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Q17:
Prove that the function $f$ given by $f(x) = \log |\cos x|$ is decreasing on $(0, \frac{\pi}{2})$ and increasing on $(\frac{3\pi}{2}, 2\pi)$.
Solution :
Step 1: Defining the Function and Its Domain
Given the function $f(x) = \log(\cos x)$ within the specified intervals. Note that for the interval $(0, \frac{\pi}{2})$, $\cos x > 0$, hence the modulus $| \cos x | = \cos x$. [By the definition of the absolute value function].
Step 2: Differentiating the Function
To determine the monotonic nature of the function, we compute the first derivative $f'(x)$ with respect to $x$ using the Chain Rule [Per the Chain Rule of Calculus]:
$f(x) = \log(\cos x)$
$f'(x) = \frac{d}{dx} [\log(\cos x)] = \frac{1}{\cos x} \cdot \frac{d}{dx}(\cos x)$
$f'(x) = \frac{1}{\cos x} \cdot (-\sin x)$
$f'(x) = -\tan x$
Step 3: Analysis of the Interval $(0, \frac{\pi}{2})$
In the first quadrant, where $0 < x < \frac{\pi}{2}$:
- The trigonometric function $\tan x$ is strictly positive ($\tan x > 0$).
- Since $f'(x) = -\tan x$, it follows that $f'(x) < 0$ for all $x \in (0, \frac{\pi}{2})$.
[By the First Derivative Test for Monotonicity: If $f'(x) < 0$ on an interval, the function is strictly decreasing on that interval.]
Step 4: Analysis of the Interval $(\frac{3\pi}{2}, 2\pi)$
In the fourth quadrant, where $\frac{3\pi}{2} < x < 2\pi$:
- The trigonometric function $\tan x$ is strictly negative ($\tan x < 0$).
- Substituting this into the derivative: $f'(x) = -(\text{negative value}) = \text{positive value}$.
- Thus, $f'(x) > 0$ for all $x \in (\frac{3\pi}{2}, 2\pi)$.
[By the First Derivative Test for Monotonicity: If $f'(x) > 0$ on an interval, the function is strictly increasing on that interval.]
Summary Table
| Interval | Sign of $\tan x$ | Sign of $f'(x) = -\tan x$ | Nature of $f(x)$ |
|---|---|---|---|
| $(0, \frac{\pi}{2})$ | Positive (+) | Negative (-) | Decreasing |
| $(\frac{3\pi}{2}, 2\pi)$ | Negative (-) | Positive (+) | Increasing |
Final Solution: The function $f(x) = \log | \cos x |$ is strictly decreasing on $(0, \frac{\pi}{2})$ because $f'(x) < 0$, and strictly increasing on $(\frac{3\pi}{2}, 2\pi)$ because $f'(x) > 0$.
More Questions from Class 12 Mathematics Application of Derivatives EXERCISE 6.2
- Q1: Show that the function given by $f(x) = 3x + 17$ is increasing on R.
- Q2: Show that the function given by $f(x) = e^{2x}$ is increasing on R.
- Q3(a): Show that the function given by $f(x) = \sin x$ is (a) increasing in $(0, \frac{\pi}{2})$
- Q3(b): Show that the function given by $f(x) = \sin x$ is (b) decreasing in $(\frac{\pi}{2}, \pi)$
- Q3(c): Show that the function given by $f(x) = \sin x$ is (c) neither increasing nor decreasing in $(0, \pi)$
- Q4(a): Find the intervals in which the function $f$ given by $f(x) = 2x^2 – 3x$ is (a) increasing
- Q4(b): Find the intervals in which the function $f$ given by $f(x) = 2x^2 – 3x$ is (b) decreasing
- Q5(a): Find the intervals in which the function $f$ given by $f(x) = 2x^3 – 3x^2 – 36x + 7$ is (a) increasing
- Q5(b): Find the intervals in which the function $f$ given by $f(x) = 2x^3 – 3x^2 – 36x + 7$ is (b) decreasing
- Q6(a): Find the intervals in which the following functions are strictly increasing or decreasing: (a) $x^2 + 2x – 5$
- Q6(b): Find the intervals in which the following functions are strictly increasing or decreasing: (b) $10 – 6x – 2x^2$
- Q6(c): Find the intervals in which the following functions are strictly increasing or decreasing: (c) $–2x^3 – 9x^2 – 12x + 1$
- Q6(d): Find the intervals in which the following functions are strictly increasing or decreasing: (d) $6 – 9x – x^2$
- Q6(e): Find the intervals in which the following functions are strictly increasing or decreasing: (e) $(x + 1)^3 (x – 3)^3$
- Q7: Show that $y = \log(1 + x) - \frac{2x}{2+x}$, $x > – 1$, is an increasing function of $x$ throughout its domain.
- Q8: Find the values of $x$ for which $y = [x(x – 2)]^2$ is an increasing function.
- Q9: Prove that $y = \frac{4\sin\theta}{(2 + \cos\theta)} - \theta$ is an increasing function of $\theta$ in $[0, \frac{\pi}{2}]$.
- Q10: Prove that the logarithmic function is increasing on $(0, \infty)$.
- Q11: Prove that the function $f$ given by $f(x) = x^2 – x + 1$ is neither strictly increasing nor decreasing on $(– 1, 1)$.
- Q12: Which of the following functions are decreasing on $(0, \frac{\pi}{2})$ ?
- Q13: On which of the following intervals is the function $f$ given by $f(x) = x^{100} + \sin x –1$ decreasing ?
- Q14: For what values of $a$ the function $f$ given by $f(x) = x^2 + ax + 1$ is increasing on $[1, 2]$?
- Q15: Let $I$ be any interval disjoint from $[–1, 1]$. Prove that the function $f$ given by $f(x) = x + \frac{1}{x}$ is increasing on $I$.
- Q16: Prove that the function $f$ given by $f(x) = \log \sin x$ is increasing on $(0, \frac{\pi}{2})$ and decreasing on $(\frac{\pi}{2}, \pi)$.
- Q18: Prove that the function given by $f(x) = x^3 – 3x^2 + 3x – 100$ is increasing in R.
- Q19: The interval in which $y = x^2 e^{–x}$ is increasing is
CBSE Solutions for Class 12 Mathematics Application of Derivatives
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