MCQ
Interval in which function $y = |x^2 -|x| -2|$ is non monotonic, can be-
- A$x \in (-2,-1)$
- B$x \in (-4,-2)$
- ✓$x \in (0,2)$
- D$x \in (2,10)$
✓
Answer
Correct option: C.
$x \in (0,2)$
c
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
$f : R \to R$ is defined as
→$f(x) = \left\{ {\begin{array}{*{20}{c}}
{{x^2} + 2mx - 1\,,}&{x \leq 0}\\
{mx - 1\,\,\,\,\,\,\,\,\,\,\,\,\,,}&{x > 0}
\end{array}} \right.$
If $f (x)$ is one-one then the set of values of $'m'$ is
Let $A = \left\{ {{x_1},{x_2},{x_3},.....,{x_7}} \right\}$ and $B = \left\{ {{y_1},{y_2},{y_3}} \right\}$ be two sets containing seven and three distinct elements respectively. Then the total number of functions $f:A \to B$ which are onto, if there exist exactly three elements $x$ in $A$ such that $f(x) = {y_2}$ , is equal to
→The value of $\int\limits_0^1 {\sqrt[3]{{2{x^3} - 3{x^2} - x + 1}}\,dx} $ is
→A biased coin with probabilty p, 0 < p < 1, of heads is tossed until a head appears for the first time. If the probability that the number of tosses required is even is $\frac{2}{5},$ then p equals:
→Let $P ( S )$ denote the power set of $S =\{1,2,3, \ldots, 10\}$. Define the relations $R_1$ and $R_2$ on $P(S)$ as $A R_1 B$ if $\left( A \cap B ^{ c }\right) \cup\left( B \cap A ^{ c }\right)=\varnothing$ and $AR _2 B$ if $A \cup B ^{ c }=$ $B \cup A ^{ c }, \forall A , B \in P ( S )$. Then :
→If $f\left( x \right) = x{e^{x\left( {1 - x} \right)}},\,x \in R$ , then $f(x)$ is
→Let $a, b$ and $c$ be positive constants. The value of $‘a’$ in terms of $‘c’$ if the value of integral $\int\limits_0^1 {(ac{x^{b + 1}} + {a^3}b{x^{3b + 5}})\,dx} $ is independent of $b$ equals
→The area of the region enclosed by the curve $f(x)=\max \{\sin x, \cos x\},-\pi \leq x \leq \pi$ and the $x$-axis is
→Derivative of ${\sec ^{ - 1}}\left\{ {{1 \over {2{x^2} - 1}}} \right\}$ w.r.t $\sqrt {1 + 3x} $ at $x = - {1 \over 3}$ is
→If $\text{f(x)}=\begin{cases}\text{mx}+1,&\text{x}\leq\frac{\pi}{2}\\\sin\text{x}+\text{n},&\text{n}>\frac{\pi}{2}\end{cases}$ is continuous at $\text{x}=\frac{\pi}{2},$ then:
→- $\text{m}=1,\text{ n}=0$
- $\text{m}=\frac{\text{n}\pi}{2}+1$
- $\text{n}=\frac{\text{m}\pi}{2}$
- $\text{m}=\text{n}=\frac{\pi}{2}$