Let f(x) = 2x + tan^{-1} x and g(x) = log_e(s | vedclass

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

Question

(B) Given $f(x) = 2x + 	an^{-1} x$ and $g(x) = \ln(\sqrt{1+x^2} + x)$ for $x \in [0, 3]$.First,find the derivatives:$f'(x) = 2 + \frac{1}{1+x^2}$$g'(

Vedclass · Accepted Answer

$\max f(x) > \max g(x)$

Vedclass · Answer

(B) Given $f(x) = 2x + 	an^{-1} x$ and $g(x) = \ln(\sqrt{1+x^2} + x)$ for $x \in [0, 3]$.First,find the derivatives:$f'(x) = 2 + \frac{1}{1+x^2}$$g'(x) = \frac{1}{\sqrt{1+x^2} + x} \cdot \left(\frac{x}{\sqrt{1+x^2}} + 1ight) = \frac{1}{\sqrt{1+x^2} + x} \cdot \frac{x + \sqrt{1+x^2}}{\sqrt{1+x^2}} = \frac{1}{\sqrt{1+x^2}}$.For $x \in [0, 3]$:$f'(x) \in [2 + \frac{1}{1+3^2}, 2 + \frac{1}{1+0^2}] = [2.1, 3]$.$g'(x) \in [\frac{1}{\sqrt{1+3^2}}, \frac{1}{\sqrt{1+0^2}}] = [\frac{1}{\sqrt{10}}, 1] \approx [0.316, 1]$.Since $f'(x) > g'(x)$ for all $x \in [0, 3]$,option $A$ is incorrect.Both $f(x)$ and $g(x)$ are strictly increasing functions on $[0, 3]$.Thus,$\max f(x) = f(3) = 6 + 	an^{-1} 3$ and $\max g(x) = g(3) = \ln(3 + \sqrt{10})$.Since $6 + 	an^{-1} 3 > 6$ and $\ln(3 + \sqrt{10}) \approx \ln(6.16)  g(3)$.Therefore,option $B$ is correct.

Let $f(x) = 2x + \tan^{-1} x$ and $g(x) = \log_e(\sqrt{1+x^2} + x)$,$x \in [0, 3]$. Then:

Similar Questions

Let $f(x) = x \cos^{-1}(-\sin |x|)$,$x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. Then which of the following is true?

If $(x-a)^{2}+(y-b)^{2}=c^{2},$ for some $c > 0,$ prove that $\frac{\left[1+\left(\frac{d y}{d x}\right)^{2}\right]^{\frac{3}{2}}}{\frac{d^{2} y}{d x^{2}}}$ is a constant independent of $a$ and $b.$

$f(x)=4 \log _{e}(x-1)-2 x^{2}+4 x+5, x>1$,which one of the following is $NOT$ correct?

Let $f: R \rightarrow R$ be defined by $f(x) = \frac{x^2-3x-6}{x^2+2x+4}$. Then which of the following statements is (are) $TRUE$?
$(A)$ $f$ is decreasing in the interval $(-2, -1)$
$(B)$ $f$ is increasing in the interval $(1, 2)$
$(C)$ $f$ is onto
$(D)$ Range of $f$ is $[-\frac{3}{2}, 2]$

If $f(x)=\int_0^x e^{t^2}(t-2)(t-3) dt$ for all $x \in(0, \infty)$,then
$(A)$ $f$ has a local maximum at $x=2$
$(B)$ $f$ is decreasing on $(2,3)$
$(C)$ there exists some $c \in(0, \infty)$ such that $f^{\prime \prime}(c)=0$
$(D)$ $f$ has a local minimum at $x=3$

Vedclass Test Series

Exam Paper Generator

Online Exam Module