Question 283 marksShort AnswerSection C

(a) Find the intervals in R\mathbb{R} for which the function f(x)=x4−2x2f(x) = x^4 - 2x^2 is increasing or decreasing.

OR

(b) Find : ∫2x+118−4x−x2dx\int \frac{2x + 1}{\sqrt{18 - 4x - x^2}} dx

Calculus (Application of Derivatives / Integrals)
Increasing/Decreasing Functions and Integration

Official answer

This question has two alternatives; either one earns full marks.


Option (a) - Monotonicity of f(x)=x4−2x2f(x) = x^4 - 2x^2:

  • f′(x)=4x3−4x=4x(x−1)(x+1)f'(x) = 4x^3 - 4x = 4x(x - 1)(x + 1).
  • Critical points: x=−1,0,1x = -1, 0, 1.
  • Sign of f' on each interval:
  • (−∞,−1)(-\infty, -1): f′<0f' < 0, so decreasing.
  • (−1,0)(-1, 0): f′>0f' > 0, so increasing.
  • (0,1)(0, 1): f′<0f' < 0, so decreasing.
  • (1,∞)(1, \infty): f′>0f' > 0, so increasing.
  • Increasing on (−1,0)(-1, 0) union (1,∞)(1, \infty); decreasing on (−∞,−1)(-\infty, -1) union (0,1)(0, 1).

Option (b) - Integral:

  • Write 2x+1=−1(−4−2x)−32x + 1 = -1(-4 - 2x) - 3, where −4−2x-4 - 2x is the derivative of 18−4x−x218 - 4x - x^2.
  • Also 18−4x−x2=22−(x+2)218 - 4x - x^2 = 22 - (x + 2)^2.
  • Integral = −218−4x−x2-2\sqrt{18 - 4x - x^2} - 3sin⁡−1((x+2)/22)3\sin^{-1}((x + 2)/\sqrt{22}) + C.
Answer keywords:increasing decreasing functionfirst derivative testsign chartcritical pointsintegration by substitutioncompleting the squareinverse sine integralmonotonicity

Marking scheme

  1. 1.Option (a): 1 mark for f′(x)=4x3−4xf'(x) = 4x^3 - 4x factored, 1 mark for critical points and sign chart, 1 mark for correct increasing/decreasing intervals.
  2. 2.Option (b): 1 mark for splitting 2x+1=−1(−4−2x)−32x+1 = -1(-4-2x) - 3, 1 mark for completing the square 22−(x+2)222 - (x+2)^2 and integrating the root part to -2 sqrt(...), 1 mark for the −3sin⁡−1((x+2)/22)-3\sin^{-1}((x+2)/\sqrt{22}) term with +C.
  3. 3.Accept open or closed interval notation at critical points; deduct for missing +C in (b).

Hint

For (a) factor f′(x)=4x(x−1)(x+1)f'(x) = 4x(x-1)(x+1) and build a sign chart. For (b) split 2x+12x+1 into a multiple of the derivative of (18−4x−x2)(18-4x-x^2) plus a constant, then complete the square.

Quick oral answer

For (a), f′(x)=4x(x−1)(x+1)f'(x) = 4x(x-1)(x+1) is positive on (−1,0)(-1,0) and (1,∞)(1, \infty) so f increases there and decreases elsewhere; for (b), split 2x+12x+1 by the derivative of the quadratic to get −218−4x−x2-2\sqrt{18-4x-x^2} - 3 sin inverse of (x+2)/22(x+2)/\sqrt{22} plus C.

Common mistakes

  1. 1.In (a), omitting x=0x = 0 as a critical point and reporting only two intervals instead of four.
  2. 2.In (b), a sign error when matching coefficients (A should be -1), which corrupts both terms.
  3. 3.In (b), forgetting the constant of integration +C or leaving the quadratic uncompleted before the inverse-sine step.

Analysis and explanation

Both alternatives sit at the heart of Unit 3 calculus but test opposite operations: differentiation for monotonicity and integration by splitting.


Option (a) - Concept:

  • A function increases where its first derivative is positive and decreases where it is negative. Factoring f′(x)=4x(x−1)(x+1)f'(x) = 4x(x-1)(x+1) exposes three sign-change points; a sign chart across them gives the four monotonic intervals.

Option (b) - Concept:

  • When the numerator is linear and the denominator is the square root of a quadratic, split the numerator into (a multiple of the derivative of the quadratic) plus (a constant). The first part integrates to 2 sqrt(quadratic); the constant part, after completing the square to 22−(x+2)222 - (x+2)^2, integrates to an inverse-sine.

Exam trap:

  • In (a), students forget x = 0 as a critical point and merge intervals wrongly. In (b), a sign slip while matching −2A=2-2A = 2 (giving A=−1A = -1) flips the whole answer.

Real-world link:

  • Monotonicity analysis identifies where profit, cost or population is rising or falling; the inverse-sine integral form appears in arc-length and probability (normal-curve related) computations.

Interesting facts

The function x4−2x2x^4 - 2x^2 has a 'W' shape with two symmetric minima at x=±1x = \pm 1 and a local maximum at x=0x = 0, a classic double-well curve also seen in physics potential-energy models.

The technique of splitting a linear numerator into 'derivative part + constant part' is one of the most frequently examined integration tricks in CBSE Applied and Core Mathematics.

Question details

Q28 of 44 questions

Board
CBSE
Class
Class 12
Subject
Applied Mathematics
Year
2026
Section
Section C
Marks
3
Type
Short Answer
Chapter
Calculus (Application of Derivatives / Integrals)
Topic
Increasing/Decreasing Functions and Integration
Bloom's level
analyze
Syllabus topic
Unit 3: Calculus — Applications of Derivatives (Monotonicity) and Integration by Substitution
Author
CBSE
Reviewed by
Padhantu Academic Team

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Frequently asked questions

How do I decide where f(x)=x4−2x2f(x) = x^4 - 2x^2 is increasing?

Differentiate to get f′(x)=4x(x−1)(x+1)f'(x) = 4x(x-1)(x+1), find the zeros x=−1,0,1x = -1, 0, 1, and test the sign of f' in each interval. Where f' is positive, namely (−1,0)(-1, 0) and (1,∞)(1, \infty), the function is increasing; where f' is negative it is decreasing.

Why do we split the numerator 2x+12x+1 in the integral?

Splitting 2x+12x+1 into a multiple of the derivative of the denominator's quadratic plus a constant turns the problem into two standard integrals: one of the form derivative/sqrt giving 2 sqrt, and one constant/a2−u2\sqrt{a^2 - u^2} giving an inverse sine after completing the square.