Basics & LogarithmsmediumJEE Main 2025Free

Basics & Logarithms — JEE Maths practice question (JEE Main 2025)

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

Question
The number of real roots of the equation x∣x−2∣+3∣x−3∣+1=0x|x-2|+3|x-3|+1=0 is
A11correct
B22
C33
D44
Solution
Step 1: Open the modulus signs by considering cases. Case I: x<2x<2. Then ∣x−2∣=−(x−2)|x-2|=-(x-2) and ∣x−3∣=−(x−3)|x-3|=-(x-3):
x(−(x−2))+3(−(x−3))+1=0x(-(x-2))+3(-(x-3))+1=0
−x2+2x−3x+9+1=0  ⇒  −x2−x+10=0  ⇒  x2+x−10=0-x^{2}+2x-3x+9+1=0\;\Rightarrow\;-x^{2}-x+10=0\;\Rightarrow\;x^{2}+x-10=0
 Step 2: Solve:
x=−1±1+402=−1±412x=\dfrac{-1\pm\sqrt{1+40}}{2}=\dfrac{-1\pm\sqrt{41}}{2}
 For x<2x<2: −1+412≈−1+6.42≈2.7\dfrac{-1+\sqrt{41}}{2}\approx\dfrac{-1+6.4}{2}\approx 2.7 (rejected, since >2>2). −1−412≈−3.7\dfrac{-1-\sqrt{41}}{2}\approx -3.7 (accepted, since <2<2). Step 3: Case II: 2≤x<32\le x<3. Then ∣x−2∣=x−2|x-2|=x-2 and ∣x−3∣=−(x−3)|x-3|=-(x-3):
x(x−2)+3(−(x−3))+1=0  ⇒  x2−2x−3x+9+1=0  ⇒  x2−5x+10=0x(x-2)+3(-(x-3))+1=0\;\Rightarrow\;x^{2}-2x-3x+9+1=0\;\Rightarrow\;x^{2}-5x+10=0
 Discriminant: 25−40=−15<025-40=-15<0. No real roots. Step 4: Case III: x≥3x\ge 3. Then ∣x−2∣=x−2|x-2|=x-2 and ∣x−3∣=x−3|x-3|=x-3:
x(x−2)+3(x−3)+1=0  ⇒  x2−2x+3x−9+1=0  ⇒  x2+x−8=0x(x-2)+3(x-3)+1=0\;\Rightarrow\;x^{2}-2x+3x-9+1=0\;\Rightarrow\;x^{2}+x-8=0
 Step 5: Solve:
x=−1±1+322=−1±332x=\dfrac{-1\pm\sqrt{1+32}}{2}=\dfrac{-1\pm\sqrt{33}}{2}
 −1+332≈−1+5.742≈2.37\dfrac{-1+\sqrt{33}}{2}\approx\dfrac{-1+5.74}{2}\approx 2.37 (rejected, since <3<3). −1−332≈−3.37\dfrac{-1-\sqrt{33}}{2}\approx -3.37 (rejected, since <3<3). Step 6: Final solution set: only x=−1−412x=\dfrac{-1-\sqrt{41}}{2}. So only 11 real root.
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