# Application of Derivatives Important Questions | Class 12 Maths

Question 1
A volleyball player serves the ball which takes a parabolic path given by the equation
$h(t)= -\frac {7}{2} t^2 + \frac {13}{2} t + 1$
, where h(t) is the height of ball at any time t (in seconds), $(t \geq 0)$.

Based on the above information, answer the following questions :
(i) Is h(t) a continuous function ? Justify.
(ii) Find the time at which the height of the ball is maximum.

(i) Since this is polynomial function, it is continous
(ii) we have
$h(t)= -\frac {7}{2} t^2 + \frac {13}{2} t + 1$
$h'(t) = -7t+ \frac {13}{2}$
For maxima
$-7t+ \frac {13}{2}=0$
or t=13/14 sec

Question 2
In order to set up a rain water harvesting system, a tank to collect rain water is to be dug. The tank should have a square base and a capacity of $250 \; m^3$. The cost of land is Rs 5,000 per square metre and cost of digging increases with depth and for the whole tank, it is Rs $40,000 h^2$, where h is the depth of the tank in metres. x is the side of the square base of the tank in metres
Based on the above information, answer the following questions :
(i) Find the total cost C of digging the tank in terms of x.
(ii) Find $\frac {dC}{dx}$
(iii)Find the value of x for which cost C is minimum
(iv)Check whether the cost function C(x) expressed in terms of x is increasing or not, where x > 0

Given
The volume of the tank = $250 \; m^3$
Now since this square base
$x^2 \times h = 250$
Solving for h:
$h = \frac {250}{ x^2}$
(i) Total cost C of digging the tank in terms of x:
Now, the cost of digging the tank increases with depth, and for the whole tank, it is $40,000 h^2$.
Cost of digging (C) = $40000 h^2$
Substituting the value of h from above
$C = 40000 \times \frac {62500}{x^4}$
$C = 25 \times 10^8 \frac {1}{x^4}$
(ii) Derivative dC/dx:
$\frac {dC}{dx} =\frac {d}{dx} (25 \times 10^8 \frac {1}{x^4})$
$= -10^10 \times \frac {1}{x^5}$
(iii) For Minima
$\frac {dC}{dx}=0$
$-10^10 \times \frac {1}{x^5}=0$
So there is no value of x that minimizes the cost. This means that the cost C keeps decreasing as x increases without bound.
So, there is no minimum cost for digging the tank; it will keep getting cheaper as the tank's side x increases.
(iv) Since derivative is negative , it is a decreasing function

Question 3
Find the maximum and minimum values of the function given by
f(x) = 5 + sin 2x

Since $-1 \geq sin 2x \geq 1$
Maximum Value is 6 and Minimum value is 4

Question 4
The median of the equilateral triangle is increasing at the rate $2 \sqrt 3 \; cm/s$ . Find the rate at which side is increasing

Median is given by
$M= \frac {\sqrt 3}{2} a$
Now
$\frac {dM}{dt} =\frac {\sqrt 3}{2} \frac {da}{dt}$
Now $\frac {dM}{dt}=2 \sqrt 3$
Therefore
$\frac {da}{dt}=4$ cm/s

Question 5
Sum of the two number is 5. if the sum of the cubes of the number is least, then find the sum of the squares

Let x and y be the two numbers
$x+ y=5$
Now
$C= x^3 + y^3$
This can be written in one variable as
$C=x^3 + (5-x)^3$
$\frac {dC}{dx} = 3x^2 - 3(5-x)^2$
Now for Minima
$\frac {dC}{dx}=0$
$3x^2 - 3(5-x)^2=0$
or x = 5/2
Now
$\frac {d^2C}{dx^2} = 6x + 6(5-x)$
At x=5/2
$\frac {d^2C}{dx^2} = 30$
Hence it is least
Now y = 5-x=5/2
Now
$x^2 + y^2 = 25/2$

Question 6
A tap is connected to such a tank whose conical part is full of water. Water is dripping out from a tap at the bottom at the uniform rate of $2 \; cm^3/s$. The semi-vertical angle of the conical tank is 45°
On the basis of given information, answer the following questions :
(i) Find the volume of water in the tank in terms of its radius r.
(ii) Find rate of change of radius at an instant when $r = 2 \sqrt 2 \; cm$.
(iii) (a) Find the rate at which the wet surface of the conical tank is decreasing at an instant when radius r = $r = 2 \sqrt 2 \; cm$.
(iii) (b) Find the rate of change of height at height is 4 cm.

Question 7
Show that the function $f(x) =\frac {16 sinx }{4+cos x} - x$ is strictly decreasing in $(\frac {\pi}{2}, \pi)$

Question 8
If the radius of a circle is increasing at a rate of 2 cm/s, how fast is the area of the circle increasing when the radius is 5 cm?

Let the radius be $r$ and the area $A$. We know $A = \pi r^2$.
Differentiating both sides with respect to time $t$, we get $\frac{dA}{dt} = 2\pi r \frac{dr}{dt}$.
Given $\frac{dr}{dt} = 2$ cm/s and $r = 5$ cm, we find
$\frac{dA}{dt} = 2\pi \times 5 \times 2 = 20\pi$ cm2/s.

Question 9
Find the maximum value of the function $f(x) = 2x^3 - 15x^2 + 36x + 1$.

First, find the derivative:
$f'(x) = 6x^2 - 30x + 36$. Set $f'(x) = 0$ to find critical points:
$6x^2 - 30x + 36 = 0$.
Solving this, $x = 2$ or $x = 3$.
To determine the maxima, check the second derivative or use the first derivative test.
Here, $f''(x) = 12x - 30$. Since $f''(2) > 0$ and $f''(3) < 0$, the function has a maximum at $x = 3$.
The maximum value is $f(3) = 2(3)^3 - 15(3)^2 + 36(3) + 1 = 28$.

Question 10
Find the equation of the tangent to the curve $y = x^2 - 4x + 3$ at $x = 2$.

First, find the derivative: $y' = 2x - 4$.
At $x = 2$, the slope of the tangent is $y'(2) = 2(2) - 4 = 0$.
The y-coordinate at this point is $y(2) = 2^2 - 4(2) + 3 = -1$.
The equation of the tangent is $y = mx + c$.
Since the slope $m = 0$ and it passes through $(2, -1)$,
$-1 = 0(2) + c$, so $c = -1$.
Thus, the equation is $y = -1$.

Question 11
Determine the intervals where the function $f(x) = x^3 - 3x^2 - 9x + 10$ is increasing or decreasing.

First, find the derivative:
$f'(x) = 3x^2 - 6x - 9$.
Set $f'(x) = 0$ to find critical points:
$3x^2 - 6x - 9 = 0$.

Solving this, we get $x = -1$ and $x = 3$.
Test intervals around these points to determine sign
The function is increasing on $(-\infty, -1)$ and $(3, \infty)$, and decreasing on $(-1, 3)$.

Question 12
Use differentials to approximate the value of $\sqrt[3]{8.1}$.

Let $y = x^{1/3}$.
At $x = 8$, $y = 2$.
The differential $dy$ is given by $dy = \frac{1}{3}x^{-2/3} dx$.
With $dx = 0.1$ (since we are approximating $\sqrt[3]{8.1}$), $dy = \frac{1}{3} \cdot 8^{-2/3} \cdot 0.1 \approx 0.0125$. Thus, $\sqrt[3]{8.1} \approx 2 + 0.0125 = 2.0125$.

Question 13
Find the area of the greatest rectangle which can be incribed in a ellipse
$\frac {x^2}{a^2} + \frac {x^2}{b^2}=1$

Let ABCD is a rectangle incribed and 2p and 2q being the side as shown below

As (p,q) lies on the ellipse
$\frac {p^2}{a^2} + \frac {q^2}{b^2}=1$
or
$q^2=b^2(1-\frac {p^2}{a^2})$
Now Area of the rectangle is given by
$A = 2p \times 2q$
$A= 4pq$
or
$A^2 = 16 p^2 q^2$
Or
$A^2= 16 p^2 b^2(1-\frac {p^2}{a^2})$
$A^2=16b^2(p^2-\frac {p^4}{a^2})$
Now
$\frac {dA^2}{dp} = 16b^2(2p-\frac {4p^3}{a^2})$
for A to be max, $A^2$ is to max
So,
$\frac {dA^2}{dp} =0$
$16b^2(2p-\frac {4p^3}{a^2})=0$
or $p^2=a^2/2$
Now
$\frac {d^2A^2}{dp^2} = 16b^2(2-\frac {12p^2}{a^2})$
Substituting the values of p
$\frac {d^2A^2}{dp^2} < 0$
Hence
Area is maximum at this value
$A=4pq$
$=4 \frac {a}{\sqrt 2} \times b^2(1-\frac {1}{2})= 2ab$

Question 14
Find the point P on the curve $y^2=4ax$ which is nearest to the point (11a,0)

Any point on the parabola is $(at^2 ,2at)$
Now distance is given by
$D^2= (at^2 -11a)^2 + (2at)^2$
$D^2 = a^2t^4 -18a^2t^2 + 121a^2$
Now D to be min, $L=D^2$ is to max
$L= a^2t^4 -18a^2t^2 + 121a^2$
$\frac {dL}{dt} = 4a^2t^3 - 36a^2 t$
For minimum of L
$\frac {dL}{dt}=0$
$4a^2t^3 - 36a^2 t=0$
or t=0,3,-3
Now $\frac {d^2L}{dt^2} = 12a^2t^2 - 36a^2$
for t=0, $\frac {d^2L}{dt^2} < 0$, hence maxima
for t=3,-3 $\frac {d^2L}{dt^2} > 0$, hence minima
Hence we have two points where the distance is minimum, the cordinates are
(9a ,6a) and (9a ,-6a)

## Multiple Choice Questions

Question 15
What is the slope of the tangent to the curve $y = x^3 - 3x + 2$ at the point where $x = 2$?
(a) 9
(b) 10
(c) 11
(d) 12

- The slope of the tangent to $y = x^3 - 3x + 2$ is given by the derivative $y' = 3x^2 - 3$.
- At $x = 2$, $y' = 3(2)^2 - 3 = 12 - 3 = 9$.

Question 16
If $y = \sin x$, then the derivative $\frac{dy}{dx}$ is:
(a) $\cos x$
(b) $-\cos x$
(c) $\sin x$
(d) $-\sin x$

The derivative of $y = \sin x$ is $\frac{dy}{dx} = \cos x$.
Answer is (a) $\cos x$

Question 17
The derivative of $f(x) = e^{2x}$ is:
(a) $2e^{2x}$
(b) $e^{2x}$
(c) $2e^x$
(d) $e^x$

The derivative of $f(x) = e^{2x}$ is obtained by using the chain rule. It is $f'(x) = 2e^{2x}$.
Answer is (a) $2e^{2x}$

Question 18
What is the equation of the normal to the curve $y = x^2$ at $x = 1$?
(a) $y + x = 0$
(b) $y - x = 0$
(c) $y - x = 2$
(d) $y + x = 2$

The slope of the normal to the curve $y = x^2$ at $x = 1$ is the negative reciprocal of the derivative at that point.
The derivative $y' = 2x$, so at $x = 1$, $y' = 2$. The slope of the normal is $-1/2$.
The equation of the normal is $y - 1 = -\frac{1}{2}(x - 1)$, which simplifies to $y + x = 2$.

Question 19
if $y = \ln(x)$, the second derivative $\frac{d^2y}{dx^2}$ is:
(a) $-\frac{1}{x^2}$
(b) $\frac{1}{x^2}$
(c) $-\frac{1}{x}$
(d) $\frac{1}{x}$

The second derivative of $y = \ln(x)$ is $\frac{d^2y}{dx^2} = -\frac{1}{x^2}$.

Question 20
The maximum value of the function $f(x) = -x^2 + 4x - 3$ is:
(a) 4
(b) -1
(c) 1
(d) -4

To find the maximum value of $f(x) = -x^2 + 4x - 3$, first find the derivative and set it to zero to find critical points.
$f'(x) = -2x + 4$. Setting $f'(x) = 0$ gives $x = 2$.
The maximum value is $f(2) = -(2)^2 + 4(2) - 3 = -4 + 8 - 3 = 1$.

Question 21
The point at which the function $f(x) = x^3 - 3x^2 + 4$ has a horizontal tangent is:
(a) $x = 0$
(b) $x = 1$
(c) $x = 2$
(d) $x = 3$

The function $f(x) = x^3 - 3x^2 + 4$ has a horizontal tangent when its derivative is zero.
$f'(x) = 3x^2 - 6x$. Setting $f'(x) = 0$ gives $x(3x - 6) = 0$. Thus, $x = 0$ or $x = 2$.

Question 22
The rate of change of the area of a circle with respect to its radius is:
(a) $2\pi r$
(b) $\pi r^2$
(c) $\pi r$
(d) $2\pi$

- The rate of change of the area $A$ of a circle with respect to its radius $r$ is given by $\frac{dA}{dr}$. Since $A = \pi r^2$, $\frac{dA}{dr} = 2\pi r$.

Question 23
The absolute minimum value of f(x)=2 sinx in $[0,\frac {3\pi}{2}]$ is
(a) 0
(b) 1
(c) -1
(d) -2

we know that
$-1 \leq sinx \leq 1$
Also sinx obtained its minimum value of -1 at $\frac {3\pi}{2}$
So absolute minimum value is -2

Question 24
The function f(x)= x -sin x decreases for
(a) all x
(b) $0 < x < \frac {\pi}{4}$
(c) $x < \frac {\pi}{2}$
(d) No value of x

f(x)= x -sin x
f'(x) =1 -cos x
for decreasing
1 -cos x < 0
or 1< cos x
Which is not possible
Hence (d)

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