NCERT Solutions for Class 12 Maths Chapter 6 - Application of Derivatives

Q. The total revenue in rupees received from the sale of x units of a product is given by R(x) = 3x2 + 36x + 5. Find the marginal revenue, when x = 5, where by marginal revenue we mean the rate of change of total revenue with respect to the number of items sold at an instant.

Ans. Since the marginal revenue is the rate of change of total revenue with respect to the number of units sold, we have $$ R(x) = 3x^2+ 36x + 5 $$ On differentialting with respect to x, we get Marginal Revenue $$(MR) = \frac{dR}{dx} = 6x + 36 \\ \text{When} \space \space x = 5, \space MR = 6(5)+ 36 =\space 66 $$ Hence, the required marginal revenue is ₹ 66.

Q. A stone is dropped into a quiet lake and wave move in a circle at a speed of 3.5 cm/sec. At the instant when the radius of the circular wave is 7.5 cm, how fast is the enclosed area increasing?

Ans.  Let r be the radius and A be the area of the circular wave at any time t. Then, area of circular wave $$ = A = \pi r^2 \space \text{and} \space \frac{dr}{dt} = 3 \cdot 5 \space cm/sec \\ \text{Now, }\space \space \space A = \pi r^2 $$ On differentiating with respect to t, we get $$ \frac{dA}{dt} =\pi\left(2r\frac{dr}{drt}\right)  \\ \Rightarrow \space \space \frac{dA}{dt} = 2\pi r \space \frac{dr}{dt} \\ \Rightarrow \frac{dA}{dt} = \space 2\pi r (3 \cdot 5)= 7\pi r \space \text{(given)} \\ \Rightarrow \left(\frac{dA}{dt}\right)_{r=7\cdot 5} = 7\pi(7\cdot 5) \\ = 52\cdot 5\pi \space cm^2 /sec.$$

Q. If the radius of a sphere is measured as 9 cm with an error of 0.03 cm,  then find the approximate error in calculating its volume.

Ans. Let r be the radius of the sphere and Δr be the error in measuring the radius. Then r = 9 cm and Δr = 0.03 cm. Now, the volume V of the sphere is given by $$\\ V = \frac{4}{3} \pi r^3 \\ or \space \space \frac{dV}{dr} = 4\pi r^2 \\ \therefore \space \space dV = \left(\frac{dV}{dr}\right) \Delta r = (4\pi r^2) \Delta r \\ = 4\pi (9)^2 (0.03) \\ = 9.72\pi \space cm^3$$

Q. Find the intervals in which the function f given by $$\textbf{f(x) = sin x + cos x, 0 ≤ x ≤ 2π}$$ is strictly increasing or strictly decreasing. 

Ans. Given, $$\text {f(x) = sin x + cos x, 0}\leq x\leq 2 \pi\\ \text {On differentiating} f(x) \text {with respect to}\space x, \text{we get} \\ \therefore f'(x)=cos \text { x – sin x }\\ = \sqrt{2}\bigg(\frac{1}{\sqrt{2}}cos \text{ x -}\frac{1}{\sqrt{2}}sin \space x \bigg)\\ =\space \sqrt{2}\bigg (sin\frac{\pi}{4} cos \text{ x -}cos\frac{\pi}{4}sin \space x \bigg)\\ = \sqrt{2}\space sin \bigg (\frac{\pi}{4} – x \bigg) \\ =-\space \sqrt{2}\space sin\bigg (x-\frac{\pi}{4}\bigg)\\ \text{For}\space f(x)\space \text{to be increasing it must be} f'(x)>0 \\ \text{Therefore, -\space} \sqrt{2}sin \bigg (x-\frac{\pi}{4}\bigg)>0 \\ \Rarr \text{sin}\bigg(x-\frac{\pi}{4}\bigg)\space <\space 0 \\ \therefore \pi<x-\frac{\pi}{4}<2\pi\\ \Rarr\frac{5\pi}{4}<x< \frac{9\pi}{4}\\ \Rarr\frac{5\pi}{4}<x< 2\pi \text{ or 2}\pi \text{< x <}\frac{9\pi}{4} \\ \Rarr\frac{5\pi}{4}<x< 2\pi \text{ or 0 } \text{<x<}\frac{\pi}{4} \\ \text{Hence, }f(x) \text{ is increasing function on } \\ 0<x< \frac{\pi}{4}\space or\space \frac{5\pi}{4}<x<2\pi \\ i.e., \bigg(0,\frac{\pi}{4}\bigg) \cup \bigg(\frac{5\pi}{4},2 \pi\bigg) \\ \text{For} f(x) \text{to be decreasing function, we must have}\\ f'(x)<0\\\Rarr \sqrt{2}sin \bigg(\frac{\pi}{4}- x\bigg)<0\\ \Rarr -\space \sqrt[]{2}sin \bigg(x-\frac{\pi}{4}\bigg) \text{\textless}\space 0 \\ \Rarr sin \bigg(x-\frac{\pi}{4}\bigg)\text{\textless}\space 0\\ \Rarr 0 \space \text{\textless}\space x\space – \frac{\pi}{4}\space\text{\textless}\space\pi \\ \Rarr \frac{\pi}{4}< \space x\space\text{\textless}\space \frac{5\pi}{4}\\ \Rarr x \space \in\space \bigg(\frac{\pi}{4},\frac{5\pi}{4}\bigg)\\ \text{Hence}\space f(x)\space\text {is decreasing on}\space\bigg(\frac{\pi}{4},\frac{5\pi}{4}\bigg). $$

Q. Find intervals in which the function given by $$\textbf{f(x) = sin 3x,\space0} ≤ \textbf{x} ≤\space\frac{\textbf{π}}{\textbf{2}}$$ is (i) increasing (ii) decreasing. 

$$\textbf{Ans. } \text{Given function is }\\ f(x)=sin \space 3x \\ \text{On differentiating f(x) w.r.t. x,we get}\\ f'(x)=3 \space cos \space 3x \\ \text{ For increasing or decreasing, it must be }f'(x)=0 \text{ we get }\\ cos\space 3x=0\\ \therefore 3x= \frac{\pi}{2}.\frac{3\pi}{2}\\ So \space x = \frac{\pi}{6}\space \text{and }\space\frac{\pi}{2}\\ \text{But it is given that}\\0\space \leq\space x \leq \frac{\pi}{2}\text{ implies }\space 0\space \leq\space 3x \leq \frac{3\pi}{2}\\ \text{The point}\frac{\pi}{6}\text{ divides the interval }\space 0\space \leq\space x \leq \frac{\pi}{2}\text{ into two disjoint intervals } \space0\space \leq\space x \leq \frac{\pi}{6}\space and \space\frac{\pi}{6}\text{\textless}\space x \space \text{\textless} \frac{\pi}{2}.\\ \text{When} \space 0\space \leq x \space \text{\textless}\frac{\pi}{6}\\f'(x)=3\space cos \space 3x \\ \text{Also}\space 0\space \leq 3x \space \text{\textless}\frac{3\pi}{6}\\0\space \leq3x\space \text{\textless} \frac{\pi}{2}\\ \Rarr f'(x)\text{\textgreater}\space 0\space \bigg(As \space cos \space \theta \space\text{\textgreater} \space 0\space for \space \theta\space \text{\textless} \frac{\pi}{2}\bigg)\\ \text{When} \space \frac{\pi}{6}\space\text{\textless}\space x \space \leq \space \frac{\pi}{2}\\ f'(x)= 3 \space cos \space 3x\\ \text{Also } \space\space \frac{3\pi}{6} \space \text{\textless}\space 3x \space \leq \frac{3\pi}{2}\\ \frac{\pi}{2}<3x\leq \frac{3\pi}{2} \\ \Rarr \space \space f'(x) \text{\textless}\space 0 \\ As \space \space cos \space \theta \space \text{\textless} \space 0 \space for \space \frac{\pi}{2} \text{\textless}\space \theta \space \leq \frac{3\pi}{2}\\ \text{Therefore,} f(x)\text{ is strictly increasing in 0 }\leq x \space \text{\textless} \frac{\pi}{6}\text{ and strictly decreasing in }\frac{\pi}{6}\space \text{\textless}\space x \space \leq \space \frac{\pi}{2}.\\ \text{ Also, the given function is continuous at }\space x =0 \space and \space x=\frac{\pi}{6}.\\ \text{ Therefore, we know that}\space f(x) \text{ is increasing on 0 }\leq\space x \leq \space \frac{\pi}{6}\text{ and decreasing on }\frac{\pi}{6}\leq \space x \space \leq\space \frac{\pi}{2}.$$

Q. Find the intervals in which $$\textbf{f(x) = (x + 1)}^\textbf{3}\textbf{(x – 3)}^\textbf{3}$$ is increasing or decreasing.

$$\textbf{Ans. } \text{Given function is}\\ f(x)=(x+1)^3(x-3)^3\\ \text{On differentiating}f(x) \space \text{with respect to} \space x, \text{we get}\\ \Rarr f'(x) = \begin{rcases}\begin{cases}3(x+1)^2\frac{d}{dx}(x+1)\end{cases}\end{rcases}(x-3)^3 +(x+1)^3.\begin{cases}3(x-3)^2\space \frac{d}{dx}(x-3)\end{cases}\begin{rcases}\text{} \end{rcases}\\ \Rarr f'(x)=3(x+1)^2\space (x-3)^3+3(x+1)^3\space (x-3)^2\\ \Rarr f'(x)=3(x+1)^2\space (x-3)^2[x+1+x-3] \\ \Rarr f'(x)=6(x+1)^2\space (x-3)^2(x-1)\\ \text{For increasing value of }f(x)\text{we must have}\\ f'(x)\text{\textgreater} \space 0 \\ \text{Therefore, }\space 6(x+1)^2(x-3)^2\space (x-1)\space \text{\textgreater}\space 0 \\ \Rarr x-1 \space \text{\textgreater}\space 0 \space and \space x \space \not =-1,3 \\ \Rarr x \space \text{\textgreater} \space 1\space and \space x \space \not =-1,3 \\ \Rarr x \space \in \space (1,3)\space \cup\space (3,\space \infty) \\ So, f(x)\text{is increasing in (1,3)}\cup\space (3, \space \infty).\\ \text{For} f(x)\space \text{to be decreasing, we must have}\\ f'(x)\space\text{\textless}\space 0 \\ \Rarr 6(x+1)^2(x-3)^2(x-1)\space \text{\textless} \space 0 \\ \Rarr x-1 \space \text{\textless} \space 0 \space and \space x \not=-1,3 \\ \Rarr x \space \text{\textless} \space 1 \space and \space x \space \not=-1,3\\ \Rarr \space x\space \in \space (- \space \infty,-1)\space \cup \space (-1,1)\\ So, f(x)\space \text{is decreasing in }\space (-\space \infty,1)\space \cup \space (-1,1).$$

Q. Show that the volume of the greatest right circular cylinder that can be inscribed in a right circular cone of height h and semi-vertical angle $$\textbf{α is}\space\frac{\textbf{4}}{\textbf{27}}\space\textbf{πh}^{\textbf{3}}\textbf{tan}^\textbf{2}\textbf{α.}\\\textbf{Also show that height of the cylinder is}\space\frac{\textbf{h}}{\textbf{3}}.$$

$$\textbf{Ans. } \text{Let r = radius of the cone, h = height of the cone and R = radius of the cylinder, H = height of the cylinder.}\\ \text{Let α be the semi-vertical angle of the cone.} \\ \text{Now, } ΔAEF \sim~ ΔADC \\ \text{ Then by property of similarity,} \\ \frac{AE}{AD} = \frac{EF}{DC}\space \\ or\space \space \frac{h-H}{h} = \frac{R}{r} \\ \text{Also, in ΔADC} \\ tan\space \alpha = \frac{r}{h} \\ or \space r = h \space tan\space \alpha$$


Q. Find the maximum area of the greatest isosceles triangle that can be inscribed in the ellipse $$\frac{\textbf{x}^\textbf{2}}{\textbf{a}^\textbf{2}} + \frac{\textbf{y}^\textbf{2}}{\textbf{b}^\textbf{2}} =\textbf{1} $$ with its vertex at one end of major axis. 

$$\textbf{Ans. } \text{Given equation of the ellipse is} \\ \frac{x^2}{a^2}+ \frac{y^2}{b^2} = 1 \text{Let ABC be an isosceles triangle having the vertex} \text{at A(a, 0), B(a cos θ, b sin θ) and C(a cos θ, – b sin θ).}\\ \therefore \space Area \space of \Delta ABC = \frac{1}{2} × BC × AM \\ = \frac{1}{2}(2bsin\space \theta)(a-acos\space \theta) \\ =\frac{1}{2}(2ab\space sin\space \theta)(1-cos\space \theta) \\ \Rightarrow \space \space A = ab\space (sin\space \theta – sin\space \theta \space cos\space \theta)$$


$$\text{Differentiate A with respect to θ, we get} \\ \frac{dA}{d\theta} = ab(cos \space \theta -cos^2 \space \theta + sin^2\space \theta) \\ = ab(cos\space \theta – cos\space 2\theta)\space\space …(i) \\ \text{For maximum or minimum value of A it must be} \\ \frac{dA}{d\theta} = 0 \\ \therefore \space ab(cos\space \theta – cos \space 2\theta) = 0 \\ \Rightarrow \space \space cos\space \theta = cos\space 2 \theta \\ \Rightarrow \space \space \theta = \frac{2\pi}{3} \\ \text{Again, differentiate (i) with respect to θ, we get} \\ \frac{d^2A}{d\theta ^2} = ab(-sin\space \theta + 2\space sin\space 2\theta) \\ \frac{d^2A}{d\theta ^2}\Bigm\vert_{\theta=\frac{2\pi}{3}} = ab\left(-sin\frac{2}{3}\pi + 2sin \frac{4}{3}\pi\right) \\ =ab\left(\frac{-\sqrt{3}}{2}-\frac{2\sqrt{3}}{2}\right)<0 \\ \text{Hence A has maximum area when } \theta = \frac{2\pi}{3}. \text{The maximum area A is given by} \\ A = ab\left(sin\frac{2\pi}{3}-sin\frac{2\pi}{3}cos\frac{2\pi}{3}\right) \\ = ab\left(\frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2}×\frac{1}{2}\right) = \frac{3\sqrt{3}}{4}ab$$

Q. The volume of spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units, find the radius of the balloon after t seconds.

$$\textbf{Ans. } \text{Let the radius of the balloon of volume V at time t seconds be r, then} \\ \frac{d}{dt}(V) = \text{constant} \\ \Rightarrow \space \space \frac{d}{dt}\left(\frac{4}{3}\pi r^3 \right) = \text{constant} \\ \Rightarrow \frac{4}{3}\pi 3 r^2. \frac{dr}{dt} = k \text{(say)} \\ \Rightarrow 4\pi r^2 \frac{dr}{dt} = k \\ \Rightarrow 4\pi r^2 dr = k \space dt\space \space \space \space …(i)\\ \text{Integrating both sides} \\ \frac{4\pi r^3}{3} =kt + c \space \space \space …(ii) \\ \text{When t = 0, r = 3 } \\ c = 4 ×\frac{27 \pi}{3} \\ c = 36\space \pi \space \space …(iii) \\ \text{Put in equation (ii),} \\ \frac{4 \pi r^3}{3} = kt+\space 36\pi \space \space …(iv) \\ \text{Also when t = 3, r = 6} \\ \Rightarrow \frac{4\pi(216)}{3} = 3k + 36\pi \\ \Rightarrow \space 3k= 288\pi – 36\pi \\ \Rightarrow 3k = 252\pi \\ \Rightarrow \space \space k = 84\pi \\ \text{Equation (iv) becomes} \\ \frac{4}{3}\pi r^3 = 84\pi t + 36 \pi \\ \Rightarrow \frac{4}{3}\pi r^3 = 12\pi(7t + 3) \\ \Rightarrow r^3 = \frac{12\pi×3(7t+3)}{4\pi} \\ \Rightarrow \space r^3 = 9(7t+3) \\ \Rightarrow \space r = \text{\{9(7t+3)}\}^{1/3}$$

Q. A point on the hypotenuse of a right triangle is at distances a and b from the sides of the triangle. Show that the minimum length of the hypotenuse is (a2/3 + b2/3)3/2 .

$$\textbf{Ans. } \text{Let ABC be a right angled triangle, Point P on AB as} \\ PL = a\space \text{and} \space PM =b \\ Let \space \space \angle BAC = \theta \\ \text{In ΔΑPL, AP = a cosec θ and in ΔBMP, BP = b sec θ …(i)} $$


$$\text{Let l be the length of the hypotenuse AB. Then} \\ \text{l = AP + BP} \\ \Rightarrow \space \space l = a\space cosec\space \theta + b \space sec\space \theta \space \space \text{[from\space (i)]}\\ \text{On differentiating l w.r.t. θ, we get} \\ \frac{dl}{d\theta} = -a\space cosec\space \theta cot\space \theta + b\space sec\space \theta\space tan\space \theta \\ \text{Again,\space differentiate } \frac{dl}{d\theta}\space \text{w.r.t.\space} \theta, \text{we\space get} \\ \frac{d^2 l}{d\theta^2} = a\space cosec^3\theta + a\space cosec\space\theta\space cot^2\space \theta + b\space sec^3 \theta + b\space sec\space \theta \space tan^2\space \theta \\ = a\space cosec\space \theta (cosec^2 \space \theta + cot^2\space\theta)+ b\space sec\space \theta(sec^2\space \theta + tan^2\space \theta) \\ = a\space cosec \space \theta (1+2\space cot^2\theta) + b\space sec\space \theta (1+2\space tan^2\space \theta)\\ \text{For maximum or minimum it must have } \frac{dl}{d\theta}=0, \\ \text{we get }\\ -a\space cosec\space \theta \space cot\space \theta + b\space sec\space \theta \space tan \space \theta = 0 \\ \Rightarrow \space -\frac{acos\space \theta}{sin^2\theta}+ \frac{bsin\space \theta}{cos^2\space \theta} = 0\\ \Rightarrow \space \space \space tan^3 \theta = \frac{a}{b} \\ \Rightarrow \space \space tan\space \theta = \left(\frac{a}{b}\right)^{1/3} \\ \text{We can write θ as sin θ =} \frac{a^{1/3}}{\sqrt{a^{2/3}+b^{2/3}}}\\ \\ and \space \space cos\space \theta = \frac{b^{1/3}}{\sqrt{a^{2/3}+b^{2/3}}}\\ \text{Since θ is acute angle, we have } \frac{d^2l}{d\theta^2}>0 \\ \text{For} \space \space \space tan\space \theta = \space \left(\frac{a}{b}\right)^{1/3} \\ \text{Thus, } l \text{ is minimum when tan θ } = \left(\frac{a}{b}\right)^{1/3} \\ \text{The minimum value of }  \text{ is given by} \\ \text{ = a cosec θ + b sec θ} \\ = a\sqrt{1+cot^2 \space \theta} + b\sqrt{1+tan^2\space \theta} \\ \Rightarrow \space \space l = \space a\sqrt{1+\left(\frac{b}{a}\right)^{2/3}} + b\sqrt{1+\left(\frac{a}{b}\right)^{2/3}} \\ \Rightarrow \space \space l = (a^{2/3} + b^{2/3})^{3/2}.$$