Let $$f:[0,\infty ) \to [0,3]$$ be a function defined by $$f(x) = \left\{ {\matrix{ {\max \{ \sin t:0 \le t \le x\} ,} & {0 \le x \le \pi } \cr {2 + \cos x,} & {x > \pi } \cr } } \right.$$

Graph of $$\max \{ \sin t:0 \le t \le x\} $$ in $$x \in [0,\pi ]$$ & graph of cos x for $$x \in [\pi ,\infty )$$ So graph of $$f(x) = \left\{ {\matrix{ {\max \{ \sin t:0 \le t \le x\} ,} & {0 \le x \le \pi } \cr {2 + \cos x,} & {x > \pi} \cr } } \right.$$ f(x) is differentiable everywhere in (0, $$\infty$$)

Limits, Continuity and Differentiability — Page 9 | JEE Mathematics

Q81

Let f : R

\to

R be a function defined as

f(x) = \left\{ \begin{array}{ll}{{\sin (a + 1)x + \sin 2x} \over {2x}},if\,x < 0 \\ b,\,if\,x\, = 0 \\ {{\sqrt {x + b{x^3}} - \sqrt x } \over {b{x^{5/2}}}},\,if\,x > 0 \end{array} \right.

If f is continuous at x = 0, then the value of a + b is equal to :

A

-

3

B

-

2

C

- {5 \over 2}

D

- {3 \over 2}

Correct Answer

Option D

Solution

Given, $f(x)=\left\{\begin{array}{cl}\dfrac{\sin (a+1) x+\sin 2 x}{2 x}, & x0\end{array}\right.$

\begin{array}{ll} \because & f(x) \text{ is continuous at } x=0 . \\\\ \therefore & \lim \limits_{x \rightarrow 0^{-}} f(x)=\lim \limits_{x \rightarrow 0^{+}} f(x)=f(0) \\\\ \because & f(0)=b \end{array}

Now, $\lim \limits_{x \rightarrow 0^{-}} f(x)=\lim \limits_{x \rightarrow 0^{-}}\left(\dfrac{\sin (a+1) x+\sin 2 x}{2 x}\right)$

\begin{aligned} \Rightarrow \quad \lim \limits_{x \rightarrow 0^{-}} f(x) & =\lim \limits_{x \rightarrow 0^{-}}\left(\frac{\sin (a+1) x}{2 x}+\frac{\sin 2 x}{2 x}\right) \\\\ & =\lim \limits_{x \rightarrow 0^{-}}\left(\frac{\sin (a+1) x}{(a+1) x} \times\left(\frac{a+1}{2}\right)+\frac{\sin 2 x}{2 x}\right) \\\\ & =\frac{a+1}{2}+1 \end{aligned}

Again, $\lim \limits_{x \rightarrow 0^{+}} f(x)=\lim \limits_{x \rightarrow 0^{+}}\left(\dfrac{\sqrt{x+b x^3}-\sqrt{x}}{b x^{5 / 2}}\right)$

=\lim \limits_{x \rightarrow 0^{+}} \frac{\left(\sqrt{x+b x^3}-\sqrt{x}\right)\left(\sqrt{x+b x^3}+\sqrt{x}\right)}{b x^{5 / 2}\left(\sqrt{x+b x^3}+\sqrt{x}\right)}

\begin{aligned} & =\lim _{x \rightarrow 0^{+}} \frac{\left(x+b x^3-x\right)}{b x^{5 / 2}\left(\sqrt{x+b x^3}+\sqrt{x}\right)} \\\\ & =\lim _{x \rightarrow 0^{+}} \frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{1+b x^2}+1\right)} \end{aligned}

\Rightarrow \quad \lim \limits_{x \rightarrow 0^{+}} f(x)=1 / 2

From Eq. (i), (ii), (iii) and (iv)

\begin{aligned} &\frac{1}{2} =b=\frac{a+1}{2}+1 \Rightarrow b=\frac{1}{2}, a=-2 \\\\ &\therefore \quad a+b =\frac{-3}{2} \end{aligned}

Q82

Let a function f : R

\to

R be defined as

f(x) = \left\{ \begin{array}{lll}{\sin x - {e^x}} & {if} & {x \le 0} \\ {a + [ - x]} & {if} & {0 < x < 1} \\ {2x - b} & {if} & {x \ge 1} \end{array} \right.

where [ x ] is the greatest integer less than or equal to x. If f is continuous on R, then (a + b) is equal to:

A 4

B 3

C 2

D 5

Correct Answer

Option B

Solution

Continuous x = 0 f(0+) = f(0 $-$ ) $\Rightarrow$ a $-$ 1 = 0 $-$ e0 $\Rightarrow$ a = 0 Continuous at x = 1 f(1+) = f(1 $-$ ) $\Rightarrow$ 2(1) $-$ b = a + ( $-$ 1) $\Rightarrow$ b = 2 $-$ a + 1 $\Rightarrow$ b = 3 $\therefore$ a + b = 3

Q83

If

f:R \to R

is given by

f(x) = x + 1

, then the value of

\mathop {\lim }\limits_{n \to \infty } {1 \over n}\left[ {f(0) + f\left( {{5 \over n}} \right) + f\left( {{{10} \over n}} \right) + ...... + f\left( {{{5(n - 1)} \over n}} \right)} \right]

is :

A

{3 \over 2}

B

{5 \over 2}

C

{1 \over 2}

D

{7 \over 2}

Correct Answer

Option D

Solution

f(0) + f\left( {{5 \over n}} \right) + f\left( {{{10} \over n}} \right) + ...... + f\left( {{{5(n - 1)} \over n}} \right)

\Rightarrow 1 + 1 + {5 \over n} + 1 + {{10} \over n} + .... + 1 + {{5(n - 1)} \over n}

\Rightarrow n + {5 \over n}{{(n - 1)n} \over 2} = {{2n + 5n - 5} \over 2} = {{7n - 5} \over 2}

\mathop {\lim }\limits_{n \to \infty } {1 \over n}\left( {{{7n - 5} \over 2}} \right) = {7 \over 2}

Q84

Let f : R

\to

R be defined as

f(x) = \left\{ \begin{array}{ll}{{{{x^3}} \over {{{(1 - \cos 2x)}^2}}}{{\log }_e}\left( {{{1 + 2x{e^{ - 2x}}} \over {{{(1 - x{e^{ - x}})}^2}}}} \right),} & {x \ne 0} \\ {\alpha ,} & {x = 0} \end{array} \right.

If f is continuous at x = 0, then

\alpha

is equal to :

A 1

B 3

C 0

D 2

Correct Answer

Option A

Solution

For continuity

\mathop {\lim }\limits_{x \to 0} {{{x^3}} \over {4{{\sin }^4}x}}(\ln (1 + 2x{e^{ - 2x}}) - 2\ln (1 - x{e^{ - x}})) = \alpha

\mathop {\lim }\limits_{x \to 0} {1 \over {4x}}[2x{e^{ - 2x}} + 2x{e^{ - x}}] = \alpha

= {1 \over 4}(4) = \alpha = 1

Q85

Let

f:[0,\infty ) \to [0,3]

be a function defined by

f(x) = \left\{ \begin{array}{ll}{\max \{ \sin t:0 \le t \le x\} ,} & {0 \le x \le \pi } \\ {2 + \cos x,} & {x > \pi } \end{array} \right.

Then which of the following is true?

A f is continuous everywhere but not differentiable exactly at one point in (0,

\infty

)

B f is differentiable everywhere in (0,

\infty

)

C f is not continuous exactly at two points in (0,

\infty

)

D f is continuous everywhere but not differentiable exactly at two points in (0,

\infty

)

Correct Answer

Option B

Solution

Graph of

\max \{ \sin t:0 \le t \le x\}

in

x \in [0,\pi ]

& graph of cos x for

x \in [\pi ,\infty )

So graph of

f(x) = \left\{ \begin{array}{ll}{\max \{ \sin t:0 \le t \le x\} ,} & {0 \le x \le \pi } \\ {2 + \cos x,} & {x > \pi} \end{array} \right.

f(x) is differentiable everywhere in (0, $\infty$ )

Q86

Let

f:\left( { - {\pi \over 4},{\pi \over 4}} \right) \to R

be defined as

f(x) = \left\{ \begin{array}{lll}{{{(1 + |\sin x|)}^{{{3a} \over {|\sin x|}}}}} & , & { - {\pi \over 4} < x < 0} \\ b & , & {x = 0} \\ {{e^{\cot 4x/\cot 2x}}} & , & {0 < x < {\pi \over 4}} \end{array} \right.

If f is continuous at x = 0, then the value of 6a + b2 is equal to :

A 1

-

e

B e

-

1

C 1 + e

D e

Correct Answer

Option C

Solution

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

\mathop {\lim }\limits_{x \to {0^ + }} x{e^{{{\cot 4x} \over {\cot 2x}}}} = {e^{{1 \over 2}}} = b

\mathop {\lim }\limits_{x \to {0^ - }} {(1 + |\sin x|)^{{{3a} \over {|\sin x|}}}} = {e^{3a}} = {e^{{1 \over 2}}}

a = {1 \over 6} \Rightarrow 6a = 1

$\therefore$

(6a + {b^2}) = (1 + e)

Q87

Let f : R

\to

R be a function such that f(2) = 4 and f'(2) = 1. Then, the value of

\mathop {\lim }\limits_{x \to 2} {{{x^2}f(2) - 4f(x)} \over {x - 2}}

is equal to :

A 4

B 8

C 16

D 12

Correct Answer

Option D

Solution

This limit can be solved using L'Hopital's Rule, which states that for the limit of the form 0/0 or ±∞/±∞, the limit can be found by taking the derivative of the numerator and the derivative of the denominator separately.

\mathop {\lim }\limits_{x \to 2} {{{x^2}f(2) - 4f(x)} \over {x - 2}}

is in the indeterminate form, and we are given that f(2) = 4 and f'(2) = 1, so we can apply L'Hopital's rule.

Taking the derivative of the numerator and the denominator, we get : Numerator: derivative of $x^2 f(2) - 4f(x)$ is $2x f(2) - 4f'(x)$ .

Denominator: derivative of $x - 2$ is $1$ .

So, the limit becomes :

\mathop {\lim }\limits_{x \to 2} {{{2xf(2) - 4f'(x)}} \over 1} = 2 \times 2 \times f(2) - 4 \times f'(2) = 16 - 4 = 12.

Therefore, Option D, 12, is the correct answer.

Q88

Let [t] denote the greatest integer less than or equal to t. Let f(x) = x

-

[x], g(x) = 1

-

x + [x], and h(x) = min{f(x), g(x)}, x

\in

[

-

2, 2]. Then h is :

A continuous in [

-

2, 2] but not differentiable at more than four points in (

-

2, 2)

B not continuous at exactly three points in [

-

2, 2]

C continuous in [

-

2, 2] but not differentiable at exactly three points in (

-

2, 2)

D not continuous at exactly four points in [

-

2, 2]

Correct Answer

Option A

Solution

min{x $-$ [x], 1 $-$ x + [x]} h(x) = min{x $-$ [x], 1 $-$ [x $-$ [x])} $\Rightarrow$ always continuous in [ $-$ 2, 2] but not differentiable at 7 points.

Q89

\mathop {\lim }\limits_{x \to 2} \left( {\sum\limits_{n = 1}^9 {{x \over {n(n + 1){x^2} + 2(2n + 1)x + 4}}} } \right)

is equal to :

A

{9 \over {44}}

B

{5 \over {24}}

C

{1 \over 5}

D

{7 \over {36}}

Correct Answer

Option A

Solution

S = \mathop {\lim }\limits_{x \to 2} \sum\limits_{n = 1}^9 {{x \over {n(n + 1){x^2} + 2(2n + 1)x + 4}}}

S = \sum\limits_{n = 1}^9 {{2 \over {4({n^2} + 3n + 2)}}} = {1 \over 2}\sum\limits_{n = 1}^9 {\left( {{1 \over {n + 1}} - {1 \over {n + 2}}} \right)}

S = {1 \over 2}\left( {{1 \over 2} - {1 \over {11}}} \right) = {9 \over {44}}

Q90

If

\alpha

,

\beta

are the distinct roots of x2 + bx + c = 0, then

\mathop {\lim }\limits_{x \to \beta } {{{e^{2({x^2} + bx + c)}} - 1 - 2({x^2} + bx + c)} \over {{{(x - \beta )}^2}}}

is equal to :

A b2 + 4c

B 2(b2 + 4c)

C 2(b2

-

4c)

D b2

-

4c

Correct Answer

Option C

Solution

\mathop {\lim }\limits_{x \to \beta } {{{e^{2({x^2} + bx + c)}} - 1 - 2({x^2} + bx + c)} \over {{{(x - \beta )}^2}}}

= \mathop {\lim }\limits_{x \to \beta } {{1\left( {1 + {{2({x^2} + bx + c)} \over {1!}} + {{{2^2}{{({x^2} + bx + c)}^2}} \over {2!}} + ...} \right) - 1 - 2({x^2} + bx + c)} \over {{{(x - \beta )}^2}}}

= \mathop {\lim }\limits_{x \to \beta } {{2{{({x^2} + bx + 1)}^2}} \over {{{(x - \beta )}^2}}}

= \mathop {\lim }\limits_{x \to \beta } {{2{{(x - \alpha )}^2}{{(x - \beta )}^2}} \over {{{(x - \beta )}^2}}}

= 2{(\beta - \alpha )^2} = 2({b^2} - 4c)