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# Test: Continuity And Differentiability (Competition Level) - 1

## 30 Questions MCQ Test Mathematics (Maths) Class 12 | Test: Continuity And Differentiability (Competition Level) - 1

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

Solution:
QUESTION: 2

### , -1 ≤ x < 0 is continuous in the interval [–1, 1], then `p' is equal to:

Solution:

lim f(x) = -½
Apply L hospital
lim(x → 0) p/2(1+px)½ + p/2(1-px)½ = -½
⇒ 2p = -1
p = -1/2

QUESTION: 3

### Let f(x) =  when – 2 ≤ x ≤ 2. Then (where [ * ] represents greatest integer function)

Solution:

At (x = 0) f(0) = |(0+½) .[0]| = 0
f(1) = |(1+½) .[1]| = 3/2
f(2) = |(2+½) .[2]| = 5
f(-1) = |(-1+½) .[-1]| = ½
LHL at x = 2
lim x-->2  f(x) = |(x+½) .[2]|
=|(5/2) * 1| = 5/2
f(2) = limx-->2 f(x) ( continuous at x=2)

QUESTION: 4

Let f(x) = sgn (x) and g(x) = x (x2 – 5x + 6). The function f(g(x)) is discontinuous at

Solution:
QUESTION: 5

If y =  where t = , then the number of points of discontinuities of y = f(x), x ∈ R is

Solution:

QUESTION: 6

The equation 2 tan x + 5x – 2 = 0 has

Solution:

Let f(x) = 2tan x + 5x – 2
Now, f(0) = – 2 < 0
and f(π/4 ) = 2 + 5π/4 – 2
= 5π/5 > 0,
the line intersects the x-axis at x = ⅖
since 0 < ⅖ <  π/4
Thus, f(x) = 0 has atleast real solution root in ( 0, π/4)

QUESTION: 7

If f(x) =  , then indicate the correct alternative(s)

Solution:
QUESTION: 8

The function f (x) = 1 + | sin x l is

Solution:
QUESTION: 9

If f(x) =  be a real valued function then

Solution:
QUESTION: 10

The function f(x) = sin-1 (cos x) is

Solution:
QUESTION: 11

Let f(x) be defined in [–2, 2] by f(x) =   then f(x)

Solution:
QUESTION: 12

If f(x) is differentiable everywhere, then

Solution:

Even if f(x) is differentiable everywhere, |f(x)| need not be differentiable everywhere. An example being where the function changes its sign from negative to positive. f(x)=x at x=0B.
∣f∣= {f if f>0                −f if f<0}
​So,∣f∣2 = {f2 if f>0        −f2 if f<0}
​Differentiating ∣f∣2,
{2ff′ if f>0     −2ff′ if f<0}
​At f=0, LHD=RHD=∣f(0)∣2=0, so function is differentiable.

QUESTION: 13

Let f(x + y) = f(x) f(y) all x and y. Suppose that f(3) = 3 and f'(0) = 11 then f'(3) is given by

Solution:

f(x + y)=f(x) f(y) wrt x we get:
f′(x+y) (1 + y′) = f(x) f′(y) (y′) + f′(x) f(y)
Now on substituting x = 0 and y = 3, we get :
f′(3) (1 + 0)=f(0) f′(3) 0+ f(3) f′(0)
f′(3) = 3 × 11 = 33

QUESTION: 14

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

Solution:
QUESTION: 15

Let f(x) = x – x2 and g(x) = . Then in the interval [0, ¥)

Solution:
QUESTION: 16

Let [x] denote the integral part of x ∈ R and g(x) = x – [x]. Let f(x) be any continuous function with f(0) = f(1) then the function h(x) = f(g(x))

Solution:
QUESTION: 17

If f (x) = [x sin p x] { where [x] denotes greatest integer function}, then f (x) is

Solution:
QUESTION: 18

If f(x) =  , then f(x) is

Solution:
QUESTION: 19

The functions defined by f(x) = max {x2, (x – 1)2, 2x (1 – x)}, 0 £ x £ 1

Solution:
QUESTION: 20

Let f(x) = x3 – x2 + x + 1 and g(x) =  then

Solution:
QUESTION: 21

Let f²(x) be continuous at x = 0 and f²(0) = 4 then value of  is

Solution:
QUESTION: 22

Let f : R → R be a function such that f , f(0) = 0 and f¢(0) = 3, then

Solution:
QUESTION: 23

Suppose that f is a differentiable function with the property that f(x + y) = f(x) + f(y) + xy and  f(h) = 3 then

Solution:
QUESTION: 24

If a differentiable function f satisfies f   x, y Î R, find f(x)

Solution:
QUESTION: 25

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

Solution:

f(x) = Min {x + 1,∣x∣ + 1}
x = 0 ⇒ ∣x∣ = x
f(x) = min {x + 1,x + 1}
= x + 1
x < 0,∣x∣ = −x
f(x) = min {x + 1,−x + 1}
−x + 1> x + 1
x < 0
⇒x < 0, f(x) = x + 1
x > 0,f (x) = x + 1
⇒ f(x) = x + 1 for all x
∴f(x) is differential everywhere (continuous and constant slope)

QUESTION: 26

The function f : R /{0} → R given by vf(x) =  can be made continuous at x = 0 by defining f(0) as

Solution:

QUESTION: 27

Function f(x) = (|x – 1| + |x – 2| + cos x) where x Î [0, 4] is not continuous at number of points

Solution:
QUESTION: 28

Let f(x + y) = f(x) f(y) for all x, y, where f(0) ≠ 0. If f'(0) = 2, then f(x) is equal to

Solution:
QUESTION: 29

A function f : R → R satisfies the equation f(x + y) = f(x) . f(y) for all x, y ∈ R, f(x) ≠ 0. Suppose that the function is differentiable at x = 0 and f'(0) = 2 then f'(x) =

Solution:
QUESTION: 30

Let f(x) = [cos x + sin x], 0 < x < 2p where [x] denotes the greatest integer less than or equal to x. the number of points of discontinuity of f(x) is

Solution: