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JEE Mathematics 2023 Question with Solution

Cover image showing Collegedunia and National Testing Agency branding with the text JEE MAIN 2023 10 Apr Shift-2 Question Paper with Solutions.

Mathematics Section-A If the coefficients of xx and x2x^2 in (1+x)p(1−x)q\left(1 + x\right)^p\left(1 - x\right)^q are 44 and −5-5 respectively, then 2p+3q2p + 3q is equal to:

  • A

    6060

  • B

    6363

  • C

    6666

  • D

    6969

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: The coefficient of xx in (1+x)p(1−x)q\left(1+x\right)^p\left(1-x\right)^q is 44 and the coefficient of x2x^2 is −5-5.

Find: The value of 2p+3q2p+3q.

Expand the two binomials up to the required terms:

(1+x)p=1+px+p(p−1)2!x2+…(1+x)^p = 1+px+\frac{p(p-1)}{2!}x^2+\ldots (1−x)q=1−qx+q(q−1)2!x2−…(1-x)^q = 1-qx+\frac{q(q-1)}{2!}x^2-\ldots

From the coefficient of xx,

p−q=4p-q=4

For the coefficient of x2x^2, combine the quadratic terms and the cross term:

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

Using p=q+4p=q+4, substitute into the quadratic relation:

(q+4)2+q2−(q−4)−q−2(4+q)q=−10(q+4)^2+q^2-(q-4)-q-2(4+q)q=-10

This simplifies as shown in the solution:

q2+8q+16−q2−q−4−q−8q−2q2=−10q^2+8q+16-q^2-q-4-q-8q-2q^2=-10 −2q=−22-2q=-22

Hence,

q=11q=11

and therefore

p=15p=15

Now calculate:

2p+3q=2(15)+3(11)2p+3q=2(15)+3(11) =30+33=30+33 =63=63

Therefore, the correct option is B.

Coefficient Matching

Given: Coefficient of xx is 44 and coefficient of x2x^2 is −5-5.

Find: 2p+3q2p+3q.

When multiplying

(1+px+p(p−1)2x2+…)(1−qx+q(q−1)2x2+…)(1+px+\frac{p(p-1)}{2}x^2+\ldots)(1-qx+\frac{q(q-1)}{2}x^2+\ldots)

the coefficient of xx comes only from pxpx and −qx-qx. So,

p−q=4p-q=4

The coefficient of x2x^2 comes from three contributions:

  • the x2x^2 term of (1+x)p\left(1+x\right)^p,
  • the x2x^2 term of (1−x)q\left(1-x\right)^q,
  • the product of the two linear terms. Thus,
p(p−1)2+q(q−1)2−pq=−5\frac{p(p-1)}{2}+\frac{q(q-1)}{2}-pq=-5

Substitute p=q+4p=q+4 and solve to get q=11q=11 and p=15p=15. Therefore,

2p+3q=2⋅15+3⋅11=632p+3q=2\cdot 15+3\cdot 11=63

So the answer is 6363.

Common mistakes

  • Ignoring the cross term px⋅(−qx)=−pqx2px\cdot (-qx)=-pqx^2 while finding the coefficient of x2x^2. This makes the quadratic equation incorrect. Always include contributions from both quadratic terms and the product of linear terms.

  • Writing the coefficient of xx as p+qp+q instead of p−qp-q. The sign of the linear term in (1−x)q\left(1-x\right)^q is negative, so the correct coefficient is p−qp-q.

  • Using q(q+1)2\frac{q(q+1)}{2} for the coefficient of x2x^2 in (1−x)q\left(1-x\right)^q. The correct term is q(q−1)2x2\frac{q(q-1)}{2}x^2 because it comes from the binomial coefficient, while the sign becomes positive for the even power.

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