Let the function f: (0, pi) rightarrow R be d | vedclass

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

Question

(B) Given $f(	heta) = (\sin 	heta + \cos 	heta)^2 + (\sin 	heta - \cos 	heta)^4$.Using $(\sin 	heta + \cos 	heta)^2 = 1 + \sin 2	heta$ and $(\

Vedclass · Accepted Answer

$0.50$

Vedclass · Answer

(B) Given $f(	heta) = (\sin 	heta + \cos 	heta)^2 + (\sin 	heta - \cos 	heta)^4$.Using $(\sin 	heta + \cos 	heta)^2 = 1 + \sin 2	heta$ and $(\sin 	heta - \cos 	heta)^2 = 1 - \sin 2	heta$,we have:$f(	heta) = (1 + \sin 2	heta) + (1 - \sin 2	heta)^2$$f(	heta) = 1 + \sin 2	heta + 1 - 2\sin 2	heta + \sin^2 2	heta = \sin^2 2	heta - \sin 2	heta + 2$.Let $x = \sin 2	heta$. Then $g(x) = x^2 - x + 2$. The derivative $g'(x) = 2x - 1$. Setting $g'(x) = 0$ gives $x = 1/2$.Thus,$\sin 2	heta = 1/2$. For $	heta \in (0, \pi)$,$2	heta \in (0, 2\pi)$.$2	heta = \pi/6$ or $2	heta = 5\pi/6$.$	heta = \pi/12$ or $	heta = 5\pi/12$.These are the points of local minima. Thus $\lambda_1 = 1/12$ and $\lambda_2 = 5/12$.The sum $\lambda_1 + \lambda_2 = 1/12 + 5/12 = 6/12 = 0.50$.

Similar Questions

It is given that at $x=1,$ the function $f(x) = x^{4}-62x^{2}+ax+9$ attains its maximum value on the interval $[0,2].$ Find the value of $a.$

$P(x) = x^4 + ax^3 + bx^2 + cx + d$ is such that $x = 0$ is the only real root of $P'(x) = 0$. If $P(-1) < P(1)$,then in the interval $[-1, 1]$:

The following figure shows the graph of a differentiable function $y=f(x)$ on the interval $[a, b]$ (not containing $0$). Let $g(x)=\frac{f(x)}{x}$. Which of the following is a possible graph of $y=g(x)$?

The number $10$ is divided into two parts such that the sum of double of the first part and the square of the second part is minimum. The two parts are respectively:

Vedclass Test Series

Exam Paper Generator

Online Exam Module