CBSE Sample Paper 2026-27 · Section A · 1 mark

Q10: Where the derivative of $y = x^4$ exceeds $y$

Sample 2026-27Ch 6 · Application of Derivatives Ch 5 · Continuity and Differentiability

The question

The values of $x$ for which the derivative of the function $y = x^4$ is greater than the value of the function are

(A) $0 < x < 4$

(B) $0 < x \le 4$

(D) $x < 0 \text{ or } x \ge 4$

Exam answer

$y = x^4 \Rightarrow \frac{dy}{dx} = 4x^3$ ⁠.

Why?

Power rule:

\frac{d}{dx}x^n = nx^{n-1}

With $n = 4$ ⁠, the derivative is $4x^3$ ⁠.

\begin{aligned} \frac{dy}{dx} > y &\Rightarrow 4x^3 > x^4 \\ &\Rightarrow 4x^3 - x^4 > 0 \\ &\Rightarrow x^3(4 - x) > 0 \\ &\Rightarrow x(x - 4) < 0 \end{aligned}

Why?

Take $x^4$ to the left, then take out the common factor $x^3$ ⁠:

\begin{aligned} 4x^3 - x^4 &= x^3 \cdot 4 - x^3 \cdot x \\ &= x^3(4 - x) \end{aligned}

$x = 0$ gives $0 > 0$ ⁠, which is false, so $x \ne 0$ and $x^2 > 0$ ⁠. Dividing $x^3(4 - x) > 0$ by $x^2$ keeps the sign and gives $x(4 - x) > 0$ ⁠.

\begin{aligned} \frac{x^3(4 - x)}{x^2} &= x(4 - x) \\ x(4 - x) &> 0 \end{aligned}

Multiplying by $-1$ turns $4 - x$ into $x - 4$ and reverses the inequality.

\begin{aligned} -x(4 - x) &< 0 \\ x(x - 4) &< 0 \end{aligned}

( $x^3$ has the sign of $x$ ⁠.) So $0 < x < 4$ ⁠.

Why?

A product of two factors is negative when they have opposite signs. $x(x - 4)$ is zero at $0$ and $4$ ⁠, and negative only between them, where $x > 0$ and $x - 4 < 0$ ⁠.

\begin{aligned} x = 4:\ 4x^3 &= 4(4)^3 = 4(64) = 256 \\ x^4 &= (4)^4 = 256 \end{aligned}

Watch out: at $x = 4$ ⁠, $4x^3 = 256 = x^4$ ⁠, equal and not greater, so (B) is wrong. (C) is where the derivative is smaller.

Answer: (A) $0 < x < 4$

Where this comes from

Chapter 6, Application of Derivatives: Comparing a derivative with the function.

Quick lesson · Worked NCERT solutions · Interactive lesson

Also uses Chapter 5, Continuity and Differentiability.

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Q10: Where the derivative of y=x4y = x^4y=x4 exceeds yyy

Q10: Where the derivative of $y = x^4$ exceeds $y$