Basics & LogarithmsmediumJEE Main 2024Free

Basics & Logarithms — JEE Maths practice question (JEE Main 2024)

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

Question
The number of distinct real roots of the equation ∣x+1∣∣x+3∣−4∣x+2∣+5=0|x+1||x+3|-4|x+2|+5=0 is
Solution
Answer: 2
Step 1: Substitute y=x+2y=x+2. Then x+1=y−1x+1=y-1 and x+3=y+1x+3=y+1, so:
∣y−1∣∣y+1∣−4∣y∣+5=0  ⇒  ∣y2−1∣−4∣y∣+5=0|y-1||y+1|-4|y|+5=0\;\Rightarrow\;|y^{2}-1|-4|y|+5=0
 Step 2: Case I: ∣y∣≥1|y|\ge 1 (so y2−1≥0y^{2}-1\ge 0, ∣y2−1∣=y2−1|y^{2}-1|=y^{2}-1):
y2−1−4∣y∣+5=0  ⇒  y2−4∣y∣+4=0  ⇒  (∣y∣−2)2=0y^{2}-1-4|y|+5=0\;\Rightarrow\;y^{2}-4|y|+4=0\;\Rightarrow\;(|y|-2)^{2}=0
 So ∣y∣=2|y|=2, giving y=±2y=\pm 2, i.e., x=0x=0 or x=−4x=-4. Both satisfy ∣y∣≥1|y|\ge 1. Accepted. Step 3: Case II: ∣y∣<1|y|<1 (so y2−1<0y^{2}-1<0, ∣y2−1∣=1−y2|y^{2}-1|=1-y^{2}):
1−y2−4∣y∣+5=0  ⇒  −y2−4∣y∣+6=0  ⇒  y2+4∣y∣−6=01-y^{2}-4|y|+5=0\;\Rightarrow\;-y^{2}-4|y|+6=0\;\Rightarrow\;y^{2}+4|y|-6=0
 Let v=∣y∣≥0v=|y|\ge 0:
v2+4v−6=0  ⇒  v=−4±16+242=−4±402=−2±10v^{2}+4v-6=0\;\Rightarrow\;v=\dfrac{-4\pm\sqrt{16+24}}{2}=\dfrac{-4\pm\sqrt{40}}{2}=-2\pm\sqrt{10}
 v=−2+10≈1.16v=-2+\sqrt{10}\approx 1.16 but we need v<1v<1. Rejected. v=−2−10<0v=-2-\sqrt{10}<0. Rejected. Step 4: Total distinct real roots: x=0x=0 and x=−4x=-4. So 22. Answer: 22
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