Previous year questions: Binomial Theorem

How to use this page

Six items are real JEE Main 2026 questions from the official National Testing Agency papers and four are from the NCERT Exemplar for Class 11 Mathematics. Each heading shows the paper and the marks. The stems were checked against the official PDF pages and the values against the official key. Most items reduce to the general term or to a coefficient sum, so write the general term before computing anything.

Q1 (JEE Main 2026, 4 Apr Shift 2, 4 marks)

In the expansion of (9x minus 1 by (3 root x)) to the eighteenth power, x greater than 0, the term independent of x is 221 times k. Find k.

Answer: The general term is 18Cr times (9x) to the (18 minus r) times (minus 1 by (3 root x)) to the r. The power of x is 18 minus r minus r by 2, that is 18 minus 3r by 2. For the term to be independent of x this power is 0, so r is 12. The coefficient is then 18C12 times 9 to the sixth power divided by 3 to the twelfth power. Since 9 to the sixth is 3 to the twelfth, the ratio is 1, and the coefficient is 18C12 equal to 18564. Dividing 18564 by 221 gives 84, so k is 84, matching the official answer.

Q2 (JEE Main 2026, 5 Apr Shift 2, 4 marks)

Find the coefficient of x squared in the expansion of (2x squared plus 1 by x) to the tenth power, x not zero.

Answer: The general term is 10Cr times (2x squared) to the (10 minus r) times (1 by x) to the r. The power of x is 20 minus 2r minus r, that is 20 minus 3r. Setting this equal to 2 gives 3r equal to 18, so r is 6. The coefficient is 10C6 times 2 to the fourth power, that is 210 times 16 equal to 3360. The official key is 3360. Checking the arithmetic a second way, 20 minus 18 is 2, so the powers match exactly and no extra factor of x remains.

Hinglish: General term me x ki power 20 − 3r hai. 20 − 3r = 2 se r = 6 milta hai. Coefficient 10C6 × 2^4 = 210 × 16 = 3360. Power check: 20 − 18 = 2, to x² hi bachta hai.

Q3 (JEE Main 2026, 5 Apr Shift 1, 4 marks)

If the sum of the coefficients of x to the seventh and x to the fourteenth in the expansion of (1 by x cubed minus x to the fourth) to the power n, with x not zero, is zero, find n.

Answer: The general term is nCr times x to the (minus 3n plus 3r) times (minus x to the fourth) to the r, so the power of x is 7r minus 3n. For x to the seventh this gives 7r minus 3n equal to 7, so r is (7 plus 3n) by 7. For x to the fourteenth the corresponding r is (14 plus 3n) by 7. Write n as 7m so both indices are integers: the first is 1 plus 3m and the second is 2 plus 3m, and they are consecutive. The coefficient sum being zero means the two binomial coefficients have opposite signs equal in size, so C(7m, 1 plus 3m) equals C(7m, 2 plus 3m). Neighbouring indices have equal nCr only when they are complementary, so (1 plus 3m) plus (2 plus 3m) equals 7m, giving 3 plus 6m equal to 7m and m equal to 3. Hence n is 21, which is the official answer.

Q4 (JEE Main 2026, 6 Apr Shift 1, 4 marks)

If the coefficients of the middle terms in the binomial expansions of (1 plus alpha x) to the twenty sixth power and (1 minus alpha x) to the twenty eighth power, with alpha not zero, are equal, find alpha.

Answer: Both indices are even, so each expansion has one middle term. For the twenty sixth power the middle term is the fourteenth and its coefficient is 26C13 times alpha to the thirteenth. For the twenty eighth power the middle term is the fifteenth and its coefficient is 28C14 times alpha to the fourteenth, because the sign from the even power of minus is positive. Equating and dividing by alpha to the thirteenth gives alpha equal to 26C13 divided by 28C14. Simplify the ratio: 26C13 is 26 factorial by 13 factorial squared and 28C14 is 28 factorial by 14 factorial squared. The ratio becomes 14 times 14 divided by 28 times 27, that is 196 by 756, which reduces to 7 by 27. The official key is 7 by 27.

Q5 (JEE Main 2026, 6 Apr Shift 2, 4 marks)

The identity (1 minus x cubed) to the tenth power equals the sum from r equal to 0 to 10 of a subscript r times x to the r times (1 minus x) to the (30 minus 2r) holds for all x. Find 9 times a9 divided by a10.

Answer: Divide both sides by (1 minus x) to the thirtieth power. The left side becomes the tenth power of (1 plus x plus x squared) divided by (1 minus x) squared, and the right side becomes the sum of a subscript r times u to the r, where u equals x divided by (1 minus x) squared. Now (1 plus x plus x squared) equals (1 minus x) squared plus 3x, so the ratio on the left is 1 plus 3u. Hence the sum of a subscript r u to the r equals (1 plus 3u) to the tenth power, which means a subscript r is 10Cr times 3 to the r.

Therefore a9 is 10C9 times 3 to the ninth and a10 is 3 to the tenth. Then 9 times a9 divided by a10 equals 9 times 10 times 3 to the ninth divided by 3 to the tenth, which is 90 by 3, that is 30. The official answer is 30.

Q6 (JEE Main 2026, 4 Apr Shift 1, 4 marks)

Let k be the smallest natural number for which the coefficient of x cubed in (1 plus x) cubed plus (1 plus x) to the fourth plus and so on up to (1 plus x) to the ninety ninth plus (1 plus kx) to the hundredth is (43n plus 101 by 4) times 100C3 for some natural number n. Taking that smallest k as p, find p plus n.

Answer: The coefficient of x cubed in (1 plus x) to the j is jC3. The hockey stick identity sums jC3 from j equal to 3 to 99 to 100C4. The last bracket contributes k cubed times 100C3. Now 100C4 equals 97 by 4 times 100C3, so the total coefficient is (97 by 4 plus k cubed) times 100C3. Equating with (43n plus 101 by 4) times 100C3 gives k cubed equal to 43n plus 1. The smallest natural k with k cubed congruent to 1 modulo 43 is 6, since 216 equals 43 times 5 plus 1, so p is 6 and n is 5. Then p plus n is 11, matching the official answer.

Q7 (NCERT Exemplar, 3 marks)

Find the coefficient of x to the fifteenth power in the expansion of (x minus x squared) to the tenth power.

Answer: Factor x out of the bracket: (x minus x squared) to the tenth is x to the tenth times (1 minus x) to the tenth. To reach x to the fifteenth the bracket must supply x to the fifth. The coefficient of x to the fifth in (1 minus x) to the tenth is 10C5 with a minus sign because the power is odd, that is minus 252. So the required coefficient is minus 252. Source: NCERT Exemplar Class 11 Mathematics, Chapter 8, Exercise 8.3, Short Answer Type, Q6.

Q8 (NCERT Exemplar, 3 marks)

Find the sixth term of the expansion of (x to the one third plus y to the one half) to the power n, given that the binomial coefficient of the third term from the end is 45.

Answer: The binomial coefficient of the third term from the end equals that of the third term from the beginning, which is nC2. Setting nC2 equal to 45 gives n(n minus 1) by 2 equal to 45, so n(n minus 1) is 90 and n is 10. The sixth term is 10C5 times x to the one third raised to the fifth times y to the one half raised to the fifth. That is 252 times x to the five thirds times y to the five halves. Source: NCERT Exemplar Class 11 Mathematics, Chapter 8, Exercise 8.3, Short Answer Type, Q8.

Q9 (NCERT Exemplar, 3 marks)

Find r if the coefficients of the (2r plus 4)th term and the (r minus 2)th term in the expansion of (1 plus x) to the eighteenth power are equal.

Answer: The coefficient of the mth term is 18C(m minus 1). Equality of the two coefficients gives 18C(2r plus 3) equal to 18C(r minus 3). Two binomial coefficients of the same row are equal when their indices are equal or complementary. Equal indices would need 2r plus 3 equal to r minus 3, giving r equal to minus 6, which is not possible for a term index. Complementary indices give (2r plus 3) plus (r minus 3) equal to 18, so 3r is 18 and r is 6. Source: NCERT Exemplar Class 11 Mathematics, Chapter 8, Exercise 8.3, Short Answer Type, Q9.

Q10 (NCERT Exemplar, 3 marks)

Find the coefficient of x to the fourth power in the expansion of (1 plus x plus x squared plus x cubed) to the eleventh power.

Answer: The bracket is the geometric sum (1 minus x to the fourth) divided by (1 minus x), so its eleventh power is (1 minus x to the fourth) to the eleventh times (1 minus x) to the minus eleventh. Expand the first factor as 1 minus 11 x to the fourth plus further terms, and the second factor as a binomial series 1 plus 11x plus 66 x squared plus 286 x cubed plus 1001 x to the fourth plus further terms. The x to the fourth coefficient gets two contributions: the constant of the first factor times the coefficient of x to the fourth in the second factor, which is 1001, and minus 11 from the x to the fourth term times the constant 1 of the second factor. The total is 1001 minus 11 equal to 990. Source: NCERT Exemplar Class 11 Mathematics, Chapter 8, Exercise 8.3, Short Answer Type, Q11.