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Functions — JEE Maths practice question

JEE Maths question with a full step-by-step solution.

Question

Let

f(x) = [\sec\{x\}]

where

[x]

and

\{x\}

denote greatest integer and fractional parts of

x

respectively, and

g(x) = 2x^2 - 3x(k+1) + k(3k+1)

. Find the number of integral values of

k

such that

g(f(x)) < 0

for all

x \in \mathbb{R}

Solution

Answer: 1

Step 1: Simplify

f(x)

Since

\{x\} \in [0,1)

, we have

\sec\{x\} \in [\sec 0, \sec 1) = [1, 1.85\ldots)

. Therefore

[\sec\{x\}] = 1

for all

x

, so

f(x) = 1

for all

x \in \mathbb{R}

. Step 2: Set up the inequality

g(f(x)) = g(1) = 2 - 3(k+1) + k(3k+1) = 3k^2 - 2k - 1 < 0

Step 3: Solve and count

(3k+1)(k-1) < 0 \implies -\frac{1}{3} < k < 1

The only integer in this interval is

k = 0

. Number of integral values = 1. Answer: 1Step 1: Simplify

f(x)

Since

\{x\} \in [0,1)

, we have

\sec\{x\} \in [\sec 0, \sec 1) = [1, 1.85\ldots)

. Therefore

[\sec\{x\}] = 1

for all

x

, so

f(x) = 1

for all

x \in \mathbb{R}

. Step 2: Set up the inequality

g(f(x)) = g(1) = 2 - 3(k+1) + k(3k+1) = 3k^2 - 2k - 1 < 0

Step 3: Solve and count

(3k+1)(k-1) < 0 \implies -\frac{1}{3} < k < 1

The only integer in this interval is

k = 0

. Number of integral values = 1. Answer: 1

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