Question

The number of solutions of the equation $x^3+2x^2+5x+2cosx=0$ in $[0,2\pi ]$ is:

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (A)

0

Let $f\left(x\right)=x^3+2x^2+5x+2\cos x$

$\Rightarrow f^{\prime }\left(x\right)=3x^2+4x+5-2sinx=2(x+\frac{2}{3}{)}^2+\frac{11}{3}-2sinx$

Now $\frac{11}{3}-2sinx>0$ $x(as-1\le sinx\le 1)$

$\Rightarrow f^{\prime }\left(x\right)>0$ $x\Rightarrow f\left(x\right)$ is an increasing function.

Now $f\left(0\right)=2$

$\Rightarrow f\left(x\right)=0$ has no solution in $[0,2\pi ]$.

Hence (A) is the correct answer.