Calculus

Calculus Level 3

Find the number of x ∈ R x \in \mathbb R for which the equation below holds true.

x 2 − x sin ⁡ x − cos ⁡ x = 0 \large x^2-x \sin x-\cos x=0

6 2 0 4

Aakhyat Singh by Aakhyat Singh , 20, India

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1 solution

Sabhrant Sachan
Sep 14, 2017

Consider f ( x ) = x 2 − x sin ⁡ x − cos ⁡ x f ′ ( x ) = 2 x − sin ⁡ x − x cos ⁡ x + sin ⁡ x = x ( 1 + 1 − cos ⁡ x ) = x ( 1 + 2 sin ⁡ 2 x 2 ) > 0 ∀ x > 0 f ( x ) is increasing ∀ x > 0 & decreasing ∀ x < 0 1st eqn Also f ( 0 ) = − 1 2nd eqn Hence from 1st & 2nd eqn f ( x ) have two solutions \text{Consider } f(x) = x^2-x\sin{x}-\cos{x} \\ \begin{aligned} f^{'}(x) & = 2x-\cancel{\sin{x}}-x\cos{x}+\cancel{\sin{x}} \\ & = x(1+1-\cos{x}) \\ & = x \left( 1+2\sin^2{\dfrac{x}{2}} \right) > 0 \hspace{4mm} \forall \hspace{4mm} x >0 \end{aligned} \\ f(x) \text{ is increasing } \forall \hspace{3mm} x > 0 \text{ \& decreasing } \forall \hspace{3mm} x<0 \hspace{ 5mm } \color{#3D99F6}{\small{\text{ 1st eqn }}} \\ \text{Also } f(0) = -1 \hspace{5mm } \color{#3D99F6}{\small{\text{ 2nd eqn }}} \\ \text{ Hence from 1st \& 2nd eqn } f(x) \text{ have two solutions}

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