The-number-of-points-in-for-which-x-2-x-sin-x-cos-x-0-is-1-6-2-4-3-2-4-0-

Question Number 17348 by Tinkutara last updated on 04/Jul/17

The number of points in (- \infty, \infty) for

which x^{2} - x \sin x - \cos x = 0 is

(1) 6

(2) 4

(3) 2

(4) 0

Answered by ajfour last updated on 04/Jul/17

(3) 2

let f (x) = x^{2} - xsin x - \cos x

f (0) = - 1

f^{'} (x) = 2 x - xcos x

= x (2 - \cos x)

as x \to - \infty, f (x) \to \infty

only decreasing for x < 0

stops decreasing when x = 0

and by this time f (x) = - 1

\Rightarrow crossed x - axis once

x = 0 is the minimum

f (x) starts increasing from

f (0) = - 1 and keeps increasing

hereafter so crosses x - axis just

once more .

Commented by Tinkutara last updated on 05/Jul/17

Thanks Sir!

Answered by mrW1 last updated on 04/Jul/17

y = f (x) = x^{2} - x \sin x - \cos x

f (- x) = {(- x)}^{2} - (- x) \sin (- x) - \cos (- x)

= x^{2} - x \sin x - \cos x

f (x) is symmetric .

f (0) = - 1

\lim_{x \to + \infty} f (x) = + \infty

\Rightarrow there is at least one point for x > 0

and f (x) = 0

f^{'} (x) = 2 x - \sin x - xcos x + \sin x = x (2 - \cos x)

for x > 0 \Rightarrow f^{'} (x) > 0

\Rightarrow f (x) is increasing for x > 0

\Rightarrow there is only one point for x \in (0, + \infty) with

f (x) = 0

due to symmetry

there are 2 point 2 for x \in (- \infty, + \infty) with

f (x) = 0

\Rightarrow answer (3) is right .

Commented by Tinkutara last updated on 05/Jul/17

Thanks Sir!

Answered by b.e.h.i.8.3.4.17@gmail.com last updated on 06/Jul/17

x^{2} - x (x - \frac{x^{3}}{6}) - (1 - \frac{x^{2}}{2}) = 0

\Rightarrow x^{2} - x^{2} + \frac{x^{4}}{6} - 1 + \frac{x^{2}}{2} = 0

\Rightarrow x^{4} + 3 x^{2} - 6 = 0 \Rightarrow x^{4} + 3 x^{2} + \frac{9}{4} = 6 + \frac{9}{4}

\Rightarrow {(x^{2} + \frac{3}{2})}^{2} = \frac{33}{4} \Rightarrow x^{2} + \frac{3}{2} = + \frac{\sqrt{33}}{2}

\Rightarrow x^{2} = \frac{\sqrt{33} - 3}{2} \Rightarrow x = \pm \sqrt{\frac{\sqrt{33} - 3}{2} .}

t h e r e i s o n l y 2 a n s w e r s . \Rightarrow (3) .

Commented by Tinkutara last updated on 07/Jul/17

Can you explain the first line ?

Commented by alex041103 last updated on 07/Jul/17

This is wrong .

He tried to use the Taylor series

expansion for \sin x and \cos x .

They have infinatly many terms, but

he uses only two of them .

This is an aproximation!

Here are the expansions for sine and cosine :

s i n x = \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} \frac{x^{2 n + 1}}{(2 n + 1)!}

c o s x = \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} \frac{x^{2 n}}{(2 n)!}