MCQ
For which value of $x$ , the function $f(x) = {x^2} - 2x$ is decreasing
- A$x > 1$
- B$x > 2$
- ✓$x < 1$
- D$x < 2$
Aliter : $f'(x) = 2x - 2 = 2(x - 1)$
To be decreasing, $2(x - 1) < 0 \Rightarrow (x - 1) < 0 \Rightarrow x < 1$.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$(1,1)(-1,1),(1,-1),(-1,-1),(-2,1)(2,-1),(-1,2)$ and $(-2,-1)$
Match List $I$ with List $II$ and select the correct answer using the code given below the lists :
| List $I$ | List $II$ |
| $P.$ $\quad$m= | $1.$ $\quad\frac{1}{2}$ |
| $Q.$ $\quad$Maximum area of $\triangle E F G$ is | $2.$ $\quad4$ |
| $R.$ $\quad y_0=$ | $3.$ $\quad2$ |
| $S.$ $\quad y_1=$ | $4.$ $\quad1$ |
Codes: $ \quad P \quad Q \quad R \quad S $
$\frac{3}{2} \cos ^{-1} \sqrt{\frac{2}{2+\pi^2}}+\frac{1}{4} \sin ^{-1} \frac{2 \sqrt{2} \pi}{2+\pi^2}+\tan ^{-1} \frac{\sqrt{2}}{\pi}$ is. . . .