Question

If $\ell = \lim_{x \to 1^+} 2^{-2^{\frac{1}{1-x}}}$ and $m = \lim_{x \to 1^+}\frac{x \sin(x-[x])}{x-1}$, where [.] denotes greatest integer function, then $\int_{\ell}^{m}\frac{\ln(x+\sqrt{1+x^2})}{\sqrt{1+x^2}}dx$ is equal to -

• A. 1
• B. \frac{1}{2}
• C. \ln\frac{1}{2}
• D. 0

Answer 1

Answer: (D)

0

Answer 2

1. Calculation of $\ell$: As $x \to 1^+$, $\frac{1}{1-x} \to -\infty$. Thus, $\ell = \lim_{u \to -\infty} 2^{-2^u} = 2^{-0} = 1$.

2. Calculation of $m$: As $x \to 1^+$, $x$ is slightly greater than 1, so $[x]=1$. Thus, $x-[x] = x-1$. $m = \lim_{x \to 1^+}\frac{x \sin(x-1)}{x-1} = \left(\lim_{x \to 1^+} x\right) \cdot \left(\lim_{x \to 1^+} \frac{\sin(x-1)}{x-1}\right) = 1 \cdot 1 = 1$.

3. Evaluation of the integral: The integral to be evaluated is $\int_{\ell}^{m}\frac{\ln(x+\sqrt{1+x^2})}{\sqrt{1+x^2}}dx$. Substituting the calculated values of $\ell=1$ and $m=1$, we get $\int_{1}^{1}\frac{\ln(x+\sqrt{1+x^2})}{\sqrt{1+x^2}}dx$. A definite integral where the upper and lower limits of integration are the same is always equal to 0.