NEET Test Series from KOTA - 10 Papers In MS WORD
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Limits, Continuity and Differentiability
79890
If \(\begin{aligned} f(x) & =\frac{4 \sin \pi x}{5 x}, \text { for } x \neq 0 \\ =2 k, & \text { for } x=0\end{aligned}\) is continuous at \(x=0\), then the value of \(k\) is
79892
If \(\begin{aligned} f(x) & =\frac{\left(\mathrm{e}^{3 x}-1\right) \sin x^0}{x^2}, & & \text { if } x \neq 0 \\ & =\frac{\pi}{60}, & & \text { if } x=0\end{aligned}\)
1 \(\lim _{x \rightarrow 0} f(x)=3\)
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, \(\mathrm{f}(\mathrm{x})=\frac{\pi}{60}, \mathrm{x}=0\) \(f(0)=\frac{\pi}{60}\) \(\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \frac{\left(e^{3 x}-1\right) \sin x^{0}}{x^{2}}\) Dividing numerator and denominator by \(\mathrm{x}^{2}\), we get \(=\frac{\lim _{x \rightarrow 0}\left(\frac{e^{3 x}-1}{3 x} \times 3\right) \frac{\sin \left(\frac{\pi x}{180}\right)^{c}}{\left(\frac{\pi x}{180}\right)^{c}} \times\left(\frac{\pi}{180}\right)}{\left(\frac{x^{2}}{x^{2}}\right)}\) \(\lim _{x \rightarrow 0} f(x)=(3)(1)\left(\frac{\pi}{180}\right)=\frac{\pi}{60} \quad\left\{\because \lim _{x \rightarrow 0} \frac{e^{x}-1}{x}=1\right\}\) \(\because\) at \(x=0, f(0)=\lim _{x \rightarrow 0} \mathrm{f}(\mathrm{x})\) Hence, \(\mathrm{f}(\mathrm{x})\) is continuous at \(\mathrm{x}=0\).
MHT CET-2020
Limits, Continuity and Differentiability
79893
If \(\begin{aligned} & f(x)=\left[\tan \left(\frac{\pi}{4}+\mathbf{x}\right)\right]^{\frac{1}{x}} \text { if } x \neq 0 \\ & =\mathbf{k} \quad \text { if } \mathbf{x}=\mathbf{0}, \\ & \end{aligned}\) is continuous at \(x=0\) then \(k=\)
79890
If \(\begin{aligned} f(x) & =\frac{4 \sin \pi x}{5 x}, \text { for } x \neq 0 \\ =2 k, & \text { for } x=0\end{aligned}\) is continuous at \(x=0\), then the value of \(k\) is
79892
If \(\begin{aligned} f(x) & =\frac{\left(\mathrm{e}^{3 x}-1\right) \sin x^0}{x^2}, & & \text { if } x \neq 0 \\ & =\frac{\pi}{60}, & & \text { if } x=0\end{aligned}\)
1 \(\lim _{x \rightarrow 0} f(x)=3\)
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, \(\mathrm{f}(\mathrm{x})=\frac{\pi}{60}, \mathrm{x}=0\) \(f(0)=\frac{\pi}{60}\) \(\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \frac{\left(e^{3 x}-1\right) \sin x^{0}}{x^{2}}\) Dividing numerator and denominator by \(\mathrm{x}^{2}\), we get \(=\frac{\lim _{x \rightarrow 0}\left(\frac{e^{3 x}-1}{3 x} \times 3\right) \frac{\sin \left(\frac{\pi x}{180}\right)^{c}}{\left(\frac{\pi x}{180}\right)^{c}} \times\left(\frac{\pi}{180}\right)}{\left(\frac{x^{2}}{x^{2}}\right)}\) \(\lim _{x \rightarrow 0} f(x)=(3)(1)\left(\frac{\pi}{180}\right)=\frac{\pi}{60} \quad\left\{\because \lim _{x \rightarrow 0} \frac{e^{x}-1}{x}=1\right\}\) \(\because\) at \(x=0, f(0)=\lim _{x \rightarrow 0} \mathrm{f}(\mathrm{x})\) Hence, \(\mathrm{f}(\mathrm{x})\) is continuous at \(\mathrm{x}=0\).
MHT CET-2020
Limits, Continuity and Differentiability
79893
If \(\begin{aligned} & f(x)=\left[\tan \left(\frac{\pi}{4}+\mathbf{x}\right)\right]^{\frac{1}{x}} \text { if } x \neq 0 \\ & =\mathbf{k} \quad \text { if } \mathbf{x}=\mathbf{0}, \\ & \end{aligned}\) is continuous at \(x=0\) then \(k=\)
79890
If \(\begin{aligned} f(x) & =\frac{4 \sin \pi x}{5 x}, \text { for } x \neq 0 \\ =2 k, & \text { for } x=0\end{aligned}\) is continuous at \(x=0\), then the value of \(k\) is
79892
If \(\begin{aligned} f(x) & =\frac{\left(\mathrm{e}^{3 x}-1\right) \sin x^0}{x^2}, & & \text { if } x \neq 0 \\ & =\frac{\pi}{60}, & & \text { if } x=0\end{aligned}\)
1 \(\lim _{x \rightarrow 0} f(x)=3\)
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, \(\mathrm{f}(\mathrm{x})=\frac{\pi}{60}, \mathrm{x}=0\) \(f(0)=\frac{\pi}{60}\) \(\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \frac{\left(e^{3 x}-1\right) \sin x^{0}}{x^{2}}\) Dividing numerator and denominator by \(\mathrm{x}^{2}\), we get \(=\frac{\lim _{x \rightarrow 0}\left(\frac{e^{3 x}-1}{3 x} \times 3\right) \frac{\sin \left(\frac{\pi x}{180}\right)^{c}}{\left(\frac{\pi x}{180}\right)^{c}} \times\left(\frac{\pi}{180}\right)}{\left(\frac{x^{2}}{x^{2}}\right)}\) \(\lim _{x \rightarrow 0} f(x)=(3)(1)\left(\frac{\pi}{180}\right)=\frac{\pi}{60} \quad\left\{\because \lim _{x \rightarrow 0} \frac{e^{x}-1}{x}=1\right\}\) \(\because\) at \(x=0, f(0)=\lim _{x \rightarrow 0} \mathrm{f}(\mathrm{x})\) Hence, \(\mathrm{f}(\mathrm{x})\) is continuous at \(\mathrm{x}=0\).
MHT CET-2020
Limits, Continuity and Differentiability
79893
If \(\begin{aligned} & f(x)=\left[\tan \left(\frac{\pi}{4}+\mathbf{x}\right)\right]^{\frac{1}{x}} \text { if } x \neq 0 \\ & =\mathbf{k} \quad \text { if } \mathbf{x}=\mathbf{0}, \\ & \end{aligned}\) is continuous at \(x=0\) then \(k=\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Limits, Continuity and Differentiability
79890
If \(\begin{aligned} f(x) & =\frac{4 \sin \pi x}{5 x}, \text { for } x \neq 0 \\ =2 k, & \text { for } x=0\end{aligned}\) is continuous at \(x=0\), then the value of \(k\) is
79892
If \(\begin{aligned} f(x) & =\frac{\left(\mathrm{e}^{3 x}-1\right) \sin x^0}{x^2}, & & \text { if } x \neq 0 \\ & =\frac{\pi}{60}, & & \text { if } x=0\end{aligned}\)
1 \(\lim _{x \rightarrow 0} f(x)=3\)
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, \(\mathrm{f}(\mathrm{x})=\frac{\pi}{60}, \mathrm{x}=0\) \(f(0)=\frac{\pi}{60}\) \(\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \frac{\left(e^{3 x}-1\right) \sin x^{0}}{x^{2}}\) Dividing numerator and denominator by \(\mathrm{x}^{2}\), we get \(=\frac{\lim _{x \rightarrow 0}\left(\frac{e^{3 x}-1}{3 x} \times 3\right) \frac{\sin \left(\frac{\pi x}{180}\right)^{c}}{\left(\frac{\pi x}{180}\right)^{c}} \times\left(\frac{\pi}{180}\right)}{\left(\frac{x^{2}}{x^{2}}\right)}\) \(\lim _{x \rightarrow 0} f(x)=(3)(1)\left(\frac{\pi}{180}\right)=\frac{\pi}{60} \quad\left\{\because \lim _{x \rightarrow 0} \frac{e^{x}-1}{x}=1\right\}\) \(\because\) at \(x=0, f(0)=\lim _{x \rightarrow 0} \mathrm{f}(\mathrm{x})\) Hence, \(\mathrm{f}(\mathrm{x})\) is continuous at \(\mathrm{x}=0\).
MHT CET-2020
Limits, Continuity and Differentiability
79893
If \(\begin{aligned} & f(x)=\left[\tan \left(\frac{\pi}{4}+\mathbf{x}\right)\right]^{\frac{1}{x}} \text { if } x \neq 0 \\ & =\mathbf{k} \quad \text { if } \mathbf{x}=\mathbf{0}, \\ & \end{aligned}\) is continuous at \(x=0\) then \(k=\)