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Functions: Let Denote Greatest Integer Fractional Parts Respectively Fi

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

Question
Let f(x)=[sec⁡{x}]f(x) = [\sec\{x\}] where [x][x] and {x}\{x\} denote greatest integer and fractional parts of xx respectively, and g(x)=2x2−3x(k+1)+k(3k+1)g(x) = 2x^2 - 3x(k+1) + k(3k+1). Find the number of integral values of kk such that g(f(x))<0g(f(x)) < 0 for all x∈Rx \in \mathbb{R}.
Solution
Answer: 1
Step 1: Simplify f(x)f(x) Since {x}∈[0,1)\{x\} \in [0,1), we have sec⁡{x}∈[sec⁡0,sec⁡1)=[1,1.85…)\sec\{x\} \in [\sec 0, \sec 1) = [1, 1.85\ldots). Therefore [sec⁡{x}]=1[\sec\{x\}] = 1 for all xx, so f(x)=1f(x) = 1 for all x∈Rx \in \mathbb{R}. Step 2: Set up the inequality
g(f(x))=g(1)=2−3(k+1)+k(3k+1)=3k2−2k−1<0g(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  ⟹  −13<k<1(3k+1)(k-1) < 0 \implies -\frac{1}{3} < k < 1
 The only integer in this interval is k=0k = 0. Number of integral values = 1. Answer: 1Step 1: Simplify f(x)f(x) Since {x}∈[0,1)\{x\} \in [0,1), we have sec⁡{x}∈[sec⁡0,sec⁡1)=[1,1.85…)\sec\{x\} \in [\sec 0, \sec 1) = [1, 1.85\ldots). Therefore [sec⁡{x}]=1[\sec\{x\}] = 1 for all xx, so f(x)=1f(x) = 1 for all x∈Rx \in \mathbb{R}. Step 2: Set up the inequality
g(f(x))=g(1)=2−3(k+1)+k(3k+1)=3k2−2k−1<0g(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  ⟹  −13<k<1(3k+1)(k-1) < 0 \implies -\frac{1}{3} < k < 1
 The only integer in this interval is k=0k = 0. Number of integral values = 1. Answer: 1
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