JEE Main 2019 — Binomial Theorem Question with Solution
From: JEE Main 2019 (Online) 10th April Morning Slot
Question
If the coefficients of x2
and x3
are both zero, in the expansion of the expression (1 + ax + bx2
) (1 – 3x)15 in
powers of x, then the ordered pair (a,b) is equal to :
Choose an option
Show full solutionCorrect option: B
Correct answer
B(28, 315)
Step-by-step explanation
(1 + ax + bx2)(1 – 3x)15
Co-eff. of x2 = 1.15C2(–3)2 + a.15C1(–3) + b.15C0
(given)
945 – 45a + b = 0 ...(i)
Now co-eff. of x3 = 0
15C3(–3)3 + a.15C2(–3)2 + b.15C1(–3) = 0
15 × 3[–3 × 7 × 13 + a × 7 × 3 – b] = 0
21a – b = 273 ...(ii)
From (i) and (ii)
a = +28, b = 315 (a, b) (28, 315)
Co-eff. of x2 = 1.15C2(–3)2 + a.15C1(–3) + b.15C0
(given)
945 – 45a + b = 0 ...(i)
Now co-eff. of x3 = 0
15C3(–3)3 + a.15C2(–3)2 + b.15C1(–3) = 0
15 × 3[–3 × 7 × 13 + a × 7 × 3 – b] = 0
21a – b = 273 ...(ii)
From (i) and (ii)
a = +28, b = 315 (a, b) (28, 315)
Practice this on the real CBT interface
Solve this JEE Main question (and the rest of the Binomial Theorem chapter) on PrepSharp's TCS iON-style CBT player — with timer, bookmarks and session analytics.
Solve interactively →About this question
This is a previous-year question from JEE Main 2019, covering the Binomial Theorem chapter of Mathematics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.