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JEE Main · 2024 · 08 Apr 2024 · Shift 2 · Basic Mathematics

The number of distinct real roots of the equation

∣ x + 1∣∣ x + 3∣ - 4∣ x + 2∣ + 5 = 0

, is ______.

Answer: 2

Solution

Solve

∣ x + 1∣∣ x + 3∣ - 4∣ x + 2∣ + 5 = 0

by cases.

x \leq - 3

:

(x + 1) (x + 3) + 4 (x + 2) + 5 = 0 ⟹ x^{2} + 8 x + 16 = 0 ⟹ (x + 4)^{2} = 0 ⟹ x = - 4

(valid).

- 3 \leq x \leq - 2

:

- (x + 1) (x + 3) + 4 (x + 2) + 5 = 0 ⟹ - x^{2} + 10 = 0 ⟹ x = \pm 10

, neither lies in

[- 3, - 2]

.

- 2 \leq x \leq - 1

:

- (x + 1) (x + 3) - 4 (x + 2) + 5 = 0 ⟹ x^{2} + 8 x + 6 = 0 ⟹ x = - 4 \pm 10

, neither lies in

[- 2, - 1]

.

x \geq - 1

:

(x + 1) (x + 3) - 4 (x + 2) + 5 = 0 ⟹ x^{2} = 0 ⟹ x = 0

(valid).
Distinct real roots:

x = - 4

and

x = 0

— total 2.

JEE Main · 2024 · 08 Apr 2024 · Shift 2 · Integer answer

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