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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Continuity and Differentiability

NTA Abhyas 2020 · Mathematics · Continuity and Differentiability

If f(x) = \ cases e ^ (1 + (1)/(x)) - a e(1)/(x) + 1 & : & x ≠ 0 b...

If f\left(x\right) = \left\{\begin{cases}\dfrac{{e}^{\left(1 + \dfrac{1}{x}\right)} - a}{e\dfrac{1}{x} + 1} & : & x \ne 0 \\ b & : & x = 0\end{cases}\right) (whereaandandbarearbitraryconstants)iscontinuousatare arbitrary constants) is continuous atx = 0,thenthevalueof, then the value of{a}^{2} is equal to

(use e=2.7e = 2 .7)

Correct answer

7.29

Step-by-step solution

At x=0x = 0,
LHL=lim⁡h→0+f(0−h)=lim⁡h→0+e(1−1h)−ae−1h+1=0−a0+1=−aLHL = \lim_{h \to {0}^{+}}f\left(0 - h\right) = \lim_{h \to {0}^{+}}\dfrac{{e}^{\left(1 - \dfrac{1}{h}\right)} - a}{{e}^{- \dfrac{1}{h}} + 1} = \dfrac{0 - a}{0 + 1} = - a
RHL=lim⁡h→0+f(0+h)=lim⁡h→0+e(1+1h)−ae1h+1=lim⁡h→0e−e−1h×a1+e−1h=eRHL = \lim_{h \to {0}^{+}}f\left(0 + h\right) = \lim_{h \to {0}^{+}}\dfrac{{e}^{\left(1 + \dfrac{1}{h}\right)} - a}{e\dfrac{1}{h} + 1} = \lim_{h \to 0}\dfrac{e - {e}^{- \dfrac{1}{h}} \times a}{1 + {e}^{- \dfrac{1}{h}}} = e
and f(0)=bf\left(0\right) = b
∵f(x)∵ f\left(x\right) is continuous at x=0x = 0
∴−a=e=b⇒a=−e∴ - a = e = b ⇒ a = - e
a2=e2{a}^{2} = {e}^{2}
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