f(x)= x+1/2 2 x in [0, ()/2] (A) f^()(x)= x- 2 x (B) The critical points of the function are x=()/6 and x=()/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 : Option 1: (A), (B) and (D) only Option 2: (A), (B) and (C) only Option 3: (A), (B), (C) and (D) Option 4: (B), (C) and (D) only

Answer: (A), (B) and (D) only. Solution: We're given f(x) = x + 1/2 2x on the interval [0, ()/2] and we need to check which statements are correct. Differentiate f(x) term by term: f'(x) = x + 1/2 * (-2 2x) f'(x) = x - 2x Option (A) is correct. Set f'(x) = 0: x - 2x = 0 Using the identity 2x = 2 x x: x - 2 x x = 0 x(1 - 2 x) = 0 Case 1: x = 0 x = ()/2 Case 2: 1 - 2 x = 0 x = 1/2 x = ()/6 The critical points are x = ()/6 and x = ()/2. Option (B) is correct. Evaluate f(x) at the critical points and endpoints: f(0) = 0 + 1/2 0 = 0 + 1/2 = 1/2 f(()/6) = ()/6 + 1/2()/3 = 1/2 + 1/2 * 1/2 = 3/4 f(()/2) = ()/2 + 1/2 = 1 + 1/2(-1) = 1/2 Maximum value = 3/4 at x = ()/6 Minimum value = 1/2 at x = 0 and x = ()/2 Option (C) is wrong since the minimum value is 1/2, not 2. Option (D) is correct. Options (A), (B) and (D) are correct.

Calculus

Application of Derivatives

Easy

core

$f(x)=\sin x+\frac{1}{2} \cos 2 x \text { in }\left[0, \frac{\pi}{2}\right]$

(A) $f^{\prime}(x)=\cos x-\sin 2 x$

(B) The critical points of the function are $x=\frac{\pi}{6}$ and $x=\frac{\pi}{2}$

(C) The minimum value of the function is $2$

(D) The maximum value of the function is $\frac{3}{4}$

Choose the correct answer from the options given below :

(A), (B) and (D) only

(A), (B) and (C) only

(A), (B), (C) and (D)

(B), (C) and (D) only

✅ Correct Option: 1

We're given $f(x) = \sin x + \dfrac{1}{2} \cos 2x$ on the interval $\left[0, \dfrac{\pi}{2}\right]$ and we need to check which statements are correct.

Differentiate $f(x)$ term by term:

$f'(x) = \cos x + \dfrac{1}{2} \cdot (-2\sin 2x)$

$f'(x) = \cos x - \sin 2x$

Option (A) is correct.

Set $f'(x) = 0$ :

$\cos x - \sin 2x = 0$

Using the identity $\sin 2x = 2\sin x \cos x$ :

$\cos x - 2\sin x \cos x = 0$

$\cos x(1 - 2\sin x) = 0$

Case 1: $\cos x = 0 \Rightarrow x = \dfrac{\pi}{2}$

Case 2: $1 - 2\sin x = 0 \Rightarrow \sin x = \dfrac{1}{2} \Rightarrow x = \dfrac{\pi}{6}$

The critical points are $x = \dfrac{\pi}{6}$ and $x = \dfrac{\pi}{2}$ .

Option (B) is correct.

Evaluate $f(x)$ at the critical points and endpoints:

$f(0) = \sin 0 + \dfrac{1}{2}\cos 0 = 0 + \dfrac{1}{2} = \dfrac{1}{2}$

$f\left(\dfrac{\pi}{6}\right) = \sin\dfrac{\pi}{6} + \dfrac{1}{2}\cos\dfrac{\pi}{3} = \dfrac{1}{2} + \dfrac{1}{2} \cdot \dfrac{1}{2} = \dfrac{3}{4}$

$f\left(\dfrac{\pi}{2}\right) = \sin\dfrac{\pi}{2} + \dfrac{1}{2}\cos\pi = 1 + \dfrac{1}{2}(-1) = \dfrac{1}{2}$

Maximum value $= \dfrac{3}{4}$ at $x = \dfrac{\pi}{6}$

Minimum value $= \dfrac{1}{2}$ at $x = 0$ and $x = \dfrac{\pi}{2}$

Option (C) is wrong since the minimum value is $\dfrac{1}{2}$ , not $2$ .

Option (D) is correct.

Options (A), (B) and (D) are correct.

2025: 22 May Shift 1

The function $f(x) = \frac{-3}{4}x^4 - 8x^3 - \frac{45}{2}x^2 + 163$ has a local maxima at

2025: 3 June Shift 1

The function $f(x) = 4 - 3x + 3x^2 - x^3$ is (Here $\mathbb{R}$ is set of real numbers)

2025: 14 May Shift 1

Match List-I with List-II

$\begin{array}{|l|l|} \hline \rule{0pt}{2.8ex}\text{List-I} & \text{List-II} \\[1.2ex] \hline \rule{0pt}{2.8ex}\text{(A) The minimum value of } f(x) = (2x - 1)^2 + 3 & \text{(I) } 4 \\[1.2ex] \hline \rule{0pt}{2.8ex}\text{(B) The maximum value of } f(x) = -|x + 1| + 4 & \text{(II) } 10 \\[1.2ex] \hline \rule{0pt}{2.8ex}\text{(C) The minimum value of } f(x) = \sin(2x) + 6 & \text{(III) } 3 \\[1.2ex] \hline \rule{0pt}{2.8ex}\text{(D) The maximum value of } f(x) = -(x - 1)^2 + 10 & \text{(IV) } 5 \\[1.2ex] \hline \end{array}$

Choose the correct answer from the options given below:

2026: 26 May Shift 2

The volume of a cube is increasing at the rate 9 $cm^3/s$ . How fast is the surface area increasing when the length of an edge is 10 cm ?

2026: 19 May Shift 1

If the sum of the lengths of hypotenuse and a side of a right angled triangle is given, then the area of the triangle is maximum when angle between them is:

2025: 22 May Shift 2

Which of the following statements are true?

(A) The function $f(x) = \frac{x^4}{4} - \frac{4}{3}x^3 + \frac{x^2}{2} + 6x$ has 3 critical points.

(B) The function $f(x) = |x| + 3$ has no minimum value.

(D) $f(x) = x^2$ has minima at $x=0$ .

Choose the correct answer from the options given below:

Solution

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