Basics & LogarithmsmediumPYQ · JEE Main · 8 Apr 2024 · Shift 1 (Morning)Free

Basics & Logarithms: Number Distinct Real Roots Equation (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

Solution

Answer: 2

Step 1: Substitute

y=x+2

. Then

x+1=y-1

and

x+3=y+1

, so:

|y-1||y+1|-4|y|+5=0\;\Rightarrow\;|y^{2}-1|-4|y|+5=0

Step 2: Case I:

|y|\ge 1

(so

y^{2}-1\ge 0

|y^{2}-1|=y^{2}-1

y^{2}-1-4|y|+5=0\;\Rightarrow\;y^{2}-4|y|+4=0\;\Rightarrow\;(|y|-2)^{2}=0

|y|=2

, giving

y=\pm 2

, i.e.,

x=0

x=-4

. Both satisfy

|y|\ge 1

. Accepted. Step 3: Case II:

|y|<1

(so

y^{2}-1<0

|y^{2}-1|=1-y^{2}

1-y^{2}-4|y|+5=0\;\Rightarrow\;-y^{2}-4|y|+6=0\;\Rightarrow\;y^{2}+4|y|-6=0

Let

v=|y|\ge 0

v^{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+\sqrt{10}\approx 1.16

but we need

v<1

. Rejected.

v=-2-\sqrt{10}<0

. Rejected. Step 4: Total distinct real roots:

x=0

and

x=-4

. So

2

. Answer:

2

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