Problem Statement
The maximum point on the curve with the equation y = x sin ( x ) y = x\sqrt{\sin(x)} y = x sin ( x ) 0 < x < π 0 < x < \pi 0 < x < π 2 tan ( x ) + x = 0 2\tan(x) + x = 0 2 tan ( x ) + x = 0
My Approach
First, let f ( x ) = x sin ( x ) f(x) = x\sqrt{\sin(x)} f ( x ) = x sin ( x )
Apply the product rule:
f ′ ( x ) = x ⋅ g ′ ( x ) + sin 1 2 ( x ) \begin{align*}
f'(x) = x \cdot g'(x) + \sin^{\frac{1}{2}}(x)
\end{align*} f ′ ( x ) = x ⋅ g ′ ( x ) + sin 2 1 ( x ) Where g ( x ) g(x) g ( x )
g ( x ) = sin 1 2 ( x ) \begin{align*}
g(x) &= \sin^{\frac{1}{2}}(x)\\
\end{align*} g ( x ) = sin 2 1 ( x ) Now, differentiate g ( x ) g(x) g ( x )
g ′ ( x ) = 1 2 sin − 1 2 ( x ) ⋅ cos ( x ) \begin{align*}
g'(x) &= \frac{1}{2} \sin^{-\frac{1}{2}}(x) \cdot \cos(x)\\
\end{align*} g ′ ( x ) = 2 1 sin − 2 1 ( x ) ⋅ cos ( x ) Substituting g ′ ( x ) g'(x) g ′ ( x ) f ′ ( x ) f'(x) f ′ ( x )
f ′ ( x ) = x ⋅ 1 2 sin − 1 2 ( x ) ⋅ cos ( x ) + sin 1 2 ( x ) = sin 1 2 ( x ) [ 1 + x 2 cos ( x ) sin − 1 ( x ) ] \begin{align*}
f'(x) &= x \cdot \frac{1}{2} \sin^{-\frac{1}{2}}(x) \cdot \cos(x) + \sin^{\frac{1}{2}}(x) \\
&= \sin ^ {\frac{1}{2}}(x)[1 + \frac{x}{2}\cos(x)\sin^{-1}(x)]
\end{align*} f ′ ( x ) = x ⋅ 2 1 sin − 2 1 ( x ) ⋅ cos ( x ) + sin 2 1 ( x ) = sin 2 1 ( x ) [ 1 + 2 x cos ( x ) sin − 1 ( x )] Given that a point A is a stationary point and thus its gradient is 0, we can form the following equation:
0 = sin 1 2 ( x ) [ 1 + x 2 cos ( x ) sin − 1 ( x ) ] \begin{align*}
0 &= \sin ^ {\frac{1}{2}}(x)[1 + \frac{x}{2}\cos(x)\sin^{-1}(x)]
\end{align*} 0 = sin 2 1 ( x ) [ 1 + 2 x cos ( x ) sin − 1 ( x )] Let's take a look at the two possible scenarios.
sin 1 2 ( x ) = 0 \sin ^ {\frac{1}{2}}(x) = 0 sin 2 1 ( x ) = 0
∴ sin ( x ) = 0 ∴ x = 0 , π \begin{align*}
\therefore\space& \sin(x) = 0 \\
\therefore\space& x = 0, \pi
\end{align*} ∴ ∴ sin ( x ) = 0 x = 0 , π However, as given by the question, x x x π \pi π
What about the other possibility?
[ 1 + x 2 cos ( x ) sin − 1 ( x ) ] = 0 [1 + \frac{x}{2}\cos(x)\sin^{-1}(x)] = 0 [ 1 + 2 x cos ( x ) sin − 1 ( x )] = 0
[ 1 + x 2 cos ( x ) sin − 1 ( x ) ] = 0 ∴ x 2 cos ( x ) sin − 1 ( x ) = − 1 \begin{align*}
[1 + \frac{x}{2}\cos(x)\sin^{-1}(x)] &= 0 \\
\therefore \frac{x}{2}\cos(x)\sin^{-1}(x) &= -1 \\
\end{align*} [ 1 + 2 x cos ( x ) sin − 1 ( x )] ∴ 2 x cos ( x ) sin − 1 ( x ) = 0 = − 1 ⟺ x 2 cos ( x ) sin ( x ) = − 1 ⟺ x 2 tan ( x ) = − 1 ⟺ x = − 2 tan ( x ) ⟺ x + 2 tan ( x ) = 0 \begin{align*}
\Longleftrightarrow\space& \frac{x}{2}\frac{\cos(x)}{\sin(x)} = -1 \\
\Longleftrightarrow\space& \frac{x}{2\tan(x)} = -1 \\
\Longleftrightarrow\space& x = -2 \tan(x) \\
\Longleftrightarrow\space& x + 2 \tan(x) = 0 \\
\end{align*} ⟺ ⟺ ⟺ ⟺ 2 x sin ( x ) cos ( x ) = − 1 2 tan ( x ) x = − 1 x = − 2 tan ( x ) x + 2 tan ( x ) = 0 Conclusion
Can't be bothered to write a conclusion today.