MCQ
જો $f(x) = {x^3} + b{x^2} + cx + d,0 < {b^2} < c$. તો $f$ એ . . .
- Aઆવૃત વિધેય છે
- Bસ્થાનીય મહતમ છે
- Cસ્થાનીય ન્યૂનતમ છે
- ✓ચુસ્ત વધતું
$\therefore$ $f'(x) = 3{x^2} + 2bx + c$
Now its discriminant $ = 4({b^2} - 3c)$
==> $4({b^2} - c) - 8c < 0,$ as ${b^2} < c$ and $c > 0$
Therefore, $f'(x) > 0$ for all $x \in R$
Hence $ f$ is strictly increasing.
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