Continuity and Differentiability Test

Result

Continuity and Differentiability Test - 3

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Question 1
1 / -0.25

A
–1

B
0

C
1

D
1/2

Solution

L 'Hopital 's rules says that the
lim x →a f(x)/g(x)
⇒f '(a)/g '(a)
Using this, we get
lim x →0 (1 −cosx)/x²
⇒−sin0/2(0)
Yet as the denominator is 0, this is impossible. So we do a second limit:
lim(x →0) sinx/2x
⇒cos0/2 = 1/2 = 0.5
So, in total lim x →0 (1 −cosx)/x²
⇒lim x →0 sinx/2x
⇒cosx/2
⇒cos0/2= 1/2
Question 2
1 / -0.25

If f (x) is a polynomial of degree m (⩾1) , then which of the following is not true ?

A
f is derivable at all x ∈R

B
f is continuous at all x ∈R

C
= 0 for all n >m

D
none of these

Solution

As all the three remaining statements are true for the given function.
Question 3
1 / -0.25

Let f and g be differentiable functions such that fog = I, the identity function. If g ’(a) = 2 and g (a) = b, then f ‘(b) =

A
1/2

B
2

C
-2

D
none of these

Solution

f(g(x)) = x
f '(g(x)) g '(x) = 1
put x = a
f '(b) g '(a) = 1
2 f '(b) = 1
f '(b) = 1/2
Question 4
1 / -0.25

A

B

C
1/x

D

Solution
Question 5
1 / -0.25

If f (x) =x² g(x) and g (x) is twice differentiable then f ’’(x) is equal to

A
x² g ′′(x)+4xg ′(x)+2g(x)

B
x² g ′′(x)+2xg ′(x)+2g(x)

C
2 g ’’(x)

D
none of these

Solution
Question 6
1 / -0.25

A
0

B

C
2/3

D
none of these

Solution
Question 7
1 / -0.25

If f(x) = | x | ∀x ∈R, then

A
f is discontinuous at x = 0

B
f is derivable at x = 0 and f ‘(0) = 1

C
f is derivable at x = 0 but f ’(0) ≠

D
none of these

Solution

The graph of f(x) = |x|
As observed from the graph, f(x) = |x| is continuous at x = 0.
As this curve is pointed at x = 0, f(x) is not derivable at x = 0.
Question 8
1 / -0.25

Differential coefficient of a function f (g (x)) w.r.t. the function g (x) is

A
f ‘(g (x))

B

C
f ‘(g (x)) g ’(x)

D
none of these

Solution
Question 9
1 / -0.25

A

B

C
y/x

D
none of these

Solution

Given x^p y^q = (x+y)^p+q Taking log on both sides we get:
_p logx+_q logy = (p+q) log (x+y). Differentiating both sides w.r.t. x we get ,
Question 10
1 / -0.25

If y = ae^mx + be^−mx , then y₂ is equal to

A
−m² y

B
my₁

C
m² y

D
none of these

Solution

y = ae^mx + be^-mx ⇒y1 = ame^mx + (-m)be^-mx ⇒y2
= am² e^mx_{+ (m}² )be^-mx ⇒y2 = m² (ae^mx + be^-mx ) ⇒y2 = m² y