NCERT Class 12 Mathematics — Application of Derivatives, Miscellaneous Exercise on Chapter 6. All 16 questions solved.

This exercise mixes all three strands of the chapter — rates of change, monotonicity, and optimisation — without telling you which is which. It is the best single preparation in the book for the long-answer question that Chapter 6 almost always supplies in the board paper.

One practical note before starting: NCERT’s answer key covers only questions 2, 3 and 4. Six of the remaining questions have definite numerical answers, and the book does not print them. They are worked out in full below.

Key insight. Every optimisation question here has the same shape: a constraint linking two variables, and a quantity to optimise. Use the constraint to eliminate one variable first, before differentiating anything. Question 6 (the tank) and question 8 (the window) are both two lines of calculus once that substitution is made, and unmanageable before it.

Question 1

Show that the function given by $f(x) = \dfrac{\log x}{x}$ has maximum at $x = e$.

Solution. By the quotient rule,

$$f'(x) = \frac{x \cdot \frac1x – \log x}{x^2} = \frac{1 – \log x}{x^2}$$

The denominator is positive on the domain $x > 0$, so $f'(x) = 0$ requires $\log x = 1$, that is $x = e$.

For $x < e$ we have $\log x < 1$, so $f’ > 0$; for $x > e$ we have $\log x > 1$, so $f’ < 0$. The derivative changes from positive to negative, so $x = e$ is a maximum, of value $\tfrac1e$.

Question 2

The two equal sides of an isosceles triangle with fixed base $b$ are decreasing at the rate of $3$ cm per second. How fast is the area decreasing when the two equal sides are equal to the base?

Solution. With equal sides $x$ and base $b$, the height is $\sqrt{x^2 – \tfrac{b^2}{4}}$, so

$$A = \frac12 b\sqrt{x^2 – \frac{b^2}{4}} = \frac{b}{4}\sqrt{4x^2 – b^2}$$

Differentiating with respect to time:

$$\frac{dA}{dt} = \frac{b}{4} \cdot \frac{8x}{2\sqrt{4x^2-b^2}}\frac{dx}{dt} = \frac{bx}{\sqrt{4x^2-b^2}}\frac{dx}{dt}$$

At the instant $x = b$, the root is $\sqrt{3b^2} = b\sqrt3$, and $\tfrac{dx}{dt} = -3$:

$$\frac{dA}{dt} = \frac{b \cdot b}{b\sqrt3}(-3) = -\frac{3b}{\sqrt3} = -\sqrt3\,b$$

Question 3

Find the intervals in which the function $f$ given by $f(x) = \dfrac{4\sin x – 2x – x\cos x}{2 + \cos x}$ is (i) increasing (ii) decreasing.

Solution. Split the fraction first — this is what makes the derivative manageable:

$$f(x) = \frac{4\sin x – x(2 + \cos x)}{2 + \cos x} = \frac{4\sin x}{2 + \cos x} – x$$

That is exactly the function of Exercise 6.2, question 9, whose derivative we found there:

$$f'(x) = \frac{\cos x\left(4 – \cos x\right)}{(2 + \cos x)^2}$$

The denominator is positive, and $4 – \cos x \ge 3 > 0$ always. So the sign of $f’$ is the sign of $\cos x$, which is positive on $\left[0, \tfrac{\pi}{2}\right)$ and on $\left(\tfrac{3\pi}{2}, 2\pi\right)$, and negative in between.

Question 4

Find the intervals in which the function $f$ given by $f(x) = x^3 + \dfrac{1}{x^3}$, $x \ne 0$, is (i) increasing (ii) decreasing.

Solution.

$$f'(x) = 3x^2 – \frac{3}{x^4} = \frac{3\left(x^6 – 1\right)}{x^4}$$

The denominator $x^4$ is positive wherever $f$ is defined, so the sign is that of $x^6 – 1$, which is positive exactly when $|x| > 1$.

Question 5

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

Solution. Put the vertex at $(a, 0)$. By symmetry the other two vertices are $(a\cos\theta,\ b\sin\theta)$ and $(a\cos\theta,\ -b\sin\theta)$, so the base is vertical with length $2b\sin\theta$ and the height is $a – a\cos\theta$:

$$A = \frac12(2b\sin\theta)\big(a – a\cos\theta\big) = ab\sin\theta(1 – \cos\theta)$$

Differentiating with respect to $\theta$:

$$\frac{dA}{d\theta} = ab\Big[\cos\theta(1 – \cos\theta) + \sin^2\theta\Big] = ab\left[\cos\theta – \cos^2\theta + 1 – \cos^2\theta\right]$$

$$= ab\left(-2\cos^2\theta + \cos\theta + 1\right) = -ab(2\cos\theta + 1)(\cos\theta – 1)$$

The factor $\cos\theta – 1$ vanishes at $\theta = 0$, which gives a degenerate triangle. The useful root is $\cos\theta = -\tfrac12$, that is $\theta = \tfrac{2\pi}{3}$, where $\sin\theta = \tfrac{\sqrt3}{2}$ and $1 – \cos\theta = \tfrac32$:

$$A = ab \cdot \frac{\sqrt3}{2} \cdot \frac32 = \frac{3\sqrt3}{4}ab$$

Question 6

A tank with rectangular base and rectangular sides, open at the top, is to be constructed so that its depth is $2$ m and volume is $8$ m³. If building of tank costs ₹70 per square metre for the base and ₹45 per square metre for the sides, what is the cost of the least expensive tank?

Solution. With base sides $\ell$ and $w$ and depth $2$, the volume constraint gives

$$2\ell w = 8 \quad\Longrightarrow\quad \ell w = 4$$

so the base area is fixed at $4$ m² and its cost is a constant $70 \times 4 = ₹280$. Only the sides can be optimised. Their total area is $2(\ell + w) \times 2 = 4(\ell + w)$, costing

$$C = 280 + 45 \cdot 4(\ell + w) = 280 + 180(\ell + w)$$

Substituting $w = \tfrac4\ell$:

$$C(\ell) = 280 + 180\left(\ell + \frac4\ell\right), \qquad \frac{dC}{d\ell} = 180\left(1 – \frac{4}{\ell^2}\right) = 0$$

giving $\ell = 2$ and hence $w = 2$ — a square base. Then $\tfrac{d^2C}{d\ell^2} = \tfrac{1440}{\ell^3} > 0$, confirming a minimum.

$$C = 280 + 180(4) = 280 + 720$$

Question 7

The sum of the perimeter of a circle and square is $k$, where $k$ is some constant. Prove that the sum of their areas is least when the side of the square is double the radius of the circle.

Solution. Let the circle have radius $r$ and the square side $s$. The constraint is

$$2\pi r + 4s = k \quad\Longrightarrow\quad s = \frac{k – 2\pi r}{4}$$

The total area is $A = \pi r^2 + s^2$. Differentiating with respect to $r$ and using the chain rule on $s$:

$$\frac{dA}{dr} = 2\pi r + 2s\frac{ds}{dr} = 2\pi r + 2s\left(-\frac{2\pi}{4}\right) = 2\pi r – \pi s$$

Setting this to zero gives $s = 2r$, and $\tfrac{d^2A}{dr^2} = 2\pi + \tfrac{\pi^2}{2} > 0$, so it is a minimum. Hence the side of the square equals twice the radius, as required.

Question 8

A window is in the form of a rectangle surmounted by a semicircular opening. The total perimeter of the window is $10$ m. Find the dimensions of the window to admit maximum light through the whole opening.

Solution. Let the semicircle have radius $r$, so the rectangle’s width is $2r$; let its height be $h$. The perimeter runs along the two sides, the base and the semicircular arc:

$$2h + 2r + \pi r = 10$$

“Maximum light” means maximum area:

$$A = 2rh + \frac12\pi r^2$$

From the constraint, $h = \dfrac{10 – 2r – \pi r}{2}$, so

$$A = r(10 – 2r – \pi r) + \frac{\pi r^2}{2} = 10r – 2r^2 – \frac{\pi r^2}{2}$$

$$\frac{dA}{dr} = 10 – 4r – \pi r = 0 \quad\Longrightarrow\quad r = \frac{10}{\pi + 4}$$

and $\tfrac{d^2A}{dr^2} = -4 – \pi < 0$, a maximum. Substituting back,

$$h = \frac{10 – r(2 + \pi)}{2} = \frac{10(\pi+4) – 10(2+\pi)}{2(\pi+4)} = \frac{10}{\pi + 4}$$

so the height of the rectangle equals the radius of the semicircle.

Question 9

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 $\left(a^{2/3} + b^{2/3}\right)^{3/2}$.

Solution. Let the right angle be at the origin with the legs along the axes, and let the hypotenuse make an angle $\theta$ with the horizontal leg. The given point $\mathrm{P}$ is at perpendicular distances $a$ and $b$ from the two legs.

The hypotenuse splits at $\mathrm{P}$ into two pieces whose lengths are $\dfrac{b}{\sin\theta}$ and $\dfrac{a}{\cos\theta}$, so

$$L(\theta) = a\sec\theta + b\operatorname{cosec}\theta$$

$$\frac{dL}{d\theta} = a\sec\theta\tan\theta – b\operatorname{cosec}\theta\cot\theta = 0$$

Rewriting in sines and cosines, $\dfrac{a\sin\theta}{\cos^2\theta} = \dfrac{b\cos\theta}{\sin^2\theta}$, so

$$\tan^3\theta = \frac{b}{a} \quad\Longrightarrow\quad \tan\theta = \left(\frac{b}{a}\right)^{1/3}$$

From this, $\sec\theta = \dfrac{\left(a^{2/3}+b^{2/3}\right)^{1/2}}{a^{1/3}}$ and $\operatorname{cosec}\theta = \dfrac{\left(a^{2/3}+b^{2/3}\right)^{1/2}}{b^{1/3}}$. Substituting:

$$L = \left(a^{2/3} + b^{2/3}\right)^{1/2}\left(a^{2/3} + b^{2/3}\right) = \left(a^{2/3} + b^{2/3}\right)^{3/2}$$

Question 10

Find the points at which the function $f$ given by $f(x) = (x-2)^4(x+1)^3$ has (i) local maxima (ii) local minima (iii) point of inflexion.

Solution. Differentiate as a product and factor out the common powers:

$$f'(x) = 4(x-2)^3(x+1)^3 + 3(x-2)^4(x+1)^2 = (x-2)^3(x+1)^2\big[4(x+1) + 3(x-2)\big]$$

$$= (x-2)^3(x+1)^2(7x – 2)$$

The critical points are $x = 2$, $x = -1$ and $x = \tfrac27$. The factor $(x+1)^2$ is never negative, so the sign of $f’$ is decided by $(x-2)^3(7x-2)$:

Interval	$(x-2)^3$	$(7x-2)$	$f’$
$x < \tfrac27$	$-$	$-$	$+$
$\tfrac27 < x < 2$	$-$	$+$	$-$
$x > 2$	$+$	$+$	$+$

So $f’$ changes from $+$ to $-$ at $x = \tfrac27$ (local maximum) and from $-$ to $+$ at $x = 2$ (local minimum). At $x = -1$ the derivative is zero but does not change sign, because the factor there is squared — that is a point of inflexion.

Question 11

Find the absolute maximum and minimum values of the function $f$ given by $f(x) = \cos^2 x + \sin x$, $x \in [0, \pi]$.

Solution. Replace $\cos^2 x$ by $1 – \sin^2 x$ so that the whole thing depends on $\sin x$ alone — this is cleaner than differentiating as written:

$$f(x) = 1 – \sin^2 x + \sin x$$

$$f'(x) = -2\sin x\cos x + \cos x = \cos x(1 – 2\sin x)$$

Setting this to zero: $\cos x = 0$ gives $x = \tfrac{\pi}{2}$, and $\sin x = \tfrac12$ gives $x = \tfrac{\pi}{6}$ or $\tfrac{5\pi}{6}$.

$$f(0) = 1, \quad f\!\left(\tfrac{\pi}{6}\right) = \tfrac34 + \tfrac12 = \tfrac54, \quad f\!\left(\tfrac{\pi}{2}\right) = 0 + 1 = 1$$

$$f\!\left(\tfrac{5\pi}{6}\right) = \tfrac34 + \tfrac12 = \tfrac54, \quad f(\pi) = 1 + 0 = 1$$

Question 12

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

Solution. Let the cone’s altitude, measured from its apex on the sphere, be $h$, and let its base radius be $\rho$. The base circle is a chord section of the sphere at distance $h – r$ from the centre, so

$$\rho^2 = r^2 – (h – r)^2 = 2rh – h^2 = h(2r – h)$$

$$V = \frac13\pi\rho^2 h = \frac13\pi h^2(2r – h)$$

$$\frac{dV}{dh} = \frac{\pi}{3}\left(4rh – 3h^2\right) = \frac{\pi h}{3}(4r – 3h)$$

This is zero at $h = \tfrac{4r}{3}$ (rejecting $h = 0$), and $\tfrac{d^2V}{dh^2} = \tfrac{\pi}{3}(4r – 6h)$, which at $h = \tfrac{4r}{3}$ equals $\tfrac{\pi}{3}(4r – 8r) < 0$: a maximum.

Question 13

Let $f$ be a function defined on $[a, b]$ such that $f'(x) > 0$ for all $x \in (a, b)$. Then prove that $f$ is an increasing function on $(a, b)$.

Solution. Take any $x_1 < x_2$ in $(a, b)$. The function $f$ is continuous on $[x_1, x_2]$ and differentiable on $(x_1, x_2)$, so the Mean Value Theorem gives a point $c \in (x_1, x_2)$ with

$$f(x_2) – f(x_1) = f'(c)\,(x_2 – x_1)$$

By hypothesis $f'(c) > 0$, and $x_2 – x_1 > 0$, so the right-hand side is positive. Hence $f(x_2) > f(x_1)$.

Since $x_1 < x_2$ were arbitrary, $f$ is increasing on $(a, b)$. The Mean Value Theorem is what converts a statement about the derivative at single points into a statement about the function across an interval.

Question 14

Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius $\mathrm{R}$ is $\dfrac{2\mathrm{R}}{\sqrt3}$. Also find the maximum volume.

Solution. Let the cylinder have height $h$ and base radius $\rho$. Its axis passes through the centre of the sphere, so by Pythagoras

$$\rho^2 + \frac{h^2}{4} = \mathrm{R}^2 \quad\Longrightarrow\quad \rho^2 = \mathrm{R}^2 – \frac{h^2}{4}$$

$$V = \pi\rho^2 h = \pi\left(\mathrm{R}^2 – \frac{h^2}{4}\right)h$$

$$\frac{dV}{dh} = \pi\left(\mathrm{R}^2 – \frac{3h^2}{4}\right) = 0 \quad\Longrightarrow\quad h = \frac{2\mathrm{R}}{\sqrt3}$$

and $\tfrac{d^2V}{dh^2} = -\tfrac{3\pi h}{2} < 0$, so it is a maximum. The volume is then

$$V = \pi\left(\mathrm{R}^2 – \frac{\mathrm{R}^2}{3}\right)\frac{2\mathrm{R}}{\sqrt3} = \pi \cdot \frac{2\mathrm{R}^2}{3} \cdot \frac{2\mathrm{R}}{\sqrt3} = \frac{4\pi\mathrm{R}^3}{3\sqrt3}$$

Question 15

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

Solution. The cone has base radius $h\tan\alpha$. Let the inscribed cylinder have height $y$ and radius $x$. By similar triangles, at height $y$ above the base the cone’s cross-sectional radius is $(h – y)\tan\alpha$, so

$$x = (h – y)\tan\alpha$$

$$V = \pi x^2 y = \pi\tan^2\alpha\,(h-y)^2 y$$

$$\frac{dV}{dy} = \pi\tan^2\alpha\Big[(h-y)^2 – 2y(h-y)\Big] = \pi\tan^2\alpha\,(h-y)(h – 3y)$$

Rejecting $y = h$ (which gives a degenerate cylinder of zero radius), the critical point is $y = \tfrac{h}{3}$ — one third of the cone’s height. There $\tfrac{dV}{dy}$ changes from $+$ to $-$, so it is a maximum, and

$$V = \pi\tan^2\alpha\left(\frac{2h}{3}\right)^2\frac{h}{3} = \pi\tan^2\alpha \cdot \frac{4h^2}{9} \cdot \frac{h}{3} = \frac{4}{27}\pi h^3\tan^2\alpha$$

Question 16

A cylindrical tank of radius $10$ m is being filled with wheat at the rate of $314$ cubic metre per hour. Then the depth of the wheat is increasing at the rate of

Solution. The radius is fixed, so only the depth varies:

$$V = \pi r^2 h = 100\pi h \quad\Longrightarrow\quad \frac{dV}{dt} = 100\pi\frac{dh}{dt}$$

$$\frac{dh}{dt} = \frac{314}{100\pi}$$

The number $314$ is chosen so that with $\pi \approx 3.14$ this is exactly $1$.

Common mistakes

• Differentiating before using the constraint. In questions 6, 7, 8, 12, 14 and 15 the constraint reduces two variables to one. Every one of these becomes a single-variable problem in one substitution.
• Missing that part of a cost is fixed. In question 6, the volume constraint pins the base area at $4$ m², so the base cost is a constant and only the sides can be optimised. Trying to minimise the total including a constant works, but seeing this first saves time.
• Treating a squared factor as a sign change. Question 10 turns on $(x+1)^2$ producing an inflexion, not an extremum.
• Rejecting the wrong root. In questions 5, 15 and 12 one critical point is geometrically degenerate — a triangle of zero height, a cylinder of zero radius. Say which one you are discarding and why.
• Assuming NCERT’s key covers this exercise. It prints answers for questions 2, 3 and 4 only. Do not read a missing answer as evidence that the question has none.
• Confusing the semicircle’s radius with the window’s width. In question 8 the rectangle’s width is $2r$, not $r$; the final answer for the width is $\tfrac{20}{\pi+4}$, twice the radius.

Practise next

• Exercise 6.3 — the optimisation drills that questions 5 to 15 here build on.
• Exercise 6.1 — the related-rates method used in questions 2 and 16.