First slide

Maxima and minima

Question

The number of solutions of the equation $x^{3} + 2 x^{2} + 5 x + 2 \cos x = 0 in [0, 2 π]$ is

Moderate

A

one
B

two
C

three
D

zero

Solution

Let $f (x) = x^{3} + 2 x^{2} + 5 x + 2 \cos x$
$∴ f^{'} (x) = 3 x^{2} + 4 x + 5 - 2 \sin x$
Now, the least value of $3 x^{2} + 4 x + 5$ is
$- \frac{D}{4 a} = - \frac{(4)^{2} - 4 (3) (5)}{4 (3)} = \frac{11}{3}$
and the greatest value of 2 sin x is 2. Therefore
$3 x^{2} + 4 x + 5 > 2 \sin x \forall x \in R$
or $f^{'} (x) = 3 x^{2} + 4 x + 5 - 2 \sin x > 0 \forall x \in R$
Thus, f(x) is strictly an increasing function.
Also, $f (0) = 2 and f (2 π) > 0$ ,
Thus, for the given interval , f (x) never becomes zero. Hence, the number of roots is zero.

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