The function $f(x) = \dfrac{2x^2 - 1}{x^4}$, $x > 0$, decreases in the interval ______.

> [!TLDR] > The function decreases where its derivative is negative. After simplifying $f'(x) = \frac{4(1 - x^2)}{x^5}$, we find $f'(x) 1$. So the interval of decrease is $(1, \infty)$. To decide where a function increases or decreases, we look at the sign of its first derivative. If $f'(x) > 0$, the function is rising; if $f'(x) < 0$, it is falling. The question asks for the interval where $f$ *decreases*, so we need $f'(x) < 0$. Let’s work through it carefully. 1. **Rewrite the function for easier differentiation.** $f(x) = \dfrac{2x^2 - 1}{x^4} = \dfrac{2x^2}{x^4} - \dfrac{1}{x^4} = 2x^{-2} - x^{-4}$. This avoids the quotient rule and makes differentiation straightforward. 2. **Differentiate term by term.** $f'(x) = 2(-2)x^{-3} - (-4)x^{-5} = -4x^{-3} + 4x^{-5}$. Factor out the common factor $4x^{-5}$: $f'(x) = 4x^{-5}(-x^{2} + 1) = \dfrac{4(1 - x^2)}{x^5}$. > [!FORMULA] > $$f'(x) = \frac{4(1 - x^2)}{x^5}$$ 3. **Analyze the sign of $f'(x)$ for $x > 0$.** Since $x > 0$, the denominator $x^5$ is always positive. The factor $4$ is also positive. So the sign of $f'(x)$ depends entirely on the numerator $(1 - x^2)$. - $1 - x^2 > 0$ when $x^2 0$, so $f$ increases. - $1 - x^2 1$, i.e., $x > 1$. Then $f'(x) < 0$, so $f$ decreases. …

The function f(x) = (2x² - 1)/(x⁴), x > 0, decreas – Class 12 Mathematics

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f(x) = \dfrac{2x^2 - 1}{x^4}

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Concept understanding — Derivative Sign Analysis

Derivative Sign Analysis: What the Slope Tells You

Imagine walking along a hilly road — sometimes uphill, sometimes downhill, occasionally flat. The derivative at any point is simply the slope of the road under your feet at that instant.

Derivative sign analysis figures out where a function is increasing, where it is decreasing, and where it has flat spots (critical points) — all from the sign of its derivative.

The Intuition First

If $f'(x)$ is positive, the function is increasing — the graph rises as you move right. If $f'(x)$ is negative, it is decreasing. If $f'(x) = 0$ , there is a horizontal tangent — a potential peak, valley, or flat inflection.

The key: a single point tells you little; you look at intervals. If $f'(x) > 0$ for all $x$ in $(a, b)$ , the function is strictly increasing on that whole interval. Same logic for negative.

Note

The analysis is local — it describes behaviour on intervals, not isolated points. A zero derivative at a single point doesn't guarantee a max or min; check the sign change across that point.

The Precise Statement

Let $f$ be differentiable on an open interval $I$ . Then:

If $f'(x) > 0$ for all $x$ in $I$ , then $f$ is strictly increasing on $I$ .
If $f'(x) < 0$ for all $x$ in $I$ , then $f$ is strictly decreasing on $I$ .
If $f'(x) = 0$ for all $x$ in $I$ , then $f$ is constant on $I$ .

Points where $f'(x) = 0$ (or where $f'$ does not exist) are critical points — the candidates for local maxima and minima.

\text{If } f'(x) > 0 \text{ on } (a,b) \implies f \text{ increasing on } (a,b)

How to Perform It (Step-by-Step)

Find the derivative $f'(x)$ .
Find critical points: solve $f'(x) = 0$ and check where $f'(x)$ is undefined (but $f$ is defined).
Plot these on a number line — they split the domain into intervals.
Pick a test point inside each interval and evaluate $f'$ ; only the sign matters.
Record the sign in each interval and interpret: + means increasing, – means decreasing.

A Concrete Example

Take $f(x) = x^3 - 3x$ .

Step 1: $f'(x) = 3x^2 - 3 = 3(x-1)(x+1)$ .

Step 2: Critical points: $x = -1$ and $x = 1$ .

Step 3: Intervals: $(-\infty, -1)$ , $(-1, 1)$ , $(1, \infty)$ .

Step 4: Test points:

$x = -2$ : $f'(-2) = 3(4-1) = 9 > 0$ .
$x = 0$ : $f'(0) = -3 < 0$ .
$x = 2$ : $f'(2) = 9 > 0$ .

Step 5: So $f$ increases on $(-\infty, -1)$ , decreases on $(-1, 1)$ , increases on $(1, \infty)$ . Thus $x = -1$ is a local maximum (sign changes + to –), and $x = 1$ is a local minimum (– to +). …

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The function decreases where its derivative is negative. After simplifying $f'(x) = \frac{4(1 - x^2)}{x^5}$ , we find $f'(x) < 0$ when $x > 1$ . So the interval of decrease is $(1, \infty)$ .

To decide where a function increases or decreases, we look at the sign of its first derivative. If $f'(x) > 0$ , the function is rising; if $f'(x) < 0$ , it is falling. The question asks for the interval where $f$ decreases, so we need $f'(x) < 0$ .

Let’s work through it carefully.

Rewrite the function for easier differentiation.

$f(x) = \dfrac{2x^2 - 1}{x^4} = \dfrac{2x^2}{x^4} - \dfrac{1}{x^4} = 2x^{-2} - x^{-4}$ .

This avoids the quotient rule and makes differentiation straightforward.
Differentiate term by term.

$f'(x) = 2(-2)x^{-3} - (-4)x^{-5} = -4x^{-3} + 4x^{-5}$ .

Factor out the common factor $4x^{-5}$ :

$f'(x) = 4x^{-5}(-x^{2} + 1) = \dfrac{4(1 - x^2)}{x^5}$ .
$f'(x) = \frac{4(1 - x^2)}{x^5}$
Analyze the sign of $f'(x)$ for $x > 0$ .

Since $x > 0$ , the denominator $x^5$ is always positive. The factor $4$ is also positive. So the sign of $f'(x)$ depends entirely on the numerator $(1 - x^2)$ .
- $1 - x^2 > 0$ when $x^2 < 1$ , i.e., $0 < x < 1$ . Then $f'(x) > 0$ , so $f$ increases.
- $1 - x^2 < 0$ when $x^2 > 1$ , i.e., $x > 1$ . Then $f'(x) < 0$ , so $f$ decreases. …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2024Set 65/2/11 markMCQ
$f(x) = x^3 - 3x^2 + 12x - 18$
›Reveal solutionSolution
The derivative $f'(x) = 3x^2 - 6x + 12$ is always positive (its discriminant is negative and leading coefficient positive), so $f(x)$ is strictly increasing on $\mathbb{R}$ . The correct option is (B).

The core question here is about monotonicity — whether a function is always increasing, always decreasing, or neither. For a polynomial, the sign of its derivative tells us everything. If the derivative is positive everywhere, the function is strictly increasing; if negative everywhere, strictly decreasing; if it changes sign, the function is neither.

Let’s see what $f'(x)$ looks like.

Find the derivative. $f(x) = x^3 - 3x^2 + 12x - 18$ Differentiating term by term:

$f'(x) = 3x^2 - 6x + 12$

Analyze the sign of $f'(x)$ . This is a quadratic: $3x^2 - 6x + 12$ . To check if it ever becomes negative or zero, compute its discriminant:

$D = (-6)^2 - 4 \cdot 3 \cdot 12 = 36 - 144 = -108$
Since $D < 0$ , the quadratic has no real roots — it never touches or crosses the x-axis.

What does a negative discriminant mean for sign? The leading coefficient $3 > 0$ , so the parabola opens upward. A quadratic that opens upward and has no real roots is always positive. Therefore, $f'(x) > 0$ for every real $x$ . …
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CBSE 2026Set V11 markMCQ
$f(x) = x^2$
(a) Both the Statements I and II are true
(b) Both the Statements I and II are false
(c) Statement I is true and Statement II is false
(d) Statement I is false and Statement II is true
›Reveal solutionSolution

Statement I is false and Statement II is true, so the answer is (d).

Statement I: For $f(x)=x^2$ , $f'(x)=2x$ . On $(0,\infty)$ we have $f'(x)=2x>0$ , so $f$ is increasing there, not decreasing. False. …
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CBSE 2026Set ANNUAL1 markMCQ
$f(x)=\int e^x (x-1)(x-2)\,dx$
(a) $(-\infty,-2)$
(b) $(-2,-1)$
(c) $(1,2)$
(d) $(2,+\infty)$
›Reveal solutionSolution
Since $f(x)=\int e^x(x-1)(x-2)\,dx$ , we get $f'(x)=e^x(x-1)(x-2)$ ; $f$ decreases where $f'(x)<0$ , i.e. on $(1,2)$ .

By the Fundamental Theorem of Calculus, if $f(x)=\int e^x(x-1)(x-2)\,dx$ , then

$f'(x)=e^x(x-1)(x-2)$

A function decreases on an interval where its derivative is negative: $f'(x)<0$ .

Since $e^x>0$ for every real $x$ , the sign of $f'(x)$ is entirely determined by the sign of $(x-1)(x-2)$ :

For $x<1$ : both factors negative $\Rightarrow$ product positive $\Rightarrow f'(x)>0$ . …
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CBSE 2025Set 65/4/11 markMCQ
$\lambda$
›Reveal solutionSolution
A function decreases everywhere when its derivative is non-positive for all $x$ . Here $f'(x) = \cos x + \sin x - \lambda$ must satisfy $\cos x + \sin x \le \lambda$ for all $x$ , which requires $\lambda \ge \sqrt{2}$ (the maximum of $\cos x + \sin x$ ).

A function decreases for all real $x$ when its rate of change is never positive. This translates to the condition $f'(x) \le 0$ for all $x \in \mathbb{R}$ . The question asks us to find which values of the parameter $\lambda$ enforce this condition.

The key insight is that we need to understand the range of the trigonometric expression in the derivative, then choose $\lambda$ large enough to dominate it everywhere.

Finding the derivative

Differentiate $f(x) = \sin x - \cos x - \lambda x + C$ :

$f'(x) = \cos x + \sin x - \lambda$

For $f$ to be decreasing everywhere, we need:

$f'(x) \le 0 \quad \text{for all } x \in \mathbb{R}$
This means:
$\cos x + \sin x - \lambda \le 0$ $\cos x + \sin x \le \lambda \quad \text{for all } x$
Finding the maximum of $\cos x + \sin x$

The condition $\cos x + \sin x \le \lambda$ for all $x$ is equivalent to requiring:

$\lambda \ge \max_{x \in \mathbb{R}} (\cos x + \sin x)$

To find this maximum, we can express the sum as a single sinusoid. Using the identity:

$\cos x + \sin x = \sqrt{2} \sin\left(x + \frac{\pi}{4}\right)$
›Proof

Derivation of the identity:

We write $\cos x + \sin x = R \sin(x + \phi)$ for some amplitude $R$ and phase $\phi$ .

Expanding: $R \sin(x + \phi) = R(\sin x \cos \phi + \cos x \sin \phi) = R \cos \phi \cdot \sin x + R \sin \phi \cdot \cos x$

Comparing coefficients:

Coefficient of $\sin x$ : $R \cos \phi = 1$

Coefficient of $\cos x$ : $R \sin \phi = 1$

Squaring and adding: $R^2(\cos^2 \phi + \sin^2 \phi) = 1 + 1 = 2$ , so $R = \sqrt{2}$ .

…
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Q.The function $f(x) = \dfrac{2x^2 - 1}{x^4}$ , $x > 0$ , decreases in the interval ______.

Concept understanding — Derivative Sign Analysis

Derivative Sign Analysis: What the Slope Tells You

The Intuition First

The Precise Statement

How to Perform It (Step-by-Step)

A Concrete Example

Method: Sign-Analysis of the Derivative for Power/Rational Functions

Steps

Common Mistakes

Q.The function f(x)=2x2−1x4f(x) = \dfrac{2x^2 - 1}{x^4}f(x)=x42x2−1​, x>0x > 0x>0, decreases in the interval ______.

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Q.The function $f(x) = \dfrac{2x^2 - 1}{x^4}$ , $x > 0$ , decreases in the interval ______.

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