MCQ
Let $h$ be a twice continuously differentiable positive function on an open interval $J.$ Let $g(x) = ln\left( {h(x)} \right)$ for each $x \in J$. Suppose ${\left( {h'(x)} \right)^2} > h''(x) h(x) $ for each $x \in J$. Then
  • A
    $g$ is increasing on $J$
  • B
    $g$ is decreasing on $J$
  • C
    $g$ is concave up on $J$
  • $g$ is concave down on $J$

Answer

Correct option: D.
$g$ is concave down on $J$
d
Given $g(x) = ln\left( {h(x)} \right)$

$g ' (x) = \frac{{h'(x)}}{{h(x)}}$

$g''(x) = \frac{{h(x)h''(x) - {{(h'(x))}^2}}}{{{h^2}(x)}}$ < 0

$ g''(x) \leq 0\Rightarrow  \,\,g (x)$  is concave down

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

Let $f$ be a differentiable function $R$ to $R$ such that $\left| {f\,(x)\, - \,f(y)} \right|\, \le \,2\,{\left| {x - y} \right|^{\frac{3}{2}}},$ for all $x,y\,\in R .$ If $f\,(0)=1$ then $\int\limits_0^1 {{f^2}\,(x)\,dx} $ is equal to
A rectangular parallelopiped is formed by planes drawn through the point $(5, 7, 9)$ and $(2, 3, 7)$ parallel to the coordinate planes. The length of an edge of this rectangular parallelopiped is:
Evaluate: $\int \sec ^2(7-4 x) d x$
Let $g(x)=f(x)+f(1-x)$ and $f^{\prime \prime}(x) > 0, x \in(0,1)$. If $g$ is decreasing in the interval $(0, \alpha)$ and increasing in the interval $(\alpha, 1)$, then $\tan ^1(2 \alpha)+\tan ^{-1}\left(\frac{1}{\alpha}\right)+\tan ^{-1}\left(\frac{\alpha+1}{\alpha}\right)$ is equal to :
If $\text{y}=\sin^{-1}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big),$ then $\frac{\text{dy}}{\text{dx}}=$
If $x\frac{{dy}}{{dx}} + y = x\frac{{f\left( {xy} \right)}}{{f'\left( {xy} \right)}}$ , then $f(xy)$ is equal to
If $y = y ( x )$ is the solution of the differential equation $2 x^{2} \frac{d y}{d x}-2 x y+3 y^{2}=0 \quad$ such that $y(e)=\frac{e}{3}$, then $y(1)$ is equal to
Let $R$ be the relation on the set of all real numbers defined by $a R b$ iff $|a-b| \leq 1$. Then, $R$ is
If $\tan^{-1}\Big\{\frac{\sqrt{1+\text{x}^2}-\sqrt{1-\text{x}^2}}{\sqrt{1+\text{x}^2}+\sqrt{1-\text{x}^2}}\Big\}=\alpha,$ then $x^2 =$
If $f(x) = \left\{ \begin{array}{l}{2^{1/x}},{\rm{for\,\,}}\,x \ne 0\\\,\,\,\,\,\,\,3,{\rm{for\,\,}}\,x = {\rm{0}}\end{array} \right.$, then