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 real solutions of the equation xx+5+2x+72=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: x5x\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+142=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: 7x<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    x25x+2x+142=0x\cdot(-(x+5))+2(x+7)-2=0\;\Rightarrow\;-x^{2}-5x+2x+14-2=0
x23x+12=0    x2+3x12=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}
357237.5525.27\dfrac{-3-\sqrt{57}}{2}\approx\dfrac{-3-7.55}{2}\approx -5.27 (accepted, since 75.27<5-7\le -5.27<-5). 3+5722.27\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
x25x2x142=0    x27x16=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: 4964=15<049-64=-15<0. No real solutions. Step 6: Total distinct real solutions: x=3,4,3572x=-3,-4,\dfrac{-3-\sqrt{57}}{2}. So 33 solutions. Answer: 33
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