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CBSE Class 12 Math 2022 Term I Solved Paper

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Question : 12 of 50
Marks: +1, -0
The function f(x)={e3x−e−5xx,if x≠0k,if x=0.f(x)=\begin{cases} \frac{e^{3x}-e^{-5x}}{x}, & \text{if } x \neq 0 \\ k, & \text{if } x=0 \end{cases}. is continuous at x=0x=0 for the value of kk, as
Solution:     👈: Video Solution
Explanation: Since, f(x)f(x) is continuous at x=0x=0, then
LHL=RHL=f(0) or LHL=RHL=k\text{LHL} = \text{RHL} = f(0) \text{ or } \text{LHL} = \text{RHL} = k
Now,
LHL=lim⁡h→0e3(0−h)−e−5(0−h)0−h\text{LHL} = \lim\limits_{h \rightarrow 0} \frac{e^{3(0-h)}-e^{-5(0-h)}}{0-h}
=lim⁡h→0e−3h−e5h−h= \lim\limits_{h \rightarrow 0} \frac{e^{-3h}-e^{5h}}{-h}
=lim⁡h→0(e−3h−1−h)+lim⁡h→0(e5h−1h)= \lim\limits_{h \rightarrow 0} \left( \frac{e^{-3h}-1}{-h} \right) + \lim\limits_{h \rightarrow 0} \left( \frac{e^{5h}-1}{h} \right)
=3lim⁡h→0(e−3h−1−3h)+5lim⁡h→0(e5h−15h)= 3 \lim\limits_{h \rightarrow 0} \left( \frac{e^{-3h}-1}{-3h} \right) + 5 \lim\limits_{h \rightarrow 0} \left( \frac{e^{5h}-1}{5h} \right)
=3×1+5×1=8= 3 \times 1 + 5 \times 1 = 8
Thus, k=8 .\text{Thus, } \quad k=8 \text{ .}
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