Differential Calculus MCQ Level - 2 - IIT JAM MCQ

= x²(log 1 – log x)
= –x²  log x

or x = e^–1/2

Hence, y is maximum at x = 1/√e

The correct answer is: 1/√e

Turning point is (1, 1) as on one side of (1, 1), the curve is decreasing while the other, is increasing.

The correct answer is: (1,1)

Correct Answer :- c

Explanation : If f'(x) = f''(x) = 0,

but we don't know the state of higher derivative. Therefore the answer is cant say.

As we can see that the denominator in f'(x) will always be positive.

Also, we notice that  i.e., f is a decreasing function and  i.e., f is an increasing function.

Hence, we conclude that there must be minima at x = 3.

The correct answer is: minimum

Let  f(x) = sin x(1 + cosx)

Now, f''(x) = –sin x – 2 sin 2x

= –1 < 0
Hence, f(x) will be maximum at x = π/3.

The correct answer is: maximum

Correct Answer :- d

Explanation : Let f(x) = (x – 1)(x – 2)2

f(x)'= (x – 2)2 + 2(x – 1)(x – 2)

= (x – 2)(3x – 4)

Putting f'(x) = (x – 2)(3x – 4) = 0

We get, x = 4/3

f(x) = x³ – 3x + 2
Putting f'(x) = 3x² – 3

We get   as the critical or turning points.
The correct answer is: –1 and +1

f(x) = x³ – 3x² + 1
⇒ 
Putting f'(x) = 0,
i.e., 3x² – 6x = 0
3x(x – 2) = 0
⇒ x = 0, 2
Now, f''(x) = 6x – 6
f''(0) = –6 < 0
and f''(0) = 6 > 0

f(4) = 17
f(0) = 1
f(2) = –3
Hence, absolute maxima will be 17 (at x = 4)
and hence absolute minima will be –3 (at x = 2).

The correct answer is: 17, –3

f(x)=x^3/5(4 – x)
= 4x^3/5 + x^8/5

Putting f'(x) = 0,

Now, either, 3 – 2x = 0  ⇒ x = 3/2 is the critical point or x = 0 (at x = 0, f'(x) doesn’t exist)

The correct answer is:  3/2 and 0

Let the two numbers be x and y
⇒ x + y = a
or      x = a – y
We need to maximise, z = xy³
or  z = (a – y)y³
= ay³ – y⁴

Putting dz/dy = 0, i.e
3ay² – 4y³ = 0
or y(3a – 4y) = 0

The correct answer is: