Limits, Continuity and Differentiability — Page 18 | JEE Mathematics

Q171

For each t

\in

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

\mathop {\lim }\limits_{x \to 1^ + } {{\left( {1 - \left| x \right| + \sin \left| {1 - x} \right|} \right)\sin \left( {{\pi \over 2}\left[ {1 - x} \right]} \right)} \over {\left| {1 - x} \right|.\left[ {1 - x} \right]}}

A equals

-

1

B equals 1

C equals 0

D does not exist

Correct Answer

Option C

Solution

\mathop {\lim }\limits_{x \to {1^ + }} {{\left( {1 - \left| x \right| + \sin \left| {1 - x} \right|} \right)\sin \left( {{\pi \over 2}\left[ {1 - x} \right]} \right)} \over {\left| {1 - x} \right|\left[ {1 - x} \right]}}

$=$

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

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

$=$

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

Q172

Let

f(2) = 4

and

f'(x) = 4.

Then

\mathop {\lim }\limits_{x \to 2} {{xf\left( 2 \right) - 2f\left( x \right)} \over {x - 2}}

is given by

A

2

B

- 2

C

- 4

D

3

Correct Answer

Option C

Solution

Given,

\mathop {\lim }\limits_{x \to 2} {{xf\left( 2 \right) - 2f\left( x \right)} \over {x - 2}}

=

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

=

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

=

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

\left[ {As\,f'\left( x \right) = \mathop {\lim }\limits_{x \to 2} {{f\left( x \right) - f\left( 2 \right)} \over {x - 2}}} \right]

=

f\left( 2 \right) - 2f'\left( x \right)

= 4 - 2 $\times$ 4 = - 4

Q173

Let a, b

\in

R, (a

\ne

0). If the function f defined as

f\left( x \right) = \left\{ \begin{array}{ll}{{{2{x^2}} \over a}\,\,,} & {0 \le x < 1} \\ {a\,\,\,,} & {1 \le x < \sqrt 2 } \\ {{{2{b^2} - 4b} \over {{x^3}}},} & {\sqrt 2 \le x < \infty } \end{array} \right.

is continuous in the interval [0,

\infty

), then an ordered pair ( a, b) is :

A

\left( {\sqrt 2 ,1 - \sqrt 3 } \right)

B

\left( { - \sqrt 2 ,1 + \sqrt 3 } \right)

C

\left( {\sqrt 2 , - 1 + \sqrt 3 } \right)

D

\left( { - \sqrt 2 ,1 - \sqrt 3 } \right)

Correct Answer

Option A

Solution

f(x) is continuous at x = 1 $\therefore$

{{2{{\left( 1 \right)}^2}} \over a} = a

$\Rightarrow$ a2 = 2 $\Rightarrow$ a = $\pm$

\sqrt 2

Also f(x) is continuous at x =

\sqrt 2

$\therefore$ a =

{{2{b^2} - 4b} \over {2\sqrt 2 }}

Now, when a =

\sqrt 2 ,

then 4 = 2b2 $-$ 4b $\Rightarrow$ b2 $-$ 2b = 2 $\Rightarrow$ b2 $-$ 2b $-$ 2 = 0 $\Rightarrow$ b =

{{2 \mp \sqrt {4 + 4.2} } \over 2}

= 1 $\pm$

\sqrt 3

$\therefore$ (a, b) =

\left( {\sqrt 2 ,1 \pm \sqrt 3 } \right)

When a = $-$

\sqrt 2

, then $-$

\sqrt 2

=

{{2{b^2} - 4b} \over {2\sqrt 2 }}

$\Rightarrow$ $-$ 4 = 2b2 $-$ 4b $\Rightarrow$ b2 $-$ 2b + 2 = 0 $\therefore$ b =

{{2 \pm \sqrt {4 - 8} } \over 2}

= 1 $\pm$ i As b

\in

Real number so, b = 1 $\pm$ i is not accepted. $\therefore$ (a, b) =

\left( {\sqrt 2 ,1 + \sqrt 3 } \right)

or

\left( {\sqrt 2 ,1 - \sqrt 3 } \right)

Q174

If the function f defined as

f\left( x \right) = {1 \over x} - {{k - 1} \over {{e^{2x}} - 1}},x \ne 0,

is continuous at x = 0, then the ordered pair (k, f(0)) is equal to :

A (3, 2)

B (3, 1)

C (2, 1)

D

\left( {{1 \over 3},\,2} \right)

Correct Answer

Option B

Solution

If the function is continuous at x = 0, then

\mathop {\lim }\limits_{x \to 0}

f(x) will exist and f(0) =

\mathop {\lim }\limits_{x \to 0}

f(x) Now,

\mathop {\lim }\limits_{x \to 0}

f(x) =

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

=

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

=

\mathop {\lim }\limits_{x \to 0} \left[ {{{\left( {1 + 2x + {{{{\left( {2x} \right)}^2}} \over {2!}} + {{{{\left( {2x} \right)}^3}} \over {3!}} + ....} \right) - 1 - kx + x} \over {\left( x \right)\left( {\left( {1 + 2x + {{{{\left( {2x} \right)}^2}} \over {2!}} + {{{{\left( {2x} \right)}^3}} \over {3!}} + ...} \right) - 1} \right)}}} \right]

=

\mathop {\lim }\limits_{x \to 0} \left[ {{{\left( {3 - k} \right)x + {{4{x^2}} \over {2!}} + {{8{x^3}} \over {3!}} + ...} \over {\left( {2{x^2} + {{4{x^3}} \over {2!}} + {{8{x^3}} \over {3!}} + ....} \right)}}} \right]

For the limit to exist, power of x in the numerator should be greater than or equal to the power of x in the denominator.

Therefore, coefficient of x in numerator is equal to zero $\Rightarrow$ 3 $-$ k = 0 $\Rightarrow$ k = 3 So the limit reduces to

\mathop {\lim }\limits_{x \to 0} {{\left( {{x^2}} \right)\left( {{4 \over {2!}} + {{8x} \over {3!}} + ...} \right)} \over {\left( {{x^2}} \right)\left( {2 + {{4x} \over {2!}} + {{8{x^2}} \over {3!}} + ...} \right)}}

=

\mathop {\lim }\limits_{x \to 0} {{{4 \over {2!}} + {{8x} \over {3!}} + ...} \over {2 + {{4x} \over {2!}} + {{8{x^2}} \over {3!}} + ...}}

= 1 Hence, f(0) = 1

Q175

Let f(x) =

\left\{ \begin{array}{ll}{{{\left( {x - 1} \right)}^{{1 \over {2 - x}}}},} & {x > 1,x \ne 2} \\ {k\,\,\,\,\,\,\,\,\,\,\,\,\,\,} & {,x = 2} \end{array} \right.

Thevaue of k for which f s continuous at x = 2 is :

A 1

B e

C e-1

D e-2

Correct Answer

Option C

Solution

Since f(x) is continuous at x = 2.

\therefore\,\,\,

\mathop {\lim }\limits_{x \to 2}

f(x) = f(2) $\Rightarrow$

\mathop {\lim }\limits_{x \to 2}

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

= k (

{{1^\infty }}

form) $\therefore$

{{e^l}}

= k Where

l

=

\mathop {\lim }\limits_{x \to 2}

(x $-$

l

$-$ 1) $\times$

{1 \over {2 - x}}

=

\mathop {\lim }\limits_{x \to 2} {{x - 2} \over {2 - x}}

=

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

$\Rightarrow$ k = e $-$ 1

Q176

Let S = {(

\lambda

,

\mu

)

\in

R

\times

R : f(t) = (|

\lambda

| e|t|

-

\mu

). sin (2|t|), t

\in

R, is a differentiable function}. Then S is a subset of :

A R

\times

[0,

\infty

)

B [0,

\infty

)

\times

R

C R

\times

(

-

\infty

, 0)

D (

-

\infty

, 0)

\times

R

Correct Answer

Option A

Solution

S = {( $\lambda$ , $\mu$ )

\in

R $\times$ R : f(t) =

\left( {\left| \lambda \right|{e^{\left| t \right|}} - \mu } \right)

sin

\left( {2\left| t \right|} \right),

t

\in

R f(t) =

\left( {\left| \lambda \right|{e^{\left| t \right|}} - \mu } \right)\sin \left( {2\left| t \right|} \right)

=

\left\{ \begin{array}{ll}{\left( {\left| \lambda \right|{e^t} - \mu } \right)\sin 2t,} & {t > 0} \\ {\left( {\left| \lambda \right|{e^{ - t}} - \mu } \right)\left( { - \sin 2t} \right),} & {t < 0} \end{array} \right.

f'(t) =

\left\{ \begin{array}{ll}{\left( {\left| \lambda \right|{e^t}} \right)\sin 2t + \left( {\left| \lambda \right|{e^t} - \mu } \right)\left( {2\cos 2t} \right),\,\,t > 0} \\ {\left| \lambda \right|{e^{ - t}}\sin 2t + \left( {\left| \lambda \right|{e^{ - t}} - \mu } \right)\left( {-2\cos 2t} \right),t < 0} \end{array} \right.

As, f(t) is differentiable $\therefore$ LHD = RHD at t = 0 $\Rightarrow$

\left| \lambda \right|

. sin2(0) +

\left( {\left| \lambda \right|{e^0} - \mu } \right)

2cos(0) =

\left| \lambda \right|{e^{ - 0}}\,

. sin 2(0) $-$ 2 cos (0)

\left( {\left| \lambda \right|{e^{ - 0}} - \mu } \right)

$\Rightarrow$

\,\,\,

0 +

\left( {\left| \lambda \right| - \mu } \right)

2 = 0 $-$ 2

\left( {\left| \lambda \right| - \mu } \right)

$\Rightarrow$ 4

\left( {\left| \lambda \right| - \mu } \right)

= 0 $\Rightarrow$

\left| \mu \right|

= $\mu$ So, S $\equiv$ ( $\lambda$ , $\mu$ ) = { $\lambda$

\in

R & $\mu$

\in

[0, $\infty$ )} Therefore set S is subset of R $\times$ [0, $\infty$ )

Q177

\mathop {\lim }\limits_{x \to {1^ - }} {{\sqrt \pi - \sqrt {2{{\sin }^{ - 1}}x} } \over {\sqrt {1 - x} }}

is equal to :

A

\sqrt {{2 \over \pi }}

B

{1 \over {\sqrt {2\pi } }}

C

\sqrt {{\pi \over 2}}

D

\sqrt \pi

Correct Answer

Option A

Solution

\mathop {\lim }\limits_{x \to {1^ - }} {{\sqrt \pi - \sqrt {2{{\sin }^{ - 1}}x} } \over {\sqrt {1 - x} }} \times {{\sqrt \pi + \sqrt {2{{\sin }^{ - 1}}} } \over {\sqrt \pi + \sqrt {2{{\sin }^{ - 1}}x} }}

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

\mathop {\lim }\limits_{x \to {1^ - }} {{2{{\cos }^{ - 1}}x} \over {\sqrt {1 - x} }}.{1 \over {2\sqrt \pi }}

Put

x = \cos \theta

\mathop {\lim }\limits_{x \to {0^ + }} {{2\theta } \over {\sqrt 2 \sin \left( {{\theta \over 2}} \right)}}.{1 \over {2\sqrt \pi }}

= \sqrt {{2 \over \pi }}

Q178

Let f be a differentiable function such that f(1) = 2 and f '(x) = f(x) for all x

\in

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

A 4e

B 2e2

C 4e2

D 2e

Correct Answer

Option A

Solution

{{f'(x)} \over {f(x)}} = 1\forall x \in R

Intergrate & use f(1) = 2 f(x) = 2ex-1 $\Rightarrow$ f '(x) = 2ex $-$ 1 h(x) = f(f(x)) $\Rightarrow$ h'(x) = f '(f(x)) f'(x) h'(1) = f '(f(1)) f'(1) = f '(2) f '(1) = 2e . 2 = 4e

Q179

Let f : R

\to

R be a function defined as

f(x) = \left\{ \begin{array}{lll}5 & ; & {x \le 1} \\ {a + bx} & ; & {1 < x < 3} \\ {b + 5x} & ; & {3 \le x < 5} \\ {30} & ; & {x \ge 5} \end{array} \right.

Then, f is

A continuous if a = 0 and b = 5

B continuous if a = –5 and b = 10

C continuous if a = 5 and b = 5

D not continuous for any values of a and b

Correct Answer

Option D

Solution

Checking if f(x) is continuous at x = 1 : f(1 $-$ ) = 5 f(1) = 5 f(1+) = a + b if f(x) is continuous at x = 1, then f(1 $-$ ) = f(1) = f(1+) $\Rightarrow$ 5 = 5 = a + b $\therefore$ a + b = 5 . . . . . . . . (1) checking if f(x) is continuous at x = 3 : f(3 $-$ ) = a + 3b f(3) = b + 15 f(3+) = b + 15 if f(x) = is continuous at x = 3 then, f(3 $-$ ) = f(3) = f(3+) $\Rightarrow$ a + 3b = b + 15 = b + 15 $\Rightarrow$ a + 2b = 15 . . . . . (2) checking if f(x) is continuous at x = 5 : f(5 $-$ ) = b + 25 f(5) = 30 f(5+) = 30 if f(x) is continuous at x = 5 then, f(5 $-$ ) = f(5) = f(5+) $\Rightarrow$ b + 25 = 30 = 30 $\Rightarrow$ b = 5 By putting this value in equation (2), we get, a + 2(5) = 15 $\Rightarrow$ a = 5 when a = 5 and b = 5 then equation (1) a + b = 5 does not satisfy. $\therefore$ f is not continuous for any value of a and b.

Q180

Let [x] denote the greatest integer less than or equal to x. Then

\mathop {\lim }\limits_{x \to 0} {{\tan \left( {\pi {{\sin }^2}x} \right) + {{\left( {\left| x \right| - \sin \left( {x\left[ x \right]} \right)} \right)}^2}} \over {{x^2}}}

A equals

\pi

+ 1

B equals 0

C does not exist

D equals

\pi

Correct Answer

Option C

Solution

R.H.L. $=$

\mathop {\lim }\limits_{x \to {0^ + }} {{\tan \left( {\pi {{\sin }^2}x} \right) + {{\left( {\left| x \right| - \sin \left( {x\left[ x \right]} \right)} \right)}^2}} \over {{x^2}}}

(as x $\to$ 0+ $\Rightarrow$ [x] $=$ 0) $=$

\mathop {\lim }\limits_{x \to {0^ + }} {{\tan \left( {\pi {{\sin }^2}x} \right) + {x^2}} \over {{x^2}}}

$=$

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

L.H.L. $=$

\mathop {\lim }\limits_{x \to {0^ + }} {{\tan \left( {\pi {{\sin }^2}x} \right) + {{\left( { - x + \sin x} \right)}^2}} \over {{x^2}}}

(as x $\to$ 0 $-$ $\Rightarrow$ [x] $=$ $-$ 1)

\mathop {\lim }\limits_{x \to {0^ + }} {{\tan \left( {\pi {{\sin }^2}x} \right)} \over {\pi {{\sin }^2}x}}\,.\,{{\pi {{\sin }^2}x} \over {{x^2}}} + {\left( { - 1 + {{\sin x} \over x}} \right)^2} \Rightarrow \pi

R.H.L. $\ne$ L.H.L.