Mathematics - Application of Derivative Question with Solution | TestHub

$f\left(x\right)=x+\ell nx-x\ell nx, x>0$

$f^{\text{'}}\left(x\right)=1+\frac{1}{x}-\ell nx-1$

$f^{\text{'}\text{'}}\left(x\right)= -\frac{1}{x^2}-\frac{1}{x}=\frac{-\left(x+1\right)}{x^2}$

(I)  $f\left(1\right) f\left(e^2\right)<0$, so true.

(II) $f^{\text{'}}\left(1\right) f^{\text{'}}\left(e\right)<0$,so true.

(III) Graph of $f^{\text{'}}\left(x\right)$ , so (III) is false as the curve does not intersect X-axis i.e, $f\text{'}\left(x\right)\ne 0$ when $x\in (0,1)$

(IV) Is false. $f\text{'}\text{'}\left(x\right)\ne 0$as tangent to the curve of $f\text{'}\left(x\right)$ is not parallel to X-axis.

As   $\underset{x\to \infty }{\lim }f\left(x\right)= \underset{x\to \infty }{\lim }x\left[1+\frac{\ell n x}{x}-\mathcal{ }\ell n x\right]= -\infty$

$\therefore$  (i) is false (ii) is true

$\underset{x\to \infty }{\lim }{f}^{\text{'}}\left(x\right)= -\infty$, so (iii) is true

$\underset{x\to \infty }{\lim }{f}^{\text{'}\text{'}}\left(x\right)=0$ , so (iv) is true.

(P)  $f^{\text{'}}\left(x\right)$, is positive in$(0, 1)$, so true.

(Q)  $f^{\text{'}}\left(x\right)<0$, for in $\left(e, e^2\right)$, so true.

As  $f^{"}\left(x\right)<0 \forall x>0$ therefore, R is false, S is true.

Alternate:

$f\left(x\right)=x+ \ell n x-x\ell n x$

$f^{\text{'}}\left(x\right)=\frac{1}{x}-\ell n x=0$ at $x=x_0$ where $x_0\in \left(1, e\right)$

$f^{\text{'}\text{'}}\left(x\right)= -\frac{1}{x^2}-\frac{1}{x}<0 \forall x>0 \Rightarrow f\left(x\right)$  concave down