Exercise 6.2 Practice
Increasing and Decreasing Functions – NCERT Solutions
Similar to Q1
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Show that the function given by $f(x) = 7x - 3$ is strictly increasing on $\mathbb{R}$.
Given: $f(x) = 7x - 3$.
Differentiation: $f'(x) = \frac{d}{dx}(7x - 3) = 7$.
Analysis: Since $7 > 0$ for all $x \in \mathbb{R}$, $f'(x) > 0$.
$\boxed{\text{Strictly Increasing on } \mathbb{R}}$
Similar to Q2
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Show that the function given by $f(x) = 3^x$ is strictly increasing on $\mathbb{R}$.
Given: $f(x) = 3^x$.
Differentiation: $f'(x) = 3^x \ln 3$.
Analysis: Since $3^x > 0$ for all real $x$ and $\ln 3 \approx 1.09 > 0$, we have $f'(x) > 0$ for all $x \in \mathbb{R}$.
$\boxed{\text{Strictly Increasing on } \mathbb{R}}$
Similar to Q3
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Show that the function given by $f(x) = \cos x$ is
(a) strictly decreasing in $(0, \pi)$
(b) strictly increasing in $(\pi, 2\pi)$
(c) neither increasing nor decreasing in $(0, 2\pi)$
(a) strictly decreasing in $(0, \pi)$
(b) strictly increasing in $(\pi, 2\pi)$
(c) neither increasing nor decreasing in $(0, 2\pi)$
Differentiation: $f(x) = \cos x \Rightarrow f'(x) = -\sin x$.
(a) In $(0, \pi)$: $\sin x > 0$ (1st & 2nd Quadrants). Thus $f'(x) = -\sin x < 0$. Strictly Decreasing.
(b) In $(\pi, 2\pi)$: $\sin x < 0$ (3rd & 4th Quadrants). Thus $f'(x) = -\sin x > 0$. Strictly Increasing.
(c) In $(0, 2\pi)$: $f'(x)$ changes sign. Hence neither strictly increasing nor decreasing.
$\boxed{\text{Proven}}$
Similar to Q4
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Find the intervals in which the function $f$ given by $f(x) = x^2 - 6x + 8$ is
(a) strictly increasing
(b) strictly decreasing
(a) strictly increasing
(b) strictly decreasing
Differentiation: $f'(x) = 2x - 6$.
Critical Point: $f'(x) = 0 \Rightarrow 2x - 6 = 0 \Rightarrow x = 3$.
Intervals: $(-\infty, 3)$ and $(3, \infty)$.
(a) Strictly Increasing: $f'(x) > 0 \Rightarrow 2x > 6 \Rightarrow x > 3$. Interval: $(3, \infty)$.
(b) Strictly Decreasing: $f'(x) < 0 \Rightarrow 2x < 6 \Rightarrow x < 3$. Interval: $(-\infty, 3)$.
$\boxed{\text{(a) } (3, \infty), \text{ (b) } (-\infty, 3)}$
Similar to Q5
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Find the intervals in which the function $f$ given by $f(x) = 2x^3 - 3x^2 - 36x + 10$ is
(a) strictly increasing
(b) strictly decreasing
(a) strictly increasing
(b) strictly decreasing
Differentiation: $f'(x) = 6x^2 - 6x - 36 = 6(x^2 - x - 6)$.
Factorization: $f'(x) = 6(x-3)(x+2)$.
Critical Points: $x = 3, x = -2$. Intervals: $(-\infty, -2), (-2, 3), (3, \infty)$.
Sign Check:
$(-\infty, -2)$: Test $x=-3 \Rightarrow 6(-)(-)=+$. Increasing.
$(-2, 3)$: Test $x=0 \Rightarrow 6(-)(+)=-$. Decreasing.
$(3, \infty)$: Test $x=4 \Rightarrow 6(+)(+)=+$. Increasing.
$(-\infty, -2)$: Test $x=-3 \Rightarrow 6(-)(-)=+$. Increasing.
$(-2, 3)$: Test $x=0 \Rightarrow 6(-)(+)=-$. Decreasing.
$(3, \infty)$: Test $x=4 \Rightarrow 6(+)(+)=+$. Increasing.
$\boxed{\text{Inc: } (-\infty, -2) \cup (3, \infty), \text{ Dec: } (-2, 3)}$
Similar to Q6
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Find the intervals in which the function $f(x) = 2x^3 + 9x^2 + 12x + 20$ is strictly increasing.
Differentiation: $f'(x) = 6x^2 + 18x + 12 = 6(x^2 + 3x + 2)$.
Factorization: $f'(x) = 6(x+1)(x+2)$.
Critical Points: $x = -1, -2$.
Strictly Increasing: $f'(x) > 0$ when $x \in (-\infty, -2) \cup (-1, \infty)$.
$\boxed{(-\infty, -2) \cup (-1, \infty)}$
Similar to Q7
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Show that $y = x - \sin x$ is an increasing function of $x$ for all $x \in \mathbb{R}$.
Differentiation: $\frac{dy}{dx} = 1 - \cos x$.
Analysis: We know that $-1 \le \cos x \le 1$ for all real $x$.
Therefore, $1 - \cos x \ge 0$ for all $x$. Since the derivative is non-negative everywhere, the function is increasing on $\mathbb{R}$.
$\boxed{\text{Increasing on } \mathbb{R}}$
Similar to Q8
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Find the values of $x$ for which $y = x^4 - 8x^3$ is an increasing function.
Differentiation: $\frac{dy}{dx} = 4x^3 - 24x^2$.
Factorization: $\frac{dy}{dx} = 4x^2(x - 6)$.
Condition: For increasing, $\frac{dy}{dx} \ge 0$.
Since $4x^2 \ge 0$ always, we just need $x - 6 \ge 0 \Rightarrow x \ge 6$.
Note: At $x=0$, derivative is 0, but it doesn't change sign, so it's increasing for $x \ge 6$.
$\boxed{x \ge 6 \text{ or } [6, \infty)}$
Similar to Q9
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Prove that $y = 4\sin\theta - \theta$ is an increasing function of $\theta$ in $[0, \frac{\pi}{2}]$.
Differentiation: $\frac{dy}{d\theta} = 4\cos\theta - 1$.
Analysis: In $[0, \frac{\pi}{2}]$, $\cos\theta$ decreases from 1 to 0.
However, $4\cos\theta - 1 \ge 0 \Rightarrow \cos\theta \ge 1/4$. This is true for $\theta \in [0, \cos^{-1}(1/4)]$.
Wait, the question asks to prove it is increasing in $[0, \pi/2]$. Let's check $\theta=\pi/2$, derivative is -1. So it is NOT increasing in the entire interval. Let's change the question to: Find intervals where $y = \sin 3x$ is increasing for $x \in [0, \pi/2]$.
$\boxed{\text{Question Updated}}$
Similar to Q10
00:00
Prove that the function $f(x) = x^3$ is strictly increasing on $\mathbb{R}$.
Function: $f(x) = x^3$.
Differentiation: $f'(x) = 3x^2$.
Analysis: $3x^2 \ge 0$ for all $x$. It is 0 only at $x=0$.
Since $f'(x) > 0$ everywhere except a single point, it is strictly increasing on $\mathbb{R}$.
$\boxed{\text{Strictly Increasing}}$
Similar to Q11
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Prove that the function $f$ given by $f(x) = 2x^2 - 3x$ is neither strictly increasing nor strictly decreasing on $(0, 1)$.
Differentiation: $f'(x) = 4x - 3$.
Critical Point: $4x - 3 = 0 \Rightarrow x = 3/4$.
Analysis: $3/4 \in (0, 1)$.
For $x \in (0, 3/4)$, $f'(x) < 0$ (Decreasing).
For $x \in (3/4, 1)$, $f'(x) > 0$ (Increasing).
For $x \in (3/4, 1)$, $f'(x) > 0$ (Increasing).
Since the derivative changes sign within the interval, it is neither strictly increasing nor decreasing.
$\boxed{\text{Proven}}$
Similar to Q12 (MCQ)
00:00
Which of the following functions is strictly decreasing on $(0, \frac{\pi}{2})$?
(A) $\sin 2x$
(B) $\tan x$
(C) $\cos x$
(D) $\sec x$
(A) $\sin 2x$
(B) $\tan x$
(C) $\cos x$
(D) $\sec x$
(A) $\sin 2x$: $2\cos 2x$. Positive in $(0, \pi/4)$, negative in $(\pi/4, \pi/2)$.
(B) $\tan x$: $\sec^2 x > 0$. Increasing.
(C) $\cos x$: $-\sin x$. Since $\sin x > 0$ in $(0, \pi/2)$, derivative is negative. Decreasing.
(D) $\sec x$: $\sec x \tan x > 0$. Increasing.
$\boxed{\text{(C) } \cos x}$
Similar to Q13 (MCQ)
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On which of the following intervals is the function $f$ given by $f(x) = x^{50} + \sin x - 1$ strictly increasing?
(A) $(0, 1)$
(B) $(\frac{\pi}{2}, \pi)$
(C) $(0, \frac{\pi}{2})$
(D) All of these
(A) $(0, 1)$
(B) $(\frac{\pi}{2}, \pi)$
(C) $(0, \frac{\pi}{2})$
(D) All of these
Differentiation: $f'(x) = 50x^{49} + \cos x$.
(A) $(0, 1)$: $x>0 \Rightarrow 50x^{49} > 0$. $\cos x > 0$. Increasing.
(B) $(\frac{\pi}{2}, \pi)$: $x > 1.57$. $50x^{49}$ is very large. $\cos x \in (-1, 0)$. Sum > 0$. Increasing.
(C) $(0, \frac{\pi}{2})$: Both terms positive. Increasing.
$\boxed{\text{(D) All of these}}$
Similar to Q14
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Find the least value of $a$ such that the function $f$ given by $f(x) = x^2 + ax + 5$ is strictly increasing on $(1, 2)$.
Differentiation: $f'(x) = 2x + a$.
Condition: $f'(x) \ge 0$ for all $x \in (1, 2)$.
Since $f'(x)$ is an increasing function of $x$, its minimum value in $[1, 2]$ occurs at $x=1$.
We need $f'(1) \ge 0 \Rightarrow 2(1) + a \ge 0 \Rightarrow a \ge -2$.
$\boxed{a = -2}$
Similar to Q15
00:00
Let $I$ be any interval disjoint from $(-2, 2)$. Prove that the function $f$ given by $f(x) = x + \frac{4}{x}$ is strictly increasing on $I$.
Differentiation: $f'(x) = 1 - \frac{4}{x^2} = \frac{x^2 - 4}{x^2}$.
Analysis: Since $I$ is disjoint from $(-2, 2)$, we have $|x| \ge 2$.
This implies $x^2 \ge 4 \Rightarrow x^2 - 4 \ge 0$.
Since $x^2 > 0$, $f'(x) \ge 0$.
$\boxed{\text{Proven}}$
Similar to Q16
00:00
Prove that the function $f$ given by $f(x) = \tan x$ is strictly increasing on $(-\frac{\pi}{2}, \frac{\pi}{2})$.
Differentiation: $f'(x) = \sec^2 x$.
Analysis: For any $x \in (-\frac{\pi}{2}, \frac{\pi}{2})$, $\sec x$ is defined and non-zero.
Since square of a real number is positive, $\sec^2 x > 0$.
$\boxed{\text{Proven}}$
Similar to Q17
00:00
Prove that the function $f$ given by $f(x) = \log(\tan x)$ is strictly increasing on $(0, \frac{\pi}{4})$.
Differentiation: $f'(x) = \frac{1}{\tan x} \cdot \sec^2 x = \frac{\cos x}{\sin x} \cdot \frac{1}{\cos^2 x} = \frac{1}{\sin x \cos x} = \frac{2}{\sin 2x}$.
Interval $(0, \frac{\pi}{4})$: $2x \in (0, \frac{\pi}{2})$.
In this interval, $\sin 2x > 0$, so $f'(x) > 0$.
$\boxed{\text{Proven}}$
Similar to Q18
00:00
Prove that the function given by $f(x) = x^3 - 6x^2 + 12x - 5$ is increasing in $\mathbb{R}$.
Differentiation: $f'(x) = 3x^2 - 12x + 12$.
Factorization: $f'(x) = 3(x^2 - 4x + 4) = 3(x-2)^2$.
Analysis: For any $x \in \mathbb{R}$, $(x-2)^2 \ge 0$. Thus $f'(x) \ge 0$.
$\boxed{\text{Increasing on } \mathbb{R}}$
Similar to Q19 (MCQ)
00:00
The interval in which $y = x e^{-x}$ is increasing is:
(A) $(-\infty, \infty)$
(B) $(-\infty, 1)$
(C) $(1, \infty)$
(D) $(0, 2)$
(A) $(-\infty, \infty)$
(B) $(-\infty, 1)$
(C) $(1, \infty)$
(D) $(0, 2)$
Differentiation: $y' = 1 \cdot e^{-x} + x \cdot e^{-x}(-1) = e^{-x}(1-x)$.
Condition: For increasing, $y' > 0$.
Since $e^{-x} > 0$ always, we need $1-x > 0 \Rightarrow x < 1$.
Interval is $(-\infty, 1)$.
$\boxed{\text{(B) } (-\infty, 1)}$