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

d) Data inadequate

View Answer

Explanation: Let, y = sin(x + α)/sin(x + β)

Then,

dy/dx = [cos(x + α)sin(x + β) – sin(x + α)cos(x + β)]/sin

^{2}(x + β)

= sin(x+β – x-α)/sin

^{2}(x + β)

Or sin(β – α)/sin

^{2}(x + β)

So, for minimum or maximum value of x we have,

dy/dx = 0

Or sin(β – α)/sin

^{2}(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) = x^{3} – 12x^{2} + 45x + 8. At which point does f(x) has its maximum?

a) 1

b) 2

c) 3

d) 4

View Answer

Explanation: We have, f(x) = x

^{3}– 12x

^{2}+ 45x + 8 ……….(1)

Differentiating both sides of (1) with respect to x we

f’(x) = 3x

^{2}– 24x + 45

3x

^{2}– 24x + 45 = 0

Or x

^{2}– 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) = x^{3} – 12x^{2} + 45x + 8. What is the maximum value of f(x)?

a) 61

b) 62

c) 63

d) 54

View Answer

Explanation: We have, f(x) = x

^{3}– 12x

^{2}+ 45x + 8 ……….(1)

Differentiating both sides of (1) with respect to x we

f’(x) = 3x

^{2}– 24x + 45

3x

^{2}– 24x + 45 = 0

Or x

^{2}– 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) = 3

^{3}– 12*3

^{2}+ 45*3 + 8 = 62.

4. Given, f(x) = x^{3} – 12x^{2} + 45x + 8. At which point does f(x) has its minimum?

a) 1

b) 7

c) 3

d) 5

View Answer

Explanation: We have, f(x) = x

^{3}– 12x

^{2}+ 45x + 8 ……….(1)

Differentiating both sides of (1) with respect to x we

f’(x) = 3x

^{2}– 24x + 45

3x

^{2}– 24x + 45 = 0

Or x

^{2}– 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) = x^{3} – 12x^{2} + 45x + 8. What is the minimum value of f(x)?

a) -1

b) 0

c) 1

d) Value does not exist

View Answer

Explanation: We have, f(x) = x

^{3}– 12x

^{2}+ 45x + 8 ……….(1)

Differentiating both sides of (1) with respect to x we

f’(x) = 3x

^{2}– 24x + 45

3x

^{2}– 24x + 45 = 0

Or x

^{2}– 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) = 5

^{3}– 12*5

^{2}+ 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

View Answer

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 = t^{4}/12 – 2t^{3}/3 + 3t^{2}/2 + t + 15. At what time is the velocity minimum?

a) 1

b) 2

c) 3

d) 4

View Answer

Explanation: Assume that the velocity of the particle at time t second is vcm/sec.

Then, v = dx/dt = 4t

^{3}/12 – 6t

^{2}/3 + 6t/2 + 1

So, v = dx/dt = t

^{3}/3 – 2t

^{2}/ + 3t + 1

Thus, dv/dt = t

^{2}– 4t + 3

And d

^{2}v/dt

^{2}= 2t – 4

For maximum and minimum value of v we have,

dv/dt = 0

Or t

^{2}– 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, [d

^{2}v/dt

^{2}]

_{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 = t^{4}/12 – 2t^{3}/3 + 3t^{2}/2 + t + 15. What is the minimum velocity?

a) 1 cm/sec

b) 2 cm/sec

c) 3 cm/sec

d) 4 cm/sec

View Answer

Explanation: Assume that the velocity of the particle at time t second is vcm/sec.

Then, v = dx/dt = 4t

^{3}/12 – 6t

^{2}/3 + 6t/2 + 1

So, v = dx/dt = t

^{3}/3 – 2t

^{2}/ + 3t + 1

Thus, dv/dt = t

^{2}– 4t + 3

And d

^{2}v/dt

^{2}= 2t – 4

For maximum and minimum value of v we have,

dv/dt = 0

Or t

^{2}– 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, [d

^{2}v/dt

^{2}]

_{t = 3}= 2*3 – 4 = 2 > 0

Thus, v is minimum at t = 3.

Putting t = 3 in (1) we get,

3

^{3}/3 – 2(3)

^{2}/ + 3(3) + 1

= 1 cm/sec.

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