Mathematics MCQ - Calculus

Calculus GATE Syllabus

Calculus: Limits, continuity and differentiability. Maxima and minima. Mean value theorem. Integration

1 If

A
2x²
B
√x
C
0
D
1

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Comment

2 The function f(x) = x³ - 6x² + 9x + 25 has

A
a maxima at x= 1 and a minima at x = 3
B
a maxima at x = 3 and a minima at x = 1
C
no maxima, but a minima at x = 1
D
a maxima at x = 1, but no minima

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Comment

f '(x) = 3(x²-4x+3) f ''(x)=6(x)-12 at x=1, f "(1) at x=3, f ''(3) = +ve

3 The value of a =

A
>0
B
2
C
0 - 1 + 100 - 10 + 1
D
undefined

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Comment

4 The interval in which the Lagrange's theorem is applicable for the function f(x) = 1/x is

A
[-3, 3]
B
[-2, 2]
C
[2, 3]
D
[-1, 1]

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Comment

Since f(x) = (1/x) is not continuous in [- 3, 3] [- 4, 2] or [- 1, 1], The point of discontinuity is '0'. Only in [2, 3] the function is continuous, and differentiable hence mean value theorem is applicable in [2, 3].

5 If f(x) = | x | , then for interval [-1, 1] ,f(x)

A
satisfied all the conditions of Rolle's Theorem
B
satisfied all the conditions of Mean Value Theorem
C
does not satisfied the -conditions of Mean Value Theorem
D
None of these

View Explanation
Comment

Since f(x) = |x| is continuous in [-1, 1] but it is not differentiable at x = 0 ε (-1, 1)

Mathematics MCQ - Calculus

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