Let f(x) = lim_{n} {rightarrow infty} left( f | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B,C) Taking the natural logarithm on both sides:$\ln f(x) = \lim_{n}$ ${\rightarrow \infty} \frac{x}{n} \sum_{r=1}^n \ln \left( \frac{x + n/r}{x^2 +

Vedclass · Accepted Answer

$B, D$

Vedclass · Answer

(B,C) Taking the natural logarithm on both sides:$\ln f(x) = \lim_{n}$ ${\rightarrow \infty} \frac{x}{n} \sum_{r=1}^n \ln \left( \frac{x + n/r}{x^2 + n^2/r^2} \cdot \frac{n}{r} \right)$$= \lim_{n \rightarrow \infty} \frac{x}{n} \sum_{r=1}^n \ln \left( \frac{1 + \frac{r}{n}x}{1 + (\frac{r}{n}x)^2} \right)$$= x \int_0^1 \ln \left( \frac{1 + tx}{1 + (tx)^2} \right) dt$Let $u = tx$,then $du = x dt$:$\ln f(x) = \int_0^x \ln \left( \frac{1+u}{1+u^2} \right) du$By the Fundamental Theorem of Calculus:$\frac{f^{\prime}(x)}{f(x)} = \ln \left( \frac{1+x}{1+x^2} \right)$For $0  1$,so $\frac{f^{\prime}(x)}{f(x)} > 0$,meaning $f(x)$ is increasing.For $x > 1$,$\frac{1+x}{1+x^2} Thus,$f(1/3) Since $f(x)$ is decreasing for $x > 1$,$f^{\prime}(2) Therefore,the correct options are $B$ and $C$.

Similar Questions

Let $f(x)$ be defined in $[-2, 2]$ by
$f(x) = \begin{cases} \max(4 - x^2, 1 + x^2), & -2 < x < 0 \\ \min(4 - x^2, 1 + x^2), & 0 < x < 2 \end{cases}$
Then $f(x)$:

Using the fact that $\sin (A+B)=\sin A \cos B+\cos A \sin B$ and the differentiation,obtain the sum formula for cosines.

Consider the function $f(x) = x \cos x - \sin x$. Identify the correct statement.

Let $f : R \to R$ be a differentiable function satisfying $f''(3) + f'(2) = 0$. Then $\mathop {\lim }\limits_{x \to 0} {\left( {\frac{{1 + f\left( {3 + x} \right) - f\left( 3 \right)}}{{1 + f\left( {2 - x} \right) - f\left( 2 \right)}}} \right)^{\frac{1}{x}}}$ is equal to

If $f(x)=x^2+g^{\prime}(1) x+g^{\prime \prime}(2)$ and $g(x)=f(1) x^2+x f^{\prime}(x)+f^{\prime \prime}(x),$ then the value of $f(4)-g(4)$ is equal to $...........$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module