If f(x) = 1 + 2 sin x + 3 cos^2 x for 0

Question

(A) Given $f(x) = 1 + 2 \sin x + 3 \cos^2 x = 1 + 2 \sin x + 3(1 - \sin^2 x) = 4 + 2 \sin x - 3 \sin^2 x$.Let $u = \sin x$. Since $0 < x < 2\pi / 3$,t

Vedclass · Accepted Answer

Minimum at $x = \pi / 2$

Vedclass · Answer

(A) Given $f(x) = 1 + 2 \sin x + 3 \cos^2 x = 1 + 2 \sin x + 3(1 - \sin^2 x) = 4 + 2 \sin x - 3 \sin^2 x$.Let $u = \sin x$. Since $0 Then $g(u) = 4 + 2u - 3u^2$.To find critical points,$g'(u) = 2 - 6u = 0 \implies u = 1/3$.Since $g''(u) = -6 Thus,$f(x)$ has a maximum at $\sin x = 1/3$,i.e.,$x = \sin^{-1}(1/3)$.At the boundaries of the interval $(0, 2\pi / 3)$:As $x 	o 0^+$,$f(x) 	o 1 + 0 + 3(1) = 4$.At $x = 2\pi / 3$,$f(2\pi / 3) = 1 + 2(\sqrt{3}/2) + 3(1/4) = 1 + \sqrt{3} + 0.75 = 1.75 + \sqrt{3} \approx 3.48$.Since $g(1/3) = 4 + 2/3 - 3(1/9) = 4 + 2/3 - 1/3 = 4.33$,the maximum is at $x = \sin^{-1}(1/3)$.Checking the options,the question asks for the behavior. Based on the derivative $f'(x) = 2 \cos x - 6 \cos x \sin x = 2 \cos x (1 - 3 \sin x)$,$f'(x) = 0$ at $x = \pi / 2$ and $x = \sin^{-1}(1/3)$.At $x = \pi / 2$,$f(\pi / 2) = 1 + 2(1) + 3(0) = 3$. Since $3 < 4$ and $3 < 4.33$,$x = \pi / 2$ is a local minimum.

If $f(x) = 1 + 2 \sin x + 3 \cos^2 x$ for $0 < x < 2\pi / 3$,then:

Similar Questions

If the function $f(x)=2 x^{3}-9 a x^{2}+12 a^{2} x+1$ attains its maximum and minimum at $p$ and $q$ respectively such that $p^{2}=q$,then $a$ equals

Let the set of all positive values of $\lambda$,for which the point of local minimum of the function $f(x) = 1 + x(\lambda^2 - x^2)$ satisfies $\frac{x^2+x+2}{x^2+5x+6} < 0$,be $(\alpha, \beta)$. Then $\alpha^2 + \beta^2$ is equal to:

Let the set of all values of $p$,for which $f(x) = (p^2 - 6p + 8)(\sin^2 2x - \cos^2 2x) + 2(2 - p)x + 7$ does not have any critical point,be the interval $(a, b)$. Then $16ab$ is equal to ..........

The number of solution$(s)$ satisfying the equation $3x^2 - 2x^3 = \log_2 (x^2 + 1) - \log_2 x$ is

Given $f(x) = -\frac{x^3}{3} + x^2 \sin(1.5a) - x \sin(a) \sin(2a) - 5 \sin^{-1}(a^2 - 8a + 17)$,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module