Sujeet Kumar

Question 1.State, by writing the first four terms, the expansion of the following, where |x| < 1.(i) (1 + x)-4Solution: (ii) (1 – x)1/3Solution: Solution: (iv) (1 + x)-1/5Solution: Solution: Question 2.State by writing first four terms, the expansion of the following, where |b| < |a|. (ii) (a + b)-4Solution: (iii) (a + b)1/4Solution: (v) … Read more

Question 1.In the following expansions, find the indicated term. Solution: Solution: Solution: Solution: Solution: Question 2.In the following expansions, find the indicated coefficients. Solution: Solution: Solution: Solution: Solution: Question 3.Find the constant term (term independent of x) in the expansion of Solution: Solution: Solution: Solution: Solution: Question 4.Find the middle terms in the expansion of … Read more

Question 1.Expand:(i) (√3 + √2)4Solution:Here, a = √3, b = √2 and n = 4.Using binomial theorem, ∴ (√3 + √2)4 = 1(9) (1) + 4(3√3) (√2) + 6(3)(2) + 4(√3) (2√2) + 1(1)(4)= 9 + 12√6 + 36 + 8√6 + 4= 49 + 20√6 (ii) (√5 – √2)5Solution:Here, a = √5, b = √2 … Read more

Prove by the method of induction, for all n ∈ N. Question 1.2 + 4 + 6 + …… + 2n = n(n + 1)Solution:Let P(n) = 2 + 4 + 6 + …… + 2n = n(n + 1), for all n ∈ N.Step I:Put n = 1L.H.S. = 2R.H.S. = 1(1 + 1) … Read more

(I) Select the correct answer from the given alternatives. Question 1.A college offers 5 courses in the morning and 3 in the evening. The number of ways a student can select exactly one course, either in the morning or in the evening is(A) 5(B) 3(C) 8(D) 15Answer:(C) 8Hint:Number of ways to select one course from … Read more

Question 1.Find the value of(a) 15C4Solution: (b) 80C2Solution: (c) 15C4 +15C5Solution: (d) 20C16 –19C16Solution: Question 2.Find n if Solution: Solution: Solution: Question 3. Solution:∴ 2r(2r – 1) (2r – 2) (2r – 3) = 14 × 12 × 10∴ 2r(2r – 1) (2r – 2) (2r – 3) = 8 × 7 × 6 × 5Comparing on both sides, we get∴ … Read more

Question 1.In how many different ways can 8 friends sit around a table?Solution:We know that ‘n’ persons can sit around a table in (n – 1)! ways.∴ 8 friends can sit around a table in 7!= 7 × 6 × 5 × 4 × 3 × 2 × 1= 5040 ways. Question 2.A party has … Read more

Question 1.Find the number of permutations of letters in each of the following words.(i) DIVYA(ii) SHANTARAM(iii) REPRESENT(iv) COMBINE(v) BAL BHARATISolution:(i) There are 5 distinct letters in the word DIVYA.∴ Number of permutations of the letters of the word DIVYA = 5! = 120 (ii) There are 9 letters in the word SHANTARAM in which ‘A’ … Read more

Question 1. ∴ (n – 3) (n – 4) (n – 5) = 120∴ (n – 3) (n – 4) (n – 5) = 6 × 5 × 4Comparing on both sides, we getn – 3 = 6∴ n = 9 Question 2. (m + n) (m + n – 1) = 56Let m + … Read more

Question 1.Evaluate:(i) 8!Solution:8!= 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1= 40320 (ii) 10!Solution:10!= 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1= 3628800 (iii) 10! – 6!Solution:10! – 6!= 10 × 9 × 8 × 7 × … Read more