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MATHEMATICS XI TEST PAPER SOL.pdf

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Ashwani Jha
Ashwani Jha

The document contains 12 math problems involving expansions of binomial expressions, arithmetic progressions, and other algebra topics. The problems are multi-step and require setting up and solving equations. No final answers are provided.

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Roll No. : Date : Time - MM - 70 2 Ans : 4 2. Ans : 4 3. Ans : 4 4. Ans : Find the number of terms in the expansion of (a + 2b ¨C 3c)n. Consider (a + 2b ¨C 3c)n = {a + (2b ¨C 3c)}n = nC0 an + nC1 an¨C1(2b ¨C 3c) + nC2 an¨C2(2b ¨C 3c)2 + ... + nCn(2b ¨C 3c)n. We notice first term gives one term, second term gives two terms, third term gives three terms and so on. ¡à Total terms 1 + 2 + ... + n + (n + 1) = Find the middle term(s) in the expansion of : (1 + x)2n Since, 2n would always be even so (n + 1)th term would be the middle term ¡à The coefficients of 2nd, 3rd and 4th terms in the expansion of (1 + x)2n are in A.P., prove that 2n2 ¨C 9n + 7 = 0 2nC1, 2nC2, 2nC3 are in A.P. ? 2. 2nC2 = 2nC1 + 2nC3 ? ? 2n2 ¨C 9n + 7 = 0 If Tr is rth term in the expansion of (1 + x)n in the ascending powers of x, prove that r (r + 1) Tr+2 = (n ¨C r + 1) (n ¨C r)x2 Tr. LHS = r (r + 1) Tr+2 = r (r + 1). nCr+1 xr+1 = r (r + 1) = RHS = (n ¨C r + 1) (n ¨C r) x2 Tr = (n ¨C r + 1) (n ¨C r)x2. nCr¨C1 xr¨C1 = (n ¨C r + 1) (n ¨C r). x2 . = LHS = RHS
4 5. Ans : 4 6. If the 3rd, 4th, 5th and 6th terms in the expansion of (x + y)n be a, b, c and d respectively, prove that If a, b, c be the three consecutive coefficients in the expansion of (1 + x)n, prove that n =
Ans : 4 7. Ans : 4 8. Using binomial theorem, prove that 6n ¨C 5n always leaves remainder 1 when divided by 25. For two numbers a and b if we can find numbers q and r such that a = bq + r, then we say that b divides a with q as quotient and r as remainder. Thus, in order to show that 6n ¨C 5n leaves remainder 1 when divided by 25, we prove that 6n ¨C 5n = 25k + 1, where k is some natural number. We have (1 + a)n = nC0 + nC1a + nC2a2 + ... nCnan For a = 5, we get (1 + 5)n = nC0 + nC15 + nC252 + ... nCn5n i.e. (6)n = 1 + 5n + 52 nC2 + 53 nC3 + ... + 5n i.e. 6n ¨C 5n = 1 + 52 (nC2 + nC3 5 + ... + 5n¨C2) or 6n ¨C 5n = 1 + 25 (nC2 + 5. nC3 + ... + 5n¨C2) or 6n ¨C 5n = 25k + 1 where k = nC2 + 5. nC3 + ... + 5n¨C2 This shows that when divided by 25, 6n ¨C 5n leaves remainder 1. If a1, a2, a3 and a4 are the coefficient of any four consecutive terms in the expansion of (1 + x)n, prove that
Ans : 4 9. Let a1, a2, a3 and a4 be the coefficient of four consecutive terms Tr+1, Tr+2, Tr+3 and Tr+4 respectively. Then a1 = coefficient of Tr+1 = nCr a2 = coefficient of Tr+2 = nCr+1 a3 = coefficient of Tr+3 = nCr+2 and a4 = coefficient of Tr+4 = nCr+3 Find the sum (33 ¨C 23) + (53 ¨C 43) + (73 ¨C 63) + ... 10 terms.
Ans : 4 10. Ans : 4 11. Ans : 4 12. Sum = (33 + 53 + 73 + ...) ¨C (23 + 43 + 63 + ...) For 33 + 53 + 73 + ... an = (2n + 1)3 = 8n3 + 12n2 + 6n + 1 ¡à S1 = 8 ¦²n3 + 12 ¦²n2 + 6 ¦²n + n ... (i) For 23 + 43 + 63 + ... an = (2n)3 = 8n3 S2 = 8¦²n3 ... (ii) ¡à Sum = S1 ¨C S2 = 12 ¦²n2 + 6 ¦²n + n [From (i), (ii)] Sum to 10 terms = = 4620 + 330 + 10 = 4960 If a, b, c be the first, third and nth term respectively of an A.P. Prove that the sum to n terms is Given first term = a, a3 = b and an = c Sn = [2a + (n ¨C 1)d] ... (i) We have first term = a b = a + 2d and a + (n ¨C 1) d = c Sn = [a + c] [From (i)] ... (ii) Also a + (n ¨C 1) = c ? n ¨C 1 = Substituting in (ii), we get Sn = If a, b, c are in G.P and . Prove that x, y, z are in A.P. ? a = kx, b = ky, c = kz Given b2 = ac ? (ky)2 = kx. kz ? k2y = kx + z ? 2y = x + z ? x, y, z in A.P. If a1, a2, a3, ... an are in A.P., where ai > 0 for all i, show that =
Ans : 4 13. The ratio of the A.M. and G.M. of two positive numbers a and b is m : n, show that:
Ans : 4 14. Given A.M. : G.M. = m : n If p, q, r are in G.P. and the equations px2 + 2qx + r = 0 and dx2 + 2ex + f = 0 have a common root then show that are in A.P.
Ans : 6 15. Ans : 4 16. Given p, q, r are in G.P. ? q2 = pr ¡­(i) Equations px2 + 2qx + r = 0 and dx2 + 2ex + f = 0 have common root ¡­(ii) Substituting for q, from (i) in px2 + 2qx + r = 0 Find the sum of, + .......................... up to n terms. If S1, S2, S3, ......, Sm are the sums of n terms of m A.P.'s whose first terms are 1, 2, 3, ......, m and common differences are 1, 3, 5, ......, 2m ¨C 1 respectively, show that S1 + S2 + S3 + ...... + Sm =
Ans : 6 17. Ans : mn (mn + 1) S1 = ¦²(1, 2, 3..............) = [2.1 + (n ¨C 1).1] S2 = ¦²(2, 5, 8..............) = [2.2 + (n ¨C 1).3] S3 = ¦²(3, 8, 13............) = [2.3 + (n ¨C 1).5] Sm = [2.m + (n ¨C 1) (2m ¨C 1)] S1 + S2 + ...... + Sm = [2 (1 + 2 + 3.......m) + (n ¨C 1) (1 + 3 + 5........+ 2m ¨C 1)] = = If A and G be A.M. and G.M., respectively between two positive numbers, prove that the numbers are A ¡À . Let the numbers be a and b. = A ? a + b = 2A ...(i) = G ? ab = G2 (a ¨C b)2 = (a + b)2 ¨C 4ab = 4A2 ¨C 4G2 ? 4(A2 ¨C G2) a ¨C b = ...(ii) From (i) and (ii), we get a = A + and b = A ¨C Hence, the numbers are A ¡À

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MATHEMATICS XI TEST PAPER SOL.pdf

  • 1. Roll No. : Date : Time - MM - 70 2 Ans : 4 2. Ans : 4 3. Ans : 4 4. Ans : Find the number of terms in the expansion of (a + 2b ¨C 3c)n. Consider (a + 2b ¨C 3c)n = {a + (2b ¨C 3c)}n = nC0 an + nC1 an¨C1(2b ¨C 3c) + nC2 an¨C2(2b ¨C 3c)2 + ... + nCn(2b ¨C 3c)n. We notice first term gives one term, second term gives two terms, third term gives three terms and so on. ¡à Total terms 1 + 2 + ... + n + (n + 1) = Find the middle term(s) in the expansion of : (1 + x)2n Since, 2n would always be even so (n + 1)th term would be the middle term ¡à The coefficients of 2nd, 3rd and 4th terms in the expansion of (1 + x)2n are in A.P., prove that 2n2 ¨C 9n + 7 = 0 2nC1, 2nC2, 2nC3 are in A.P. ? 2. 2nC2 = 2nC1 + 2nC3 ? ? 2n2 ¨C 9n + 7 = 0 If Tr is rth term in the expansion of (1 + x)n in the ascending powers of x, prove that r (r + 1) Tr+2 = (n ¨C r + 1) (n ¨C r)x2 Tr. LHS = r (r + 1) Tr+2 = r (r + 1). nCr+1 xr+1 = r (r + 1) = RHS = (n ¨C r + 1) (n ¨C r) x2 Tr = (n ¨C r + 1) (n ¨C r)x2. nCr¨C1 xr¨C1 = (n ¨C r + 1) (n ¨C r). x2 . = LHS = RHS
  • 2. 4 5. Ans : 4 6. If the 3rd, 4th, 5th and 6th terms in the expansion of (x + y)n be a, b, c and d respectively, prove that If a, b, c be the three consecutive coefficients in the expansion of (1 + x)n, prove that n =
  • 3. Ans : 4 7. Ans : 4 8. Using binomial theorem, prove that 6n ¨C 5n always leaves remainder 1 when divided by 25. For two numbers a and b if we can find numbers q and r such that a = bq + r, then we say that b divides a with q as quotient and r as remainder. Thus, in order to show that 6n ¨C 5n leaves remainder 1 when divided by 25, we prove that 6n ¨C 5n = 25k + 1, where k is some natural number. We have (1 + a)n = nC0 + nC1a + nC2a2 + ... nCnan For a = 5, we get (1 + 5)n = nC0 + nC15 + nC252 + ... nCn5n i.e. (6)n = 1 + 5n + 52 nC2 + 53 nC3 + ... + 5n i.e. 6n ¨C 5n = 1 + 52 (nC2 + nC3 5 + ... + 5n¨C2) or 6n ¨C 5n = 1 + 25 (nC2 + 5. nC3 + ... + 5n¨C2) or 6n ¨C 5n = 25k + 1 where k = nC2 + 5. nC3 + ... + 5n¨C2 This shows that when divided by 25, 6n ¨C 5n leaves remainder 1. If a1, a2, a3 and a4 are the coefficient of any four consecutive terms in the expansion of (1 + x)n, prove that
  • 4. Ans : 4 9. Let a1, a2, a3 and a4 be the coefficient of four consecutive terms Tr+1, Tr+2, Tr+3 and Tr+4 respectively. Then a1 = coefficient of Tr+1 = nCr a2 = coefficient of Tr+2 = nCr+1 a3 = coefficient of Tr+3 = nCr+2 and a4 = coefficient of Tr+4 = nCr+3 Find the sum (33 ¨C 23) + (53 ¨C 43) + (73 ¨C 63) + ... 10 terms.
  • 5. Ans : 4 10. Ans : 4 11. Ans : 4 12. Sum = (33 + 53 + 73 + ...) ¨C (23 + 43 + 63 + ...) For 33 + 53 + 73 + ... an = (2n + 1)3 = 8n3 + 12n2 + 6n + 1 ¡à S1 = 8 ¦²n3 + 12 ¦²n2 + 6 ¦²n + n ... (i) For 23 + 43 + 63 + ... an = (2n)3 = 8n3 S2 = 8¦²n3 ... (ii) ¡à Sum = S1 ¨C S2 = 12 ¦²n2 + 6 ¦²n + n [From (i), (ii)] Sum to 10 terms = = 4620 + 330 + 10 = 4960 If a, b, c be the first, third and nth term respectively of an A.P. Prove that the sum to n terms is Given first term = a, a3 = b and an = c Sn = [2a + (n ¨C 1)d] ... (i) We have first term = a b = a + 2d and a + (n ¨C 1) d = c Sn = [a + c] [From (i)] ... (ii) Also a + (n ¨C 1) = c ? n ¨C 1 = Substituting in (ii), we get Sn = If a, b, c are in G.P and . Prove that x, y, z are in A.P. ? a = kx, b = ky, c = kz Given b2 = ac ? (ky)2 = kx. kz ? k2y = kx + z ? 2y = x + z ? x, y, z in A.P. If a1, a2, a3, ... an are in A.P., where ai > 0 for all i, show that =
  • 6. Ans : 4 13. The ratio of the A.M. and G.M. of two positive numbers a and b is m : n, show that:
  • 7. Ans : 4 14. Given A.M. : G.M. = m : n If p, q, r are in G.P. and the equations px2 + 2qx + r = 0 and dx2 + 2ex + f = 0 have a common root then show that are in A.P.
  • 8. Ans : 6 15. Ans : 4 16. Given p, q, r are in G.P. ? q2 = pr ¡­(i) Equations px2 + 2qx + r = 0 and dx2 + 2ex + f = 0 have common root ¡­(ii) Substituting for q, from (i) in px2 + 2qx + r = 0 Find the sum of, + .......................... up to n terms. If S1, S2, S3, ......, Sm are the sums of n terms of m A.P.'s whose first terms are 1, 2, 3, ......, m and common differences are 1, 3, 5, ......, 2m ¨C 1 respectively, show that S1 + S2 + S3 + ...... + Sm =
  • 9. Ans : 6 17. Ans : mn (mn + 1) S1 = ¦²(1, 2, 3..............) = [2.1 + (n ¨C 1).1] S2 = ¦²(2, 5, 8..............) = [2.2 + (n ¨C 1).3] S3 = ¦²(3, 8, 13............) = [2.3 + (n ¨C 1).5] Sm = [2.m + (n ¨C 1) (2m ¨C 1)] S1 + S2 + ...... + Sm = [2 (1 + 2 + 3.......m) + (n ¨C 1) (1 + 3 + 5........+ 2m ¨C 1)] = = If A and G be A.M. and G.M., respectively between two positive numbers, prove that the numbers are A ¡À . Let the numbers be a and b. = A ? a + b = 2A ...(i) = G ? ab = G2 (a ¨C b)2 = (a + b)2 ¨C 4ab = 4A2 ¨C 4G2 ? 4(A2 ¨C G2) a ¨C b = ...(ii) From (i) and (ii), we get a = A + and b = A ¨C Hence, the numbers are A ¡À