MCQ
What are the minimum and maximum values of the function ${x^5} - 5{x^4} + 5{x^3} - 10$
- ✓$-37, -9$
- B$10, 0$
- CIt has $2 $ min. and $1 $ max. values
- DIt has $2$ max. and $ 1$ min. values
$\therefore$ $\frac{{dy}}{{dx}} = 5{x^4} - 20{x^3} + 15{x^2}$$ = \,\,5{x^2}({x^2} - 4x + 3)$
$ = \,\,5{x^2}(x - 3)\,(x - 1)$
$\frac{{dy}}{{dx}} = 0$, gives $x = 0,\,1,\,3$
Now, $\frac{{{d^2}y}}{{d{x^2}}} = 20{x^3} - 60{x^2} + 30x$ = $10x(2{x^2} - 6x + 3)$
and $\frac{{{d^3}y}}{{d{x^3}}} = 10(6{x^2} - 12x + 3)$
For $x = 0$: $\frac{{dy}}{{dx}} = 0,\,\frac{{{d^2}y}}{{d{x^2}}} = 0,\,\frac{{{d^3}y}}{{d{x^3}}} \ne 0$
$\therefore$ Neither minimum nor maximum
For$x = 1,\,\frac{{{d^2}y}}{{d{x^2}}} = - 10 = {\rm{negative}}$.
$\therefore$ Maximum value ${y_{{\rm{max}}{\rm{.}}}} = - 9$
For $x = 3,\,\frac{{{d^2}y}}{{d{x^2}}} = 90 = {\rm{positive}}$
$\therefore$ Minimum value ${y_{{\rm{min}}{\rm{.}}}} = - 37$.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
A coin is tossed n times. The probability of geting at least once is greater than 0.8. Then, the least value of n, is:
$\left|\begin{array}{ccc}\cos ^{2} x & 1+\sin ^{2} x & \sin 2 x \\ 1+\cos ^{2} x & \sin ^{2} x & \sin 2 x \\ \cos ^{2} x & \sin ^{2} x & 1+\sin 2 x\end{array}\right|$.
Then the ordered pair $( m , M )$ is equal to