Download App

1. relation and function — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths1. relation and function1 Mark
MCQ
If $f(x) = 2{x^6} + 3{x^4} + 4{x^2}$ then $f'(x)$ is
  • A
    Even function
  • ✓
    An odd function
  • C
    Neither even nor odd
  • D
    None of these
✓

Answer

Correct option: B.
An odd function
b
(b) $f(x) = 2{x^6} + 3{x^4} + 4{x^2}$

$f( - x) = 2{( - x)^6} + 3{( - x)^4} + 4{( - x)^2} = f(x)$

==> $f(x)$ is an even function and derivative of an even function is always odd.

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 1. relation and function→1. relation and function — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

$\int_1^e {\frac{1}{x}\,dx} $ is equals to
→
Statement $- 1:$ The function $x^2 (e^x + e^{-x})$ is increasing for all $x > 0.$

Statement $-2:$ The functions $x^2e^x$ and $x^2e^{-x}$ are increasing for all $x > 0$ and the sum of two increasing functions in any interval $(a, b)$ is an increasing function in $(a, b).$

→
Statement $-1 :$  The point $A(1, 0, 7)$ is the mirror image of the point $B( 1, 6, 3)$ in the line:  $\frac{x}{1} = \frac{{y - 1}}{2} = \frac{{z - 2}}{3}$

Statement $-2 :$ The line $\frac{x}{1} = \frac{{y - 1}}{2} = \frac{{z - 2}}{3}$ bisects the line joining $A(1, 0, 7)$ and $B( 1, 6, 3)$

→
The sum of two forces is $18\, N$  and resultant whose direction is at right angles to the smaller force is $12\,N.$  The magnitude of the two forces are
→
Let $y=y(x), x>1$, be the solution of the differential equation $(x-1) \frac{d y}{d x}+2 x y=\frac{1}{x-1}$, with $y(2)=\frac{1+e^{4}}{2 e^{4}}$. If $y(3)=\frac{e^{\alpha}+1}{\beta e^{\alpha}}$. then the value of $\alpha+\beta$ is equal to
→
In a triangle $ABC,$  right angled at the vertex $A,$  if the position vectors of $A, B$  and $C$  are respectively $3\hat i\, + \hat j\, - \hat k,\,\, - \hat i\, + 3\hat j\, + p\hat k$  and $5\hat i\, + q\hat j\, - 4\hat k,\,$ then the point $(p, q)$  lies on a line
→
If f(x) = |x2 - 9x + 20|, then f(x) is equal to:
  1. -2x + 9 for all $\text{x}\in\text{R}$
  2. 2x - 9 if 4 < x < 5
  3. -2x + 9, if 4 < x < 5
  4. None of these.
→
The solution of the equation $\frac{{dy}}{{dx}} + \sqrt {\frac{{1 - {y^2}}}{{1 - {x^2}}}} = 0$ is
→
The solution of the differential equation $(1 + \cos x)dy = (1 - \cos x)dx$ is
→
If $A = \left[ {\begin{array}{*{20}{c}}\alpha &2\\2&\alpha \end{array}} \right]$ and $|{A^3}|$=125, then $\alpha = $
→