Limits, Continuity and Differentiability — Page 17 | JEE Mathematics

Q161

For

\alpha, \beta, \gamma \in \mathbf{R}

, if

\lim \limits_{x \rightarrow 0} \dfrac{x^2 \sin \alpha x+(\gamma-1) \mathrm{e}^{x^2}}{\sin 2 x-\beta x}=3

, then

\beta+\gamma-\alpha

is equal to :

A

-

1

B 4

C 6

D 7

Correct Answer

Option D

Solution

\begin{aligned} & \text{ At } x \rightarrow 0 \\ & \sin 2 x-\beta x \rightarrow 0 \\ & \Rightarrow \quad \frac{0}{0} \text{ form } \\ & \Rightarrow \quad(\gamma-1) e^0+0 \sin (\alpha x) \rightarrow 0 \\ & \Rightarrow \quad(\gamma-1)=0 \\ & \Rightarrow \quad \gamma=1 \\ & \Rightarrow \quad \lim _{x \rightarrow 0} \frac{x^2 \sin (\alpha x)}{(\sin 2 x-\beta x)}=3 \end{aligned}

$\Rightarrow \lim _{x \rightarrow 0} \dfrac{x^2\left[\alpha x-\dfrac{(\alpha x)^3}{3!}+\dfrac{(\alpha x)^5}{5!}-\cdots\right]}{\left[(2 x)-\dfrac{(2 x)^3}{3!}+\dfrac{(2 x)^5}{5!}\right]-\beta x}$ $\Rightarrow \lim _{x \rightarrow 0} \dfrac{\alpha x^3-\dfrac{\alpha^3 x^5}{3!}+\dfrac{\alpha^5 x^7}{5!}-\cdots}{x(2-\beta)-\dfrac{8 x^3}{6}+\dfrac{2^5 \cdot x^5}{5!}-\cdots}=3$

\begin{aligned} \Rightarrow & 2-\beta=0 \text{ and } \frac{\alpha}{\frac{-8}{6}}=3 \\ \Rightarrow & \beta=2 \\ & \alpha=3\left(-\frac{8}{6}\right)=-4 \\ \Rightarrow & \gamma=1, \beta=2, \alpha=-4 \\ \Rightarrow & \beta+\gamma-\alpha=7 \end{aligned}

Q162

\lim \limits_{x \rightarrow 0^{+}} \dfrac{\tan \left(5(x)^{\dfrac{1}{3}}\right) \log _e\left(1+3 x^2\right)}{\left(\tan ^{-1} 3 \sqrt{x}\right)^2\left(e^{5(x)^{\dfrac{4}{3}}}-1\right)}

is equal to

A

\dfrac{5}{3}

B 1

C

\dfrac{1}{3}

D

\dfrac{1}{15}

Correct Answer

Option C

Solution

$\lim _{x \rightarrow 0^{+}} \dfrac{\tan \left(5(x)^{\dfrac{1}{3}}\right) \ln \left(1+3 x^3\right)}{\left(\tan ^{-1}(3 \sqrt{x})\right)^2\left(e^{5 x^{\dfrac{4}{3}}}-1\right)}$

\mathop {\lim }\limits_{x \to {0^ + }} {{{{\tan \left( {5{{(x)}^{{1 \over 3}}}} \right)} \over {5{{(x)}^{{1 \over 3}}}}}{{\ln (1 + 3{x^2})} \over {3{x^2}}} \times 5{{(x)}^{{1 \over 3}}}(3{x^2})} \over {{{{{\left( {{{\tan }^2}\left( {3\sqrt x } \right)} \right)}^2}} \over {{{\left( {3\sqrt x } \right)}^2}}}{{\left( {{e^{5x{4 \over 3}}} - 1} \right)} \over {5{x^{{1 \over 3}}}}} \times 9x \times 5{x^{{4 \over 3}}}}}

\mathop {\lim }\limits_{x \to {0^ + }} {{{{\tan \left( {5{{(x)}^{{1 \over 3}}}} \right)} \over {5{{(x)}^{{1 \over 3}}}}}{{\ln (1 + 3{x^3})} \over {3{x^2}}} \times 15{x^{{7 \over 3}}}} \over {{{{{\left( {{{\tan }^2}\left( {3\sqrt x } \right)} \right)}^2}} \over {{{\left( {3\sqrt x } \right)}^2}}}{{\left( {{e^{5(x){4 \over 3}}} - 1} \right)} \over {5{x^{{1 \over 3}}}}} \times 45{x^{{7 \over 3}}}}}

= {1 \over 3}

Q163

If\,\mathop {\lim }\limits_{x \to 0} {{\cos (2x) + a\cos (4x) - b} \over {{x^4}}}is\,finite,\,then\,(a + b)\,is\,equal\,to:

A 0

B

\dfrac{3}{4}

C -1

D

\dfrac{1}{2}

Correct Answer

Option D

Solution

To find the value of $a + b$ for which the following limit is finite: $\lim\limits_{x \to 0} \dfrac{\cos(2x) + a \cos(4x) - b}{x^4}$ we start by expanding the cosine functions using their Taylor series: $\cos(2x) = 1 - \dfrac{(2x)^2}{2!} + \dfrac{(2x)^4}{4!} - \ldots$ $\cos(4x) = 1 - \dfrac{(4x)^2}{2!} + \dfrac{(4x)^4}{4!} - \ldots$ Substitute these expansions into the limit expression: $\lim\limits_{x \to 0} \dfrac{\left(1 - \dfrac{4x^2}{2!} + \dfrac{(2x)^4}{4!} - \ldots \right) + a\left(1 - \dfrac{(4x)^2}{2!} + \dfrac{(4x)^4}{4!} - \ldots\right) - b}{x^4}$ To ensure that the limit is finite, the linear and quadratic terms must sum to zero.

Let's equate the constant terms first: For the limit to be finite: $1 + a - b = 0$ Equating the coefficients of $x^2$ : $-2 - 8a = 0$ Solving the second equation, we have: $8a = -2 \\ a = \dfrac{-1}{4}$ Substitute $a$ back into the first equation to find $b$ : $1 + \dfrac{-1}{4} - b = 0 \\ b = 1 - \dfrac{1}{4} \\ b = \dfrac{3}{4}$ Thus, the sum of $a$ and $b$ is: $a + b = \dfrac{-1}{4} + \dfrac{3}{4} = \dfrac{1}{2}$ Therefore, the correct value of $a + b$ is $\dfrac{1}{2}$ .

Q164

Let

\quad f(x)= \begin{cases}(1+a x)^{1 / x} & , x0\end{cases}

be continuous at

x=0

. Then

e^a b c

is equal to:

A 64

B 48

C 36

D 72

Correct Answer

Option B

Solution

$f(x)= \begin{cases}(1+a x)^{1 / x} & , x0\end{cases}$

\begin{aligned} &\text{ Hence, } f(0)=1+6\\ &\mathrm{RHL}=\lim _{x \rightarrow 0^{+}} \frac{(x+4)^{1 / 2}-2}{(x+c)^{1 / 3}-2} \quad\left[\text{ For } \frac{0}{0} \text{ form, } c=8\right] \end{aligned}

\begin{aligned} & =\lim _{x \rightarrow 0} \frac{2\left(1+\frac{x}{4}\right)^{1 / 2}-2}{\left(1+\frac{x}{8}\right)^{1 / 3}-2} \\ & =\lim _{x \rightarrow 0} \frac{1+\frac{x}{8}-1}{1+\frac{x}{8} \cdot \frac{1}{3}-1}=\frac{\frac{1}{8}}{\frac{1}{8} \cdot \frac{1}{3}}=3 \\ & \therefore 1+b=3 \\ & \Rightarrow b=2 \end{aligned}

\begin{aligned} & \mathrm{LHL}=\lim _{x \rightarrow 0}(1+a x)^{1 / x}=\lim _{x \rightarrow 0} e^{\frac{1+a x-1}{x}}=e^a=3 \\ & \therefore \quad e^a \cdot b c=3 \cdot 2.8=48 \end{aligned}

Q165

Let

f

be a differentiable function on

\mathbf{R}

such that

f(2)=1, f^{\prime}(2)=4

. Let

\lim \limits_{x \rightarrow 0}(f(2+x))^{3 / x}=\mathrm{e}^\alpha

. Then the number of times the curve

y=4 x^3-4 x^2-4(\alpha-7) x-\alpha

meets

x

-axis is :

A 3

B 1

C 2

D 0

Correct Answer

Option C

Solution

\begin{aligned} &\begin{aligned} & \lim _{x \rightarrow 0}(f(2+x))^{3 / x}=\left(1^{\infty} \text{ form }\right) \\ & e^{\lim _{x \rightarrow 0 x}^3(f(2+x)-1)}=e^{\lim _{x \rightarrow} 3 f^{\prime}(2+x)} \\ & =e^{3 f^{\prime}(2)} \\ & =e^{12} \\ & \Rightarrow \alpha=12 \\ & y=4 x^3-4 x^2-4(12-7) x-12 \\ & y=4 x^3-4 x^2-20 x-12 \\ & y=4\left(x^3-x^2-5 x-3\right) \\ & =4(x+1)^2(x-3) \end{aligned}\\ &\text{ It meets the } x \text{-axis at two points } \end{aligned}

Q166

Let

f: \mathbb{R} \rightarrow \mathbb{R}

be a continuous function satisfying

f(0)=1

and

f(2 x)-f(x)=x

for all

x \in \mathbb{R}

. If

\lim \limits_{n \rightarrow \infty}\left\{f(x)-f\left(\dfrac{x}{2^n}\right)\right\}=G(x)

, then

\sum\limits_{r=1}^{10} G\left(r^2\right)

is equal to

A 215

B 420

C 385

D 540

Correct Answer

Option C

Solution

\begin{aligned} & f(0)=1, f(2 x)-f(x)=x \\ & \text{ Replace } x \rightarrow \frac{x}{2} \\ & f(x)-f\left(\frac{x}{2}\right)=\frac{x}{2}\quad\text{..... (1)} \end{aligned}

Again, Replace $x \rightarrow \dfrac{x}{2}$ $f\left(\dfrac{x}{2}\right)-f\left(\dfrac{x}{2^2}\right)=\dfrac{x}{2^2}\quad\text{...... (2)}$ : $\qquad$ : $\qquad$ : : $\qquad$ : $\qquad$ : : $\qquad$ : $\qquad$ :

\begin{aligned} & f\left(\frac{x}{2^{n-1}}\right)-f\left(\frac{x}{2^n}\right)=\frac{x}{2^n} \\ & \text{ Adding }(1)+(2)+(3)+\ldots+(\mathrm{n}) \\ & \text{ We get } f(x)-f\left(\frac{x}{2^n}\right)=\frac{x}{2}+\frac{x}{2^2}+\ldots .+\frac{x}{2^n} \\ & \lim _{n \rightarrow \infty}\left(f(x)-f\left(\frac{x}{2^n}\right)\right)=\lim _{n \rightarrow \infty}\left(\frac{x}{2}+\frac{x}{2^n}+\ldots+\frac{x}{2^n}\right) \end{aligned}

\begin{aligned} & f(x)-f(0)=\frac{\frac{x}{2}}{\frac{1}{2}} \\ & \Rightarrow G(x)=x \\ & \Rightarrow G\left(r^2\right)=r^2 \\ & \Rightarrow \sum_{r=1}^{10} G\left(r^2\right)=\sum_{r=1}^{10} G\left(r^2\right) \\ & =\frac{(10)(11)(21)}{6}=(55) 7 \\ & \Rightarrow 385 \end{aligned}

Q167

If

\lim \limits_{x \rightarrow 1^{+}} \dfrac{(x-1)(6+\lambda \cos (x-1))+\mu \sin (1-x)}{(x-1)^3}=-1

, where

\lambda, \mu \in \mathbb{R}

, then

\lambda+\mu

is equal to

A 20

B 19

C 18

D 17

Correct Answer

Option C

Solution

\begin{aligned} &\lim _{x \rightarrow 1^{+}} \frac{(x-1)(6+\lambda \cos (x-1))+\mu \sin (1-x)}{(x-1)^3}=-1\\ &\text{ Let } x-1=t\\ &\lim _{t \rightarrow 0^{+}} \frac{6 t+\lambda t \cos t-\mu \sin t}{t^3}=-1 \end{aligned}

\begin{aligned} & =\lim _{t \rightarrow 0^{+}} \frac{6 t+\lambda t\left(1-\frac{t^2}{2!}+\frac{t^4}{4!}+\cdots\right)-\mu\left(t-\frac{t^3}{3!}+\cdots\right)}{t^3}=-1 \\ & =\lim _{t \rightarrow 0^{+}} \frac{t(6+\lambda-\mu)+t^3\left(-\frac{\lambda}{2}+\frac{\mu}{6}\right)+\cdots}{t^3}=-1 \end{aligned}

\begin{aligned} &\begin{aligned} \therefore\quad & \lambda-\mu+6=0 \quad\text{..... (i)}\\ & \frac{\mu}{6}-\frac{\lambda}{2}=-1 \quad\text{..... (ii)} \end{aligned}\\ &\text{ Solving (i) and (ii) }\\ &\begin{aligned} & \lambda=6, \mu=12 \\ & \lambda+\mu=18 \end{aligned} \end{aligned}

Q168

If the function f(x) =

\left\{ \begin{array}{ll}{ - x} & {x < 1} \\ {a + {{\cos }^{ - 1}}\left( {x + b} \right),} & {1 \le x \le 2} \end{array} \right.

is differentiable at x = 1, then

{a \over b}

is equal to :

A

{{\pi - 2} \over 2}

B

{{ - \pi - 2} \over 2}

C

{{\pi + 2} \over 2}

D

- 1 - {\cos ^{ - 1}}\left( 2 \right)

Correct Answer

Option C

Solution

As f(x) is differentiable at x = 1 $\therefore$

\mathop {\lim }\limits_{x \to {1^ - }} \left( { - x} \right) = \mathop {\lim }\limits_{x \to {1^ + }} \left( {a + {{\cos }^{ - 1}}\left( {x + b} \right)} \right) = f(1)

$\Rightarrow$ $-$ 1 $=$ a + cos $-$ 1(1 + b) $\Rightarrow$ cos $-$ 1 (1 + b) $=$ $-$ 1 $-$ a . . . . . .(

1) As f(x) is differentiable, so, 2 .

H .

D $=$ R .

H .

D Here, L .

H .

D $=$

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

{{f\left( {1 - h} \right) - f\left( 1 \right)} \over { - h}}

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

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

$=$

= \mathop {\lim }\limits_{x \to 0} {h \over { - h}}

$=$ $-$ 1 R. H. D

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

= \mathop {\lim }\limits_{h \to 0} {{a + {{\cos }^{ - 1}}\left( {1 + h + b} \right) - \left[ {a + {{\cos }^{ - 1}}\left( {1 + b} \right)} \right]} \over h}

= \mathop {\lim }\limits_{h \to 0} {{{{\cos }^{ - 1}}\left( {1 + h + b} \right) - {{\cos }^{ - 1}}\left( {1 + b} \right)} \over h}\left[ {{0 \over 0}\,\,form\left. \, \right]} \right.

= \mathop {\lim }\limits_{h \to 0} {{ - 1} \over {\sqrt {1 - {{\left( {1 + h + b} \right)}^2}} }}

[ Using L' Hospital Rule]

= {{ - 1} \over {\sqrt {1 - {{\left( {1 + b} \right)}^2}} }}

$\therefore$

- 1 = {{ - 1} \over {\sqrt {1 - {{\left( {1 + b} \right)}^2}} }}

\Rightarrow 1 - {\left( {1 + b} \right)^2} = 1

$\Rightarrow$

{\left( {1 + b} \right)^2} = 0

$\Rightarrow$

b = - 1

putting value of b in equation (1), we get,

{\cos ^{ - 1}}\left( {1 - 1} \right) = - 1 - a

$\Rightarrow$

{\pi \over 2} = - 1 - a

$\Rightarrow$

a = - 1 - {\pi \over 2}

$\therefore$

{a \over b} = {{ - 1 - {\pi \over 2}} \over { - 1}} = 1 + {\pi \over 2} = {{\pi + 2} \over 2}

Q169

\mathop {\lim }\limits_{x \to 3}

{{\sqrt {3x} - 3} \over {\sqrt {2x - 4} - \sqrt 2 }}

is equal to :

A

\sqrt 3

B

{1 \over {\sqrt 2 }}

C

{{\sqrt 3 } \over 2}

D

{1 \over {2\sqrt 2 }}

Correct Answer

Option B

Solution

Given,

\mathop {\lim }\limits_{x \to 3}

{{\sqrt {3x} - 3} \over {\sqrt {2x - 4} - \sqrt 2 }}

Here if you put x = 3 in

{{\sqrt {3x} - 3} \over {\sqrt {2x - 4 - \sqrt 2 } }}

you will get

{0 \over 0}

form. So, we can apply L' Hospital rule

\therefore\,\,\,

\mathop {\lim }\limits_{x \to 3} {{\sqrt {3x} - 3} \over {\sqrt {2x - 4} - \sqrt 2 }}

=

\mathop {\lim }\limits_{x \to 3}

{{\sqrt 3 .{1 \over {2\sqrt x }}} \over {{2 \over {2\sqrt {2x - 4} }}}}

(applying L' Hospital rule) =

{{\sqrt 3 .{1 \over {2\sqrt 3 }}} \over {{1 \over {\sqrt 6 - 4}}}}

=

{1 \over 2}

$\times$

\sqrt 2

=

{1 \over {\sqrt 2 }}

Q170

\mathop {\lim }\limits_{x \to 0} \,\,{{{{\left( {27 + x} \right)}^{{1 \over 3}}} - 3} \over {9 - {{\left( {27 + x} \right)}^{{2 \over 3}}}}}

equals.

A

{1 \over 3}

B

-

{1 \over 3}

C

-

{1 \over 6}

D

{1 \over 6}

Correct Answer

Option C

Solution

Given,

\mathop {lim}\limits_{x \to 0} \,{{{{\left( {27 + x} \right)}^{{1 \over 3}}} - 3} \over {9 - {{\left( {27 + x} \right)}^{{2 \over 3}}}}}

=

\mathop {\lim }\limits_{x \to 0} \,{{3\left[ {{{\left( {1 + {x \over {27}}} \right)}^{{1 \over 3}}} - 1} \right]} \over {9\left[ {1 - {{\left( {1 + {x \over {27}}} \right)}^{{2 \over 3}}}} \right]}}

=

\mathop {\lim }\limits_{x \to 0} \,\,{{\left( {1 + {x \over {3 \times 27}}} \right) - 1} \over {3\left[ {1 - \left( {1 + {{2x} \over {3 \times 27}}} \right)} \right]}}

=

\mathop {\lim }\limits_{x \to 0} \,\,{{{x \over {81}}} \over {3\left( { - {{2x} \over {81}}} \right)}} = - {1 \over 6}