Question

prove that y=4 sin x/2+cos x -x is an increasing function of x in [0,pi/2]

Answer 1

Answer:

Step-by-step explanation:

Given:

$\tt y=\dfrac{4\:sin\:x}{2+cos\:x} -x$

To Prove:

y is an increasing function of x in the interval [0, π/2]

Proof:

$\tt y=\dfrac{4\:sin\:x}{2+cos\:x} -x$

Differentiating with respect to x,

$\tt \dfrac{dy}{dx } =\dfrac{d}{dx}\bigg(\dfrac{4\:sin\:x}{2+cos\:x} -x\bigg)$

By quotient rule,

$\tt \bigg(\dfrac{u}{v} \bigg)'=\dfrac{vu'-uv'}{v^2}$

Also,

$\tt \dfrac{d}{dx}(sin\:x)=cos\:x$

$\tt \dfrac{d}{dx} (cos\:x)=-sin\:x$

$\tt \dfrac{d}{dx} (x)=1$

Hence,

$\tt \dfrac{dy}{dx} =\dfrac{(2+cos\:x)\times 4\:cos\:x-(4\:sin\:x)\times -sin\:x}{(2+cos\:x)^2} -1$

$\tt \dfrac{dy}{dx} =\dfrac{(8\:cos\:x+4\:cos^2\:x)+(4\:sin^2\:x)}{(2+cos\:x)^2} -1$

We know that,

sin²x + cos²x = 1

Therefore,

$\tt \dfrac{dy}{dx} =\dfrac{8\:cos\:x+4}{(2+cos\:x)^2} -1$

Now cross multiplying,

$\tt \dfrac{dy}{dx} =\dfrac{8\:cos\:x+4-(2+cos\:x)^2}{(2+cos\:x)^2}$

$\tt \dfrac{dy}{dx} =\dfrac{8\:cos\:x+4-(4+4\:cos\:x+cos^2\:x)}{(2+cos\:x)^2}$

$\tt \dfrac{dy}{dx} =\dfrac{8\:cos\:x+4-4-4\:cos\:x-cos^2\:x}{(2+cos\:x)^2}$

$\tt \dfrac{dy}{dx} =\dfrac{4\:cos\:x-cos^2\:x}{(2+cos\:x)^2}$

$\tt \dfrac{dy}{dx} =\dfrac{cos\:x(4-cos\:x)}{(2+cos\:x)^2}$

Here,

$\tt \dfrac{cos\:x(4-cos\:x)}{(2+cos\:x)^2}\geq 0$

∵ cos x is positive in [0, π/2]

4 - cos x ≥ 0

(2 + cos x)² ≥ 0

Therefore y is an increasing function on the interval [0, π/2].