QBManager

Limits, Continuity and Differentiability

79890 If $\begin{aligned} f (x) & = \frac{4 \sin π x}{5 x}, for x \neq 0 \\ = 2 k, & for x = 0 \end{aligned}$
is continuous at $x = 0$ , then the value of $k$ is

1

\frac{4 π}{5}

2

\frac{π}{5}

3

\frac{2 π}{5}

4

\frac{π}{10}

Explanation:

(C) : Given, $f (x) = \frac{4 \sin π x}{5 x}, x \neq 0$
$f (x) = 2 k, at x = 0$
$f (0) = 2 k$
Since, $f (x)$ is continuous at $x = 0$ ,
$lim_{x \to 0} f (x) = f (0)$
$lim_{x \to 0} \frac{4 \sin π x}{5 x} = 2 k$
$lim_{x \to 0} (\frac{4 \sin π x}{π x}) \cdot \frac{π}{5} = 2 k$
Or
(1). $\frac{4 π}{5} = 2 k$
$k = \frac{2 π}{5}$

MHT CET-2020

Limits, Continuity and Differentiability

79891 If $f (x) = \frac{1 - \sin x + \cos x}{1 + \sin x + \cos x}$ , for $x \neq π$ is
continuous at $x = π$ , then $f (π) =$

1 -1

2 2

3 0

4 1

Explanation:

(A) : Given,
$f (x)$ is continuous at $x = π$
$f (π) = lim_{x \to π} f (x) = lim_{x \to π} \frac{1 - \sin x + \cos x}{1 + \sin x + \cos x}$
Apply L-Hospital rule,
$= lim_{x \to π} \frac{- \cos x - \sin x}{\cos x - \sin x}$
$= \frac{- \cos π - \sin π}{\cos π - \sin π} = \frac{- (- 1) - 0}{- 1 - 0} = \frac{1}{- 1} = - 1$

MHT CET-2020

Limits, Continuity and Differentiability

79892 If
$\begin{aligned} f (x) & = \frac{(e^{3 x} - 1) \sin x^{0}}{x^{2}}, & if x \neq 0 \\ = \frac{π}{60}, & if x = 0 \end{aligned}$

1

lim_{x \to 0} f (x) = 3

2 f has removable discontinuity at

x = 0

3

f

is continuous at

x = 0

4 f has irremovable discontinuity at

x = 0

Explanation:

(C) : Given, $f (x) = \frac{π}{60}, x = 0$
$f (0) = \frac{π}{60}$
$lim_{x \to 0} f (x) = lim_{x \to 0} \frac{(e^{3 x} - 1) \sin x^{0}}{x^{2}}$
Dividing numerator and denominator by $x^{2}$ , we get
$= \frac{lim_{x \to 0} (\frac{e^{3 x} - 1}{3 x} \times 3) \frac{\sin {(\frac{π x}{180})}^{c}}{{(\frac{π x}{180})}^{c}} \times (\frac{π}{180})}{(\frac{x^{2}}{x^{2}})}$
$lim_{x \to 0} f (x) = (3) (1) (\frac{π}{180}) = \frac{π}{60} {∵ lim_{x \to 0} \frac{e^{x} - 1}{x} = 1}$
$∵$ at $x = 0, f (0) = lim_{x \to 0} f (x)$
Hence, $f (x)$ is continuous at $x = 0$ .

MHT CET-2020

Limits, Continuity and Differentiability

79893 If
$\begin{aligned} f (x) = {[\tan (\frac{π}{4} + x)]}^{\frac{1}{x}} if x \neq 0 \\ = k if x = 0, \end{aligned}$
is continuous at $x = 0$ then $k =$

1 e

2

\sqrt{e}

3

e^{4}

4

e^{2}

Explanation:

(D) : Given, $f (x) = {[\tan (\frac{π}{4} + x)]}^{\frac{1}{x}}$ if $x \neq 0$
$= k if x = 0$
Given $f (x)$ is continuous at $x = 0$
$∴ lim_{x \to 0} f (x) = f (0)$
$lim_{x \to 0} {[\tan (\frac{π}{4} + x)]}^{\frac{1}{x}} = k$
$lim_{x \to 0} {(\frac{1 + \tan x}{1 - \tan x})}^{\frac{1}{x}} = k$
$[∵ lim_{x \to 0} [f (x)]^{g (x)} = e^{lim_{x \to 0} [f (x) - 1] g (x)}]$
$∴ e^{lim_{x \to 0} (\frac{1 + \tan x}{1 - \tan x} - 1) \frac{1}{x}} = k$
$e^{2 lim_{x \to 0} \frac{\frac{\tan x}{x}}{(1 - \tan 0^{\circ})}} = k$
$e^{2 (\frac{1}{1 - 0})} = k$
$k = e^{2}$

MHT CET-2020

Limits, Continuity and Differentiability

79890 If $\begin{aligned} f (x) & = \frac{4 \sin π x}{5 x}, for x \neq 0 \\ = 2 k, & for x = 0 \end{aligned}$
is continuous at $x = 0$ , then the value of $k$ is

1

\frac{4 π}{5}

2

\frac{π}{5}

3

\frac{2 π}{5}

4

\frac{π}{10}

Explanation:

(C) : Given, $f (x) = \frac{4 \sin π x}{5 x}, x \neq 0$
$f (x) = 2 k, at x = 0$
$f (0) = 2 k$
Since, $f (x)$ is continuous at $x = 0$ ,
$lim_{x \to 0} f (x) = f (0)$
$lim_{x \to 0} \frac{4 \sin π x}{5 x} = 2 k$
$lim_{x \to 0} (\frac{4 \sin π x}{π x}) \cdot \frac{π}{5} = 2 k$
Or
(1). $\frac{4 π}{5} = 2 k$
$k = \frac{2 π}{5}$

MHT CET-2020

Limits, Continuity and Differentiability

79891 If $f (x) = \frac{1 - \sin x + \cos x}{1 + \sin x + \cos x}$ , for $x \neq π$ is
continuous at $x = π$ , then $f (π) =$

1 -1

2 2

3 0

4 1

Explanation:

(A) : Given,
$f (x)$ is continuous at $x = π$
$f (π) = lim_{x \to π} f (x) = lim_{x \to π} \frac{1 - \sin x + \cos x}{1 + \sin x + \cos x}$
Apply L-Hospital rule,
$= lim_{x \to π} \frac{- \cos x - \sin x}{\cos x - \sin x}$
$= \frac{- \cos π - \sin π}{\cos π - \sin π} = \frac{- (- 1) - 0}{- 1 - 0} = \frac{1}{- 1} = - 1$

MHT CET-2020

Limits, Continuity and Differentiability

79892 If
$\begin{aligned} f (x) & = \frac{(e^{3 x} - 1) \sin x^{0}}{x^{2}}, & if x \neq 0 \\ = \frac{π}{60}, & if x = 0 \end{aligned}$

1

lim_{x \to 0} f (x) = 3

2 f has removable discontinuity at

x = 0

3

f

is continuous at

x = 0

4 f has irremovable discontinuity at

x = 0

Explanation:

(C) : Given, $f (x) = \frac{π}{60}, x = 0$
$f (0) = \frac{π}{60}$
$lim_{x \to 0} f (x) = lim_{x \to 0} \frac{(e^{3 x} - 1) \sin x^{0}}{x^{2}}$
Dividing numerator and denominator by $x^{2}$ , we get
$= \frac{lim_{x \to 0} (\frac{e^{3 x} - 1}{3 x} \times 3) \frac{\sin {(\frac{π x}{180})}^{c}}{{(\frac{π x}{180})}^{c}} \times (\frac{π}{180})}{(\frac{x^{2}}{x^{2}})}$
$lim_{x \to 0} f (x) = (3) (1) (\frac{π}{180}) = \frac{π}{60} {∵ lim_{x \to 0} \frac{e^{x} - 1}{x} = 1}$
$∵$ at $x = 0, f (0) = lim_{x \to 0} f (x)$
Hence, $f (x)$ is continuous at $x = 0$ .

MHT CET-2020

Limits, Continuity and Differentiability

79893 If
$\begin{aligned} f (x) = {[\tan (\frac{π}{4} + x)]}^{\frac{1}{x}} if x \neq 0 \\ = k if x = 0, \end{aligned}$
is continuous at $x = 0$ then $k =$

1 e

2

\sqrt{e}

3

e^{4}

4

e^{2}

Explanation:

(D) : Given, $f (x) = {[\tan (\frac{π}{4} + x)]}^{\frac{1}{x}}$ if $x \neq 0$
$= k if x = 0$
Given $f (x)$ is continuous at $x = 0$
$∴ lim_{x \to 0} f (x) = f (0)$
$lim_{x \to 0} {[\tan (\frac{π}{4} + x)]}^{\frac{1}{x}} = k$
$lim_{x \to 0} {(\frac{1 + \tan x}{1 - \tan x})}^{\frac{1}{x}} = k$
$[∵ lim_{x \to 0} [f (x)]^{g (x)} = e^{lim_{x \to 0} [f (x) - 1] g (x)}]$
$∴ e^{lim_{x \to 0} (\frac{1 + \tan x}{1 - \tan x} - 1) \frac{1}{x}} = k$
$e^{2 lim_{x \to 0} \frac{\frac{\tan x}{x}}{(1 - \tan 0^{\circ})}} = k$
$e^{2 (\frac{1}{1 - 0})} = k$
$k = e^{2}$

MHT CET-2020

Limits, Continuity and Differentiability

79890 If $\begin{aligned} f (x) & = \frac{4 \sin π x}{5 x}, for x \neq 0 \\ = 2 k, & for x = 0 \end{aligned}$
is continuous at $x = 0$ , then the value of $k$ is

1

\frac{4 π}{5}

2

\frac{π}{5}

3

\frac{2 π}{5}

4

\frac{π}{10}

Explanation:

(C) : Given, $f (x) = \frac{4 \sin π x}{5 x}, x \neq 0$
$f (x) = 2 k, at x = 0$
$f (0) = 2 k$
Since, $f (x)$ is continuous at $x = 0$ ,
$lim_{x \to 0} f (x) = f (0)$
$lim_{x \to 0} \frac{4 \sin π x}{5 x} = 2 k$
$lim_{x \to 0} (\frac{4 \sin π x}{π x}) \cdot \frac{π}{5} = 2 k$
Or
(1). $\frac{4 π}{5} = 2 k$
$k = \frac{2 π}{5}$

MHT CET-2020

Limits, Continuity and Differentiability

79891 If $f (x) = \frac{1 - \sin x + \cos x}{1 + \sin x + \cos x}$ , for $x \neq π$ is
continuous at $x = π$ , then $f (π) =$

1 -1

2 2

3 0

4 1

Explanation:

(A) : Given,
$f (x)$ is continuous at $x = π$
$f (π) = lim_{x \to π} f (x) = lim_{x \to π} \frac{1 - \sin x + \cos x}{1 + \sin x + \cos x}$
Apply L-Hospital rule,
$= lim_{x \to π} \frac{- \cos x - \sin x}{\cos x - \sin x}$
$= \frac{- \cos π - \sin π}{\cos π - \sin π} = \frac{- (- 1) - 0}{- 1 - 0} = \frac{1}{- 1} = - 1$

MHT CET-2020

Limits, Continuity and Differentiability

79892 If
$\begin{aligned} f (x) & = \frac{(e^{3 x} - 1) \sin x^{0}}{x^{2}}, & if x \neq 0 \\ = \frac{π}{60}, & if x = 0 \end{aligned}$

1

lim_{x \to 0} f (x) = 3

2 f has removable discontinuity at

x = 0

3

f

is continuous at

x = 0

4 f has irremovable discontinuity at

x = 0

Explanation:

(C) : Given, $f (x) = \frac{π}{60}, x = 0$
$f (0) = \frac{π}{60}$
$lim_{x \to 0} f (x) = lim_{x \to 0} \frac{(e^{3 x} - 1) \sin x^{0}}{x^{2}}$
Dividing numerator and denominator by $x^{2}$ , we get
$= \frac{lim_{x \to 0} (\frac{e^{3 x} - 1}{3 x} \times 3) \frac{\sin {(\frac{π x}{180})}^{c}}{{(\frac{π x}{180})}^{c}} \times (\frac{π}{180})}{(\frac{x^{2}}{x^{2}})}$
$lim_{x \to 0} f (x) = (3) (1) (\frac{π}{180}) = \frac{π}{60} {∵ lim_{x \to 0} \frac{e^{x} - 1}{x} = 1}$
$∵$ at $x = 0, f (0) = lim_{x \to 0} f (x)$
Hence, $f (x)$ is continuous at $x = 0$ .

MHT CET-2020

Limits, Continuity and Differentiability

79893 If
$\begin{aligned} f (x) = {[\tan (\frac{π}{4} + x)]}^{\frac{1}{x}} if x \neq 0 \\ = k if x = 0, \end{aligned}$
is continuous at $x = 0$ then $k =$

1 e

2

\sqrt{e}

3

e^{4}

4

e^{2}

Explanation:

(D) : Given, $f (x) = {[\tan (\frac{π}{4} + x)]}^{\frac{1}{x}}$ if $x \neq 0$
$= k if x = 0$
Given $f (x)$ is continuous at $x = 0$
$∴ lim_{x \to 0} f (x) = f (0)$
$lim_{x \to 0} {[\tan (\frac{π}{4} + x)]}^{\frac{1}{x}} = k$
$lim_{x \to 0} {(\frac{1 + \tan x}{1 - \tan x})}^{\frac{1}{x}} = k$
$[∵ lim_{x \to 0} [f (x)]^{g (x)} = e^{lim_{x \to 0} [f (x) - 1] g (x)}]$
$∴ e^{lim_{x \to 0} (\frac{1 + \tan x}{1 - \tan x} - 1) \frac{1}{x}} = k$
$e^{2 lim_{x \to 0} \frac{\frac{\tan x}{x}}{(1 - \tan 0^{\circ})}} = k$
$e^{2 (\frac{1}{1 - 0})} = k$
$k = e^{2}$

MHT CET-2020

Limits, Continuity and Differentiability

79890 If $\begin{aligned} f (x) & = \frac{4 \sin π x}{5 x}, for x \neq 0 \\ = 2 k, & for x = 0 \end{aligned}$
is continuous at $x = 0$ , then the value of $k$ is

1

\frac{4 π}{5}

2

\frac{π}{5}

3

\frac{2 π}{5}

4

\frac{π}{10}

Explanation:

(C) : Given, $f (x) = \frac{4 \sin π x}{5 x}, x \neq 0$
$f (x) = 2 k, at x = 0$
$f (0) = 2 k$
Since, $f (x)$ is continuous at $x = 0$ ,
$lim_{x \to 0} f (x) = f (0)$
$lim_{x \to 0} \frac{4 \sin π x}{5 x} = 2 k$
$lim_{x \to 0} (\frac{4 \sin π x}{π x}) \cdot \frac{π}{5} = 2 k$
Or
(1). $\frac{4 π}{5} = 2 k$
$k = \frac{2 π}{5}$

MHT CET-2020

Limits, Continuity and Differentiability

79891 If $f (x) = \frac{1 - \sin x + \cos x}{1 + \sin x + \cos x}$ , for $x \neq π$ is
continuous at $x = π$ , then $f (π) =$

1 -1

2 2

3 0

4 1

Explanation:

(A) : Given,
$f (x)$ is continuous at $x = π$
$f (π) = lim_{x \to π} f (x) = lim_{x \to π} \frac{1 - \sin x + \cos x}{1 + \sin x + \cos x}$
Apply L-Hospital rule,
$= lim_{x \to π} \frac{- \cos x - \sin x}{\cos x - \sin x}$
$= \frac{- \cos π - \sin π}{\cos π - \sin π} = \frac{- (- 1) - 0}{- 1 - 0} = \frac{1}{- 1} = - 1$

MHT CET-2020

Limits, Continuity and Differentiability

79892 If
$\begin{aligned} f (x) & = \frac{(e^{3 x} - 1) \sin x^{0}}{x^{2}}, & if x \neq 0 \\ = \frac{π}{60}, & if x = 0 \end{aligned}$

1

lim_{x \to 0} f (x) = 3

2 f has removable discontinuity at

x = 0

3

f

is continuous at

x = 0

4 f has irremovable discontinuity at

x = 0

Explanation:

(C) : Given, $f (x) = \frac{π}{60}, x = 0$
$f (0) = \frac{π}{60}$
$lim_{x \to 0} f (x) = lim_{x \to 0} \frac{(e^{3 x} - 1) \sin x^{0}}{x^{2}}$
Dividing numerator and denominator by $x^{2}$ , we get
$= \frac{lim_{x \to 0} (\frac{e^{3 x} - 1}{3 x} \times 3) \frac{\sin {(\frac{π x}{180})}^{c}}{{(\frac{π x}{180})}^{c}} \times (\frac{π}{180})}{(\frac{x^{2}}{x^{2}})}$
$lim_{x \to 0} f (x) = (3) (1) (\frac{π}{180}) = \frac{π}{60} {∵ lim_{x \to 0} \frac{e^{x} - 1}{x} = 1}$
$∵$ at $x = 0, f (0) = lim_{x \to 0} f (x)$
Hence, $f (x)$ is continuous at $x = 0$ .

MHT CET-2020

Limits, Continuity and Differentiability

79893 If
$\begin{aligned} f (x) = {[\tan (\frac{π}{4} + x)]}^{\frac{1}{x}} if x \neq 0 \\ = k if x = 0, \end{aligned}$
is continuous at $x = 0$ then $k =$

1 e

2

\sqrt{e}

3

e^{4}

4

e^{2}

Explanation:

(D) : Given, $f (x) = {[\tan (\frac{π}{4} + x)]}^{\frac{1}{x}}$ if $x \neq 0$
$= k if x = 0$
Given $f (x)$ is continuous at $x = 0$
$∴ lim_{x \to 0} f (x) = f (0)$
$lim_{x \to 0} {[\tan (\frac{π}{4} + x)]}^{\frac{1}{x}} = k$
$lim_{x \to 0} {(\frac{1 + \tan x}{1 - \tan x})}^{\frac{1}{x}} = k$
$[∵ lim_{x \to 0} [f (x)]^{g (x)} = e^{lim_{x \to 0} [f (x) - 1] g (x)}]$
$∴ e^{lim_{x \to 0} (\frac{1 + \tan x}{1 - \tan x} - 1) \frac{1}{x}} = k$
$e^{2 lim_{x \to 0} \frac{\frac{\tan x}{x}}{(1 - \tan 0^{\circ})}} = k$
$e^{2 (\frac{1}{1 - 0})} = k$
$k = e^{2}$

MHT CET-2020

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