Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Binomial Theorem question

2023 · 10 Apr · Shift 2 · Q25
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Mathematics
  4. /Binomial Theorem
  5. /2023 · 10 Apr · Shift 2 · Q25

Binomial Theorem question

2023 · 10 Apr · Shift 2 · Q25

JEE MainMathematicsBinomial TheoremMCQ+4 / −1
If the coefficients of xxx and x2x^{2}x2 in (1+x)p(1−x)q(1+x)^{\mathrm{p}}(1-x)^{\mathrm{q}}(1+x)p(1−x)q are 4 and −-− 5 respectively, then 2p+3q2 p+3 q2p+3q is equal to :
  1. A
    66
  2. B
    60
  3. C
    69
  4. D
    63
View written solutionFree

Correct answer: D

  1. Expand up to the x2x^2x2 term

We have

(1+x)p(1−x)q.(1+x)^p(1-x)^q.(1+x)p(1−x)q.

Using binomial expansion up to second degree:

(1+x)p=1+px+p(p−1)2x2+⋯(1+x)^p = 1+px+\frac{p(p-1)}{2}x^2+\cdots(1+x)p=1+px+2p(p−1)​x2+⋯ (1−x)q=1−qx+q(q−1)2x2+⋯(1-x)^q = 1-qx+\frac{q(q-1)}{2}x^2+\cdots(1−x)q=1−qx+2q(q−1)​x2+⋯

Now multiply:

(1+x)p(1−x)q=(1+px+p(p−1)2x2)(1−qx+q(q−1)2x2).(1+x)^p(1-x)^q = \left(1+px+\frac{p(p-1)}{2}x^2\right) \left(1-qx+\frac{q(q-1)}{2}x^2\right).(1+x)p(1−x)q=(1+px+2p(p−1)​x2)(1−qx+2q(q−1)​x2).
  1. Coefficient of xxx

The coefficient of xxx is

p−q.p-q.p−q.

Given that it is 444, so

p−q=4.(1)p-q=4. \qquad (1)p−q=4.(1)
  1. Coefficient of x2x^2x2

The coefficient of x2x^2x2 comes from:

  • p(p−1)2\dfrac{p(p-1)}{2}2p(p−1)​ from the first expansion,
  • q(q−1)2\dfrac{q(q-1)}{2}2q(q−1)​ from the second,
  • and the product of pxpxpx and −qx-qx−qx, i.e. −pq-pq−pq.

So,

p(p−1)2+q(q−1)2−pq=−5.\frac{p(p-1)}{2}+\frac{q(q-1)}{2}-pq=-5.2p(p−1)​+2q(q−1)​−pq=−5.

Simplify:

p2−p+q2−q−2pq2=−5\frac{p^2-p+q^2-q-2pq}{2}=-52p2−p+q2−q−2pq​=−5 (p−q)2−(p+q)2=−5.\frac{(p-q)^2-(p+q)}{2}=-5.2(p−q)2−(p+q)​=−5.

Using (1)(1)(1), p−q=4p-q=4p−q=4, hence

16−(p+q)2=−5.\frac{16-(p+q)}{2}=-5.216−(p+q)​=−5.

So,

16−(p+q)=−1016-(p+q)=-1016−(p+q)=−10 p+q=26.(2)p+q=26. \qquad (2)p+q=26.(2)
  1. Solve for ppp and qqq

From

p−q=4,p-q=4,p−q=4, p+q=26,p+q=26,p+q=26,

adding gives

2p=30  ⟹  p=15.2p=30 \implies p=15.2p=30⟹p=15.

Then

q=11.q=11.q=11.
  1. Find 2p+3q2p+3q2p+3q
2p+3q=2(15)+3(11)=30+33=63.2p+3q=2(15)+3(11)=30+33=63.2p+3q=2(15)+3(11)=30+33=63.
  1. Compare with options

The value is

63,63,63,

which corresponds to Option D.

  1. Compare with stored correct answer

Stored correct answer: D

Our derived answer is also D, so they agree.

PreviousNext

More from Binomial Theorem

  • The mean of the coefficients of x,x2,…,x7 in the binomial expansion of (2+x)9 is ​.2023 · Numerical
  • The number of integral terms in the expansion of (321​+541​)680 is equal to ​.2023 · Numerical
  • The sum of the coefficients of three consecutive terms in the binomial expansion of (1+x)n+2, which are in the ratio 1:3:5, is equal to :2023 · MCQ
  • If the 1011th  term from the end in the binominal expansion of (54x​−2x5​)2022 is 1024 times 1011th  R term from the beginning, then ∣x∣ is equal to2023 · MCQ
  • Fractional part of the number 1542022​ is equal to2023 · MCQ
  • Let α be the constant term in the binomial expansion of (x​−x23​6​)n,n≤15. If the sum of the coefficients of the remaining terms in the expansion is 649 and the coefficient of…2023 · Numerical
  • The coefficient of x5 in the expansion of (2x3−3x21​)5 is :2023 · MCQ
  • The remainder, when 7103 is divided by 17, is ​2023 · Numerical