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# Competition Level Test: Continuity And Differentiability- 2

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

Competition Level Test: Continuity And Differentiability- 2 for JEE 2023 is part of Mathematics (Maths) Class 12 preparation. The Competition Level Test: Continuity And Differentiability- 2 questions and answers have been prepared according to the JEE exam syllabus.The Competition Level Test: Continuity And Differentiability- 2 MCQs are made for JEE 2023 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Competition Level Test: Continuity And Differentiability- 2 below.
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Competition Level Test: Continuity And Differentiability- 2 - Question 1

### The function is continuous at

Detailed Solution for Competition Level Test: Continuity And Differentiability- 2 - Question 1 Competition Level Test: Continuity And Differentiability- 2 - Question 2

### If the function is continuous at x = 0 then a =

Detailed Solution for Competition Level Test: Continuity And Differentiability- 2 - Question 2 Competition Level Test: Continuity And Differentiability- 2 - Question 3

### is continuous at x = 0 then

Detailed Solution for Competition Level Test: Continuity And Differentiability- 2 - Question 3 Competition Level Test: Continuity And Differentiability- 2 - Question 4

The function is discontinuous at the points

Detailed Solution for Competition Level Test: Continuity And Differentiability- 2 - Question 4

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

Competition Level Test: Continuity And Differentiability- 2 - Question 5

The values of a and b if f is continuous at x = 0, where Detailed Solution for Competition Level Test: Continuity And Differentiability- 2 - Question 5 Competition Level Test: Continuity And Differentiability- 2 - Question 6 is continuous at then k =

Detailed Solution for Competition Level Test: Continuity And Differentiability- 2 - Question 6  Competition Level Test: Continuity And Differentiability- 2 - Question 7 so that f(x) is continuous at then

Detailed Solution for Competition Level Test: Continuity And Differentiability- 2 - Question 7 By L-Hospital rule  Competition Level Test: Continuity And Differentiability- 2 - Question 8 where [.] denotes greatest integer function and the function is continuous then

Detailed Solution for Competition Level Test: Continuity And Differentiability- 2 - Question 8 Competition Level Test: Continuity And Differentiability- 2 - Question 9 is continuous everywhere. Then the  equation whose roots are a and b is

Detailed Solution for Competition Level Test: Continuity And Differentiability- 2 - Question 9  Competition Level Test: Continuity And Differentiability- 2 - Question 10 where [x] is the greatest integer function. The function f (x) is

Detailed Solution for Competition Level Test: Continuity And Differentiability- 2 - Question 10 Competition Level Test: Continuity And Differentiability- 2 - Question 11

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

Detailed Solution for Competition Level Test: Continuity And Differentiability- 2 - Question 11

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

Competition Level Test: Continuity And Differentiability- 2 - Question 12

If the function is continuous for every x ∈ R then

Detailed Solution for Competition Level Test: Continuity And Differentiability- 2 - Question 12

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)

Competition Level Test: Continuity And Differentiability- 2 - Question 13 Detailed Solution for Competition Level Test: Continuity And Differentiability- 2 - Question 13

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

Competition Level Test: Continuity And Differentiability- 2 - Question 14

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

Detailed Solution for Competition Level Test: Continuity And Differentiability- 2 - Question 14 f (x) is continuous at x = π, - π

Competition Level Test: Continuity And Differentiability- 2 - Question 15 Detailed Solution for Competition Level Test: Continuity And Differentiability- 2 - Question 15  Competition Level Test: Continuity And Differentiability- 2 - Question 16 then which is correct

Detailed Solution for Competition Level Test: Continuity And Differentiability- 2 - Question 16 Competition Level Test: Continuity And Differentiability- 2 - Question 17

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

Detailed Solution for Competition Level Test: Continuity And Differentiability- 2 - Question 17 Competition Level Test: Continuity And Differentiability- 2 - Question 18 Detailed Solution for Competition Level Test: Continuity And Differentiability- 2 - Question 18

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  Competition Level Test: Continuity And Differentiability- 2 - Question 19

The set of all points where the function is differentiable is

Detailed Solution for Competition Level Test: Continuity And Differentiability- 2 - Question 19 is not differentiable only at x = 0

Competition Level Test: Continuity And Differentiability- 2 - Question 20

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

Detailed Solution for Competition Level Test: Continuity And Differentiability- 2 - Question 20 Competition Level Test: Continuity And Differentiability- 2 - 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

Detailed Solution for Competition Level Test: Continuity And Differentiability- 2 - Question 21

f (x + 2y) = f (x) + f (2 y) + 4xy for x, y ∈ R putting x = y = 0, we get f (0) = 0  Competition Level Test: Continuity And Differentiability- 2 - Question 22

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

Detailed Solution for Competition Level Test: Continuity And Differentiability- 2 - Question 22 replacing x and y both by 1, we get    Competition Level Test: Continuity And Differentiability- 2 - Question 23

The function is not differentiable at

Detailed Solution for Competition Level Test: Continuity And Differentiability- 2 - Question 23

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

Competition Level Test: Continuity And Differentiability- 2 - Question 24 then set of all points where f is differentiable is

Detailed Solution for Competition Level Test: Continuity And Differentiability- 2 - Question 24

The function is clearly differentiable except possible at x = 2, 3 Competition Level Test: Continuity And Differentiability- 2 - Question 25 Competition Level Test: Continuity And Differentiability- 2 - Question 26

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

Detailed Solution for Competition Level Test: Continuity And Differentiability- 2 - Question 26 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

Competition Level Test: Continuity And Differentiability- 2 - 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

Detailed Solution for Competition Level Test: Continuity And Differentiability- 2 - Question 27  Competition Level Test: Continuity And Differentiability- 2 - Question 28

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

Detailed Solution for Competition Level Test: Continuity And Differentiability- 2 - Question 28

Take  f (x) = ax+ b

Competition Level Test: Continuity And Differentiability- 2 - Question 29

Let f (x) be differentiable function such that and y. If Detailed Solution for Competition Level Test: Continuity And Differentiability- 2 - Question 29 Competition Level Test: Continuity And Differentiability- 2 - 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?

Detailed Solution for Competition Level Test: Continuity And Differentiability- 2 - Question 30 ## Mathematics (Maths) Class 12

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