The function $$f(x) = \left\{ {\matrix{ {{\pi \over 4} + {{\tan }^{ - 1}}x,} & {\left| x \right| \le 1} \cr {{1 \over 2}\left( {\left| x \right| - 1} \right),} & {\left| x \right| > 1} \cr

$$f\left( x \right) = \left\{ {\matrix{ {{\pi \over 4} + {{\tan }^{ - 1}}x,} & {x \in \left[ { - 1,1} \right]} \cr {{1 \over 2}\left( {x - 1} \right),} & {x > 1} \cr {{1 \over 2}\left( { - x - 1} \right),} & {x < - 1} \cr } } \right.$$ At x = 1 L.H.L = $$\mathop {\lim }\limits_{x \to {1^ - }} \left( {{\pi \over 4} + {{\tan }^{ - 1}}x} \right)$$ = $${{\pi \over 4} + {\pi \over 4}}$$ = $${{\pi \over 2}}$$ f(1) = $${{\pi \over 4} + {{\tan }^{ - 1}}x}$$ = $${{\pi \over 4} + {\pi \o

If the function $$f\left( x \right) = \left\{ {\matrix{ {{k_1}{{\left( {x - \pi } \right)}^2} - 1,} & {x \le \pi } \cr {{k_2}\cos x,} & {x > \pi } \cr } } \right.$$ is twice differentiable,

Given, $$f\left( x \right) = \left\{ {\matrix{ {{k_1}{{\left( {x - \pi } \right)}^2} - 1,} & {x \le \pi } \cr {{k_2}\cos x,} & {x > \pi } \cr } } \right.$$ Differentiating one time, $$f'\left( x \right) = \left\{ {\matrix{ {2{k_1}\left( {x - \pi } \right),} & {x \le \pi } \cr { - {k_2}\sin x,} & {x > \pi } \cr } } \right.$$ Differentiating one more time, $$f''\left( x \right) = \left\{ {\matrix{ {2{k_1},} & {x \le \pi } \cr { - {k_2}\cos x,} & {x > \pi } \cr } }

Limits, Continuity and Differentiability — Page 6 | JEE Mathematics

Q51

For each x

\in

R, let [x] be the greatest integer less than or equal to x. Then

\mathop {\lim }\limits_{x \to {0^ - }} \,\,{{x\left( {\left[ x \right] + \left| x \right|} \right)\sin \left[ x \right]} \over {\left| x \right|}}

is equal to :

A

-

sin 1

B 1

C sin 1

D 0

Correct Answer

Option A

Solution

\mathop {\lim }\limits_{x \to {0^ - }} {{x\left( {\left[ x \right] + \left| x \right|} \right)\sin \left[ x \right]} \over {\left| x \right|}}

= \mathop {\lim }\limits_{h \to 0} {{\left( {0 - h} \right)\left( {\left[ {0 - h} \right] + \left| {0 - h} \right|} \right)\sin \left[ {0 - h} \right]} \over {\left| {0 - h} \right|}}

= \mathop {\lim }\limits_{h \to 0} {{ - h\left( { - 1 + h} \right)\sin \left( { - 1} \right)} \over h}

= \mathop {\lim }\limits_{h \to 0} \left( { - 1 + h} \right)\sin \left( 1 \right) = - \sin 1

Q52

Let

f:\left( {0,\infty } \right) \to \left( {0,\infty } \right)

be a differentiable function such that f(1) = e and

\mathop {\lim }\limits_{t \to x} {{{t^2}{f^2}(x) - {x^2}{f^2}(t)} \over {t - x}} = 0

. If f(x) = 1, then x is equal to :

A

{1 \over e}

B e

C

{1 \over 2e}

D 2e

Correct Answer

Option A

Solution

\mathop {\lim }\limits_{t \to x} {{{t^2}{f^2}(x) - {x^2}{f^2}(t)} \over {t - x}} = 0

(Using L'Hospital's Rule)

\Rightarrow \mathop {\lim }\limits_{t \to x} {{2t{f^2}(x) - 2{x^2}f(t).f'(t)} \over 1} = 0

$\Rightarrow$ 2xf2(x) - 2x2.f(x).f'(x) = 0 $\Rightarrow$ 2xf(x){f(x) - xf'(x)} = 0 [x $\ne$ 0, f(x) $\ne$ 0 as given function

f:\left( {0,\infty } \right) \to \left( {0,\infty } \right)

only takes positive value as input and output]

\Rightarrow f(x) = xf'(x)

\Rightarrow {{f'(x)} \over {f(x)}} = {1 \over x}

Integrating w.r.t x, we get

\Rightarrow ln\,f(x) = ln\,x + ln\,C

\Rightarrow f(x) = Cx

$\because$ f(1) = e

\Rightarrow C = e;\,so\,f(x) = ex

When f(x) = 1 = ex

\Rightarrow x = {1 \over e}

Q53

The function

f(x) = \left\{ \begin{array}{ll}{{\pi \over 4} + {{\tan }^{ - 1}}x,} & {\left| x \right| \le 1} \\ {{1 \over 2}\left( {\left| x \right| - 1} \right),} & {\left| x \right| > 1} \end{array} \right.

is :

A continuous on R–{–1} and differentiable on R–{–1, 1}

B both continuous and differentiable on R–{1}

C both continuous and differentiable on R–{–1}

D continuous on R–{1} and differentiable on R–{–1, 1}

Correct Answer

Option D

Solution

f\left( x \right) = \left\{ \begin{array}{ll}{{\pi \over 4} + {{\tan }^{ - 1}}x,} & {x \in \left[ { - 1,1} \right]} \\ {{1 \over 2}\left( {x - 1} \right),} & {x > 1} \\ {{1 \over 2}\left( { - x - 1} \right),} & {x < - 1} \end{array} \right.

At x = 1 L.H.L =

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

=

{{\pi \over 4} + {\pi \over 4}}

=

{{\pi \over 2}}

f(1) =

{{\pi \over 4} + {{\tan }^{ - 1}}x}

=

{{\pi \over 4} + {\pi \over 4}}

=

{{\pi \over 2}}

R.H.L =

\mathop {\lim }\limits_{x \to {1^ + }} \left( {{1 \over 2}\left( {x - 1} \right)} \right)

= 0 As L.H.L $\ne$ R.H.L so function is discontinuous $\Rightarrow$ non differentiable. At x = -1 L.H.L =

\mathop {\lim }\limits_{x \to - {1^ - }} \left( {{1 \over 2}\left( { - x - 1} \right)} \right)

=

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

= 0 f(-1) =

{\pi \over 4} + {\tan ^{ - 1}}\left( { - 1} \right)

=

{\pi \over 4} - {\pi \over 4}

= 0 R.H.L =

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

=

{\pi \over 4} + {\tan ^{ - 1}}\left( { - 1} \right)

=

{\pi \over 4} - {\pi \over 4}

= 0 As L.H.L = f(-1) = R.H.L so function is continuous.

f'\left( x \right) = \left\{ \begin{array}{ll}{{1 \over {1 + {x^2}}},} & {x \in \left[ { - 1,1} \right]} \\ {{1 \over 2},} & {x > 1} \\ { - {1 \over 2},} & {x < - 1} \end{array} \right.

For differentiability at x = –1 L.H.D =

{ - {1 \over 2}}

R.H.D. =

{{1 \over 2}}

So, non differentiable at x = –1

Q54

If the function

f\left( x \right) = \left\{ \begin{array}{ll}{{k_1}{{\left( {x - \pi } \right)}^2} - 1,} & {x \le \pi } \\ {{k_2}\cos x,} & {x > \pi } \end{array} \right.

is twice differentiable, then the ordered pair (k1, k2) is equal to :

A

\left( {{1 \over 2},-1} \right)

B (1, 1)

C (1, 0)

D

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

Correct Answer

Option D

Solution

Given,

f\left( x \right) = \left\{ \begin{array}{ll}{{k_1}{{\left( {x - \pi } \right)}^2} - 1,} & {x \le \pi } \\ {{k_2}\cos x,} & {x > \pi } \end{array} \right.

Differentiating one time,

f'\left( x \right) = \left\{ \begin{array}{ll}{2{k_1}\left( {x - \pi } \right),} & {x \le \pi } \\ { - {k_2}\sin x,} & {x > \pi } \end{array} \right.

Differentiating one more time,

f''\left( x \right) = \left\{ \begin{array}{ll}{2{k_1},} & {x \le \pi } \\ { - {k_2}\cos x,} & {x > \pi } \end{array} \right.

As f''(x) is differentiable so f''( $\pi$ +) = f''( $\pi$ -) $\Rightarrow$ -k2(-1) = 2k1 $\Rightarrow$ 2k1 = k2 $\therefore$ (k1, k2) =

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

Q55

If

\alpha

is positive root of the equation, p(x) = x2 - x - 2 = 0, then

\mathop {\lim }\limits_{x \to {\alpha ^ + }} {{\sqrt {1 - \cos \left( {p\left( x \right)} \right)} } \over {x + \alpha - 4}}

is equal to :

A

{1 \over \sqrt2}

B

{1 \over 2}

C

{3 \over \sqrt2}

D

{3 \over 2}

Correct Answer

Option C

Solution

{x^2} - x - 2 = 0

roots are 2 & $-$ 1 $\Rightarrow$ $\alpha$ = 2 (given $\alpha$ is positive) Now

\mathop {\lim }\limits_{x \to {2^ + }} {{\sqrt {1 - \cos ({x^2} - x - 2)} } \over {(x - 2)}}

= \mathop {\lim }\limits_{x \to {2^ + }} {{\sqrt {2{{\sin }^2}{{({x^2} - x - 2)} \over 2}} } \over {(x - 2)}}

= \mathop {\lim }\limits_{x \to {2^ + }} {{\sqrt 2 \sin \left( {{{(x - 2)(x + 1)} \over 2}} \right)} \over {(x - 2)}}

= {3 \over {\sqrt 2 }}

Q56

Let ƒ be any function continuous on [a, b] and twice differentiable on (a, b). If for all x

\in

(a, b), ƒ'(x) > 0 and ƒ''(x) < 0, then for any c

\in

(a, b),

{{f(c) - f(a)} \over {f(b) - f(c)}}

is greater than :

A 1

B

{{b - c} \over {c - a}}

C

{{b + a} \over {b - a}}

D

{{c - a} \over {b - c}}

Correct Answer

Option D

Solution

It is clear from graph that, slope of AC

>

slope of CB $\Rightarrow$

{{f\left( c \right) - f\left( a \right)} \over {c - a}}

>

{{f\left( b \right) - f\left( c \right)} \over {b - c}}

$\Rightarrow$

{{f\left( c \right) - f\left( a \right)} \over {f\left( b \right) - f\left( c \right)}}

> {{c - a} \over {b - c}}

Q57

\mathop {\lim }\limits_{x \to 0} {{x\left( {{e^{\left( {\sqrt {1 + {x^2} + {x^4}} - 1} \right)/x}} - 1} \right)} \over {\sqrt {1 + {x^2} + {x^4}} - 1}}

A is equal to 0.

B is equal to

\sqrt e

.

C is equal to 1.

D does not exist.

Correct Answer

Option C

Solution

\mathop {\lim }\limits_{x \to 0} {{x\left( {{e^{\left( {\sqrt {1 + {x^2} + {x^4}} - 1} \right)/x}} - 1} \right)} \over {\sqrt {1 + {x^2} + {x^4}} - 1}}

=

\mathop {\lim }\limits_{x \to 0} {{x\left[ {{e^{{{\left( {\sqrt {1 + {x^2} + {x^4}} - 1} \right)\left( {\sqrt {1 + {x^2} + {x^4}} + 1} \right)} \over {x\left( {\sqrt {1 + {x^2} + {x^4}} + 1} \right)}}}} - 1} \right] \times \left( {\sqrt {1 + {x^2} + {x^4}} + 1} \right)} \over {\left( {\sqrt {1 + {x^2} + {x^4}} - 1} \right)\left( {\sqrt {1 + {x^2} + {x^4}} + 1} \right)}}

=

\mathop {\lim }\limits_{x \to 0} {{x\left[ {{e^{{{\left( {1 + {x^2} + {x^4}} \right) - 1} \over {x\left( {\sqrt {1 + {x^2} + {x^4}} + 1} \right)}}}} - 1} \right] \times \left( {\sqrt {1 + {x^2} + {x^4}} + 1} \right)} \over {\left( {1 + {x^2} + {x^4} - 1} \right)}}

=

\mathop {\lim }\limits_{x \to 0} {{\left[ {{e^{{{{x^2} + {x^4}} \over {x\left( {\sqrt {1 + 0 + 0} + 1} \right)}}}} - 1} \right] \times \left( {\sqrt {1 + 0 + 0} + 1} \right)} \over {x + {x^3}}}

=

2\mathop {\lim }\limits_{x \to 0} {{\left[ {{e^{{{{x^2} + {x^4}} \over {2x}}}} - 1} \right]} \over {x + {x^3}}}

=

2\mathop {\lim }\limits_{x \to 0} {{\left[ {{e^{{{x + {x^3}} \over 2}}} - 1} \right]} \over {{{x + {x^3}} \over 2} \times 2}}

=

2 \times {1 \over 2} \times 1

= 1 Note : As from formula,

\mathop {\lim }\limits_{x \to 0} {{\left[ {{e^{{{x + {x^3}} \over 2}}} - 1} \right]} \over {{{x + {x^3}} \over 2}}}

= 1

Q58

Let f : R

\to

R be a function defined by f(x) = max {x, x2}. Let S denote the set of all points in R, where f is not differentiable. Then :

A {0, 1}

B {0}

C

\phi

(an empty set)

D {1}

Correct Answer

Option A

Solution

From graph you can see, (1) when x < 0 then y = x2 is greater than y = x.

That is why for f(x) that curved part is chosen. (2) when 0 $\le$ x < 1 then y = x is greater than y = x2.

That is why for f(x) part of that straight line is chosen. (3) when x $\ge$ 1 then y = x2 is greater than y = x.

That is why for f(x) that curved part is chosen.

Here on the graph of f(x) there is two sharp corner at x = 0 and x = 1.

As we know no function is differentiable at the sharp corner.

So f(x) is not differentiable at those two sharp corner.

Q59

\mathop {\lim }\limits_{x \to 0} {\left( {\tan \left( {{\pi \over 4} + x} \right)} \right)^{{1 \over x}}}

is equal to :

A 2

B 1

C

e

D

e

2

Correct Answer

Option D

Solution

\mathop {\lim }\limits_{x \to 0} {\left( {\tan \left( {{\pi \over 4} + x} \right)} \right)^{{1 \over x}}}

This is 1 $\infty$ form. =

{e^{\mathop {\lim }\limits_{x \to 0} \left[ {\tan \left( {{\pi \over 4} + x} \right) - 1} \right] \times {1 \over x}}}

=

{e^{\mathop {\lim }\limits_{x \to 0} \left[ {{{1 + \tan x} \over {1 - \tan x}} - 1} \right] \times {1 \over x}}}

=

{e^{\mathop {\lim }\limits_{x \to 0} \left[ {{{2\tan x} \over {x\left( {1 - \tan x} \right)}}} \right]}}

=

{e^{2\mathop {\lim }\limits_{x \to 0} \left[ {{{\tan x} \over x} \times {1 \over {\left( {1 - \tan x} \right)}}} \right]}}

=

{e^{2\mathop {\lim }\limits_{x \to 0} \left[ {1 \times {1 \over {\left( {1 - 0} \right)}}} \right]}}

= e2

Q60

Let [t] denote the greatest integer

\le

t. If for some

\lambda

\in

R - {1, 0},

\mathop {\lim }\limits_{x \to 0} \left| {{{1 - x + \left| x \right|} \over {\lambda - x + \left[ x \right]}}} \right|

= L, then L is equal to :

A 1

B 2

C 0

D

{1 \over 2}

Correct Answer

Option B

Solution

Here

\mathop {\lim }\limits_{x \to 0} \left| {{{1 - x + \left| x \right|} \over {\lambda - x + [x]}}} \right| = L

Here L.H.L.

\mathop {\lim }\limits_{h \to 0^-} \left| {{{1 + h + h} \over {\lambda + h - 1}}} \right| = \left| {{1 \over {\lambda - 1}}} \right|

R.H.L. =

\mathop {\lim }\limits_{h \to 0^+} \left| {{{1 - h + h} \over {\lambda + h + 0}}} \right| = \left| {{1 \over \lambda }} \right|

$\because$ Limit exists. Hence L.H.L. = R.H.L. $\Rightarrow$

\left| {\lambda - 1} \right| = \left| \lambda \right|

$\Rightarrow$

\lambda = {1 \over 2}

$\therefore$ L =

{1 \over {\left| \lambda \right|}}

= 2