Basics & LogarithmsmediumFree

Basics & Logarithms — JEE Maths practice question

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

Question
If the solution of the inequality 1<3x2−7x+8x2+1≤21 < \dfrac{3x^2-7x+8}{x^2+1} \leq 2 is [α,β][\alpha, \beta], then mark the incorrect option:
Aα\alpha is the least natural number
Bβ−α\beta - \alpha is a prime number
Cβ+α\beta + \alpha is a prime number
Dβ+3α\beta + 3\alpha is a prime numbercorrect
Solution
Step 1: Solve the left inequality
3x2−7x+8>x2+1  ⟹  2x2−7x+7>03x^2-7x+8 > x^2+1 \implies 2x^2-7x+7 > 0
 Discriminant =49−56<0= 49-56 < 0, so this holds for all x∈Rx \in \mathbb{R}. Step 2: Solve the right inequality
3x2−7x+8≤2(x2+1)  ⟹  x2−7x+6≤0  ⟹  (x−1)(x−6)≤03x^2-7x+8 \leq 2(x^2+1) \implies x^2-7x+6 \leq 0 \implies (x-1)(x-6) \leq 0
x∈[1,6]  ⟹  α=1,β=6x \in [1, 6] \implies \alpha = 1, \beta = 6
 Step 3: Test the options α=1\alpha = 1 (least natural number): true. β−α=5\beta-\alpha = 5 (prime): true. β+α=7\beta+\alpha = 7 (prime): true. β+3α=9=32\beta+3\alpha = 9 = 3^2 (not prime): false. Answer: (4)
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