# Mathematics Questions and Answers – Calculus Application – Maxima and Minima – 2

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This set of Mathematics Aptitude Test for Engineering Entrance Exams focuses on “Calculus Application – Maxima and Minima – 2”.

1. What will be the nature of the equation sin(x + α)/sin(x + β)?
a) Possess only minimum value
b) Possess only maximum value
c) Does not possess a maximum or minimum value

Explanation: Let, y = sin(x + α)/sin(x + β)
Then,
dy/dx = [cos(x + α)sin(x + β) – sin(x + α)cos(x + β)]/sin2(x + β)
= sin(x+β – x-α)/sin2(x + β)
Or sin(β – α)/sin2(x + β)
So, for minimum or maximum value of x we have,
dy/dx = 0
Or sin(β – α)/sin2(x + β) = 0
Or sin(β – α) = 0 ……….(1)
Clearly, equation (1) is independent of x; hence, we cannot have a real value of x as root of equation (1).
Therefore, y has neither a maximum or minimum value.

2. Given, f(x) = x3 – 12x2 + 45x + 8. At which point does f(x) has its maximum?
a) 1
b) 2
c) 3
d) 4

Explanation: We have, f(x) = x3 – 12x2 + 45x + 8 ……….(1)
Differentiating both sides of (1) with respect to x we
f’(x) = 3x2 – 24x + 45
3x2 – 24x + 45 = 0
Or x2 – 8x + 15 = 0
Or (x – 3)(x – 5) = 0
Thus, either x – 3 = 0 i.e., x = 3 or x – 5 = 0 i.e., x = 5
Therefore, f’(x) = 0 for x = 3 and x = 5.
If h be a positive quantity, however small, then,
f’(3 – h) = 3*(3 – h – 3)(3 – h – 5) = 3h(h + 2) = positive.
f’(3 + h) = 3*(3 + h – 3)(3 + h – 5) = 3h(h – 2) = negative.
Clearly, f’(x) changes sign from positive on the left to negative on the right of the point x = 3.
So, f(x) has maximum at 3.

3. Given, f(x) = x3 – 12x2 + 45x + 8. What is the maximum value of f(x)?
a) 61
b) 62
c) 63
d) 54

Explanation: We have, f(x) = x3 – 12x2 + 45x + 8 ……….(1)
Differentiating both sides of (1) with respect to x we
f’(x) = 3x2 – 24x + 45
3x2 – 24x + 45 = 0
Or x2 – 8x + 15 = 0
Or (x – 3)(x – 5) = 0
Thus, either x – 3 = 0 i.e., x = 3 or x – 5 = 0 i.e., x = 5
Therefore, f’(x) = 0 for x = 3 and x = 5.
If h be a positive quantity, however small, then,
f’(3 – h) = 3*(3 – h – 3)(3 – h – 5) = 3h(h + 2) = positive.
f’(3 + h) = 3*(3 + h – 3)(3 + h – 5) = 3h(h – 2) = negative.
Clearly, f’(x) changes sign from positive on the left to negative on the right of the point x = 3.
So, f(x) has maximum at 3.
Putting, x = 3 in (1)
Thus, its maximum value is,
f(3) = 33 – 12*32 + 45*3 + 8 = 62.
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4. Given, f(x) = x3 – 12x2 + 45x + 8. At which point does f(x) has its minimum?
a) 1
b) 7
c) 3
d) 5

Explanation: We have, f(x) = x3 – 12x2 + 45x + 8 ……….(1)
Differentiating both sides of (1) with respect to x we
f’(x) = 3x2 – 24x + 45
3x2 – 24x + 45 = 0
Or x2 – 8x + 15 = 0
Or (x – 3)(x – 5) = 0
Thus, either x – 3 = 0 i.e., x = 3 or x – 5 = 0 i.e., x = 5
Therefore, f’(x) = 0 for x = 3 and x = 5.
If h be a positive quantity, however small, then,
f’(5 – h) = 3*(5 – h – 3)(5 – h – 5) = -3h(2 – h) = negative.
f’(5 + h) = 3*(5 + h – 3)(5 + h – 5) = 3h(2 + h) = positive.
Clearly, f’(x) changes sign from negative on the left to positive on the right of the point x = 5.
So, f(x) has minimum at 5.

5. Given, f(x) = x3 – 12x2 + 45x + 8. What is the minimum value of f(x)?
a) -1
b) 0
c) 1
d) Value does not exist

Explanation: We have, f(x) = x3 – 12x2 + 45x + 8 ……….(1)
Differentiating both sides of (1) with respect to x we
f’(x) = 3x2 – 24x + 45
3x2 – 24x + 45 = 0
Or x2 – 8x + 15 = 0
Or(x – 3)(x – 5) = 0
Thus, either x – 3 = 0 i.e., x = 3 or x – 5 = 0 i.e., x = 5
Therefore, f’(x) = 0 for x = 3 and x = 5.
If h be a positive quantity, however small, then,
f’(5 – h) = 3*(5 – h – 3)(5 – h – 5) = -3h(2 – h) = negative.
f’(5 + h) = 3*(5 + h – 3)(5 + h – 5) = 3h(2 + h) = positive.
Clearly, f’(x) changes sign from negative on the left to positive on the right of the point x = 5.
So, f(x) has minimum at 5.
Putting, x = 5 in (1)
Thus, its maximum value is,
f(3) = 53 – 12*52 + 45*5 + 8 = 58.

6. At which point does f(x) = |x – 1| has itslocal minimum?
a) They are unequal
b) They are equal
c) Depend on the numbers
d) Can’t be predicted

Explanation: The given function is f(x) = ∣x − 1∣, x ∈ R.
It is known that a function f is differentiable at point x = c in its domain if both
$$\lim\limits_{h \rightarrow 0^-}$$ hf(c + h) – f(c)
And
$$\lim\limits_{h \rightarrow 0^+}$$ hf(c + h) – f(c) are finite and equal.
To check the differentiability of the function at x = 1,
LHS,
Consider the left hand limit of f at x=1
$$\lim\limits_{h \rightarrow 0^-}\frac{|1+h-1|-|1-1|}{h}$$
= $$\lim\limits_{h \rightarrow 0^-}\frac{|h|}{h}$$
= $$\lim\limits_{h \rightarrow 0^-}\frac{-h}{h}$$
= −1
RHS,
Consider the right hand limit of f at x − 1
$$\lim\limits_{h \rightarrow 0^+}\frac{|1+h-1|-|1-1|}{h}$$
= $$\lim\limits_{h \rightarrow 0^+}\frac{|h|}{h}$$
= 1
Since the left and right hand limits of f at x = 1 are not equal, f is not differentiable at x = 1.
As, LHS = -1 and RHS = 1, it is clear that, f’(1) < 0 on the left of x = 1 and f’(x) > 0 on the right of the point x = 1.
Hence, f’(x) changes sign, from negative on the left to positive on the right of the point x = 1.
Therefore, f(x) has a local minima at x = 1.

7. A particle is moving in a straight line and its distance x from a fixed point on the line at any time t seconds is given by, x = t4/12 – 2t3/3 + 3t2/2 + t + 15. At what time is the velocity minimum?
a) 1
b) 2
c) 3
d) 4

Explanation: Assume that the velocity of the particle at time t second is vcm/sec.
Then, v = dx/dt = 4t3/12 – 6t2/3 + 6t/2 + 1
So, v = dx/dt = t3/3 – 2t2/ + 3t + 1
Thus, dv/dt = t2 – 4t + 3
And d2v/dt2 = 2t – 4
For maximum and minimum value of v we have,
dv/dt = 0
Or t2 – 4t + 3 = 0
Or (t – 1)(t – 3) = 0
Thus, t – 1 = 0 i.e., t = 1 Or t – 3 = 0 i.e., t = 3
Now, [d2v/dt2]t = 3 = 2*3 – 4 = 2 > 0
Thus, v is minimum at t = 3.

8. A particle is moving in a straight line and its distance x from a fixed point on the line at any time t seconds is given by, x = t4/12 – 2t3/3 + 3t2/2 + t + 15. What is the minimum velocity?
a) 1 cm/sec
b) 2 cm/sec
c) 3 cm/sec
d) 4 cm/sec

Explanation: Assume that the velocity of the particle at time t second is vcm/sec.
Then, v = dx/dt = 4t3/12 – 6t2/3 + 6t/2 + 1
So, v = dx/dt = t3/3 – 2t2/ + 3t + 1
Thus, dv/dt = t2 – 4t + 3
And d2v/dt2 = 2t – 4
For maximum and minimum value of v we have,
dv/dt = 0
Or t2 – 4t + 3 = 0
Or (t – 1)(t – 3) = 0
Thus, t – 1 = 0 i.e., t = 1 Or t – 3 = 0 i.e., t = 3
Now, [d2v/dt2]t = 3 = 2*3 – 4 = 2 > 0
Thus, v is minimum at t = 3.
Putting t = 3 in (1) we get,
33/3 – 2(3)2/ + 3(3) + 1
= 1 cm/sec.

Sanfoundry Global Education & Learning Series – Mathematics – Class 12. 