Basics & LogarithmsmediumPYQ · JEE Main · 5 Apr 2024 · Shift 2 (Afternoon)Free

Basics & Logarithms: Number Real Solutions Equation (JEE Main 2024)

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

Question
The number of real solutions of the equation x∣x+5∣+2∣x+7∣−2=0x|x+5|+2|x+7|-2=0 is
Solution
Answer: 3
Step 1: Split into cases based on the sign of x+5x+5 and x+7x+7. Case I: x≥−5x\ge -5. Then ∣x+5∣=x+5|x+5|=x+5 and ∣x+7∣=x+7|x+7|=x+7 (since −5>−7-5>-7):
x(x+5)+2(x+7)−2=0  ⇒  x2+5x+2x+14−2=0  ⇒  x2+7x+12=0x(x+5)+2(x+7)-2=0\;\Rightarrow\;x^{2}+5x+2x+14-2=0\;\Rightarrow\;x^{2}+7x+12=0
 Step 2: Factor:
(x+3)(x+4)=0  ⇒  x=−3 or x=−4(x+3)(x+4)=0\;\Rightarrow\;x=-3\text{ or }x=-4
 Both are ≥−5\ge -5, accepted. Step 3: Case II: −7≤x<−5-7\le x<-5. Then ∣x+5∣=−(x+5)|x+5|=-(x+5) and ∣x+7∣=x+7|x+7|=x+7:
x⋅(−(x+5))+2(x+7)−2=0  ⇒  −x2−5x+2x+14−2=0x\cdot(-(x+5))+2(x+7)-2=0\;\Rightarrow\;-x^{2}-5x+2x+14-2=0
−x2−3x+12=0  ⇒  x2+3x−12=0-x^{2}-3x+12=0\;\Rightarrow\;x^{2}+3x-12=0
 Step 4: Solve:
x=−3±9+482=−3±572x=\dfrac{-3\pm\sqrt{9+48}}{2}=\dfrac{-3\pm\sqrt{57}}{2}
 −3−572≈−3−7.552≈−5.27\dfrac{-3-\sqrt{57}}{2}\approx\dfrac{-3-7.55}{2}\approx -5.27 (accepted, since −7≤−5.27<−5-7\le -5.27<-5). −3+572≈2.27\dfrac{-3+\sqrt{57}}{2}\approx 2.27 (rejected, outside range). Step 5: Case III: x<−7x<-7. Then ∣x+5∣=−(x+5)|x+5|=-(x+5) and ∣x+7∣=−(x+7)|x+7|=-(x+7):
x⋅(−(x+5))+2⋅(−(x+7))−2=0x\cdot(-(x+5))+2\cdot(-(x+7))-2=0
−x2−5x−2x−14−2=0  ⇒  −x2−7x−16=0  ⇒  x2+7x+16=0-x^{2}-5x-2x-14-2=0\;\Rightarrow\;-x^{2}-7x-16=0\;\Rightarrow\;x^{2}+7x+16=0
 Discriminant: 49−64=−15<049-64=-15<0. No real solutions. Step 6: Total distinct real solutions: x=−3,−4,−3−572x=-3,-4,\dfrac{-3-\sqrt{57}}{2}. So 33 solutions. Answer: 33
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