MCQ
The solution set of $\frac{x^{2}}{x^{2}+1}<0$ is
  • A
    $0$
  • B
    $(-1,1)$
  • $\phi$
  • D
    $\mathrm{R}$

Answer

Correct option: C.
$\phi$
c

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

$\sin 12^\circ \sin 24^\circ \sin 48^\circ \sin 84^\circ = $
Let $\alpha, \beta$ be the roots of the equation $x^{2}-\sqrt{2} x+\sqrt{6}=0$ and $\frac{1}{\alpha^{2}}+1, \frac{1}{\beta^{2}}+1$ be the roots of the equation $x^{2}+a x+b=0$. Then the roots of the equation $x ^{2}-( a + b -2) x +( a + b +2)$ $=0$ are$...$
The standard deviation of $25$ numbers is $40$. If each of the numbers is increased by $5$, then the new standard deviation will be
The number of triangles that can be formed by choosing the vertices from a set of $12$ points,$7$ of which lie on the same straight line, is
If $\sum_{i=1}^{5}(x_i-10)=5$ and $\sum_{i=1}^{5}(x_i-10)^2=5$ then standard deviation of observations $2x_1 + 7, 2x_2 + 7, 2x_3 + 7, 2x_4 + 7$ and $2x_5 + 7$ is equal to-
A card is drawn at random from a pack of cards. The probability of this card being a red or a queen is
If $a, b, c$ are in $A.P.$ and $a^2, b^2, c^2$ are in $G.P.$ such that $ a <  b$ $ < c$ and $a+b+c\,= \frac{3}{4}$ , then the value of $a$ is
The value of $\sum\limits_{r = 1}^{15} {{r^2}\,\left( {\frac{{^{15}{C_r}}}{{^{15}{C_{r - 1}}}}} \right)} $ is equal to
Let $f(x)=x^6-2 x^3+x^3+x^2-x-1$ and $g(x)=x^4-x^3-x^2-1$ be two polynomials. Let $a, b, c$ and $d$ be the roots of $g(x)=0$. Then, the value of $f(a)+f(b)+f(c)+f(d)$ is
If $\int \frac{\cos \theta}{5+7 \sin \theta-2 \cos ^{2} \theta} d \theta=A \log _{e}|B(\theta)|+C$ where $C$ is a constant of integration, then $\frac{ B (\theta)}{ A }$ can be