Let h(x) = f(x) - {f(x)}^2 + {f(x)}^3 for eve | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given $h(x) = f(x) - \{f(x)\}^2 + \{f(x)\}^3$.Differentiating with respect to $x$,we get:$h'(x) = f'(x) - 2f(x)f'(x) + 3\{f(x)\}^2f'(x)$Factoring

Vedclass · Accepted Answer

Both $(A)$ and $(C)$

Vedclass · Answer

(D) Given $h(x) = f(x) - \{f(x)\}^2 + \{f(x)\}^3$.Differentiating with respect to $x$,we get:$h'(x) = f'(x) - 2f(x)f'(x) + 3\{f(x)\}^2f'(x)$Factoring out $f'(x)$:$h'(x) = f'(x) [1 - 2f(x) + 3\{f(x)\}^2]$Let $y = f(x)$. Then the expression becomes:$h'(x) = f'(x) (3y^2 - 2y + 1)$Consider the quadratic expression $Q(y) = 3y^2 - 2y + 1$. The discriminant $D = (-2)^2 - 4(3)(1) = 4 - 12 = -8$.Since $D  0$,the quadratic $3y^2 - 2y + 1$ is always positive for all real $y$.Thus,$h'(x) = f'(x) 	imes (	ext{a positive quantity})$.This implies that the sign of $h'(x)$ is the same as the sign of $f'(x)$.Therefore,$h(x)$ is increasing whenever $f(x)$ is increasing $(f'(x) > 0 \implies h'(x) > 0)$ and $h(x)$ is decreasing whenever $f(x)$ is decreasing $(f'(x) Hence,both $(A)$ and $(C)$ are correct.

Let $h(x) = f(x) - \{f(x)\}^2 + \{f(x)\}^3$ for every real number $x$,then

Similar Questions

On which of the following intervals is the function $f(x) = x^{100} + \sin x - 1$ strictly increasing?

If $f(x) = \frac{\sin x + b \cos x}{\sin x + 4 \cos x}$ is a strictly increasing function,then:

Let $f(x) = \int\limits_1^x {\left( {t\ln(t) - \frac{{\ln(t)}}{t}} \right)dt}$ for $x > 1$. Then:

For the function $f(x) = \cos x - x + 1, x \in R$,consider the following two statements:
$(S1)$ $f(x) = 0$ for only one value of $x$ in $[0, \pi]$.
$(S2)$ $f(x)$ is decreasing in $[0, \frac{\pi}{2}]$ and increasing in $[\frac{\pi}{2}, \pi]$.

The set of all points,for which $f(x) = x^2 e^{-x}$ strictly increases,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module