Basics & LogarithmsmediumJEE Main 2023Free

Basics & Logarithms — JEE Maths practice question (JEE Main 2023)

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

Question
The number of integral solutions xx of log⁡(x+7/2)(x−72x−3)2≥0\log_{(x+7/2)}\left(\dfrac{x-7}{2x-3}\right)^{2}\ge 0 is
A88
B66correct
C77
D44
Solution
Step 1: Domain conditions: For the base: x+72>0x+\dfrac{7}{2}>0 and x+72≠1x+\dfrac{7}{2}\neq 1, i.e., x>−72x>-\dfrac{7}{2} and x≠−52x\neq -\dfrac{5}{2}. For the argument: x−72x−3≠0\dfrac{x-7}{2x-3}\neq 0 (so the squared expression is positive). So x≠7x\neq 7 and x≠32x\neq\dfrac{3}{2}. Step 2: Combine: x∈(−72,∞)∖{−52,32,7}x\in\left(-\dfrac{7}{2},\infty\right)\setminus\left\{-\dfrac{5}{2},\dfrac{3}{2},7\right\}. Step 3: log⁡ab≥0\log_{a}b\ge 0 holds when a>1a>1 with b≥1b\ge 1, or 0<a<10<a<1 with 0<b≤10<b\le 1. So we have two cases for (x−72x−3)2\left(\dfrac{x-7}{2x-3}\right)^{2}. Step 4: Case I: x+72>1x+\dfrac{7}{2}>1, i.e., x>−52x>-\dfrac{5}{2}. Then need (x−72x−3)2≥1\left(\dfrac{x-7}{2x-3}\right)^{2}\ge 1:
(x−7)2≥(2x−3)2(x-7)^{2}\ge(2x-3)^{2}
 Expanding: x2−14x+49≥4x2−12x+9x^{2}-14x+49\ge 4x^{2}-12x+9, so 3x2+2x−40≤03x^{2}+2x-40\le 0. Solve: discriminant =4+480=484=222=4+480=484=22^{2}, roots x=−2±226x=\dfrac{-2\pm 22}{6}, giving x=206=103x=\dfrac{20}{6}=\dfrac{10}{3} or x=−4x=-4. So −4≤x≤103-4\le x\le\dfrac{10}{3}. Combined with x>−52x>-\dfrac{5}{2}: x∈(−52,103]x\in\left(-\dfrac{5}{2},\dfrac{10}{3}\right] excluding 32\dfrac{3}{2}. Step 5: Case II: 0<x+72<10<x+\dfrac{7}{2}<1, i.e., −72<x<−52-\dfrac{7}{2}<x<-\dfrac{5}{2}. Then need 0<(x−72x−3)2≤10<\left(\dfrac{x-7}{2x-3}\right)^{2}\le 1: (x−7)2≤(2x−3)2(x-7)^{2}\le(2x-3)^{2}, giving 3x2+2x−40≥03x^{2}+2x-40\ge 0, so x≤−4x\le -4 or x≥103x\ge\dfrac{10}{3}. Combined with −72<x<−52-\dfrac{7}{2}<x<-\dfrac{5}{2}: empty (since −72>−4-\dfrac{7}{2}>-4). Wait — actually −72=−3.5-\dfrac{7}{2}=-3.5 and −4<−3.5-4<-3.5, so the range (−72,−52)(-\dfrac{7}{2},-\dfrac{5}{2}) contains no points with x≤−4x\le -4. So Case II gives no solutions. Step 6: Final solution set: x∈(−52,103]∖{32}x\in\left(-\dfrac{5}{2},\dfrac{10}{3}\right]\setminus\left\{\dfrac{3}{2}\right\}. Integer values: −2,−1,0,1,2,3-2,-1,0,1,2,3 (excluding 32\dfrac{3}{2}, which isn't an integer anyway). Step 7: Count: 66 integers. Answer: (2) 66
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