CBSE Class 12 Mathematics Application of Derivatives Class Test Set 01

This class test on Chapter 6 Application of Derivatives for Class 12 Mathematics (CBSE) has 6 questions carrying 1 to 5 marks each, on increasing functions, minimum values and maxima–minima word problems with cones, cylinders and spheres. Complete worked solutions follow every question.

Class Test Questions with Solutions

Question 1. Find the intervals in which the function f given by f(x) = 2x² − 3x is strictly increasing.   [1]
Answer:
Given: f(x) = 2x² − 3x
⇒ f′(x) = 4x − 3   ……(i)
Put f′(x) = 0
⇒ 4x − 3 = 0
⇒ \( x = \frac{3}{4} \)
Therefore, we have two disjoint sub-intervals \( \left(-\infty, \frac{3}{4}\right) \) and \( \left(\frac{3}{4}, \infty\right) \)
f′(x) > 0 when \( x \in \left(\frac{3}{4}, \infty\right) \)
so f is strictly increasing when \( x \in \left(\frac{3}{4}, \infty\right) \)

Question 2. Find the minimum value of f(x), where f(x) = (2x² − 3) + 3(3 − x) + 4.   [2]
Answer:
Given f(x) = (2x² − 3) + 3(3 − x) + 4   …..(i)
f′(x) = 4x − 3
for maxima or minima f′(x) = 0
⇒ 4x − 3 = 0
⇒ \( x = \frac{3}{4} \)
f″(x) = 4
\( f''\left(\frac{3}{4}\right) > 0 \)
So, \( x = \frac{3}{4} \) is the point of minima.
Now, \( f\left(\frac{3}{4}\right) = \left(2\times\frac{9}{16} - 3\right) + 3\left(3 - \frac{3}{4}\right) + 4 \)
\( f\left(\frac{3}{4}\right) = \frac{-15}{8} + 9 - \frac{9}{4} + 4 \)
\( f\left(\frac{3}{4}\right) = \frac{-33}{8} + 13 = \frac{71}{8} \) (min. value)

Question 3. Prove that \( y = \frac{4\sin\theta}{2 + \cos\theta} - \theta \) is an increasing function of θ in \( \left[0, \frac{\pi}{2}\right] \).   [3]
Answer:
\( y = \frac{4\sin\theta}{2 + \cos\theta} - \theta \)
\( y' = \frac{4(2 + \cos\theta)\cos\theta - 4\sin\theta(-\sin\theta)}{(2 + \cos\theta)^2} - 1 \)
\( y' = \frac{4[2\cos\theta + \cos^2\theta + \sin^2\theta]}{(2 + \cos\theta)^2} - 1 \)
\( = \frac{4(1 + 2\cos\theta)}{(2 + \cos\theta)^2} - 1 = \frac{\cos\theta(4 - \cos\theta)}{(2 + \cos\theta)^2} \)   ...(1)
∵ cos θ ≥ 0 if \( \theta \in \left[0, \frac{\pi}{2}\right] \),
4 − cos θ > 0 as cos θ ∈ [−1, 1]
and (2 + cos θ)² > 0 as it is a square
∴ y′ ≥ 0 in \( \left[0, \frac{\pi}{2}\right] \)
⇒ y is an increasing function.   ...(2)

Question 4. Show that the right circular cone of least curved surface and given volume has an altitude equal to √2 times the radius of the base.   [5]
Answer:
Let r be radius, ℓ be slant height, h be height of cone of given volume V
⇒ \( V = \frac{1}{3}\pi r^2 h \Rightarrow r^2 = \frac{3V}{\pi h} \)
Let S be curved surface area
\( S = \pi r\ell = \pi r\sqrt{r^2 + h^2} \)
\( S^2 = \pi^2 r^2(r^2 + h^2) = \frac{\pi^2\, 3V}{\pi h}\left[\frac{3V}{\pi h} + h^2\right] \)   \( \left(\because\; r^2 = \frac{3V}{\pi h}\right) \)
\( S^2 = 3V\left[\frac{3V}{\pi h^2} + h\right]\pi \)
Let S² = z, then S is max. or min. according as z is max. or min.
\( z = 3V\left[\frac{3V}{\pi h^2} + h\right]\pi \)
\( \frac{dz}{dh} = 3V\left[\frac{-6V}{\pi h^3} + 1\right]\pi \)
for max. or min. \( \frac{dz}{dh} = 0 \Rightarrow V = \frac{\pi h^3}{6} \)
\( \frac{d^2z}{dh^2} = 3\pi V\left[\frac{18V}{\pi h^4}\right] > 0 \)
Hence z is minimum when \( V = \frac{\pi h^3}{6} \)
Now \( \frac{\pi h^3}{6} = \frac{\pi r^2 h}{3} \Rightarrow h^2 = 2r^2 \)
⇒ h = √2 r

Question 5. Show that semi-vertical angle of right circular cone of given surface area and maximum volume is \( \sin^{-1}\left(\frac{1}{3}\right) \).   [5]
Answer:
Let x be the radius and y be the height of the cone and semi-vertical angle be θ.
Total surface area of cone \( S = \pi x\sqrt{x^2 + y^2} + \pi x^2 \)
⇒ \( x\sqrt{x^2 + y^2} + x^2 = \frac{S}{\pi} = k \)
⇒ \( x\sqrt{x^2 + y^2} = k - x^2 \)
⇒ x²(x² + y²) = (k − x²)²
⇒ x⁴ + x²y² = k² + x⁴ − 2kx²
⇒ x²y² = k² − 2kx²
⇒ \( x^2 = \frac{k^2}{y^2 + 2k} \)   …(1)
Volume of cone \( V = \frac{1}{3}\pi x^2 y \)
\( = \frac{1}{3}\pi\left(\frac{k^2}{y^2 + 2k}\right)y \)
\( = \frac{1}{3}\pi k^2\left(\frac{y}{y^2 + 2k}\right) \)
\( \frac{dV}{dy} = \frac{1}{3}\pi k^2\,\frac{d}{dy}\left(\frac{y}{y^2 + 2k}\right) \)
\( = \frac{1}{3}\pi k^2\left[\frac{y^2 + 2k - 2y^2}{(y^2 + 2k)^2}\right] = \frac{1}{3}\pi k^2\left[\frac{2k - y^2}{(y^2 + 2k)^2}\right] \)
\( \frac{dV}{dy} = 0 \) for maxima and minima
⇒ y² = 2k ⇒ \( y = \sqrt{2k} \) (height cannot be negative)
Since \( \frac{dV}{dy} > 0 \) for \( y < \sqrt{2k} \) and \( \frac{dV}{dy} < 0 \) for \( y > \sqrt{2k} \), the volume is maximum at \( y = \sqrt{2k} \)
from equation (1) \( x^2 = \frac{k}{4} \Rightarrow x = \frac{\sqrt{k}}{2} \)
Now semi-vertical angle of cone
\( \sin\theta = \frac{x}{\sqrt{x^2 + y^2}} = \frac{\sqrt{k}}{2}\times\sqrt{\frac{4}{9k}} = \frac{1}{3} \)
\( \theta = \sin^{-1}\left(\frac{1}{3}\right) \)

Question 6. Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius R is \( \frac{2R}{\sqrt{3}} \). Also find the maximum volume.   [5]
Answer:

Given: Radius of the sphere = R
Let h be the height and x be the diameter of the base of inscribed cylinder.
Then, h² + x² = (2R)²
⇒ h² + x² = 4R²   ..... (i)
∴ Volume of the cylinder
\( V = \pi\times\left(\frac{x}{2}\right)^2\times h \)
\( = \frac{1}{4}\pi x^2 h \)   …(ii)
Putting the value of x² from (i) into (ii), we get
\( V = \frac{1}{4}\pi h(4R^2 - h^2) = \pi R^2 h - \frac{1}{4}\pi h^3 \)
On differentiating w.r.t. h, we get,
\( \frac{dV}{dh} = \pi R^2 - \frac{3}{4}\pi h^2 \)
Putting \( \frac{dV}{dh} = 0 \Rightarrow R^2 = \frac{3}{4}h^2 \)
or \( h = \frac{2R}{\sqrt{3}} \); put in equation (i): \( x^2 = 4R^2 - \frac{4R^2}{3} \)
Also, \( \frac{d^2V}{dh^2} = \frac{-3}{4}\times 2\pi h \)
At \( h = \frac{2R}{\sqrt{3}};\quad \frac{d^2V}{dh^2} = \frac{-3}{4}\times 2\pi\times\left(\frac{2R}{\sqrt{3}}\right) = -\sqrt{3}\pi R < 0 \)
⇒ V is maximum at \( h = \frac{2R}{\sqrt{3}} \)
∴ Maximum volume at \( h = \frac{2R}{\sqrt{3}} \) is:
\( V = \frac{1}{4}\pi\left(\frac{2R}{\sqrt{3}}\right)\left(4R^2 - \frac{4R^2}{3}\right) \)   \( \left(\because\; x^2 = 4R^2 - \frac{4R^2}{3}\right) \)
\( = \frac{\pi R}{2\sqrt{3}}\left(\frac{8R^2}{3}\right) \)
\( = \frac{4\pi R^3}{3\sqrt{3}} \) cubic units
Thus, the volume of the cylinder is maximum when \( h = \frac{2R}{\sqrt{3}} \).