Given f(x) = 4x^3 - 6x^2 cos 2a + 3x sin 2a s | vedclass

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

Question

(D) The function $f(x)$ is defined only if the argument of the square root is non-negative,i.e.,$\ln(2a - a^2) \ge 0$.This implies $2a - a^2 \ge 1$,wh

Vedclass · Accepted Answer

$f'(1/2) > 0$

Vedclass · Answer

(D) The function $f(x)$ is defined only if the argument of the square root is non-negative,i.e.,$\ln(2a - a^2) \ge 0$.This implies $2a - a^2 \ge 1$,which simplifies to $a^2 - 2a + 1 \le 0$,or $(a - 1)^2 \le 0$.Since the square of a real number cannot be negative,we must have $(a - 1)^2 = 0$,which means $a = 1$.Substituting $a = 1$ into the expression for $f(x)$,we get $f(x) = 4x^3 - 6x^2 \cos 2 + 3x \sin 2 \sin 6$.Now,find the derivative $f'(x) = 12x^2 - 12x \cos 2 + 3 \sin 2 \sin 6$.Evaluating at $x = 1/2$,we get $f'(1/2) = 12(1/4) - 12(1/2) \cos 2 + 3 \sin 2 \sin 6 = 3 - 6 \cos 2 + 3 \sin 2 \sin 6$.Using the identity $\sin 2 \sin 6 = \frac{1}{2}(\cos(2-6) - \cos(2+6)) = \frac{1}{2}(\cos 4 - \cos 8)$,we have $f'(1/2) = 3 - 6 \cos 2 + \frac{3}{2}(\cos 4 - \cos 8)$.Since $\cos 2 \approx -0.416$,$-6 \cos 2 \approx 2.496$. Thus $f'(1/2) = 3 + 2.496 + \dots > 0$.

Given $f(x) = 4x^3 - 6x^2 \cos 2a + 3x \sin 2a \sin 6a + \sqrt{\ln(2a - a^2)}$,then:

Similar Questions

Differentiate $(x^{2}-5x+8)(x^{3}+7x+9)$ using the product rule.

If $y=\tan ^{-1}(\sin \sqrt{x})+\operatorname{cosec}^{-1}\left(e^{2 x+1}\right)$, then $\frac{d y}{d x}=$

The derivative of $F[f\{ \phi (x)\} ]$ with respect to $x$ is:

If $y = \sec(x^{\circ})$,then $\frac{dy}{dx} = $

The first derivative of the function $(\sin 2x \cos 2x \cos 3x + \log_2 2^{x+3})$ with respect to $x$ at $x = \pi$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module