Question 34 of 82intermediate🔧 ApplyLong Answer5 marks

Prove that the function ff given by f(x)=log⁡sin⁡xf(x) = \log \sin x is increasing on (0, π2)\left(0,\, \dfrac{\pi}{2}\right) and decreasing on (π2, π)\left(\dfrac{\pi}{2},\, \pi\right).

Correct Answer

We are given the function f(x)=log⁡sin⁡xf(x) = \log \sin x. The function is defined for x∈(0,π)x \in (0, \pi) because sin⁡x>0\sin x > 0 in this interval. To determine whether ff is increasing or decreasing, we find its derivative.

Differentiating f(x)f(x) with respect to xx, we get:

f′(x)=1sin⁡x⋅cos⁡x=cot⁡xf'(x) = \frac{1}{\sin x} \cdot \cos x = \cot x

Now consider the interval (0,π/2)(0, \pi/2). For any x∈(0,π/2)x \in (0, \pi/2), we have sin⁡x>0\sin x > 0 and cos⁡x>0\cos x > 0. Therefore,

cot⁡x=cos⁡xsin⁡x>0\cot x = \frac{\cos x}{\sin x} > 0

Hence f′(x)>0f'(x) > 0 on (0,π/2)(0, \pi/2). This implies that ff is increasing on (0,π/2)(0, \pi/2).

Next, consider the interval (π/2,π)(\pi/2, \pi). For any x∈(π/2,π)x \in (\pi/2, \pi), sin⁡x>0\sin x > 0 but cos⁡x<0\cos x < 0. Therefore,

cot⁡x=cos⁡xsin⁡x<0\cot x = \frac{\cos x}{\sin x} < 0

Hence f′(x)<0f'(x) < 0 on (π/2,π)(\pi/2, \pi). This implies that ff is decreasing on (π/2,π)(\pi/2, \pi).

Thus, we have proved that f(x)=log⁡sin⁡xf(x) = \log \sin x is increasing on (0,π/2)(0, \pi/2) and decreasing on (π/2,π)(\pi/2, \pi).

Exercise: EXERCISE 6.2 | Q: 16 | (Page 13)
For More Understanding

Explanation

This proof follows the standard method shown in the textbook (e.g., Example 9 and Example 13) where the sign of the derivative determines monotonicity. The derivative of log⁡sin⁡x\log \sin x is obtained using the chain rule and the derivative of the logarithmic function, which is implicitly known from the curriculum (as referenced in Question 10). The analysis of cot⁡x\cot x's sign on the given intervals is straightforward from trigonometric properties.

Solution Steps

  1. Step 1: Identify the domain of f(x)=log⁡sin⁡xf(x) = \log \sin x, which is (0,π)(0, \pi) because sin⁡x>0\sin x > 0 there.

  2. Step 2: Compute the derivative:

    f′(x)=1sin⁡x⋅cos⁡x=cot⁡xf'(x) = \frac{1}{\sin x} \cdot \cos x = \cot x

  3. Step 3: For x∈(0,π/2)x \in (0, \pi/2), sin⁡x>0\sin x > 0 and cos⁡x>0⇒cot⁡x>0⇒f′(x)>0⇒f\cos x > 0 \Rightarrow \cot x > 0 \Rightarrow f'(x) > 0 \Rightarrow f is increasing.

  4. Step 4: For x∈(π/2,π)x \in (\pi/2, \pi), sin⁡x>0\sin x > 0 and cos⁡x<0⇒cot⁡x<0⇒f′(x)<0⇒f\cos x < 0 \Rightarrow \cot x < 0 \Rightarrow f'(x) < 0 \Rightarrow f is decreasing.

  5. Step 5: Conclude the result.

    ∴ f(x)=log⁡sin⁡xf(x) = \log \sin x is increasing on (0,π/2)(0, \pi/2) and decreasing on (π/2,π)(\pi/2, \pi).

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