Daily Question 113

Daily Question 113

$\textbf{Q113.}$ If $f(x)=\dfrac{x^2}{4+(\ln x)(\ln x)\ldots\infty}$ $\forall$ $x\in[1,\infty)$, then

(A) $f(x)=\dfrac{x^2}{5}, x\in[1,e)$

(B) $f(x)=\dfrac{x^2}{4}, x\in \{e\}$

(C) $f(x)=\dfrac{x^2}{4}, x > e$

(D) $\displaystyle\int\limits_{1}^{2}f(x)\mathrm{d}x=\dfrac{7}{12}$

$\textbf{Ans.} (D)$

$\textbf{Sol.}$

Given $f(x)=\dfrac{x^2}{4+(\ln x)(\ln x)\ldots\infty}$ $\forall$ $x\in[1,\infty)$

Let $h(x)=4+(\ln x)(\ln x)\ldots\infty$

$h(x)=\begin{cases} 4, & 1 \le x < e\\ 5, & x=e\\ \infty & x > e\\ \end{cases}$

For $x = e,$ $f(x)=\dfrac{x^2}{5}$

For $1 < x < e,$ $f(x)=\dfrac{x^2}{4}$

Now, $\displaystyle\int\limits_{1}^{2}f(x)\mathrm{d}x=\int\limits_{1}^{2}\dfrac{x^2}{4}\mathrm{d}x$

$\implies\displaystyle\int\limits_{1}^{2}f(x)\mathrm{d}x=\left[\dfrac{x^3}{12}\right]_{1}^{2}$

$\implies\displaystyle\int\limits_{1}^{2}f(x)\mathrm{d}x=\dfrac{7}{12}$

Daily Question 112

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