Question

⇒ Give necessary calculations​

Answer 1

★ Concept :-

Here the concept of differentiation had been used. We see that we are given a function f(x). We are also given a condition for the value of x. When we see at the options given, we find that we have to find the maximum and minimum value of the function to choose the correct option. Firstly we shall find the first derivative of the function. Then we can simplify it and thus we can find the answer.

$\;\;\underline{\underline{\sf{\bf{Let's\;\:do\;\:it\;\:!}}}}$

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★ Solution :-

Given,

» Function = 1 + 2 sin x + 3 cos²x

» Range : 0 ≤ x ≤ 2π/3

This function can be written as,

» f(x) = 1 + 2 sin x + 3 cos²x

We know that a function f(x) at a point x is always 0. So,

» f(x) = 1 + 2 sin x + 3 cos²x = 0

Now we have to find the first derivative of the function. And first derivative is found using dy/dx that is rate of change of y with respect to x where y is the function.

First derivative of f(x) = f'(x)

So,

$\;\;\tt{\rightarrow\;\;f'(x)\;=\;\dfrac{d}{dx}(1\:+\:2\sin x\:+\:3\cos^{2}x)}$

We know that derivatives are distributive over addition. So this can be written as,

$\;\;\tt{\rightarrow\;\;f'(x)\;=\;\dfrac{d}{dx}(1)\:+\:\dfrac{d}{dx}(2\sin x)\:+\:\dfrac{d}{dx}(3\cos^{2}x)}$

• Since 1 is constant so it's derivative will be 0

• Here since 2 and 3 are constant terms but they are multiplied with functions. Then we have to take them out of the derivative. So,

$\;\;\tt{\rightarrow\;\;f'(x)\;=\;0\:+\:2\bigg(\dfrac{d}{dx}(\sin x)\bigg)\:+\:3\bigg(\dfrac{d}{dx}(3\cos^{2}x)\bigg)}$

• Derivative of sin x is cos x

• Derivative of cos²x is 2 cos x (- sin x)

By applying this, we get

$\;\;\tt{\rightarrow\;\;f'(x)\;=\;0\:+\:2\bigg(\dfrac{d}{dx}(\sin x)\bigg)\:+\:3\bigg(\dfrac{d}{dx}(\cos^{2}x)\bigg)}$

$\;\;\tt{\rightarrow\;\;f'(x)\;=\;0\:+\:2(\cos x)\:+\:3(2\cos x(-\sin x)}$

Now here we can take out the -ve sign from third term. Then,

$\;\;\tt{\rightarrow\;\;f'(x)\;=\;0\:+\:2(\cos x)\:+\:3(-2\cos x(\sin x)}$

$\;\;\tt{\rightarrow\;\;f'(x)\;=\;0\:+\:2(\cos x)\:-\:3(2\cos x\sin x)}$

$\;\;\tt{\rightarrow\;\;f'(x)\;=\;0\:+\:2\cos x\:-\:6\cos x\sin x}$

$\;\;\tt{\rightarrow\;\;f'(x)\;=\;2\cos x\:-\:6\cos x\sin x}$

Now since f(x) = 0 so f'(x) = 0 .

$\;\;\tt{\rightarrow\;\;2\cos x\:-\:6\cos x\sin x\;=\;0}$

Taking the common term, we get

$\;\;\tt{\rightarrow\;\;2\cos x(1\:-\:3\sin x)\;=\;0}$

Now transposing 2 to another side, we get

$\;\;\tt{\rightarrow\;\;\cos x(1\:-\:3\sin x)\;=\;\dfrac{0}{2}}$

$\;\;\tt{\rightarrow\;\;\cos x(1\:-\:3\sin x)\;=\;0}$

Now here we see that either cos x = 0 or (1 - 3 sin x) = 0 . So,

>> cos x = 0 or (1 - 3 sin x) = 0

>> cos x = 0 or 3 sin x = 1

>> cos x = 0 or sin x = 1/3

$\;\;\bf{\Longrightarrow\;\;\cos x\;=\;0\;\;\;\tt{0}\;\;\;\bf{x\;=\;\sin^{-1}\bigg(\dfrac{1}{3}\bigg)}}$

We know that,

• cos x = 0 when x = 90°

• sin x = ⅓ when x ≈ 19.47°

This means that,

$\;\;\bf{\Longrightarrow\;\;x\;=\;\cos^{-1}\bigg(0\bigg)\;\;\;\tt{0}\;\;\;\bf{x\;=\;\sin^{-1}\bigg(\dfrac{1}{3}\bigg)}}$

$\;\;\bf{\Longrightarrow\;\;x\;=\;90^{\circ}\;\;\;\tt{0}\;\;\;\bf{x\;=\;19.47^{\circ}}}$

We know that π = 180° (in Trignometry). And 90° = π/2 so,

$\;\;\bf{\Longrightarrow\;\;x\;=\;\dfrac{\pi}{2}\;\;\;\tt{0}\;\;\;\bf{x\;=\;19.47^{\circ}}}$

Now if we apply the value of x = π/2 in the given function, then

>> f(π/2) = 1 + 2 sin 90° + 3 cos² 90°

>> f(90°) = 1 + 2(1) + 3(0)

• Since, cos 90° = 0

• And, sin 90° = 0

So,

>> f(90°) = 1 + 2

>> f(90°) = 3

Clearly we get that,

The function is minimum at x = 90° .

So the option A) min. at x = 90° is correct.

$\;\underline{\boxed{\tt{Required\;\:Answer\;:\;\bf{\purple{min.\;\:at\;\:x\;=\;90^{\circ}}}}}}$