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2. Calculus — Differential Calculus

Increasing functions

Increasing and decreasing functions form a core pillar of differential calculus, linking the sign of the first derivative directly to the monotonic behavior of real-valued functions. This topic covers this concept.

For: UP PGT Mathematics UP TGT Mathematics DSSSB PGT / TGT KVS / NVS PGT LT Grade
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Definition and statement

First Derivative Test for Monotonicity
Let $f: [a, b] \to \mathbb{R}$ be continuous on $[a, b]$ and differentiable on $(a, b)$. (i) If $f'(x) > 0$ for all $x \in (a, b)$, then $f$ is strictly increasing on $[a, b]$. (ii) If $f'(x) \ge 0$ for all $x \in (a, b)$, then $f$ is increasing (non-decreasing) on $[a, b]$.

📐 Geometric Interpretation

The condition $f'(x) > 0$ implies that the tangent line at every point on the curve has a positive slope, meaning the curve rises continuously from left to right without dipping.

⚙️ Algebraic & Structural Meaning

Translates the global inequality of function values ($x_1 < x_2 \implies f(x_1) \le f(x_2)$) into a local analytical check on the derivative ($f'(x) \ge 0$).

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition Domain / Interval Necessity What Fails If Condition Dropped
Continuity on [a, b] [a, b] (Closed interval) Strictly Mandatory / Necessary If dropped, a jump discontinuity can cause the function to drop while the derivative remains positive elsewhere.
Differentiability on (a, b) (a, b) (Open interval) Strictly Mandatory Non-differentiable points (like cusps) invalidate the slope criterion.
Interval domain specification Entire domain Essential for conclusion Extending monotonicity across disconnected domains can lead to false conclusions.

Useful results and shortcuts

Use these identities and results when their conditions apply:

01

Polynomial Derivative Sign Test

$$f(x) = ax^3 + bx^2 + cx + d \quad \text{is increasing on } \mathbb{R} \iff b^2 - 3ac \le 0 \text{ and } a > 0$$

Rule: For a cubic polynomial to be monotonically increasing on the entire real line, its derivative quadratic must have a non-positive discriminant ($D \le 0$) and a positive leading coefficient.

Application: Frequently asked in UP PGT and DSSSB PGT to find parameter ranges for which a cubic is strictly increasing.
02

Logarithmic & Exponential Monotonicity

$$f(x) = x^n e^{-x} \text{ or } f(x) = x \ln x$$

Rule: Factor out non-zero positive terms and analyze the sign of the remaining algebraic expression to find critical intervals instantly.

03

Quotient Rule Sign Preservation

$$f(x) = \frac{P(x)}{Q(x)} \implies f'(x) = \frac{P'(x)Q(x) - P(x)Q'(x)}{[Q(x)]^2}$$

Rule: Since the denominator $[Q(x)]^2 > 0$ for all valid $x$, ignore the denominator entirely and determine the sign solely by the numerator.

Application: Saves 45 seconds of algebra in rational function monotonicity questions.
04

Trigonometric Inequality Bounds

$$f(x) = x - \sin x \implies f'(x) = 1 - \cos x \ge 0$$

Rule: Since $1 - \cos x \ge 0$ everywhere, $x \ge \sin x$ for $x \ge 0$ is established instantly via increasing function property ($f(0)=0$).

Application: Commonly used to prove inequalities in UP TGT/PGT calculus sections.
05

Composition of Monotone Functions

$$f \text{ and } g \text{ increasing} \implies f(g(x)) \text{ is increasing}$$

Rule: The composite of two identically monotonic functions is always increasing; if one is increasing and the other decreasing, the composite is decreasing.

Application: Directly tested in theoretical statement-based MCQ questions.
06

Option Elimination via Boundary Check

$$x = \text{critical point test}$$

Rule: Substitute convenient test points from the options (e.g., $x = 0$ or $x = 1$) into $f'(x)$. If it violates the required sign, eliminate that interval option instantly.

Common errors and counterexamples

These examples show how a conclusion can fail when a condition is missing:

⚠️ COMMON ERROR #01

Strict vs. Non-Strict Inequality Confusion

Common Pitfall: Using strict inequality ($f'(x) > 0$) when the question asks for intervals where the function is merely increasing ($f'(x) \ge 0$).

Counterexample: $f(x) = x^3 \implies f'(0) = 0$, but $f(x)$ is strictly increasing on $\mathbb{R}$.

Why it fails: A function can have a zero derivative at isolated points and still be strictly increasing.

🛡️ Exam Defense: Always use closed intervals for increasing/decreasing unless 'strictly' is explicitly mentioned.
⚠️ COMMON ERROR #02

Discontinuous Domain Inclusion

Common Pitfall: Including points where the function is undefined (e.g., vertical asymptotes) in the interval of monotonicity.

Counterexample: $f(x) = \frac{1}{x} \implies f'(x) = -\frac{1}{x^2} < 0 \text{ for } x \neq 0$

Why it fails: The function is decreasing on $(-\infty, 0)$ and $(0, \infty)$, but NOT decreasing on $(-\infty, \infty)$ because $f(-1) < f(1)$ is false.

🛡️ Exam Defense: Never bridge across points of discontinuity with a union symbol if the derivative changes behavior or function is undefined.

Comparison

Concept / TheoremHypotheses / AssumptionsKey Formula / OutcomeGeometric MeaningExam Identification Clue
Strictly IncreasingContinuous on $[a,b]$, Differentiable on $(a,b)$$f'(x) > 0 \forall x \in (a,b)$Tangent has strictly positive slope; curve always risesKeywords: 'strictly increasing', 'one-to-one'
Increasing (Non-decreasing)Continuous on $[a,b]$, Differentiable on $(a,b)$$f'(x) \ge 0 \forall x \in (a,b)$Tangent slope is non-negative; curve never fallsKeywords: 'increasing', 'monotonic non-decreasing'
Strictly DecreasingContinuous on $[a,b]$, Differentiable on $(a,b)$$f'(x) < 0 \forall x \in (a,b)$Tangent has strictly negative slope; curve always fallsKeywords: 'strictly decreasing'
Constant FunctionContinuous on $[a,b]$, Differentiable on $(a,b)$$f'(x) = 0 \forall x \in (a,b)$Horizontal tangent; flat line parallel to x-axisKeywords: 'independent of x', 'zero derivative everywhere'

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

Find the interval in which the function $f(x) = 2x^3 - 9x^2 + 12x + 15$ is strictly increasing.

(A) $(-\infty, 1) \cup (2, \infty)$(B) $(1, 2)$(C) $(-\infty, 1]$(D) $[2, \infty)$
✓ Correct Answer: (A)
📚 Method A: Formal Textbook Derivation
Step 1: Compute $f'(x) = 6x^2 - 18x + 12$. Step 2: Set $f'(x) > 0 \implies 6(x^2 - 3x + 2) > 0$. Step 3: Factorize $6(x-1)(x-2) > 0$. Step 4: By wavy curve method, $x \in (-\infty, 1) \cup (2, \infty)$.
⚡ Method B: 25-Second Exam Shortcut
Find roots $1$ and $2$ of derivative. Since coefficient of $x^2$ is positive ($6 > 0$), the function increases outside the roots: $(-\infty, 1) \cup (2, \infty)$.
💡 Tip: Always check whether the question asks for 'strictly increasing' (open intervals) or 'increasing' (closed intervals).
CANONICAL PROBLEM 02 Dual-Method Breakdown

For what value of $a$ is the function $f(x) = \sin x - \cos x - ax + b$ strictly decreasing for all $x \in \mathbb{R}$?

(A) $a \ge \sqrt{2}$(B) $a \le -\sqrt{2}$(C) $a \ge 2$(D) $a \le 0$
✓ Correct Answer: (A)
📚 Method A: Formal Textbook Derivation
Step 1: Differentiate: $f'(x) = \cos x + \sin x - a$. Step 2: For $f$ to be decreasing, $f'(x) \le 0$ for all $x \implies \cos x + \sin x \le a$. Step 3: Maximum value of $\cos x + \sin x$ is $\sqrt{2}$. Step 4: Hence $a \ge \sqrt{2}$.
⚡ Method B: 25-Second Exam Shortcut
Recognize $A \cos x + B \sin x$ bounded by $-\sqrt{A^2+B^2}$ to $\sqrt{A^2+B^2}$. For upper bound inequality, $a \ge \sqrt{1^2 + 1^2} = \sqrt{2}$.
💡 Tip: A classic UP PGT repeating pattern combining trigonometric range bounds with monotonicity.

Practice questions

Choose an answer to check it and see the solution:

Q1
MCQ 1: The function $f(x) = x^3 - 3x^2 + 3x - 100$ is:
A strictly increasing on $\mathbb{R}$
B strictly decreasing on $\mathbb{R}$
C increasing only for $x > 1$
D decreasing for $x < 1$
Derivation & Analysis:

$f'(x) = 3x^2 - 6x + 3 = 3(x-1)^2 \ge 0$ for all $x \in \mathbb{R}$. Since $f'(x) = 0$ only at $x = 1$ (isolated point), the function is strictly increasing on the entire real line.

Q2
MCQ 2: The interval in which $f(x) = x^e e^{-x}$ is strictly increasing is:
A $(0, e)$
B $(e, \infty)$
C $(-\infty, 0)$
D $(0, \infty)$
Derivation & Analysis:

$f'(x) = e x^{e-1} e^{-x} - x^e e^{-x} = x^{e-1} e^{-x}(e - x)$. For $f'(x) > 0$, since $x^{e-1} e^{-x} > 0$ for $x > 0$, we require $e - x > 0 \implies x < e$. Thus, on $(0, e)$.នុ

Q3
MCQ 3: If $f(x) = \tan x - x$, then $f(x)$ is:
A increasing for all $x in (-\pi/2, \pi/2)$
B decreasing for all $x \in (-\pi/2, \pi/2)$
C neither increasing nor decreasing
D constant
Derivation & Analysis:

$f'(x) = \sec^2 x - 1 = \tan^2 x \ge 0$ for all $x$ in the domain. Hence, the function is increasing.

Q4
MCQ 4: The function $f(x) = \frac{x}{\ln x}$ is strictly increasing on:
A $(0, 1)$
B $(1, e)$
C $(e, \infty)$
D $(0, \infty)$
Derivation & Analysis:

$f'(x) = \frac{\ln x - 1}{(\ln x)^2}$. For $f'(x) > 0$, we need $\ln x > 1 \implies x > e$. Thus, it is strictly increasing on $(e, \infty)$.

Q5
MCQ 5: Which of the following functions is strictly decreasing on $(0, \pi/2)$?
A $\sin x$
B $\tan x$
C $\cos x$
D $\sec x$
Derivation & Analysis:

For $f(x) = \cos x$, $f'(x) = -\sin x$. On $(0, \pi/2)$, $\sin x > 0$, so $f'(x) < 0$, making $\cos x$ strictly decreasing.

Q6
MCQ 6: The values of $a$ for which $f(x) = ax^2 + 2x + 1$ is strictly increasing on $(1, 2)$ are:
A $a \ge -1$
B $a \ge -1/2$
C $a \ge -1/4$
D $a \ge 0$
Derivation & Analysis:

$f'(x) = 2ax + 2$. For $f'(x) > 0$ on $(1, 2)$, since it's a linear function, it suffices that $f'(1) \ge 0 \implies 2a + 2 \ge 0 \implies a \ge -1$. Wait, let's test: if $a = -1$, $f'(x) = -2x + 2$, which is negative on $(1, 2)$. Correct condition for $2ax + 2 > 0$ on $(1, 2)$: minimum value of $2ax+2$ on $(1, 2)$ must be $>0$. Since $a$ could be negative, $2a(1) + 2 > 0 \implies a > -1$. More precisely, if $a < 0$, $f'$ decreases, so $f'(2) = 4a + 2 > 0 \implies a > -1/2$. Combining gives $a \ge -1/2$.

Q7
MCQ 7: If $f(x) = x^3 + av^2 + gx + s$ is an increasing function on $\mathbb{R}$, then:
A $a^2 - 3g \le 0$
B $a^2 - 3g \ge 0$
C $a^2 - g \le 0$
D $a^2 - 3g > 0$
Derivation & Analysis:

$f'(x) = 3x^2 + 2ax + g$. For $f'(x) \ge 0$ for all $x$, discriminant $D \le 0 \implies (2a)^2 - 4(3)(g) \le 0 \implies 4a^2 - 12g \le 0 \implies a^2 - 3g \le 0$.

Q8
MCQ 8: Assertion (A): $f(x) = \ln(1+x) - \frac{x}{1+x}$ is increasing for $x > 0$. Reason (R): $f'(x) > 0$ for $x > 0$.
A Both (A) and (R) are true and (R) is the correct explanation of (A)
B Both (A) and (R) are true but (R) is not the correct explanation of (A)
C (A) is true but (R) is false
D (A) is false but (R) is true
Derivation & Analysis:

$f'(x) = \frac{1}{1+x} - \frac{1}{(1+x)^2} = \frac{x}{(1+x)^2} > 0$ for $x > 0$. Thus both assertion and reason are correct and (R) explains (A).

Q9
MCQ 9: The function $f(x) = x^nV$ (let's say $x^3 - 3x$) has a local max/min, but for strict monotonicity on $(1, \infty)$, consider $f(x) = 2x^2 - \ln |x|$. The interval where it is decreasing is:
A $(-1/2, 0) \cup (0, 1/2)$
B $(-\infty, -1/2) \cup (1/2, \infty)$
C $(0, 1/2)$
D $(-1/2, 1/2)$
Derivation & Analysis:

$f'(x) = 4x - \frac{1}{x} = \frac{4x^2 - 1}{x}$. For $f'(x) < 0$, numerator and denominator must have opposite signs. Roots are $-1/2, 0, 1/2$. Testing intervals yields $(-\infty, -1/2) \cup (0, 1/2)$.

Q10
MCQ 10: If $f(x) = x^3 + kx^2 + hx + 6$ is strictly increasing on $\mathbb{R}$, what is the relation between $k$ and $h$?
A $k^2 < 3h$
B $k^2 > 3h$
C $k^2 \le 3h$
D $k^2 \ge 3h$
Derivation & Analysis:

$f'(x) = 3x^2 + 2kx + h$. For strict monotonicity on $\mathbb{R}$, $f'(x) > 0$ for all $x$ except possibly at a single point, so $4k^2 - 12h < 0 \implies k^2 < 3h$.

Common questions

Yes. For example, $f(x) = x^3$ has $f'(0) = 0$, yet the function is strictly increasing across $x = 0$ because the derivative is positive everywhere else in any neighborhood around zero.

Confusing open and closed intervals in the final answer options, or failing to exclude points where the function is undefined (such as vertical asymptotes in rational or logarithmic functions).

By leveraging polynomial discriminant shortcuts ($D \le 0$ for quadratic derivatives) and testing boundary values from the options rather than solving inequalities from scratch.

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