Basics & LogarithmseasyJEE 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 real solutions of the equation x(x2+3∣x∣+5∣x−1∣+6∣x−2∣)=0x\left(x^{2}+3|x|+5|x-1|+6|x-2|\right)=0 is
Solution
Answer: 1
Step 1: A product is zero when at least one factor is zero. So either x=0x=0 or:
x2+3∣x∣+5∣x−1∣+6∣x−2∣=0x^{2}+3|x|+5|x-1|+6|x-2|=0
 Step 2: Examine the second factor. All terms are non-negative: x2≥0x^{2}\ge 0, 3∣x∣≥03|x|\ge 0, 5∣x−1∣≥05|x-1|\ge 0, 6∣x−2∣≥06|x-2|\ge 0. Step 3: For the sum to be zero, each term must be zero simultaneously: x2=0⇒x=0x^{2}=0\Rightarrow x=0, but then 5∣x−1∣=5≠05|x-1|=5\neq 0. So the second factor cannot be zero anywhere. Step 4: Therefore, the only solution is x=0x=0.
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