MCQ
Which of the following function is even function
  • A
    $f(x) = \frac{{{a^x} + 1}}{{{a^x} - 1}}$
  • $f(x) = x\left( {\frac{{{a^x} - 1}}{{{a^x} + 1}}} \right)$
  • C
    $f(x) = \frac{{{a^x} - {a^{ - x}}}}{{{a^x} + {a^{ - x}}}}$
  • D
    $f(x) = \sin x$

Answer

Correct option: B.
$f(x) = x\left( {\frac{{{a^x} - 1}}{{{a^x} + 1}}} \right)$
b
(b) In $(a)$, $f( - x) = \frac{{{a^{ - x}} + 1}}{{{a^{ - x}} - 1}} = \frac{{1 + {a^x}}}{{1 - {a^x}}} = - \frac{{{a^x} + 1}}{{{a^x} - 1}} = - f(x)$

So, it is an odd function.

In $(b)$, $f( - x) = ( - x)\frac{{{a^{ - x}} - 1}}{{{a^{ - x}} + 1}} = - x\frac{{1 - {a^x}}}{{1 + {a^x}}} = x\frac{{{a^x} - 1}}{{{a^x} + 1}} = f(x)$

So, it is an even function.

In $(c)$, $f( - x) = - \sin \left[ {\log (x + \sqrt {1 + {x^2}} )} \right]$

So, it is an odd function.

In $(d)$, $f( - x) = \sin ( - x) = - \sin x = - f(x)$

So, it is an odd function.

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

The solution for $x$ ofthe equation $\mathop \smallint \limits_{\sqrt 2 }^x \frac{{dt}}{{t\sqrt {{t^2} - 1} }} = \frac{\pi }{2}$ is
Let $X Y$ be the diameter of a semi-circle with centre $O$. Let $A$ be a variable point on the semi-circle and $B$ another point on the semi-circle such that $A B$ is parallel to $X Y$.The value of $\angle B O Y$ for which the inradius of $\triangle A O B$ is maximum, is
The area of the region enclosed between the parabolas $y ^{2}=2 x -1$ and $y ^{2}=4 x -3$ is
If $A$ and $B$ are two independent events with $P(A)=\frac{1}{3}$ and $P(B)=\frac{1}{4}$, then $P\left(B^{\prime} \mid A\right)$ is equal to
For the function $\text{f}(\text{x})=\text{x}+1\text{x},\text{x}\in[1,3],$ the value of c for the Lagrange's mean value theorem is:
  1. 1
  2. $\sqrt3$
  3. 2
  4. none of these
If $f(x)=\begin{vmatrix}
x^3-x &2e^{2x}  &sin x^2 \\ 
cos (2x) &x+x^2  &e^{-x} \\ 
tan 3x & ln (1-2x) &x^2+x+1 
\end{vmatrix}$ then $f'(0)$ equals to
Evaluate: $\int \frac{d x}{\sqrt{1-2 x-x^2}}$
For the equation ${\cos ^{ - 1}}\left| x \right| + {\cos ^{ - 1}}\left| {2x} \right| = \pi $ , the number of real solution $(s)$ is
$Let\,\,f(x) = \left\{ {\begin{array}{*{20}{c}}
  {\operatorname{sgn} ({x^2} - 3x + 2)\,\,\,;\,x \in Q} \\ 
  {0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,;\,x \notin Q} 
\end{array}} \right.$ then number of points where $f(x)$ is continuous is (where sgn $(x)$ denotes signum function of $x$)
Area of the region bounded by the curve $y=\cos x$ between $x=0$ and $x=\pi$ is