The value of k for which the function $$f\left( x \right) = \left\{ {\matrix{ {{{\left( {{4 \over 5}} \right)}^{{{\tan \,4x} \over {\tan \,5x}}}}\,\,,} & {0 < x < {\pi \over 2}} \cr {k + {2

$f(x)=\left(\frac{4}{5}\right)^{\frac{\tan 4 x}{\tan 5 x}}$ $\lim\limits_{x \rightarrow \frac{x}{2}}\left(\frac{4}{5}\right)^{\frac{\tan 4 x}{\tan 5 x}}=k+\frac{2}{5}$ $\Rightarrow \lim\limits_{x \rightarrow \frac{x}{2}}\left(\frac{4}{5}\right)^{\tan 4 x \cot 5 x}=k+\frac{2}{5}$ $\Rightarrow\left(\frac{4}{5}\right)^{\lim\limits_{x \rightarrow \frac{x}{2}}(\tan 4 x \cdot \cot (5 x))}=k+\frac{2}{5}$ $\Rightarrow\left(\frac{4}{5}\right)^{0 \times \cos \left(2 x+\frac{x}{2}\right)}=k+\frac{2}{5}$ $\

Limits, Continuity and Differentiability — Page 4 | JEE Mathematics

Q31

Consider the function,

f\left( x \right) = \left| {x - 2} \right| + \left| {x - 5} \right|,x \in R

Statement - 1 :

f'\left( 4 \right) = 0

Statement - 2 :

f

is continuous in [2, 5], differentiable in (2, 5) and

f

(2) =

f

(5)

A Statement - 1 is false, statement - 2 is true

B Statement - 1 is true, statement - 2 is true; statement - 2 is a correct explanation for statement - 1

C Statement - 1 is true, statement - 2 is true; statement - 2 is not a correct explanation for statement - 1

D Statement - 1 is true, statement - 2 is false

Correct Answer

Option C

Solution

f\left( x \right) = \left| {x - 2} \right| = \left\{ \begin{array}{ll}{x - 2\,\,\,,} & {x - 2 \ge 0} \\ {2 - x\,\,\,,} & {x - 2 \le 0} \end{array} \right.

\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left\{ \begin{array}{ll}{x - 2\,\,\,,} & {x \ge 2} \\ {2 - x\,\,\,,} & {x \le 2} \end{array} \right.

Similarly,

f\left( x \right) = \left| {x - 5} \right| = \left\{ \begin{array}{ll}{x - 5\,\,\,,} & {x \ge 5} \\ {5 - x\,\,\,,} & {x \le 5} \end{array} \right.

$\therefore$

f\left( x \right) = \left| {x - 2} \right| + \left| {x - 5} \right|

= \left\{ {x - 2 + 5 - x = 3,2 \le x \le 5} \right\}

Thus

\,\,\,\,\,\,\,\,\,f\left( x \right) = 3,2 \le x \le 5

\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,f'\left( x \right) = 0,2 < x < 5

\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,f'\left( 4 \right) = 0

$\therefore$ Statement

1

is true. As

f\left( 2 \right) = 0 + \left| {2 - 5} \right| = 3

and

f\left( 5 \right) = \left| {5 - 2} \right| + 0 = 3

$\therefore$ Statement -

2

is also true but not a correct explanation for statement

1.

Q32

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

is equal to

A

- {1 \over 4}

B

{1 \over 2}

C 1

D 2

Correct Answer

Option D

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

Q33

If the function.

g\left( x \right) = \left\{ \begin{array}{ll}{k\sqrt {x + 1} ,} & {0 \le x \le 3} \\ {m\,x + 2,} & {3 < x \le 5} \end{array} \right.

is differentiable, then the value of

k+m

is :

A

{{10} \over 3}

B

4

C

2

D

{{16} \over 5}

Correct Answer

Option C

Solution

Since

g(x)

is differentiable, - it will be continuous at

x=3

$\therefore$

\mathop {\lim }\limits_{x \to {3^ - }} g\left( x \right) = \mathop {\lim }\limits_{x \to {3^ + }} g\left( x \right)

2k = 3m + 2\,\,\,\,\,...\left( 1 \right)

Also

g(x)

is differentiable at

x=0

$\therefore$

\mathop {\lim }\limits_{x \to {3^ - }} g'\left( x \right) = \mathop {\lim }\limits_{x \to {3^ + }} g'\left( x \right)

{K \over {2\sqrt {3 + 1} }} = m

k=4m

\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 2 \right)

Solving

(1)

and

(2)

, we get

m = {2 \over 5},\,\,k = {8 \over 5}

$\therefore$

k+m=2

Q34

For

x \in \,R,\,\,f\left( x \right) = \left| {\log 2 - \sin x} \right|\,\,

and

\,\,g\left( x \right) = f\left( {f\left( x \right)} \right),\,\,

then :

A

g

is not differentiable at

x=0

B

g'\left( 0 \right) = \cos \left( {\log 2} \right)

C

g'\left( 0 \right) = - \cos \left( {\log 2} \right)

D

g

is differentiable at

x=0

and

g'\left( 0 \right) = - \sin \left( {\log 2} \right)

Correct Answer

Option B

Solution

g\left( x \right) = f\left( {f\left( x \right)} \right)

In the neighbourhood of

x=0,

f\left( x \right) = \left| {\log 2 - \sin \,x} \right| = \left( {\log 2 - \sin x} \right)

$\therefore$

g\left( x \right) = \left| {\log 2 - \sin \left. {\left| {\log 2 - \sin x} \right|} \right|} \right.

= \left( {\log 2 - \sin \left( {\log 2 - \sin x} \right)} \right)

$\therefore$

g(x)

is differentiable and

g'\left( x \right) = - \cos \left( {\log 2 - \sin x} \right)\left( { - \cos x} \right)

\Rightarrow g'\left( 0 \right) = \cos \left( {\log 2} \right)

Q35

Let

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

then

log

p

is equal to :

A

{1 \over 2}

B

{1 \over 4}

C

2

D

1

Correct Answer

Option A

Solution

\ln \,P = \mathop {\lim }\limits_{x \to {0^ + }} {1 \over {2x}}\ln \left( {1 + {{\tan }^2}\sqrt x } \right)

\mathop {\lim }\limits_{x \to {0^ + }} {1 \over x}\ln \left( {\sec \sqrt x \,\,\,\,\,\,} \right)

Applying

L

Hospital's rule :

= \mathop {\lim }\limits_{x \to {0^ + }} {{\sec \sqrt x \tan \sqrt x } \over {\sec \sqrt x .2\sqrt x }}

= \mathop {\lim }\limits_{x \to {0^ + }} {{\tan \sqrt x } \over {2\sqrt x }}

= {1 \over 2}

Q36

The value of k for which the function

f\left( x \right) = \left\{ \begin{array}{ll}{{{\left( {{4 \over 5}} \right)}^{{{\tan \,4x} \over {\tan \,5x}}}}\,\,,} & {0 < x < {\pi \over 2}} \\ {k + {2 \over 5}\,\,\,,} & {x = {\pi \over 2}} \end{array} \right.

is continuous at x =

{\pi \over 2},

is :

A

{{17} \over {20}}

B

{{2} \over {5}}

C

{{3} \over {5}}

D

-

{{2} \over {5}}

Correct Answer

Option C

Solution

$f(x)=\left(\dfrac{4}{5}\right)^{\dfrac{\tan 4 x}{\tan 5 x}}$ $\lim\limits_{x \rightarrow \dfrac{x}{2}}\left(\dfrac{4}{5}\right)^{\dfrac{\tan 4 x}{\tan 5 x}}=k+\dfrac{2}{5}$ $\Rightarrow \lim\limits_{x \rightarrow \dfrac{x}{2}}\left(\dfrac{4}{5}\right)^{\tan 4 x \cot 5 x}=k+\dfrac{2}{5}$ $\Rightarrow\left(\dfrac{4}{5}\right)^{\lim\limits_{x \rightarrow \dfrac{x}{2}}(\tan 4 x \cdot \cot (5 x))}=k+\dfrac{2}{5}$ $\Rightarrow\left(\dfrac{4}{5}\right)^{0 \times \cos \left(2 x+\dfrac{x}{2}\right)}=k+\dfrac{2}{5}$ $\Rightarrow\left(\dfrac{4}{5}\right)^0=k+\dfrac{2}{5}$ $\Rightarrow k+\dfrac{2}{5}=1 \Rightarrow k=1-\dfrac{2}{5} \Rightarrow k=\dfrac{3}{5}$

Q37

\mathop {\lim }\limits_{x \to {\pi \over 2}} {{\cot x - \cos x} \over {{{\left( {\pi - 2x} \right)}^3}}}

equals

A

{1 \over {16}}

B

{1 \over 8}

C

{1 \over {4}}

D

{1 \over {24}}

Correct Answer

Option A

Solution

\mathop {\lim }\limits_{x \to {\pi \over 2}} {{\cot x - \cos x} \over {{{\left( {\pi - 2x} \right)}^3}}}

=

\mathop {\lim }\limits_{x \to {\pi \over 2}} {1 \over 8}.{{\cos x\left( {1 - \sin x} \right)} \over {\sin x{{\left( {{\pi \over 2} - x} \right)}^3}}}

Put

{{\pi \over 2} - x}

= t $\Rightarrow$ x $\to$

{{\pi \over 2}}

t $\to$ 0 =

\mathop {\lim }\limits_{t \to 0} {1 \over 8}.{{\cos \left( {{\pi \over 2} - t} \right)\left( {1 - \sin \left( {{\pi \over 2} - t} \right)} \right)} \over {\sin \left( {{\pi \over 2} - t} \right){{\left( {{\pi \over 2} - {\pi \over 2} + t} \right)}^3}}}

=

\mathop {{1 \over 8}\lim }\limits_{t \to 0} {{\sin t\left( {1 - \cos t} \right)} \over {\cos t{{\left( t \right)}^3}}}

=

\mathop {{1 \over 8}\lim }\limits_{t \to 0} {{\sin t\left( {2{{\sin }^2}{t \over 2}} \right)} \over {\cos t{{\left( t \right)}^3}}}

=

\mathop {{1 \over 4}\lim }\limits_{t \to 0} {{\sin t\left( {{{\sin }^2}{t \over 2}} \right)} \over {\cos t{{\left( t \right)}^3}}}

=

\mathop {\lim }\limits_{t \to 0} \left( {{{\sin t} \over t}} \right){\left( {{{\sin {t \over 2}} \over {{t \over 2}}}} \right)^2}{1 \over {\cos t}}{1 \over 4}

=

{1 \over 4} \times {1 \over 4}

=

{1 \over {16}}

Q38

Let the function

f(x)=\left(x^2-1\right)\left|x^2-a x+2\right|+\cos |x|

be not differentiable at the two points

x=\alpha=2

and

x=\beta

. Then the distance of the point

(\alpha, \beta)

from the line

12 x+5 y+10=0

is equal to :

A 5

B 2

C 4

D 3

Correct Answer

Option D

Solution

$\cos |\mathrm{x}|$ is always differentiable $\therefore$ we have to check only for $\left|\mathrm{x}^2-\mathrm{ax}+2\right|$ $\therefore$ Not differentiable at

x^2-a x+2=0

One root is given, $\alpha=2$

\begin{aligned} \therefore \quad 4 & -2 a+2=0 \\ & a=3 \end{aligned}

$\therefore$ other root $\beta=1$ but for $x=1 f(x)$ is differentiable (Drop)

Q39

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

equals :

A

{1 \over 4}

B 1

C

{1 \over 2}

D

-

{1 \over 2}

Correct Answer

Option C

Solution

Let, L =

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

=

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

(say) $\Rightarrow$ K =

{{x\left[ {{{2\tan x} \over {1 - {{\left( {\tan x} \right)}^2}}}} \right] - 2x\tan x} \over {{{\left( {1 - \left( {1 - 2{{\sin }^2}x} \right)} \right)}^2}}}

=

{{2x\tan x - \left[ {2x\tan x - 2x{{\tan }^3}x} \right]} \over {4{{\sin }^4}x \times (1 - {{\tan }^2}x)}}

=

{{2x{{\tan }^3}x} \over {4{{\sin }^4}x \times \left( {1 - {{\tan }^2}x} \right)}}

=

{{2x{{\tan }^3}x} \over {4{{\sin }^4}x \times \left( {{{{{\cos }^2}x - {{\sin }^2}x} \over {{{\cos }^2}x}}} \right)}}

=

{{2x{{{{\sin }^3}x} \over {{{\cos }^3}x}}} \over {4{{\sin }^4}x \times \left( {{{{{\cos }^2}x - {{\sin }^2}x} \over {{{\cos }^2}x}}} \right)}}

$\Rightarrow$ K =

{x \over {2\sin x \times ({{\cos }^2}x - {{\sin }^2}x)\cos x}}

\therefore\,\,\,

L =

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

{x \over {2\sin x}} \times \mathop {\lim }\limits_{x \to 0} {1 \over {\cos x({{\cos }^2}x - {{\sin }^2}x)}}

=

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

{x \over {2\sin x}} \times \mathop {\lim }\limits_{x \to 0} {1 \over {\cos 0\left( {{{\cos }^2}0 - {{\sin }^2}0} \right)}}

=

{1 \over 2}

Q40

Let S = { t

\in R:f(x) = \left| {x - \pi } \right|.\left( {{e^{\left| x \right|}} - 1} \right)

\sin \left| x \right|

is not differentiable at t}, then the set S is equal to

A {0,

\pi

}

B

\phi

(an empty set)

C {0}

D {

\pi

}

Correct Answer

Option B

Solution

Check differtiability at x = $\pi$ and x = 0 at x = 0 : We have L. H. D =

\mathop {\lim }\limits_{h \to 0} {{f\left( {0 - h} \right) - f\left( 0 \right)} \over { - h}}

=

\mathop {\lim }\limits_{h \to 0} {{\left| { - h - \pi } \right|\left( {{e^{\left| { - h} \right|}} - 1} \right)\sin \left| h \right| - 0} \over { - h}} = 0

R. H. D =

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

= \mathop {\lim }\limits_{h \to 0} {{\left| {h - \pi } \right|\left( {{e^{\left| h \right|}} - 1} \right)\sin \left| h \right| - o} \over h}

= 0

\therefore,\,\,

LHD = RHD Therefore, function is differentiable at x = $\pi$ . at x = $\pi$ : L. H. D =

\mathop {\lim }\limits_{h \to 0} {{f\left( {\pi - h} \right) - f\left( \pi \right)} \over { - h}}

=

\mathop {\lim }\limits_{h \to 0} {{\left| {\pi - h - \pi } \right|\left( {{e^{\left| {\pi - h} \right|}} - 1} \right)\sin \left| {\pi - h} \right| - 0.} \over { - h}}

= 0 RH. D =

\mathop {\lim }\limits_{h \to 0} {{f\left( {\pi + h} \right) - f\left( \pi \right)} \over h}

=

\mathop {\lim }\limits_{h \to 0} {{\left| {\pi + h - \pi } \right|\left( {{e^{\left| {\pi + h} \right|}} - 1} \right)\sin \left| {\pi + h} \right| - 0} \over h}

= 0

\therefore\,\,\,

L.

H.

D = R H D Therefore, function is differentiable at x = $\pi$ , Since, the function f(x) is differentiable at all the points including 0 and $\pi$ .

So, f(x) is differentiable everywhere.

Therefore, set S is empty set.

S = $\phi$