Application of DerivativesmediumComprehensionFree

Application of Derivatives: Function Defined

JEE Maths reading comprehension with full step-by-step solutions.

Passage

f : \mathbb{R} \to \mathbb{R}

is a function defined by

f(x) = 2\cos^{2}2x + \frac34\sin 4x + ax ,

where

a \in \mathbb{R}

Question 1 · Single correct

The complete set of values of

a

for which

f(x)

is strictly increasing for all

x \in \mathbb{R}

(-\infty, -5]

[5, \infty)

correct

(-5, \infty)

(-\infty, 5]

Solution

Step 1: Simplify the first term with the double-angle identity

2\cos^{2}\theta = 1 + \cos2\theta

, taking

\theta = 2x

2\cos^{2}2x = 1 + \cos4x .

Step 2: Rewrite

f

f(x) = 1 + \cos4x + \frac34\sin4x + ax .

Step 3: Differentiate.

f'(x) = -4\sin4x + 3\cos4x + a .

Step 4: Write down the condition. For

f

to be strictly increasing everywhere we need

f'(x) \ge 0

for all

x

, i.e.

a \ \ge\ 4\sin4x - 3\cos4x \quad \text{for every } x .

Step 5: So

a

must be at least the *maximum* of the right-hand side. For an expression

p\sin\theta + q\cos\theta

the maximum is

\sqrt{p^{2}+q^{2}}

, so with

p = 4

q = -3

\max\left(4\sin4x - 3\cos4x\right) = \sqrt{16+9} = 5 .

Step 6: Hence

a \ge 5 , \quad \text{i.e.} \quad a \in [5, \infty) .

Answer: (2).

Question 2 · Single correct

The complete set of values of

a

for which

f(x)

does not have any critical point is

(-\infty, -5]

[5, \infty)

(-5, 5)

(-\infty, -5) \cup (5, \infty)

correct

Solution

Step 1: Recall the derivative from the previous part.

f'(x) = -4\sin4x + 3\cos4x + a .

Step 2: Say what "no critical point" means. A critical point is a solution of

f'(x) = 0

, so we need

-4\sin4x + 3\cos4x + a \ne 0 \quad \text{for every } x ,

that is,

a \ne 4\sin4x - 3\cos4x \quad \text{for every } x .

Step 3: Find the full range of the right-hand side. An expression

p\sin\theta + q\cos\theta

takes every value in

\left[-\sqrt{p^{2}+q^{2}},\ \sqrt{p^{2}+q^{2}}\right]

, and

4x

runs over all reals as

x

does, so

4\sin4x - 3\cos4x \ \text{takes every value in } [-5,\ 5] .

Step 4: So

a

must avoid that whole closed interval.

a \notin [-5, 5] \implies a \in (-\infty, -5) \cup (5, \infty) .

Step 5: Note the contrast with the previous part - there the condition was

a \ge 5

, an inequality; here the endpoints

\pm5

are themselves excluded, because at

a = \pm5

the derivative touches zero and a critical point appears. Answer: (4).

Stuck on this comprehension?Ask your doubt on WhatsApp

More comprehension in Application of Derivatives

Application of Derivatives · hard

Let

f(x)

be a thrice differentiable function on the interval

[a, e]

. Define the auxiliary functions:

G(x) = [f'(x)]^2 + f(x)f''(x), \qquad H(x) = e^{-x}(f'(x) - f(x))

It is given that

f(a) = f(e) = 0

. For constants

b, c, d

satisfying

a < b < c < d < e

, the function takes the values

f(b) = 2

f(c) = -1

, and

f(d) = 2

Solve more, learn faster

Sign up free to practice comprehension sets and explore doMath — timed drills, mastery sprints, bookmarks, and chapter-wise progress tracking.