Sample 2023-24 Class 12 Maths Q19 Solution: Minimum at x = 1 when f′(x) = (x − 1)³(x − 3)²

The question

In the following questions, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct answer out of the following choices.

(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.

Let $f(x)$ be a polynomial function of degree 6 such that $\frac{d}{dx}(f(x)) = (x - 1)^3(x - 3)^2$ ⁠, then

ASSERTION (A): $f(x)$ has a minimum at $x = 1$ ⁠.

REASON (R): When $\frac{d}{dx}(f(x)) < 0$ ⁠, $\forall\ x \in (a - h, a)$ and $\frac{d}{dx}(f(x)) > 0$ ⁠, $\forall\ x \in (a, a + h)$ ⁠; where ' $h$ ' is an infinitesimally small positive quantity, then $f(x)$ has a minimum at $x = a$ ⁠, provided $f(x)$ is continuous at $x = a$ ⁠.

Exam answer

$f'(x) = (x - 1)^3(x - 3)^2$ ⁠, and $(x - 3)^2 > 0$ near $x = 1$ ⁠.

Why?

A square is positive except where it is 0, and $(x - 3)^2 = 0$ only at $x = 3$ ⁠, which is far from $x = 1$ ⁠.

So near $x = 1$ ⁠, the sign of $f'(x)$ is the sign of $(x - 1)^3$ ⁠. An odd power keeps the sign of $x - 1$ ⁠: left of 1, $x - 1 < 0$ and its cube is negative; right of 1, $x - 1 > 0$ and its cube is positive.

For $x \in (1 - h, 1)$ ⁠: $(x - 1)^3 < 0$ ⁠, so $f'(x) < 0$ ⁠.

For $x \in (1, 1 + h)$ ⁠: $(x - 1)^3 > 0$ ⁠, so $f'(x) > 0$ ⁠.

$f$ is a polynomial, so it is continuous at $x = 1$ ⁠. By the first derivative test, $f$ has a minimum at $x = 1$ ⁠. So Assertion (A) is true.

Why?

First derivative test: if $f'$ changes sign from negative to positive as $x$ passes through $a$ ⁠, then $f$ decreases and then increases. So $f$ has a local minimum at $a$ ⁠.

Watch out: at $x = 3$ the factor $(x - 3)^2$ does not change sign, so $f'$ does not change sign there, and $f$ has no extreme value at 3.

Reason (R) is the first derivative test, so it is true, and it is exactly the argument above, so it explains (A).

Why?

(R) states the test at $x = a$ ⁠: $f'$ negative just left of $a$ ⁠, positive just right of $a$ ⁠, and $f$ continuous at $a$ ⁠. With $a = 1$ it gives (A) directly.

A is true, R is true and R explains A, which is code (a).

Answer: (a) Both (A) and (R) are true and (R) is the correct explanation of (A).

Where this comes from

Chapter 6, Application of Derivatives: First derivative test for a local minimum.

Question type: When the second derivative test fails

How to spot it: click to see

Type 8. When the second derivative test fails: the first derivative test

Spot it: $f''(c) = 0$ at a critical point, or $f$ has a corner there (like $|x|$ at 0), or the question says "first derivative test".

Here: The Reason is this test itself. Near $x = 1$ ⁠, $(x - 3)^2$ is positive, so $f'$ has the sign of $(x - 1)^3$ ⁠: $-$ just left of 1 and $+$ just right, so a minimum.

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Q19: Minimum at x=1x = 1x=1 when f′(x)=(x−1)3(x−3)2f'(x) = (x - 1)^3(x - 3)^2f′(x)=(x−1)3(x−3)2

Q19: Minimum at $x = 1$ when $f'(x) = (x - 1)^3(x - 3)^2$