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Calculus

GATE CSE 2014 Set 2 · Q5

MCQ1 markEasy

Time

A non-zero polynomial f(x)f(x) of degree 3 has roots at x=1,x=2x = 1, x = 2 and x=3x = 3. Which one of the following must be TRUE?

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Solution

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Official answer: A) f(0)f(4) < 0

Calculus

Since f(x)f(x) is a polynomial of degree 3 with roots at x=1,2,3x = 1, 2, 3, it can be written as:

f(x)=c(x−1)(x−2)(x−3)f(x) = c(x - 1)(x - 2)(x - 3)

where cc is a non-zero real constant (c≠0c \neq 0).

Evaluating $f(x)$ at $x = 0$ and $x = 4$:

f(0)=c(0−1)(0−2)(0−3)=−6cf(0) = c(0 - 1)(0 - 2)(0 - 3) = -6c
f(4)=c(4−1)(4−2)(4−3)=6cf(4) = c(4 - 1)(4 - 2)(4 - 3) = 6c

Now, compute their product:

f(0)f(4)=(−6c)(6c)=−36c2f(0)f(4) = (-6c)(6c) = -36c^2

Since c≠0c \neq 0, c2>0c^2 > 0, which means:

f(0)f(4)=−36c2<0f(0)f(4) = -36c^2 < 0

Also, f(0)+f(4)=−6c+6c=0f(0) + f(4) = -6c + 6c = 0, which rules out (C) and (D). Therefore, option (A) must be TRUE.

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This Calculus previous year question is Q5 of the GATE CSE 2014 Set 2 paper. Find more GATE CSE PYQs on Calculus in Engineering Mathematics, or browse all GATE CSE PYQs.

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Calculus in GATE CSE

Calculus has 40 GATE CSE PYQs across 25 papers from 2011 to 2026. 14 of them are 1-mark MCQs like this one, and GATE CSE 2014 Set 2 has 2.

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A non-zero polynomial f(x) of degree 3 has roots at x = 1, x = 2 and x = 3. Which one of the following must be TRUE?

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