If f(x) = sinx+ 1/2 cos2x in[ 0, π/2 ], then:

Options:

(A) f′(x) = cosx−sin2x
(B) The critical points of the function are x = π/6 and x = π/2
(C) The minimum value of the function is 2
(D) The maximum value of the function is 3/4

Choose the correct answer from the options given below:

The correct answer is (A), (B), and (D) only.

Let's analyze the given function and evaluate each statement step-by-step to understand why (A), (B), and (D) are the correct statements, while (C) is incorrect.

The given function is defined on the closed interval [0, π/2] as follows:
$$f(x)=\sin x+\frac{1}{2}\cos 2x$$

Step 1: Check Statement (A)
To find the derivative of $f\left(x\right)$, we differentiate each term with respect to $x$:
$$f^{\prime }(x)=\frac{d}{dx}(\sin x+\frac{1}{2}\cos 2x)$$
Using the derivative formulas $\frac{d}{dx}(\sin x)=\cos x$ and $\frac{d}{dx}(\cos 2x)=-2\sin 2x$, we obtain:
$$f^{\prime }(x)=\cos x+\frac{1}{2}(-2\sin 2x)=\cos x-\sin 2x$$
Thus, Statement (A) is correct.

Step 2: Check Statement (B)
Critical points occur where $f^{\prime }\left(x\right)=0$ within the interval [0, π/2].
Setting the derivative to zero:
$$\cos x-\sin 2x=0$$
We use the double-angle identity $\sin 2x=2\sin x\cos x$:
$$\cos x-2\sin x\cos x=0$$
Factor out $\cos x$:
$$\cos x(1-2\sin x)=0$$
This gives us two cases to solve within the interval $[0,\pi /2]$:
1) $\cos x=0\Rightarrow x=\pi /2$
2) $1-2\sin x=0\Rightarrow \sin x=1/2\Rightarrow x=\pi /6$
Both critical points $x=\pi /6$ and $x=\pi /2$ lie in the interval [0, π/2].
Thus, Statement (B) is correct.

Step 3: Evaluate the function at critical points and boundary points to check Statements (C) and (D)
We need to compute the function values at the boundary point $x=0$, and at the critical points $x=\pi /6$ and $x=\pi /2$:

• At $x=0$:
$$f(0)=\sin 0+\frac{1}{2}\cos 0=0+\frac{1}{2}(1)=\frac{1}{2}$$
• At $x=\pi /6$:
$$f(\pi /6)=\sin (\pi /6)+\frac{1}{2}\cos (\pi /3)=\frac{1}{2}+\frac{1}{2}(\frac{1}{2})=\frac{1}{2}+\frac{1}{4}=\frac{3}{4}$$
• At $x=\pi /2$:
$$f(\pi /2)=\sin (\pi /2)+\frac{1}{2}\cos \pi =1+\frac{1}{2}(-1)=1-\frac{1}{2}=\frac{1}{2}$$

Comparing these values:
- The maximum value of the function on the interval is $3/4$ (at $x=\pi /6$). Hence, Statement (D) is correct.
- The minimum value of the function on the interval is $1/2$ (at $x=0$ and $x=\pi /2$). Thus, Statement (C) is incorrect because the minimum value is not 2.

Therefore, statements (A), (B), and (D) are correct, which matches the option (A), (B), and (D) only.