QBManager

Limits, Continuity and Differentiability

79894 If $f(x)=\frac{\left(e^{2 x}-1\right) \sin x^{0}}{x^{2}}, x \neq 0$ is continuous at $x$ $=0$, then $\mathbf{f}(0)=$

1 $\frac{90}{\pi}$

2 $\frac{\pi}{180}$

3 $\frac{\pi}{90}$

4 $\frac{180}{\pi}$

MHT CET-2020

Limits, Continuity and Differentiability

79895 If the function
$\begin{array}{cc}f(x)=\frac{1-\sin 2 x+\cos 2 x}{1+\sin 2 x+\cos 2 x}, & \text { if } x \neq \frac{\pi}{2} \\ f(x)=k, & \text { if } x=\frac{\pi}{2}\end{array}$
is continuous at $\mathrm{x}=\frac{\boldsymbol{\pi}}{\boldsymbol{2}}$, then $\mathrm{k}=$

1 2

2 0

3 -1

4 1

MHT CET-2020

Limits, Continuity and Differentiability

79896 If $\begin{aligned} f(x) & =\frac{(81)^x-9^x}{(k)^x-1}, \text { if } x \neq 0 \\ & =2, \quad \text { if } x=0\end{aligned}$
is continuous at $x=0$, then the value of $k$ is

1 9

2 2

3 3

4 4

MHT CET-2020

Limits, Continuity and Differentiability

79897 If the function
$\begin{aligned} f(x) & =\frac{\log 10+\log (0.1+2 x)}{2 x}, \text { if } x \neq 0 \\ & =k . \quad \text { if } x=0\end{aligned}$
is continuous at $x=0$, then $k+2=$

1 2

2 12

3 11

4 10

nda

Limits, Continuity and Differentiability

79898 If $f(x)=\frac{\sqrt{2}-\sqrt{1+\sin x}}{\cos ^{2} x}, x \neq \frac{\pi}{2}$ is continuous at $\mathrm{x}=\frac{\pi}{2}$, then $f (\frac{π}{2}) =$

1

\frac{1}{4 \sqrt{2}}

2

\frac{1}{2 \sqrt{2}}

3

\frac{1}{\sqrt{2}}

4

\sqrt{2}

Explanation:

(A) : Given, $f (x) = \frac{\sqrt{2} - \sqrt{1 + \sin x}}{\cos^{2} x}, x \neq \frac{π}{2}$
$f (x)$ is continuous at $x = \frac{π}{2}$
$∴ f (\frac{π}{2}) = lim_{x \to \frac{π}{2}} \frac{\sqrt{2} - \sqrt{1 + \sin x}}{\cos^{2} x}$
$= lim_{x \to \frac{π}{2}} \frac{\sqrt{2} - \sqrt{1 + \sin x}}{\cos^{2} x} \times \frac{\sqrt{2} + \sqrt{1 + \sin x}}{\sqrt{2} + \sqrt{1 + \sin x}}$
$= lim_{x \to \frac{π}{2}} \frac{2 - (1 + \sin x)}{(1 - \sin^{2} x) (\sqrt{2} + \sqrt{1 + \sin x})}$
$= lim_{x \to \frac{π}{2}} \frac{1 - \sin x}{(1 - \sin x) (1 + \sin x) (\sqrt{2} + \sqrt{1 + \sin x})}$
$= lim_{x \to \frac{π}{2}} \frac{1}{(1 + \sin x) (\sqrt{2} + \sqrt{1 + \sin x})}$
$= \frac{1}{(1 + 1) (\sqrt{2} + \sqrt{2})} = \frac{1}{2 (2 \sqrt{2})} = \frac{1}{4 \sqrt{2}}$

MHT CET-2019

Limits, Continuity and Differentiability

79894 If $f (x) = \frac{(e^{2 x} - 1) \sin x^{0}}{x^{2}}, x \neq 0$ is continuous at $x$ $= 0$ , then $f (0) =$

1

\frac{90}{π}

2

\frac{π}{180}

3

\frac{π}{90}

4

\frac{180}{π}

Explanation:

(C) : Given,
$f (x) = \frac{(e^{2 x} - 1) \sin x^{\circ}}{x^{2}}, x \neq 0$
For continuous at $x = 0$
$f (0) = lim_{x \to 0} f (x)$
$\begin{array}{r} f (0) = lim_{x \to 0} \frac{(e^{2 x} - 1) \sin \frac{x π}{180}}{x^{2}} [∵ x^{0} = (\frac{x π}{180})] \\ = lim_{x \to 0} (\frac{e^{2 x} - 1}{x}) (\frac{\sin \frac{π x}{180}}{x}) \end{array}$
$= (2 lim_{x \to 0} \frac{e^{2 x} - 1}{2 x}) \frac{π}{180} [lim_{x \to 0} \frac{\frac{\sin x π}{180}}{\frac{x π}{180}}] = 2 (\log e) \times (\frac{π}{180}) = \frac{π}{90}$

MHT CET-2020

Limits, Continuity and Differentiability

79895 If the function
$\begin{array}{cc} f (x) = \frac{1 - \sin 2 x + \cos 2 x}{1 + \sin 2 x + \cos 2 x}, & if x \neq \frac{π}{2} \\ f (x) = k, & if x = \frac{π}{2} \end{array}$
is continuous at $x = \frac{\boldsymbol π}{\boldsymbol 2}$ , then $k =$

1 2

2 0

3 -1

4 1

Explanation:

(C) : Given, $f (x) = k$ , at $x = \frac{π}{2}$
$f (\frac{π}{2}) = k$
For continuous at $x = \frac{π}{2}, f (\frac{π}{2}) = lim_{x \to \frac{π}{2}} f (x)$
$k = lim_{x \to \frac{π}{2}} \frac{1 - \sin 2 x + \cos 2 x}{1 + \sin 2 x + \cos 2 x}$
$k = lim_{x \to \frac{π}{2}} \frac{(1 + \cos 2 x) - \sin 2 x}{(1 + \cos 2 x) + \sin 2 x}$
$k = lim_{x \to \frac{π}{2}} \frac{2 \cos^{2} x - 2 \sin x \cos x}{2 \cos^{2} x + 2 \sin x \cos x}$
$k = lim_{x \to \frac{π}{2}} \frac{2 \cos x (\cos x - \sin x)}{2 \cos x (\cos x + \sin x)} \Rightarrow k = lim_{x \to \frac{π}{2}} \frac{\cos x - \sin x}{\cos x + \sin x}$
$k = \frac{0 - 1}{0 + 1}$
$k = \frac{1}{1}$
Since, $f (x)$ is continuous at $x = \frac{π}{2}$ therefore, we get $k = - 1$

MHT CET-2020

Limits, Continuity and Differentiability

79896 If $\begin{aligned} f (x) & = \frac{(81)^{x} - 9^{x}}{(k)^{x} - 1}, if x \neq 0 \\ = 2, if x = 0 \end{aligned}$
is continuous at $x = 0$ , then the value of $k$ is

1 9

2 2

3 3

4 4

Explanation:

(C)Given, $f (x) = 2$ at $x = 0$
$f (0) = 2$
For continuous at $x = 0$ ,
$lim_{x \to 0} f (x) = f (0)$
$lim_{x \to 0} \frac{(81)^{x} - (9)^{x}}{k^{x} - 1} = 2$
$lim_{x \to 0} \frac{9^{x} (9^{x} - 1)}{k^{x} - 1} = 2 \Rightarrow \frac{(lim_{x \to 0} 9^{x}) (lim_{x \to 0} \frac{9^{x} - 1}{x})}{lim_{x \to 0} \frac{k^{x} - 1}{x}} = 2$
$\frac{\log 9}{\log k} = 2 \Rightarrow \log 9 = 2 \log k$
$\Rightarrow \log k^{2} = \log 9 \Rightarrow k^{2} = 9 \Rightarrow k = 3$

MHT CET-2020

Limits, Continuity and Differentiability

79897 If the function
$\begin{aligned} f (x) & = \frac{\log 10 + \log (0.1 + 2 x)}{2 x}, if x \neq 0 \\ = k . if x = 0 \end{aligned}$
is continuous at $x = 0$ , then $k + 2 =$

1 2

2 12

3 11

4 10

Explanation:

(B) : Given,
$f (x) = \frac{\log 10 + \log (0.1 + 2 x)}{2 x}, if x \neq 0$
$= k, if x = 0$
For continuous at $x = 0$
$f (0) = lim_{x \to 0} f (x)$
$k = lim_{x \to 0} \frac{\log 10 + \log (0.1 + 2 x)}{2 x}$
$k = lim_{x \to 0} \frac{\log 10 \times (0.1 + 2 x)}{2 x}$
$k = lim_{x \to 0} \frac{\log (1 + 20 x)}{2 x}$
$k = lim_{x \to 0} \frac{10 \times \log (1 + 20 x)}{20 x}$
$k = 10 lim_{x \to 0} \frac{\log (1 + 20 x)}{20 x}$
Then,
$k + 2 = 10 + 2 = 12$

nda

Limits, Continuity and Differentiability

79898 If $f (x) = \frac{\sqrt{2} - \sqrt{1 + \sin x}}{\cos^{2} x}, x \neq \frac{π}{2}$ is continuous at $x = \frac{π}{2}$ , then $f (\frac{π}{2}) =$

1

\frac{1}{4 \sqrt{2}}

2

\frac{1}{2 \sqrt{2}}

3

\frac{1}{\sqrt{2}}

4

\sqrt{2}

Explanation:

(A) : Given, $f (x) = \frac{\sqrt{2} - \sqrt{1 + \sin x}}{\cos^{2} x}, x \neq \frac{π}{2}$
$f (x)$ is continuous at $x = \frac{π}{2}$
$∴ f (\frac{π}{2}) = lim_{x \to \frac{π}{2}} \frac{\sqrt{2} - \sqrt{1 + \sin x}}{\cos^{2} x}$
$= lim_{x \to \frac{π}{2}} \frac{\sqrt{2} - \sqrt{1 + \sin x}}{\cos^{2} x} \times \frac{\sqrt{2} + \sqrt{1 + \sin x}}{\sqrt{2} + \sqrt{1 + \sin x}}$
$= lim_{x \to \frac{π}{2}} \frac{2 - (1 + \sin x)}{(1 - \sin^{2} x) (\sqrt{2} + \sqrt{1 + \sin x})}$
$= lim_{x \to \frac{π}{2}} \frac{1 - \sin x}{(1 - \sin x) (1 + \sin x) (\sqrt{2} + \sqrt{1 + \sin x})}$
$= lim_{x \to \frac{π}{2}} \frac{1}{(1 + \sin x) (\sqrt{2} + \sqrt{1 + \sin x})}$
$= \frac{1}{(1 + 1) (\sqrt{2} + \sqrt{2})} = \frac{1}{2 (2 \sqrt{2})} = \frac{1}{4 \sqrt{2}}$

MHT CET-2019

Limits, Continuity and Differentiability

79894 If $f (x) = \frac{(e^{2 x} - 1) \sin x^{0}}{x^{2}}, x \neq 0$ is continuous at $x$ $= 0$ , then $f (0) =$

1

\frac{90}{π}

2

\frac{π}{180}

3

\frac{π}{90}

4

\frac{180}{π}

Explanation:

(C) : Given,
$f (x) = \frac{(e^{2 x} - 1) \sin x^{\circ}}{x^{2}}, x \neq 0$
For continuous at $x = 0$
$f (0) = lim_{x \to 0} f (x)$
$\begin{array}{r} f (0) = lim_{x \to 0} \frac{(e^{2 x} - 1) \sin \frac{x π}{180}}{x^{2}} [∵ x^{0} = (\frac{x π}{180})] \\ = lim_{x \to 0} (\frac{e^{2 x} - 1}{x}) (\frac{\sin \frac{π x}{180}}{x}) \end{array}$
$= (2 lim_{x \to 0} \frac{e^{2 x} - 1}{2 x}) \frac{π}{180} [lim_{x \to 0} \frac{\frac{\sin x π}{180}}{\frac{x π}{180}}] = 2 (\log e) \times (\frac{π}{180}) = \frac{π}{90}$

MHT CET-2020

Limits, Continuity and Differentiability

79895 If the function
$\begin{array}{cc} f (x) = \frac{1 - \sin 2 x + \cos 2 x}{1 + \sin 2 x + \cos 2 x}, & if x \neq \frac{π}{2} \\ f (x) = k, & if x = \frac{π}{2} \end{array}$
is continuous at $x = \frac{\boldsymbol π}{\boldsymbol 2}$ , then $k =$

1 2

2 0

3 -1

4 1

Explanation:

(C) : Given, $f (x) = k$ , at $x = \frac{π}{2}$
$f (\frac{π}{2}) = k$
For continuous at $x = \frac{π}{2}, f (\frac{π}{2}) = lim_{x \to \frac{π}{2}} f (x)$
$k = lim_{x \to \frac{π}{2}} \frac{1 - \sin 2 x + \cos 2 x}{1 + \sin 2 x + \cos 2 x}$
$k = lim_{x \to \frac{π}{2}} \frac{(1 + \cos 2 x) - \sin 2 x}{(1 + \cos 2 x) + \sin 2 x}$
$k = lim_{x \to \frac{π}{2}} \frac{2 \cos^{2} x - 2 \sin x \cos x}{2 \cos^{2} x + 2 \sin x \cos x}$
$k = lim_{x \to \frac{π}{2}} \frac{2 \cos x (\cos x - \sin x)}{2 \cos x (\cos x + \sin x)} \Rightarrow k = lim_{x \to \frac{π}{2}} \frac{\cos x - \sin x}{\cos x + \sin x}$
$k = \frac{0 - 1}{0 + 1}$
$k = \frac{1}{1}$
Since, $f (x)$ is continuous at $x = \frac{π}{2}$ therefore, we get $k = - 1$

MHT CET-2020

Limits, Continuity and Differentiability

79896 If $\begin{aligned} f (x) & = \frac{(81)^{x} - 9^{x}}{(k)^{x} - 1}, if x \neq 0 \\ = 2, if x = 0 \end{aligned}$
is continuous at $x = 0$ , then the value of $k$ is

1 9

2 2

3 3

4 4

Explanation:

(C)Given, $f (x) = 2$ at $x = 0$
$f (0) = 2$
For continuous at $x = 0$ ,
$lim_{x \to 0} f (x) = f (0)$
$lim_{x \to 0} \frac{(81)^{x} - (9)^{x}}{k^{x} - 1} = 2$
$lim_{x \to 0} \frac{9^{x} (9^{x} - 1)}{k^{x} - 1} = 2 \Rightarrow \frac{(lim_{x \to 0} 9^{x}) (lim_{x \to 0} \frac{9^{x} - 1}{x})}{lim_{x \to 0} \frac{k^{x} - 1}{x}} = 2$
$\frac{\log 9}{\log k} = 2 \Rightarrow \log 9 = 2 \log k$
$\Rightarrow \log k^{2} = \log 9 \Rightarrow k^{2} = 9 \Rightarrow k = 3$

MHT CET-2020

Limits, Continuity and Differentiability

79897 If the function
$\begin{aligned} f (x) & = \frac{\log 10 + \log (0.1 + 2 x)}{2 x}, if x \neq 0 \\ = k . if x = 0 \end{aligned}$
is continuous at $x = 0$ , then $k + 2 =$

1 2

2 12

3 11

4 10

Explanation:

(B) : Given,
$f (x) = \frac{\log 10 + \log (0.1 + 2 x)}{2 x}, if x \neq 0$
$= k, if x = 0$
For continuous at $x = 0$
$f (0) = lim_{x \to 0} f (x)$
$k = lim_{x \to 0} \frac{\log 10 + \log (0.1 + 2 x)}{2 x}$
$k = lim_{x \to 0} \frac{\log 10 \times (0.1 + 2 x)}{2 x}$
$k = lim_{x \to 0} \frac{\log (1 + 20 x)}{2 x}$
$k = lim_{x \to 0} \frac{10 \times \log (1 + 20 x)}{20 x}$
$k = 10 lim_{x \to 0} \frac{\log (1 + 20 x)}{20 x}$
Then,
$k + 2 = 10 + 2 = 12$

nda

Limits, Continuity and Differentiability

79898 If $f (x) = \frac{\sqrt{2} - \sqrt{1 + \sin x}}{\cos^{2} x}, x \neq \frac{π}{2}$ is continuous at $x = \frac{π}{2}$ , then $f (\frac{π}{2}) =$

1

\frac{1}{4 \sqrt{2}}

2

\frac{1}{2 \sqrt{2}}

3

\frac{1}{\sqrt{2}}

4

\sqrt{2}

Explanation:

(A) : Given, $f (x) = \frac{\sqrt{2} - \sqrt{1 + \sin x}}{\cos^{2} x}, x \neq \frac{π}{2}$
$f (x)$ is continuous at $x = \frac{π}{2}$
$∴ f (\frac{π}{2}) = lim_{x \to \frac{π}{2}} \frac{\sqrt{2} - \sqrt{1 + \sin x}}{\cos^{2} x}$
$= lim_{x \to \frac{π}{2}} \frac{\sqrt{2} - \sqrt{1 + \sin x}}{\cos^{2} x} \times \frac{\sqrt{2} + \sqrt{1 + \sin x}}{\sqrt{2} + \sqrt{1 + \sin x}}$
$= lim_{x \to \frac{π}{2}} \frac{2 - (1 + \sin x)}{(1 - \sin^{2} x) (\sqrt{2} + \sqrt{1 + \sin x})}$
$= lim_{x \to \frac{π}{2}} \frac{1 - \sin x}{(1 - \sin x) (1 + \sin x) (\sqrt{2} + \sqrt{1 + \sin x})}$
$= lim_{x \to \frac{π}{2}} \frac{1}{(1 + \sin x) (\sqrt{2} + \sqrt{1 + \sin x})}$
$= \frac{1}{(1 + 1) (\sqrt{2} + \sqrt{2})} = \frac{1}{2 (2 \sqrt{2})} = \frac{1}{4 \sqrt{2}}$

MHT CET-2019

Limits, Continuity and Differentiability

79894 If $f (x) = \frac{(e^{2 x} - 1) \sin x^{0}}{x^{2}}, x \neq 0$ is continuous at $x$ $= 0$ , then $f (0) =$

1

\frac{90}{π}

2

\frac{π}{180}

3

\frac{π}{90}

4

\frac{180}{π}

Explanation:

(C) : Given,
$f (x) = \frac{(e^{2 x} - 1) \sin x^{\circ}}{x^{2}}, x \neq 0$
For continuous at $x = 0$
$f (0) = lim_{x \to 0} f (x)$
$\begin{array}{r} f (0) = lim_{x \to 0} \frac{(e^{2 x} - 1) \sin \frac{x π}{180}}{x^{2}} [∵ x^{0} = (\frac{x π}{180})] \\ = lim_{x \to 0} (\frac{e^{2 x} - 1}{x}) (\frac{\sin \frac{π x}{180}}{x}) \end{array}$
$= (2 lim_{x \to 0} \frac{e^{2 x} - 1}{2 x}) \frac{π}{180} [lim_{x \to 0} \frac{\frac{\sin x π}{180}}{\frac{x π}{180}}] = 2 (\log e) \times (\frac{π}{180}) = \frac{π}{90}$

MHT CET-2020

Limits, Continuity and Differentiability

79895 If the function
$\begin{array}{cc} f (x) = \frac{1 - \sin 2 x + \cos 2 x}{1 + \sin 2 x + \cos 2 x}, & if x \neq \frac{π}{2} \\ f (x) = k, & if x = \frac{π}{2} \end{array}$
is continuous at $x = \frac{\boldsymbol π}{\boldsymbol 2}$ , then $k =$

1 2

2 0

3 -1

4 1

Explanation:

(C) : Given, $f (x) = k$ , at $x = \frac{π}{2}$
$f (\frac{π}{2}) = k$
For continuous at $x = \frac{π}{2}, f (\frac{π}{2}) = lim_{x \to \frac{π}{2}} f (x)$
$k = lim_{x \to \frac{π}{2}} \frac{1 - \sin 2 x + \cos 2 x}{1 + \sin 2 x + \cos 2 x}$
$k = lim_{x \to \frac{π}{2}} \frac{(1 + \cos 2 x) - \sin 2 x}{(1 + \cos 2 x) + \sin 2 x}$
$k = lim_{x \to \frac{π}{2}} \frac{2 \cos^{2} x - 2 \sin x \cos x}{2 \cos^{2} x + 2 \sin x \cos x}$
$k = lim_{x \to \frac{π}{2}} \frac{2 \cos x (\cos x - \sin x)}{2 \cos x (\cos x + \sin x)} \Rightarrow k = lim_{x \to \frac{π}{2}} \frac{\cos x - \sin x}{\cos x + \sin x}$
$k = \frac{0 - 1}{0 + 1}$
$k = \frac{1}{1}$
Since, $f (x)$ is continuous at $x = \frac{π}{2}$ therefore, we get $k = - 1$

MHT CET-2020

Limits, Continuity and Differentiability

79896 If $\begin{aligned} f (x) & = \frac{(81)^{x} - 9^{x}}{(k)^{x} - 1}, if x \neq 0 \\ = 2, if x = 0 \end{aligned}$
is continuous at $x = 0$ , then the value of $k$ is

1 9

2 2

3 3

4 4

Explanation:

(C)Given, $f (x) = 2$ at $x = 0$
$f (0) = 2$
For continuous at $x = 0$ ,
$lim_{x \to 0} f (x) = f (0)$
$lim_{x \to 0} \frac{(81)^{x} - (9)^{x}}{k^{x} - 1} = 2$
$lim_{x \to 0} \frac{9^{x} (9^{x} - 1)}{k^{x} - 1} = 2 \Rightarrow \frac{(lim_{x \to 0} 9^{x}) (lim_{x \to 0} \frac{9^{x} - 1}{x})}{lim_{x \to 0} \frac{k^{x} - 1}{x}} = 2$
$\frac{\log 9}{\log k} = 2 \Rightarrow \log 9 = 2 \log k$
$\Rightarrow \log k^{2} = \log 9 \Rightarrow k^{2} = 9 \Rightarrow k = 3$

MHT CET-2020

Limits, Continuity and Differentiability

79897 If the function
$\begin{aligned} f (x) & = \frac{\log 10 + \log (0.1 + 2 x)}{2 x}, if x \neq 0 \\ = k . if x = 0 \end{aligned}$
is continuous at $x = 0$ , then $k + 2 =$

1 2

2 12

3 11

4 10

Explanation:

(B) : Given,
$f (x) = \frac{\log 10 + \log (0.1 + 2 x)}{2 x}, if x \neq 0$
$= k, if x = 0$
For continuous at $x = 0$
$f (0) = lim_{x \to 0} f (x)$
$k = lim_{x \to 0} \frac{\log 10 + \log (0.1 + 2 x)}{2 x}$
$k = lim_{x \to 0} \frac{\log 10 \times (0.1 + 2 x)}{2 x}$
$k = lim_{x \to 0} \frac{\log (1 + 20 x)}{2 x}$
$k = lim_{x \to 0} \frac{10 \times \log (1 + 20 x)}{20 x}$
$k = 10 lim_{x \to 0} \frac{\log (1 + 20 x)}{20 x}$
Then,
$k + 2 = 10 + 2 = 12$

nda

Limits, Continuity and Differentiability

79898 If $f (x) = \frac{\sqrt{2} - \sqrt{1 + \sin x}}{\cos^{2} x}, x \neq \frac{π}{2}$ is continuous at $x = \frac{π}{2}$ , then $f (\frac{π}{2}) =$

1

\frac{1}{4 \sqrt{2}}

2

\frac{1}{2 \sqrt{2}}

3

\frac{1}{\sqrt{2}}

4

\sqrt{2}

Explanation:

(A) : Given, $f (x) = \frac{\sqrt{2} - \sqrt{1 + \sin x}}{\cos^{2} x}, x \neq \frac{π}{2}$
$f (x)$ is continuous at $x = \frac{π}{2}$
$∴ f (\frac{π}{2}) = lim_{x \to \frac{π}{2}} \frac{\sqrt{2} - \sqrt{1 + \sin x}}{\cos^{2} x}$
$= lim_{x \to \frac{π}{2}} \frac{\sqrt{2} - \sqrt{1 + \sin x}}{\cos^{2} x} \times \frac{\sqrt{2} + \sqrt{1 + \sin x}}{\sqrt{2} + \sqrt{1 + \sin x}}$
$= lim_{x \to \frac{π}{2}} \frac{2 - (1 + \sin x)}{(1 - \sin^{2} x) (\sqrt{2} + \sqrt{1 + \sin x})}$
$= lim_{x \to \frac{π}{2}} \frac{1 - \sin x}{(1 - \sin x) (1 + \sin x) (\sqrt{2} + \sqrt{1 + \sin x})}$
$= lim_{x \to \frac{π}{2}} \frac{1}{(1 + \sin x) (\sqrt{2} + \sqrt{1 + \sin x})}$
$= \frac{1}{(1 + 1) (\sqrt{2} + \sqrt{2})} = \frac{1}{2 (2 \sqrt{2})} = \frac{1}{4 \sqrt{2}}$

MHT CET-2019

Limits, Continuity and Differentiability

79894 If $f (x) = \frac{(e^{2 x} - 1) \sin x^{0}}{x^{2}}, x \neq 0$ is continuous at $x$ $= 0$ , then $f (0) =$

1

\frac{90}{π}

2

\frac{π}{180}

3

\frac{π}{90}

4

\frac{180}{π}

Explanation:

(C) : Given,
$f (x) = \frac{(e^{2 x} - 1) \sin x^{\circ}}{x^{2}}, x \neq 0$
For continuous at $x = 0$
$f (0) = lim_{x \to 0} f (x)$
$\begin{array}{r} f (0) = lim_{x \to 0} \frac{(e^{2 x} - 1) \sin \frac{x π}{180}}{x^{2}} [∵ x^{0} = (\frac{x π}{180})] \\ = lim_{x \to 0} (\frac{e^{2 x} - 1}{x}) (\frac{\sin \frac{π x}{180}}{x}) \end{array}$
$= (2 lim_{x \to 0} \frac{e^{2 x} - 1}{2 x}) \frac{π}{180} [lim_{x \to 0} \frac{\frac{\sin x π}{180}}{\frac{x π}{180}}] = 2 (\log e) \times (\frac{π}{180}) = \frac{π}{90}$

MHT CET-2020

Limits, Continuity and Differentiability

79895 If the function
$\begin{array}{cc} f (x) = \frac{1 - \sin 2 x + \cos 2 x}{1 + \sin 2 x + \cos 2 x}, & if x \neq \frac{π}{2} \\ f (x) = k, & if x = \frac{π}{2} \end{array}$
is continuous at $x = \frac{\boldsymbol π}{\boldsymbol 2}$ , then $k =$

1 2

2 0

3 -1

4 1

Explanation:

(C) : Given, $f (x) = k$ , at $x = \frac{π}{2}$
$f (\frac{π}{2}) = k$
For continuous at $x = \frac{π}{2}, f (\frac{π}{2}) = lim_{x \to \frac{π}{2}} f (x)$
$k = lim_{x \to \frac{π}{2}} \frac{1 - \sin 2 x + \cos 2 x}{1 + \sin 2 x + \cos 2 x}$
$k = lim_{x \to \frac{π}{2}} \frac{(1 + \cos 2 x) - \sin 2 x}{(1 + \cos 2 x) + \sin 2 x}$
$k = lim_{x \to \frac{π}{2}} \frac{2 \cos^{2} x - 2 \sin x \cos x}{2 \cos^{2} x + 2 \sin x \cos x}$
$k = lim_{x \to \frac{π}{2}} \frac{2 \cos x (\cos x - \sin x)}{2 \cos x (\cos x + \sin x)} \Rightarrow k = lim_{x \to \frac{π}{2}} \frac{\cos x - \sin x}{\cos x + \sin x}$
$k = \frac{0 - 1}{0 + 1}$
$k = \frac{1}{1}$
Since, $f (x)$ is continuous at $x = \frac{π}{2}$ therefore, we get $k = - 1$

MHT CET-2020

Limits, Continuity and Differentiability

79896 If $\begin{aligned} f (x) & = \frac{(81)^{x} - 9^{x}}{(k)^{x} - 1}, if x \neq 0 \\ = 2, if x = 0 \end{aligned}$
is continuous at $x = 0$ , then the value of $k$ is

1 9

2 2

3 3

4 4

Explanation:

(C)Given, $f (x) = 2$ at $x = 0$
$f (0) = 2$
For continuous at $x = 0$ ,
$lim_{x \to 0} f (x) = f (0)$
$lim_{x \to 0} \frac{(81)^{x} - (9)^{x}}{k^{x} - 1} = 2$
$lim_{x \to 0} \frac{9^{x} (9^{x} - 1)}{k^{x} - 1} = 2 \Rightarrow \frac{(lim_{x \to 0} 9^{x}) (lim_{x \to 0} \frac{9^{x} - 1}{x})}{lim_{x \to 0} \frac{k^{x} - 1}{x}} = 2$
$\frac{\log 9}{\log k} = 2 \Rightarrow \log 9 = 2 \log k$
$\Rightarrow \log k^{2} = \log 9 \Rightarrow k^{2} = 9 \Rightarrow k = 3$

MHT CET-2020

Limits, Continuity and Differentiability

79897 If the function
$\begin{aligned} f (x) & = \frac{\log 10 + \log (0.1 + 2 x)}{2 x}, if x \neq 0 \\ = k . if x = 0 \end{aligned}$
is continuous at $x = 0$ , then $k + 2 =$

1 2

2 12

3 11

4 10

Explanation:

(B) : Given,
$f (x) = \frac{\log 10 + \log (0.1 + 2 x)}{2 x}, if x \neq 0$
$= k, if x = 0$
For continuous at $x = 0$
$f (0) = lim_{x \to 0} f (x)$
$k = lim_{x \to 0} \frac{\log 10 + \log (0.1 + 2 x)}{2 x}$
$k = lim_{x \to 0} \frac{\log 10 \times (0.1 + 2 x)}{2 x}$
$k = lim_{x \to 0} \frac{\log (1 + 20 x)}{2 x}$
$k = lim_{x \to 0} \frac{10 \times \log (1 + 20 x)}{20 x}$
$k = 10 lim_{x \to 0} \frac{\log (1 + 20 x)}{20 x}$
Then,
$k + 2 = 10 + 2 = 12$

nda

Limits, Continuity and Differentiability

79898 If $f (x) = \frac{\sqrt{2} - \sqrt{1 + \sin x}}{\cos^{2} x}, x \neq \frac{π}{2}$ is continuous at $x = \frac{π}{2}$ , then $f (\frac{π}{2}) =$

1

\frac{1}{4 \sqrt{2}}

2

\frac{1}{2 \sqrt{2}}

3

\frac{1}{\sqrt{2}}

4

\sqrt{2}

Explanation:

(A) : Given, $f (x) = \frac{\sqrt{2} - \sqrt{1 + \sin x}}{\cos^{2} x}, x \neq \frac{π}{2}$
$f (x)$ is continuous at $x = \frac{π}{2}$
$∴ f (\frac{π}{2}) = lim_{x \to \frac{π}{2}} \frac{\sqrt{2} - \sqrt{1 + \sin x}}{\cos^{2} x}$
$= lim_{x \to \frac{π}{2}} \frac{\sqrt{2} - \sqrt{1 + \sin x}}{\cos^{2} x} \times \frac{\sqrt{2} + \sqrt{1 + \sin x}}{\sqrt{2} + \sqrt{1 + \sin x}}$
$= lim_{x \to \frac{π}{2}} \frac{2 - (1 + \sin x)}{(1 - \sin^{2} x) (\sqrt{2} + \sqrt{1 + \sin x})}$
$= lim_{x \to \frac{π}{2}} \frac{1 - \sin x}{(1 - \sin x) (1 + \sin x) (\sqrt{2} + \sqrt{1 + \sin x})}$
$= lim_{x \to \frac{π}{2}} \frac{1}{(1 + \sin x) (\sqrt{2} + \sqrt{1 + \sin x})}$
$= \frac{1}{(1 + 1) (\sqrt{2} + \sqrt{2})} = \frac{1}{2 (2 \sqrt{2})} = \frac{1}{4 \sqrt{2}}$

MHT CET-2019

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