Basics & LogarithmsmediumJEE Main 2021Free

Basics & Logarithms — JEE Maths practice question (JEE Main 2021)

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

Question
The number of elements in the set {x∈R:(∣x∣−3)∣x+4∣=6}\{x\in\mathbb{R}:(|x|-3)|x+4|=6\} is equal to
A33
B22correct
C44
D11
Solution
Step 1: If x≠−4x\neq -4, divide both sides by ∣x+4∣|x+4|:
∣x∣−3=6∣x+4∣|x|-3=\dfrac{6}{|x+4|}
∣x∣−3=6∣x+4∣|x|-3=\dfrac{6}{|x+4|}
 Step 2: From the graph (book approach): plot y=∣x∣−3y=|x|-3 (V-shape, vertex at (0,−3)(0,-3)) and y=6∣x+4∣y=\dfrac{6}{|x+4|} (hyperbola-type, always positive, with vertical asymptote at x=−4x=-4). Step 3: The curves can intersect. The equation ∣x∣−3=6∣x+4∣|x|-3=\dfrac{6}{|x+4|} means (∣x∣−3)∣x+4∣=6(|x|-3)|x+4|=6. For x≠−4x\neq -4: multiply both sides: (∣x∣−3)∣x+4∣=6(|x|-3)|x+4|=6. Step 4: From graph analysis, the two curves intersect at 22 points. Verification algebraically: Case x≥0x\ge 0: (x−3)(x+4)=6  ⇒  x2+x−12=6  ⇒  x2+x−18=0(x-3)(x+4)=6\;\Rightarrow\;x^{2}+x-12=6\;\Rightarrow\;x^{2}+x-18=0. x=−1+732≈3.77x=\dfrac{-1+\sqrt{73}}{2}\approx 3.77 (accepted). −1−732<0\dfrac{-1-\sqrt{73}}{2}<0 (rejected). Case −3≤x<0-3\le x<0: (−x−3)(x+4)=6  ⇒  −x2−7x−12=6  ⇒  x2+7x+18=0(-x-3)(x+4)=6\;\Rightarrow\;-x^{2}-7x-12=6\;\Rightarrow\;x^{2}+7x+18=0. Discriminant =49−72<0=49-72<0. No solution. Case x<−3x<-3: (−x−3)∣x+4∣(-x-3)|x+4|. Sub-case −4<x<−3-4<x<-3: (−x−3)(x+4)=6(-x-3)(x+4)=6. Same as above: x2+7x+18=0x^{2}+7x+18=0. No real solution. Sub-case x<−4x<-4: (−x−3)(−(x+4))=6  ⇒  (−x−3)(−x−4)=6  ⇒  x2+7x+12=6  ⇒  x2+7x+6=0(-x-3)(-(x+4))=6\;\Rightarrow\;(-x-3)(-x-4)=6\;\Rightarrow\;x^{2}+7x+12=6\;\Rightarrow\;x^{2}+7x+6=0. (x+1)(x+6)=0  ⇒  x=−1(x+1)(x+6)=0\;\Rightarrow\;x=-1 (rejected, not <−4<-4) or x=−6x=-6 (accepted). Step 5: Solutions: x=−1+732x=\dfrac{-1+\sqrt{73}}{2} and x=−6x=-6. So 22 elements. Answer: (2) 22
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