NEET Test Series from KOTA - 10 Papers In MS WORD
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Limits, Continuity and Differentiability
79899
If the function \(\begin{aligned} \mathbf{f}(\mathbf{x}) & =\left(1+3 \tan ^2 x\right)^{\frac{\cot ^2 x}{4}}, & & \mathbf{x} \neq 0 \\ & =\mathbf{k}, & & x=0\end{aligned}\) is continuous at \(\mathrm{x}=0\), then \(\mathrm{k}=\)
79901
If \(f(x)\) is continuous at \(x=3\), where \(\begin{aligned} \mathbf{f}(\mathbf{x}) & =\mathbf{a x}+1, & & \text { for } \mathbf{x} \leq \mathbf{3} \\ & =\mathbf{b x}+3, & & \text { for } \mathbf{x}>3\end{aligned}\) then
1 \(\mathrm{a}-\mathrm{b}=\frac{2}{3}\)
2 \(\mathrm{a}+\mathrm{b}=\frac{-2}{3}\)
3 \(\mathrm{a}-\mathrm{b}=\frac{-2}{3}\)
4 \(a+b=\frac{2}{3}\)
Explanation:
(A) : Given, \(\mathrm{f}(\mathrm{x})=\mathrm{ax}+1\), for \(\mathrm{x} \leq 3\) \(=\mathrm{bx}+3\), for \(\mathrm{x}>3\) \(\therefore \lim _{x \rightarrow 3^{-}} f(x)=\lim _{x \rightarrow 3^{-}} a x+1=3 a+1\) \(\lim _{x \rightarrow 3^{+}} f(x)=\lim _{x \rightarrow 3^{+}} b x+3=3 b+3\) \(\mathrm{f}(\mathrm{x})\) is continuous at \(\mathrm{x}=3\) \(\therefore \quad \lim _{x \rightarrow 3^{-}} \mathrm{f}(\mathrm{x})=\lim _{\mathrm{x} 3^{+}} \mathrm{f}(\mathrm{x})\) \(3 a^{x \rightarrow 3^{-}}+1=3 b^{x \rightarrow 3^{+}}+3\) \(3 a-3 b=2\) \(\mathrm{a}-\mathrm{b}=\frac{2}{3}\)
MHT CET-2019
Limits, Continuity and Differentiability
79902
If the function \(\begin{aligned} \mathbf{f}(\mathbf{x}) & =\frac{\left(\mathrm{e}^{\mathrm{kx}}-1\right) \tan \mathrm{kx}}{4 \mathrm{x}^2}, & & \mathbf{x} \neq 0 \\ & =16, & & \mathbf{x}=0\end{aligned}\) is continuous at \(x=0\), then \(k=\)
79899
If the function \(\begin{aligned} \mathbf{f}(\mathbf{x}) & =\left(1+3 \tan ^2 x\right)^{\frac{\cot ^2 x}{4}}, & & \mathbf{x} \neq 0 \\ & =\mathbf{k}, & & x=0\end{aligned}\) is continuous at \(\mathrm{x}=0\), then \(\mathrm{k}=\)
79901
If \(f(x)\) is continuous at \(x=3\), where \(\begin{aligned} \mathbf{f}(\mathbf{x}) & =\mathbf{a x}+1, & & \text { for } \mathbf{x} \leq \mathbf{3} \\ & =\mathbf{b x}+3, & & \text { for } \mathbf{x}>3\end{aligned}\) then
1 \(\mathrm{a}-\mathrm{b}=\frac{2}{3}\)
2 \(\mathrm{a}+\mathrm{b}=\frac{-2}{3}\)
3 \(\mathrm{a}-\mathrm{b}=\frac{-2}{3}\)
4 \(a+b=\frac{2}{3}\)
Explanation:
(A) : Given, \(\mathrm{f}(\mathrm{x})=\mathrm{ax}+1\), for \(\mathrm{x} \leq 3\) \(=\mathrm{bx}+3\), for \(\mathrm{x}>3\) \(\therefore \lim _{x \rightarrow 3^{-}} f(x)=\lim _{x \rightarrow 3^{-}} a x+1=3 a+1\) \(\lim _{x \rightarrow 3^{+}} f(x)=\lim _{x \rightarrow 3^{+}} b x+3=3 b+3\) \(\mathrm{f}(\mathrm{x})\) is continuous at \(\mathrm{x}=3\) \(\therefore \quad \lim _{x \rightarrow 3^{-}} \mathrm{f}(\mathrm{x})=\lim _{\mathrm{x} 3^{+}} \mathrm{f}(\mathrm{x})\) \(3 a^{x \rightarrow 3^{-}}+1=3 b^{x \rightarrow 3^{+}}+3\) \(3 a-3 b=2\) \(\mathrm{a}-\mathrm{b}=\frac{2}{3}\)
MHT CET-2019
Limits, Continuity and Differentiability
79902
If the function \(\begin{aligned} \mathbf{f}(\mathbf{x}) & =\frac{\left(\mathrm{e}^{\mathrm{kx}}-1\right) \tan \mathrm{kx}}{4 \mathrm{x}^2}, & & \mathbf{x} \neq 0 \\ & =16, & & \mathbf{x}=0\end{aligned}\) is continuous at \(x=0\), then \(k=\)
79899
If the function \(\begin{aligned} \mathbf{f}(\mathbf{x}) & =\left(1+3 \tan ^2 x\right)^{\frac{\cot ^2 x}{4}}, & & \mathbf{x} \neq 0 \\ & =\mathbf{k}, & & x=0\end{aligned}\) is continuous at \(\mathrm{x}=0\), then \(\mathrm{k}=\)
79901
If \(f(x)\) is continuous at \(x=3\), where \(\begin{aligned} \mathbf{f}(\mathbf{x}) & =\mathbf{a x}+1, & & \text { for } \mathbf{x} \leq \mathbf{3} \\ & =\mathbf{b x}+3, & & \text { for } \mathbf{x}>3\end{aligned}\) then
1 \(\mathrm{a}-\mathrm{b}=\frac{2}{3}\)
2 \(\mathrm{a}+\mathrm{b}=\frac{-2}{3}\)
3 \(\mathrm{a}-\mathrm{b}=\frac{-2}{3}\)
4 \(a+b=\frac{2}{3}\)
Explanation:
(A) : Given, \(\mathrm{f}(\mathrm{x})=\mathrm{ax}+1\), for \(\mathrm{x} \leq 3\) \(=\mathrm{bx}+3\), for \(\mathrm{x}>3\) \(\therefore \lim _{x \rightarrow 3^{-}} f(x)=\lim _{x \rightarrow 3^{-}} a x+1=3 a+1\) \(\lim _{x \rightarrow 3^{+}} f(x)=\lim _{x \rightarrow 3^{+}} b x+3=3 b+3\) \(\mathrm{f}(\mathrm{x})\) is continuous at \(\mathrm{x}=3\) \(\therefore \quad \lim _{x \rightarrow 3^{-}} \mathrm{f}(\mathrm{x})=\lim _{\mathrm{x} 3^{+}} \mathrm{f}(\mathrm{x})\) \(3 a^{x \rightarrow 3^{-}}+1=3 b^{x \rightarrow 3^{+}}+3\) \(3 a-3 b=2\) \(\mathrm{a}-\mathrm{b}=\frac{2}{3}\)
MHT CET-2019
Limits, Continuity and Differentiability
79902
If the function \(\begin{aligned} \mathbf{f}(\mathbf{x}) & =\frac{\left(\mathrm{e}^{\mathrm{kx}}-1\right) \tan \mathrm{kx}}{4 \mathrm{x}^2}, & & \mathbf{x} \neq 0 \\ & =16, & & \mathbf{x}=0\end{aligned}\) is continuous at \(x=0\), then \(k=\)
79899
If the function \(\begin{aligned} \mathbf{f}(\mathbf{x}) & =\left(1+3 \tan ^2 x\right)^{\frac{\cot ^2 x}{4}}, & & \mathbf{x} \neq 0 \\ & =\mathbf{k}, & & x=0\end{aligned}\) is continuous at \(\mathrm{x}=0\), then \(\mathrm{k}=\)
79901
If \(f(x)\) is continuous at \(x=3\), where \(\begin{aligned} \mathbf{f}(\mathbf{x}) & =\mathbf{a x}+1, & & \text { for } \mathbf{x} \leq \mathbf{3} \\ & =\mathbf{b x}+3, & & \text { for } \mathbf{x}>3\end{aligned}\) then
1 \(\mathrm{a}-\mathrm{b}=\frac{2}{3}\)
2 \(\mathrm{a}+\mathrm{b}=\frac{-2}{3}\)
3 \(\mathrm{a}-\mathrm{b}=\frac{-2}{3}\)
4 \(a+b=\frac{2}{3}\)
Explanation:
(A) : Given, \(\mathrm{f}(\mathrm{x})=\mathrm{ax}+1\), for \(\mathrm{x} \leq 3\) \(=\mathrm{bx}+3\), for \(\mathrm{x}>3\) \(\therefore \lim _{x \rightarrow 3^{-}} f(x)=\lim _{x \rightarrow 3^{-}} a x+1=3 a+1\) \(\lim _{x \rightarrow 3^{+}} f(x)=\lim _{x \rightarrow 3^{+}} b x+3=3 b+3\) \(\mathrm{f}(\mathrm{x})\) is continuous at \(\mathrm{x}=3\) \(\therefore \quad \lim _{x \rightarrow 3^{-}} \mathrm{f}(\mathrm{x})=\lim _{\mathrm{x} 3^{+}} \mathrm{f}(\mathrm{x})\) \(3 a^{x \rightarrow 3^{-}}+1=3 b^{x \rightarrow 3^{+}}+3\) \(3 a-3 b=2\) \(\mathrm{a}-\mathrm{b}=\frac{2}{3}\)
MHT CET-2019
Limits, Continuity and Differentiability
79902
If the function \(\begin{aligned} \mathbf{f}(\mathbf{x}) & =\frac{\left(\mathrm{e}^{\mathrm{kx}}-1\right) \tan \mathrm{kx}}{4 \mathrm{x}^2}, & & \mathbf{x} \neq 0 \\ & =16, & & \mathbf{x}=0\end{aligned}\) is continuous at \(x=0\), then \(k=\)
