Limits, Continuity and Differentiability Questions — JEE Mathematics

Q1

For all twice differentiable functions f : R

\to

R, with f(0) = f(1) = f'(0) = 0

A f''(x)

\ne

0, at every point x

\in

(0, 1)

B f''(x) = 0, for some x

\in

(0, 1)

C f''(0) = 0

D f''(x) = 0, at every point x

\in

(0, 1)

Correct Answer

Option B

Solution

f : R $\to$ R, with f(0) = f(1) = 0 and f'(0) = 0 $\because$ f(x) is differentiable and continuous and f(0) = f(1) = 0 Applying Rolle’s theorem in [0, 1] for function f(x) f'(c) = 0, c

\in

(0, 1) Now again $\because$ f'(c) = 0, f'(0) = 0 again applying Rolles theorem in [0, c] for function f'(x) f''(c1) = 0 for some c1

\in

(0, c)

\in

(0, 1)

Q2

\mathop {\lim }\limits_{x \to 0} {{\left( {1 - \cos 2x} \right)\left( {3 + \cos x} \right)} \over {x\tan 4x}}

is equal to

A 2

B

{1 \over 2}

C 4

D 3

Correct Answer

Option A

Solution

Multiply and divide by

x

in the given expression, we get

\mathop {\lim }\limits_{x \to 0} {{\left( {1 - \cos 2x} \right)} \over {{x^2}}}{{\left( {3 + \cos x} \right)} \over 1}.{x \over {\tan \,4x}}

= \mathop {\lim }\limits_{x \to 0} {{2{{\sin }^2}x} \over {{x^2}}}.{{3 + \cos x} \over 1}.{x \over {\tan \,4x}}

= 2\mathop {\lim }\limits_{x \to 0} {{{{\sin }^2}x} \over {{x^2}}}.\mathop {\lim }\limits_{x \to 0} 3 + \cos x.\mathop {\lim }\limits_{x \to 0} {x \over {\tan 4x}}

= 2.4{1 \over 4}\mathop {\lim }\limits_{x \to 0} {{4x} \over {\tan 4x}} = 2.4.{1 \over 4} = 2

Q3

Let f : (

-

1, 1)

\to

R be a function defined by f(x) = max

\left\{ { - \left| x \right|, - \sqrt {1 - {x^2}} } \right\}.

If K be the set of all points at which f is not differentiable, then K has exactly -

A one element

B three elements

C five elements

D two elements

Correct Answer

Option B

Solution

f : ( $-$ 1, 1) $\to$ R f(x) = max { $-$

\left| x \right|, - \sqrt {1 - {x^2}}

} Non-derivable at 3 points in ( $-$ 1, 1)

Q4

\mathop {\lim }\limits_{x \to \pi /4} {{{{\cot }^3}x - \tan x} \over {\cos \left( {x + {\pi \over 4}} \right)}}

is :

A

8\sqrt 2

B 4

C

4\sqrt 2

D 8

Correct Answer

Option D

Solution

\mathop {\lim }\limits_{x \to \pi /4} {{{{\cot }^3}x - \tan x} \over {\cos \left( {x + {\pi \over 4}} \right)}}

=

\mathop {\lim }\limits_{x \to {\pi \over 4}} {{{1 \over {{{\tan }^3}x}} - \tan x} \over {\cos \left( {x + {\pi \over 4}} \right)}}

=

\mathop {\lim }\limits_{x \to {\pi \over 4}} {{1 - {{\tan }^4}x} \over {\left( {{{\tan }^3}x} \right)\cos \left( {x + {\pi \over 4}} \right)}}

=

\mathop {\lim }\limits_{x \to {\pi \over 4}} {{\left( {1 - {{\tan }^2}x} \right)\left( {1 + {{\tan }^2}x} \right)} \over {\left( {{{\tan }^3}x} \right)\cos \left( {x + {\pi \over 4}} \right)}}

=

\mathop {\lim }\limits_{x \to {\pi \over 4}} {{\left( {1 - {{\tan }^2}x} \right)\left( {1 + {{\tan }^2}{\pi \over 4}} \right)} \over {\left( {{{\tan }^3}{\pi \over 4}} \right)\cos \left( {x + {\pi \over 4}} \right)}}

=

\mathop {\lim }\limits_{x \to {\pi \over 4}} {{\left( {1 - {{\tan }^2}x} \right)\left( {1 + 1} \right)} \over {1.\cos \left( {x + {\pi \over 4}} \right)}}

=

\mathop {\lim }\limits_{x \to {\pi \over 4}} {{2\left( {1 - {{\tan }^2}x} \right)} \over {\left( {{{\cos x - \sin x} \over {\sqrt 2 }}} \right)}}

=

\mathop {\lim }\limits_{x \to {\pi \over 4}} {{2\sqrt 2 \left( {1 - {{{{\sin }^2}x} \over {{{\cos }^2}x}}} \right)} \over {\left( {\cos x - \sin x} \right)}}

=

\mathop {\lim }\limits_{x \to {\pi \over 4}} {{2\sqrt 2 \left( {{{\cos }^2}x - {{\sin }^2}x} \right)} \over {{{\cos }^2}x\left( {\cos x - \sin x} \right)}}

=

\mathop {\lim }\limits_{x \to {\pi \over 4}} {{2\sqrt 2 \left( {\cos x + \sin x} \right)\left( {\cos x - \sin x} \right)} \over {{{\cos }^2}x\left( {\cos x - \sin x} \right)}}

=

\mathop {\lim }\limits_{x \to {\pi \over 4}} {{2\sqrt 2 \left( {\cos x + \sin x} \right)} \over {{{\cos }^2}x}}

=

\mathop {\lim }\limits_{x \to {\pi \over 4}} {{2\sqrt 2 \left( {\cos {\pi \over 4} + \sin {\pi \over 4}} \right)} \over {{{\cos }^2}{\pi \over 4}}}

=

{{2\sqrt 2 \left( {{1 \over {\sqrt 2 }} + {1 \over {\sqrt 2 }}} \right)} \over {{1 \over 2}}}

=

{{2\sqrt 2 \left( {{2 \over {\sqrt 2 }}} \right)} \over {{1 \over 2}}}

= 8 Other Method :

\mathop {\lim }\limits_{x \to \pi /4} {{{{\cot }^3}x - \tan x} \over {\cos \left( {x + {\pi \over 4}} \right)}}

This is in

{0 \over 0}

form so L' Hospital rule is applicable. =

\mathop {\lim }\limits_{x \to {\pi \over 4}} {{3{{\cot }^3}x\left( { - \cos e{c^2}x} \right) - {{\sec }^2}x} \over { - \sin \left( {x + {\pi \over 4}} \right)}}

=

{{3 \times 1\left( { - {{\left( {\sqrt 2 } \right)}^2}} \right) - {{\left( {\sqrt 2 } \right)}^2}} \over { - 1}}

=

{{ - 6 - 2} \over { - 1}}

= 8

Q5

\mathop {\lim }\limits_{x \to 0} {{x\cot \left( {4x} \right)} \over {{{\sin }^2}x{{\cot }^2}\left( {2x} \right)}}

is equal to :

A 0

B 4

C 1

D 2

Correct Answer

Option C

Solution

\mathop {\lim }\limits_{x \to 0} {{x{{\tan }^2}2x} \over {\tan 4x{{\sin }^2}x}} = \mathop {\lim }\limits_{x \to 0} {{x\left( {{{{{\tan }^2}2x} \over {4{x^2}}}} \right)4{x^2}} \over {\left( {{{\tan 4x} \over {4x}}} \right)4x\left( {{{{{\sin }^2}x} \over {{x^2}}}} \right){x^2}}} = 1

Q6

\text{ Let the function } f(x)=\left\{\begin{array}{cl} \frac{\log _{e}(1+5 x)-\log _{e}(1+\alpha x)}{x} & ;\text{ if } x \neq 0 \\ 10 & ; \text{ if } x=0 \end{array} \text{ be continuous at } x=0 .\right.

Then

\alpha

is equal to

A 10

B

-

10

C 5

D

-

5

Correct Answer

Option D

Solution

f(x)

is continuous at

x = 0

$\therefore$

f(0) = \mathop {\lim }\limits_{x \to 0} f(x)

\Rightarrow 10 = \mathop {\lim }\limits_{x \to 0} {{{{\log }_e}(1 + 5x) - {{\log }_e}(1 + \alpha x)} \over x}

= \mathop {\lim }\limits_{x \to 0} {{\log (1 + 5x)} \over {5x}} \times 5 - {{{{\log }_e}(1 + \alpha x)} \over {\alpha x}} \times \alpha

= 1 \times 5 - \alpha

\Rightarrow \alpha = 5 - 10 = - 5

Q7

\mathop {\lim }\limits_{t \to 0} {\left( {{1^{{1 \over {{{\sin }^2}t}}}} + {2^{{1 \over {{{\sin }^2}t}}}}\, + \,...\, + \,{n^{{1 \over {{{\sin }^2}t}}}}} \right)^{{{\sin }^2}t}}

is equal to

A

{{n(n + 1)} \over 2}

B n

C n

^2

+ n

D n

^2

Correct Answer

Option B

Solution

\begin{aligned} & \lim _{t \rightarrow 0}\left(1^{\operatorname{cosec}^2 t}+2^{\operatorname{cosec}^2 t}+\ldots \ldots . .+n^{\operatorname{cosec}^2 t}\right)^{\sin ^2 t} \\\\ & =\lim _{t \rightarrow 0} n\left(\left(\frac{1}{n}\right)^{\operatorname{cosec}^2 t}+\left(\frac{2}{n}\right)^{\operatorname{cosec}^2 t}+\ldots \ldots . .+1\right)^{\sin ^2 t} \\\\ & =\mathrm{n} \end{aligned}

Q8

If f(x + y) = f(x).f(y)

\forall

x, y and f(5) = 2, f'(0) = 3, then f'(5) is

A 0

B 1

C 6

D 2

Correct Answer

Option C

Solution

f\left( {x + y} \right) = f\left( x \right) \times f\left( y \right)

Differeniate with respect to

x,

treating

y

as constant

f'\left( {x + y} \right) = f'\left( x \right)f\left( y \right)

Putting

x=0

and

y=x

, we get

f'\left( x \right) = f'\left( 0 \right)f\left( x \right);

\Rightarrow f'\left( 5 \right) = 3f\left( 5 \right) = 3 \times 2 = 6.

Q9

If

\mathop {\lim }\limits_{x \to 0} {{\log \left( {3 + x} \right) - \log \left( {3 - x} \right)} \over x}

= k, the value of k is

A

- {2 \over 3}

B 0

C

- {1 \over 3}

D

{2 \over 3}

Correct Answer

Option D

Solution

\mathop {\lim }\limits_{x \to 0} {{\log \left( {3 + x} \right) - \log \left( {3 - x} \right)} \over x} = K

(by

L'

Hospital rule)

\mathop {\lim }\limits_{x \to 0} {{{1 \over {3 + x}} - {{ - 1} \over {3 - x}}} \over 1} = K

$\therefore$

{2 \over 3} = K

Q10

Let

f(a) = g(a) = k

and their nth derivatives

{f^n}(a)

,

{g^n}(a)

exist and are not equal for some n. Further if

\mathop {\lim }\limits_{x \to a} {{f(a)g(x) - f(a) - g(a)f(x) + f(a)} \over {g(x) - f(x)}} = 4

then the value of k is

A 0

B 4

C 2

D 1

Correct Answer

Option B

Solution

\mathop {\lim }\limits_{x \to a} {{f\left( a \right)g'\left( x \right) - g\left( a \right)f'\left( x \right)} \over {g'\left( x \right) - f'\left( x \right)}}

\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,

(By

L'

Hospital rule)

\mathop {\lim }\limits_{x \to a} {{k\,\,g'\left( x \right) - k\,\,f'\left( x \right)} \over {g'\left( x \right) - f'\left( x \right)}} = 4

$\therefore$

k=4.