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# Test: Continuity And Differentiability With Solution (Competition Level) - 2

## 30 Questions MCQ Test Mathematics (Maths) Class 12 | Test: Continuity And Differentiability With Solution (Competition Level) - 2

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This mock test of Test: Continuity And Differentiability With Solution (Competition Level) - 2 for JEE helps you for every JEE entrance exam. This contains 30 Multiple Choice Questions for JEE Test: Continuity And Differentiability With Solution (Competition Level) - 2 (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Continuity And Differentiability With Solution (Competition Level) - 2 quiz give you a good mix of easy questions and tough questions. JEE students definitely take this Test: Continuity And Differentiability With Solution (Competition Level) - 2 exercise for a better result in the exam. You can find other Test: Continuity And Differentiability With Solution (Competition Level) - 2 extra questions, long questions & short questions for JEE on EduRev as well by searching above.
QUESTION: 1

Solution:

QUESTION: 2

Solution:

QUESTION: 3

### is continuous at x = 0 then

Solution:

QUESTION: 4

The function  is discontinuous at the points

Solution:

f (x) is discontinuous when x2 - 3|x| + 2 = 0
⇒ |x|2 - 3|x| + 2 = 0 ⇒ |x| = 1, 2

QUESTION: 5

The values of a and b if f is continuous at x = 0, where

Solution:

QUESTION: 6

is continuous at then k =

Solution:

QUESTION: 7

so that f(x) is continuous at then

Solution:

By L-Hospital rule

QUESTION: 8

where [.] denotes greatest integer function and the function is continuous then

Solution:

QUESTION: 9

is continuous everywhere. Then the  equation whose roots are a and b is

Solution:

QUESTION: 10

where [x] is the greatest integer function. The function f (x) is

Solution:

QUESTION: 11

The function  is continuous at exactly two  points then the possible values of ' a ' are

Solution:

f (x) is continuous when x2 - ax + 3 =2 - x
⇒ x2 - a -1 x + 1 = 0. This must have two distinct roots ⇒ Δ > 0 ⇒ (a -1)2 - 4 > 0

QUESTION: 12

If the function  is continuous for every x ∈ R then

Solution:

x2 + kx + 1>0 and x2 - k must not have any real root ;
∴ k2 - 4 < 0 &k < 0
⇒ k ∈ [-2, 2] and k < 0 ⇒ k ∈ [-2, 0)

QUESTION: 13

Solution:

|x| is not differentiable at x = 0
|x| is continuous at x = 0

QUESTION: 14

The function f (x) = cos-1 (cos x) is

Solution:

f (x) is continuous at x = π, - π

QUESTION: 15

Solution:

QUESTION: 16

then which is correct

Solution:

QUESTION: 17

Let f (x) = |x - 1| + |x + 1|

Solution:

QUESTION: 18

Solution:

Since g(x) = |x| is a continuous function and   so f is continuous function. In particular f is continuous at a = 1 and x = 4) f is clearly not differentiable at x = 4) Since g(x) = |x| is not differentiable at x = 0. Now

QUESTION: 19

The set of all points where the function  is differentiable is

Solution:

is not differentiable only at x = 0

QUESTION: 20

If   then derivative of f(x) at x = 0 is

Solution:

QUESTION: 21

If f : R → R be a differentiable function, such that f (x + 2y) = f (x) + f (2y) + 4xy for all x, y ∈ R then

Solution:

f (x + 2y) = f (x) + f (2 y) + 4xy for x, y ∈ R putting x = y = 0, we get f (0) = 0

QUESTION: 22

Let f be a differentiable function satisfying the condition for all
, then f ' (x) is equal to

Solution:

replacing x and y both by 1, we get

QUESTION: 23

The function is not differentiable at

Solution:

By verification f ' (2 -) ≠ f ' (2 +)
∴ f(x) is not differentiable at x = 2

QUESTION: 24

then set of all points where f is differentiable is

Solution:

The function is clearly differentiable except possible at x = 2, 3

QUESTION: 25

Solution:
QUESTION: 26

Let h(x) = min {x, x2} for  Then which of the following is correct

Solution:

From the graph it is clear that h is continuous. Also h is differentiable except possible at x = 0 & 1

so h is not differentiable at 1
similarly h' (0 +)= 0 but h ' (0 - ) = 1

QUESTION: 27

If f (x + y) = 2f (x) f (y) for all x, y ∈ R where f ' (0) = 3 and f (4) = 2, then f ' (4) is equal to

Solution:

QUESTION: 28

If  and f ' (0) = -1, f (0) = 1, then f (2) =

Solution:

Take  f (x) = ax+ b

QUESTION: 29

Let f (x) be differentiable function such that and y. If

Solution:

QUESTION: 30

Let f : R → R be a function defined by f (x) = min {x + 1, |x| + 1}, Then which of the following is true?

Solution: