Download App

6. Application of derivatives — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths6. Application of derivatives1 Mark
MCQ
Function $f\left( x \right) = \frac{{\left| {x - 1} \right|}}{{{x^2}}}$ is monotonic decreasing in
  • A
    $\left( { - \infty ,\infty } \right)$
  • B
    $(0,1)$
  • C
    $\left( {2,\infty } \right)$
  • ✓
    $\left( {0,1} \right) \cup \left( {2,\infty } \right)$
✓

Answer

Correct option: D.
$\left( {0,1} \right) \cup \left( {2,\infty } \right)$
d
$f(x)=\left\{\begin{array}{ll}\frac{-(x-1)}{x^{2}} & x<1 \\ \frac{(x-1)}{x^{2}} & x \geq 1\end{array}\right.$

$f^{\prime}(x)=\left\{\begin{array}{ll}\frac{(x-2)}{x^{3}} & x<1 \\ \frac{-(x-2)}{x^{3}} & x \geq 1\end{array}\right.$

$x<1,$ if $f^{\prime}(x)<0$ (for $f(x)$ to be monotonically decreasing $\Rightarrow \frac{(x-2)}{x^{3}}<0 \Rightarrow x \in(0,2)$

But $x<1 \Rightarrow x \in(0,1)$

For $x \geq 1,$ if $f^{\prime}(x)<0 \Rightarrow \frac{-(x-2)}{x^{3}}<0$

$\Rightarrow \frac{(x-2)}{x^{3}}>0 \Rightarrow x \in(-\infty, 0) \cup(2, \infty)$

But, $x \geq 1 \Rightarrow x \in(2, \infty)$

Hence, $x \in(0,1) \cup(2, \infty)$

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

Explore more

More "M.C.Q (1 Marks)" from 6. Application of derivatives→6. Application of derivatives — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

Solution of the differential equation $xdy = (y + xy^3 (1 + log_ex))\ dx$ is (Where $C$ is arbitary constant)
→
The number of real solutions of ${\tan ^{ - 1}}\sqrt {x(x + 1)} + {\sin ^{ - 1}}\sqrt {{x^2} + x + 1} = \frac{\pi }{2}$ is
→
If $a_i^2 + b_i^2 + c_i^2 = 1,\,\,(i = 1,2,3)$ and ${a_i}{a_j} + {b_i}{b_j} + {c_i}{c_j} = 0$ $(i \ne j,i,j = 1,2,3)$ then the value of ${\left| {\,\begin{array}{*{20}{c}}{{a_1}}&{{a_2}}&{{a_3}}\\{{b_1}}&{{b_2}}&{{b_3}}\\{{c_1}}&{{c_2}}&{{c_3}}\end{array}\,} \right|^2}$ is
→
If $\sqrt{1-\text{x}^6}+\sqrt{1-\text{y}^6}=\text{a}^3(\text{x}^3-\text{y})^3,$ then $\frac{\text{dy}}{\text{dx}}$ is equal to:
  1. $\frac{\text{x}^2}{\text{y}^2}\sqrt{\frac{1-\text{y}^6}{1-\text{x}^6}}$
  2. $\frac{\text{y}^2}{\text{x}^2}\sqrt{\frac{1-\text{y}^6}{1+\text{x}^6}}$
  3. $\frac{\text{x}^2}{\text{y}^2}\sqrt{\frac{1-\text{x}^6}{1-\text{y}^6}}$
  4. $\text{None of these.}$
→
Let f : R → R be given by f(x) = [x2] + [x + 1] - 3 where [x] denotes the greatest integer less than or equal to x. Then, f(x) is:
  1. Many-one and onto.
  2. Many-one and into.
  3. One-one and into.
  4. One-one and onto.
→
If the area of the region

$\left\{(\mathrm{x}, \mathrm{y}): \frac{\mathrm{a}}{\mathrm{x}^2} \leq \mathrm{y} \leq \frac{1}{\mathrm{x}}, 1 \leq \mathrm{x} \leq 2,0<\mathrm{a}<1\right\}$ is

$\left(\log _e 2\right)-\frac{1}{7}$ then the value of $7 a-3$ is equal to:

→
If the mirror image of the point $\mathrm{P}(3,4,9)$ in the line $\frac{x-1}{3}=\frac{y+1}{2}=\frac{z-2}{1}$ is $(\alpha, \beta, \gamma)$, then $14(\alpha+\beta+\gamma)$ is :
→
A boy tosses fair coin 3 times. If he gets $₹ 2 X$ for $X$ heads, then his expected gain (in ₹) equals to
→
The value of $\tan\Big\{\cos^{-1}\frac{1}{5\sqrt2}-\sin^{-1}\frac{4}{\sqrt{17}}\Big\}$ is:
  1. $\frac{\sqrt{29}}{3}$
  2. $\frac{29}{3}$
  3. $\frac{\sqrt3}{29}$
  4. $\frac{3}{29}$
→
Function

$f\left( x \right) = \int_1^x {\left\{ {2\left( {t - 1} \right){{\left( {t - 2} \right)}^3} + 3{{\left( {t - 1} \right)}^2}{{\left( {t - 2} \right)}^2}} \right\}} dt$ is maximum when $x$ is equal to

→