Rightarrow \mathrm{AC}=\mathrm{AP}+\mathrm{PC} \Rightarrow A C=a \sec \theta+b \operatorname{cosec} \theta

\frac{d(A C)}{d \theta}=-b \operatorname{cosec} \theta \cot \theta+a \sec \theta \tan \theta

\therefore \frac{d(A C)}{d \theta}=0 \Rightarrow a \sec \theta \tan \theta=b \operatorname{cosec} \theta \cot \theta

\Rightarrow \frac{a}{\cos \theta} \cdot \frac{\sin \theta}{\cos \theta}=\frac{b}{\sin \theta} \cdot \frac{\cos \theta}{\sin \theta}

\Rightarrow a \sin ^3 \theta=b \cos ^3 \theta \Rightarrow \tan \theta=\left(\frac{b}{a}\right)^{\frac{1}{3}}

\therefore \sin \theta=\frac{(b)^{\frac{1}{3}}}{\sqrt{a^{\frac{2}{3}}+b^{\frac{2}{3}}}} \text { and } \cos \theta=\frac{(a)^{\frac{1}{3}}}{\sqrt{a^{\frac{2}{3}}+b^{\frac{2}{3}}}}

It can be clearly shown that \frac{d^2(A C)}{d \theta^2}>0 when \tan \theta=\left(\frac{b}{a}\right)^{\frac{1}{3}}

Therefore, by second derivative test, the length of the hypotenuse is minimum when \tan \theta=\left(\frac{b}{a}\right)^{\frac{1}{3}}.

Now, if \tan \theta=\left(\frac{b}{a}\right)^{\frac{1}{3}}, we have:

A C=\frac{b \sqrt{a^{\frac{2}{3}}+b^{\frac{2}{3}}}}{b^{\frac{2}{3}}}+\frac{b \sqrt{a^{\frac{2}{3}}+b^{\frac{2}{3}}}}{a^{\frac{2}{3}}}

AC=\sqrt{a^{\frac{2}{3}}+b^{\frac{2}{3}}}\left(a^{\frac{2}{3}}+b^{\frac{2}{3}}\right)=\left(a^{\frac{2}{3}}+b^{\frac{2}{3}}\right)^{\frac{2}{3}}

Hence, the minimum length of the hypotenuses is \left(a^{\frac{2}{3}}+b^{\frac{2}{3}}\right)^{\frac{2}{3}}.

Question 13: Find the points at which the function f given by f(x)=(x-2)^2(x+1)^3 has

(i) local maxima

(ii) local minima

(iii) point of inflexion.

Answer: The given function is f(x)=(x-2)^2(x+1)^3

\Rightarrow 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)

Now, f^{\prime}(x)=0

\Rightarrow x=-1 and x=\frac{2}{7} 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 the right of \frac{2}{7}, f^{\prime}(x)<0.

Thus, x=\frac{2}{7} is the point of local maxima.

Now, for values of \mathrm{x} close to 2 and to the left of 2, f^{\prime}(x)<0 Also, for values of \mathrm{x} close to 2 and to the right of 2, f^{\prime}(x)>0.

Thus, x=2 is the point of local minima.

Now, as the value of x varies through -1, f^{\prime}(x) does not changes its sign.

Thus, x=-1 is the point of inflexion.

Question 14: Find the absolute maximum and minimum values of the function \mathrm{f} given by
f(x)=\cos ^2 x+\sin x, \quad x \in[0, \pi]

Answer: f(x)=\cos ^2 x+\sin x

\Rightarrow f^{\prime}(x)=2 \cos x(-\sin x)+\cos x

\Rightarrow f^{\prime}(x)=-2 \sin x \cos x+\cos x

\text { Now, } f^{\prime}(x)=0

\Rightarrow 2 \sin x \cos x=\cos x \Rightarrow \cos x(2 \sin x-1)=0

\Rightarrow \sin x=\frac{1}{2} \text { or } \cos x=0

\Rightarrow x=\frac{\pi}{6} \text { or } \frac{\pi}{2} \text { as } x \in[0, \pi]

Now, evaluating the value of f at critical points x=\frac{\pi}{2} and x=\frac{\pi}{6} and at the end points of the interval [0, \pi] (i.e., at x=0 and x=\pi ),

we have:

f\left(\frac{\pi}{6}\right)=\cos ^2 \frac{\pi}{6}+\sin \frac{\pi}{6}=\left(\frac{\sqrt{3}}{2}\right)^2+\frac{1}{2}=\frac{5}{4}

f(0)=\cos ^2 0+\sin 0=1+0=1

f(\pi)=\cos ^2 \pi+\sin \pi=(-1)^2+0=1

f\left(\frac{\pi}{2}\right)=\cos ^2 \frac{\pi}{2}+\sin \frac{\pi}{2}=0+1=1

Hence, the absolute maximum value of f is \frac{5}{4} occurring at x=\frac{\pi}{6} and the absolute minimum value of f is 1 occurring at x=0, \frac{\pi}{2} and \pi.

Question 15: Show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is \frac{4 r}{3}.

Answer : Let O be the centre and r be the radius of the given sphere, O M =x

class 12 maths chapter 6 Miscellaneous ncert solution

In ΔOBM, using pythagoras theorem

OB^2=OM^2+BM^2

\Rightarrow r^2 = x^2 + BM^2

\Rightarrow BM^2 = r^2-x^2

AM= h = r+x

BM is the radius of cone

Volume of cone V = \frac{1}{3}\pi r^2 h

\Rightarrow V = \frac{1}{3}\pi(r^2-x^2)(r+x)--(i)

\Rightarrow V = \frac{1}{3}\pi(r-x)(r+x)(r+x)

\Rightarrow V = \frac{1}{3}\pi (r-x)(r+x)^2

Differentiating with respect to x

\frac{dV}{dx} = \frac{1}{3}\pi [(r-x)2(r+x)+(r+x)^2(-1)]

= \frac{1}{3}\pi (R+x)[2(R-x)-(R+x)]

= \frac{1}{3}\pi(r+x)[2r-2x-r-x]

=\frac{1}{3}\pi (r+x)[r-3x]—(ii)

For max and minima \frac{dV}{dx}=0

\frac{1}{3}\pi (r+x)[r-3x]=0

Hence r+x=0 or r-3x = 0

x = -r or x = r/3

x =-r Not possible

Agian differentiate with respect to x of eq (ii)

\frac{d^2V}{dx^2} = \frac{1}{3}\pi[(r+x)(-2)+(r-2x).1]

= \frac{1}{3}\pi[-2r-2x+r-2x]

= \frac{1}{3}\pi[-r-4x]

At x = r/3

\frac{d^V}{dx^2}=\frac{1}{3}\pi[-r-4\times \frac{r}{3}]

= \frac{1}{3}[-7r/3]<0

Hence volume of cone is max when x=r/3

Height of cone = r+x

= r+\frac{r}{3}=\frac{4r}{3}

Question 16: Let f be a function defined on [a, b] such that f^{\prime}(x)>0, for all x \in(a, b). Then prove that f is an increasing function on (a, b).

Answer : We have to prove that function is always increasing, i.e.

f\left(x_2\right)>f\left(x_1\right) for all x_2>x_1 . \quad\left[\right. Where \left.x_1, x_2 \in[a, b]\right]

Let x_1 and x_2 be two numbers in the interval [a, b], i . e . x_1, x_2 \in[a, b] and x_2>x_1.

Let the interval \left[x_1, x_2\right].

Function f is continuous as well as differential in \left[x_1, x_2\right] as it is continuous and differential in [a, b].

Using Mean value theorem, there exists c \in\left[x_1, x_2\right], such that

f'(c)=\frac{f\left(x_2\right)-f\left(x_1\right)}{x_2-x_1}

Given that f^{\prime}(x)>0 for all x \in[a, b].

Therefore, f^{\prime}(c)>0 for all x \in\left[x_1, x_2\right].

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

\Rightarrow f\left(x_2\right)-f\left(x_1\right)>0 \quad \Rightarrow f\left(x_2\right)>f\left(x_1\right)

Now, for the two point x_1, x_2 \in[a, b], where x_2>x_1, we have f\left(x_2\right)>f\left(x_1\right).

Hence, the functions f is increasing in the interval [a, b].

Question 17: Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius R is \frac{2 R}{\sqrt{3}}. Also find the maximum volume.

Answer : Consider O be the centre and R be the radius of the given sphere, O M =x

class 12 maths chapter 6 Miscellaneous ncert solution

In ΔOBM, using pythagoras theorem

OB^2=OM^2+BM^2

\Rightarrow R^2 = x^2 + BM^2

\Rightarrow BM^2 = R^2-x^2

MN= h =2x

BM is the radius of cone

Volume of cone V = \pi r^2 h

\Rightarrow V = \pi(R^2-x^2)2x--(i)

Differentiating with respect to x

\frac{dV}{dx} = \pi [(R^2-x^2).2+2x(-2x)]

= \pi[2R^2-2x^2-4x^2]

\frac{dV}{dx} = \pi[2R^2-6x^2]--(ii)

For max and minima \frac{dV}{dx}=0

\pi[2R^2-6x^2]=0

\Rightarrow 2R^2 = 6x^2

\Rightarrow x^2 = \frac{R^2}{3}

\Rightarrow x =\frac{R}{\sqrt{3}}

Again differentiate of eq (ii)

\frac{d^2V}{dx^2}= \pi[0-12x]

At x=\frac{R}{\sqrt{3}}

\frac{d^2V}{dx^2}= \pi[-12(\frac{R}{\sqrt{3}})]<0

Therefore volume of cylinde is max when x = \frac{R}{\sqrt{3}}

Height of cylinder =2x= \frac{2R}{\sqrt{3}}

Question 18: Show that the height of the cylinder of greatest volume which can be inscribed in a right circular cone of height h and having semi vertical angle \alpha is one third that of the cone and the greatest volume of the cylinder is \frac{4}{27}\pi h^2\tan^3\alpha

Solution: Let radius of cone = r and Radius of cylinder= x

class 12 maths chapter 6 Miscellaneous ncert solution

Semivertical angle of cone =\alpha and height of cone = h

In \triangle APQ

\tan\alpha = \frac{x}{AP}

\Rightarrow AP = x\cot\alpha

Therefore, Height of cylinder (QM) = AO - AP

= h-x\cot\alpha ---(i)

Volume of cylinder = \pi R^2 H

V = \pi x^2(h-x\cot\alpha)

\Rightarrow V = \pi(hx^2-x^3\cot\alpha)---(ii)

Differentiating with respect to ‘x’

\frac{dV}{dx}=\pi(h2x-3x^2\cot\alpha)

\Rightarrow \frac{dV}{dx}=\pi(2xh-3x^2\cot\alpha)

For max and minima \frac{dV}{dx}=0

\pi(2xh-3x^2\cot\alpha)=0

\Rightarrow 2xh = 3x^2\cot\alpha

\Rightarrow 2h=3x\cot\alpha

\Rightarrow x = \frac{2}{3}\tan\alpha

Again differentiate with respect to x

\frac{d^2V}{dx^2}= \pi(2h-6x\cot\alpha)

At x = \frac{2}{3}h\tan\alpha

\frac{d^2V}{dx^2} = \pi(2h-6(\frac{2}{3}h\tan\alpha)\cot\alpha)

= \pi(2h-4h)=\pi\times -2h<0

Therefore the volume of cylinder is max when x=\frac{2}{3}h\tan\alpha

Now from eq (i)

Height of cylinder = h-\frac{2}{3}h\tan\alpha\cot\alpha

=h-\frac{2}{3}=\frac{1}{3}h

Volume of cylinder =\pi x^2(h-x\cot\alpha)

= \pi (\frac{2}{3})\tan\alpha)^2\times \frac{h}{3}

= \pi\frac{4}{9}h^2\tan^2\alpha\frac{h}{3}

Volume of cylinder = \frac{4}{27}\pi h^3\tan^2\alpha

Question 19: A cylindrical tank of radius 10 \mathrm{~m} is being filled with wheat at the rate of 314 cubic meter per hour. Then the depth of wheat is increasing at the rate of:

(A) 1 \mathrm{~m} / \mathrm{h}

(B) 0.1 \mathrm{~m} / \mathrm{h}

(C) 1.1 \mathrm{~m} / \mathrm{h}

(D) 0.5 \mathrm{~m} / \mathrm{h}

Solution: Option (A) is correct.

Consider y be the depth of the wheat in the cylindrical tank whose radius is 10 \mathrm{~m} at time t.

\mathrm{V}= Volume of wheat in cylindrical tank at time t,

So, {t=\pi(10)^2 y=100 \pi y} cubic meter

We are given that \frac{d \mathrm{~V}}{d t}=314 cubic meter / \mathrm{hr}

So, \frac{d}{d t} 100 \pi y=314

100 \pi y=314

100(3.14) y=314

Therefore, y=1 \mathrm{~m} / \mathrm{h}

Question 20: The slope of the tangent to the curve x=t^2+3 t-8, y=2 t^2-2 t-5 at the point (2, \left.-1\right) is:

(A) \frac{22}{7}

(B) \frac{6}{7}

(C) \frac{7}{6}

(D) \frac{-6}{7}

Solution: Option (B) is correct.

Solution: Equation of the curves are x=t^2+3 t-8 \ldots(1) and y=2 \frac{d x}{d t}=2 t+3 and \frac{d y}{d t}=4 t-2 Slope of the tangent to the given curve at point (\mathrm{x}, \mathrm{y}) is

(x, y)=\frac{d y}{d x}=\frac{d y / d t}{d x / d t}=\frac{4 t-2}{2 t+3}

At the given point (2,-1)=x=2 and y=-1

At x=2, from equation (1), 2=t^2+3 t-8

t^2+3 t-10=0

(t+5)(t-2)=0

t=-5, t=2

At y=-1, From eq(2) -1=2t^2-2t-5

2t^2-2t-4=0

\Rightarrow t^2 - t -2 =0

(t-2)(t+1)=0

t=2, t=-1

Here, common value of t in the two sets of values is 2 .
Again, from equation (3),

Slope of the tangent to the given curve at point (2,-1)=\frac{4(2)-2}{2(2)+3}=\frac{6}{7}

Question 21: The line y=m x+1 is a tangent to the curve y^2=4 x if the value of m is:

(A) 1

(B) 2

(C) 3

(D) 1 / 2

Solution: Option (A) is correct.

Equation of the curve is y^2=4 x

2 y \frac{d y}{d x}=4.1

\frac{d y}{d x}=\frac{2}{y}

Slope of the tangent to the given curve at point (x, y)

\frac{d y}{d x}=\frac{2}{y}

\frac{2}{y}= m

y=\frac{2}{m} \ldots(ii)

Now y=m x+1

\frac{2}{m}=m x+1

m x=\frac{2}{m}-1

x=\frac{2-m}{m} .

Putting the values of x and y in equation (1), \frac{4}{m^2}=\frac{4(2-m)}{m^2}

2-m=1 \Rightarrow m=1

Question 22: The normal at the point (1,1) on the curve 2 y+x^2=3 is:

(A) x+y=0

(B) x-y=0

(C) x+y+1=0

(D) x-y=1

Solution: Option (B) is correct.

Equation of the given curve is 2 y+x^2=3

2 \frac{d y}{d x}+2 x=0

\frac{d y}{d x}=-x

Slope of the tangent to the given curve at point (1,1) is

\frac{d y}{d x}=-x=-1=m

\text { (say) }

\text { Slope of the normal }=\frac{-1}{m}=\frac{-1}{-1}=1

Equation of the normal at (1,1) is y-1=1(x-1)

y-1=x-1

x-y=0

Question 23: The normal to the curve x^2=4 y passing through (1,2) is:

(A) x+y=3

(B) x-y=3

(C) x+y=1

(D) x-y=1

Solution: Option (A) is correct.

Equation of the curve is x^2=4 y

2 x=4 \frac{d y}{d x}

\frac{d y}{d x}=\frac{x}{2}--(i)

Slope of the normal at (x, y) is

-\frac{d x}{d y}=\frac{-2}{x}--(ii)

Again slope of normal at given point (1,2)=\frac{y-2}{x-1}--(iii)

From equation (2) and (3), we have \frac{-2}{x}=\frac{y-2}{x-1}

-2 x+2=x y-2 x

x y=2

y=\frac{2}{x}

From equation (1),

x^2=\frac{8}{x}

x^3=8

x=2

y=\frac{2}{x}=\frac{2}{2}=1

Now, at point (2,1), slope of the normal from equation (2)=\frac{-2}{x}=\frac{-2}{2}=-1

Equation of the normal is y-1=-1(x-2)

y-1=-x+2

or x+y=3

Question 24: The points on the curve 9 y^2=x^3= where the normal to the curve make equal intercepts with axes are:

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

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

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

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

Solution: Option (A) is correct.

Equation of the curve is 9 y^2=x^3--(i)

18 y \frac{d y}{d x}=3 x^2

\frac{d y}{d x}=\frac{3 x^2}{18 y}=\frac{x^2}{6 y}

Slope of the tangent to curve (1) at any point (x, y) is

\frac{dy}{dx} =\frac{x^2}{6y}

Slope of the normal =-\frac{1}{dy/dx}.

Slopes of lines with equal intercepts on the axis are =\pm1

So -6y = \pm x^2

if -6y = x^2

y=-\frac{x^2}{6}--(ii)

From (i) and (ii) we have

x=4 and y=-8/3

If -6y = -x^2

y =\frac{x^2}{6}--(iii)

we have, x=4 and y = 8/3

Required points are = (4, \pm 8/3)

application of derivatives multiple choice

Case study application of derivative 5
Indian Railways is the largest rail network in Asia and world’s second largest. No doubt

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