| Statement-1: The value of\[\underset{x\to \infty }{\mathop{\lim }}\,\left( (x+2){{\tan }^{-1}}(x+2)-(x{{\tan }^{-1}}x) \right)\] is equal to\[\pi \]. |
| Statement-2: If \[f(x)\] is derivable in \[[a,b]\] then \[\exists \] at least one \[c\in (a,b)\] such that \[f'(c)=\frac{f(b)-f(a)}{b-a}\] |
A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.
B) Statement-1 is true, Statement-2 is true and Statement-2 is NOT the correct explanation for Statement-1.
C) Statement-1 is true, Statement-2 is false.
D) Statement-1 is false, Statement-2 is true
Correct Answer: A
Solution :
Let \[f(x)=x\,{{\tan }^{-1}}x\] By LMVT, \[f(x+2)\,-f(x)\,=2f'(y)\,\] for some \[y\in \,(x,\,\,x+2)\] As \[x\to \infty ,\] we have \[y\to \infty \] Hence \[\underset{x\to \infty }{\mathop{Lim}}\,[f(x+2)-f(x)]\,=\underset{y\to \infty }{\mathop{Lim}}\,2f'(x)\] \[=\underset{y\to \infty }{\mathop{Lim}}\,\,\left[ \frac{2y}{1+{{y}^{2}}}\,+2{{\tan }^{-1}}\,y \right]\,=\pi \]
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