The number of solutions of the equation $x^{3}$ + $x^{2}$ + 4x + 2sinx = 0 in 0 ≤ x ≤ 2 $π$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B
1

$x^{3}$ + $(x + 2)^{2}$ + 2sinx = 4

Clearly x = 0 satisfies the equations

If 0 < x ≤ $π$ , $x^{3}$ + $(x + 2)^{2}$ + 2sinx > 4

If $π$ < x ≤ 2 $π$ , $x^{3}$ + $(x + 2)^{2}$ + 2sinx > 27 + 25 - 2

So x = 0 is the only solution.

Suggest Corrections

Similar questions

x^{3} + x^{2} + 3 x + 2 s i n x = 0

- 2 π \leq x \leq 4 π

sin 2 x + 2 sin x - cos x - 1 = 0

0 \leq x \leq 2 π

x^{3} + x^{2} + 2 x + sin x = 0

(- 2 π, 2 π)

| x^{2} + 4 x + 3 | + 2 x + 5 = 0

s i n x + s i n y + s i n z = - 3 f o r 0 \leq x \leq 2 π, 0 \leq y \leq 2 π, 0 \leq z \leq 2 π

Join BYJU'S Learning Program