If g(x) = 2f(2x^3 - 3x^2) + f(6x^2 - 4x^3 - 3 | vedclass

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

Question

(B) Given $f''(x) > 0$,$f'(x)$ is a strictly increasing function.$g'(x) = 2f'(2x^3 - 3x^2) \cdot (6x^2 - 6x) + f'(6x^2 - 4x^3 - 3) \cdot (12x - 12x^2)

Vedclass · Accepted Answer

$\left( - \frac{1}{2},0 \right) \cup \left( 1,\infty \right)$

Vedclass · Answer

(B) Given $f''(x) > 0$,$f'(x)$ is a strictly increasing function.$g'(x) = 2f'(2x^3 - 3x^2) \cdot (6x^2 - 6x) + f'(6x^2 - 4x^3 - 3) \cdot (12x - 12x^2)$$g'(x) = 12(x^2 - x) f'(2x^3 - 3x^2) - 12(x^2 - x) f'(6x^2 - 4x^3 - 3)$$g'(x) = 12x(x - 1) [f'(2x^3 - 3x^2) - f'(6x^2 - 4x^3 - 3)]$For $g'(x) > 0$,we analyze the sign of $12x(x - 1)$ and the difference $[f'(2x^3 - 3x^2) - f'(6x^2 - 4x^3 - 3)]$.Since $f'(x)$ is increasing,$f'(A) > f'(B) \iff A > B$.Let $A = 2x^3 - 3x^2$ and $B = 6x^2 - 4x^3 - 3$.$A - B = (2x^3 - 3x^2) - (6x^2 - 4x^3 - 3) = 6x^3 - 9x^2 + 3 = 3(2x^3 - 3x^2 + 1) = 3(x - 1)^2(2x + 1)$.Thus,$f'(A) - f'(B) > 0$ when $3(x - 1)^2(2x + 1) > 0$,which implies $x > -\frac{1}{2}$ (excluding $x = 1$).Now,$g'(x) > 0$ when $12x(x - 1)$ and $(f'(A) - f'(B))$ have the same sign.Case $I$: $x(x - 1) > 0$ and $x > -\frac{1}{2}$. This occurs when $x \in (1, \infty)$.Case $II$: $x(x - 1) Combining these,$g'(x) > 0$ for $x \in (-\frac{1}{2}, 0) \cup (1, \infty)$.

If $g(x) = 2f(2x^3 - 3x^2) + f(6x^2 - 4x^3 - 3)$,$\forall x \in R$ and $f''(x) > 0$,$\forall x \in R$,then $g'(x) > 0$ for $x$ belonging to

Similar Questions

The sum of the lengths of the hypotenuse and one side of a right-angled triangle is constant. The area of the triangle will be maximum if the angle between them is:

Given $f(x) = e^{\sin x} + e^{\cos x}$. The global maximum value of $f(x)$

$x$ and $y$ are two positive integers such that $2x + 3y = 50$. If $x^2 y^3$ is maximum for $x = \alpha$ and $y = \beta$, then $\frac{\alpha}{2} + \frac{\beta}{5} =$

The maximum area of a parallelogram inscribed in a triangle $ABC$ (given that $A$ is also one of the vertices of the parallelogram) is equal to (where symbols have their usual meaning in $\Delta ABC$):

If the lengths of three sides of a trapezium other than the base are equal to $10 \ cm$,find the maximum area of the trapezium.

Vedclass Test Series

Exam Paper Generator

Online Exam Module