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BITSAT Mathematics Sequences and Series 2024 BITSAT 2024 (Memory Based Paper 1)

BITSAT Mathematics Question (2024) — Solution

Question

If y= ^ -1 1 x^2+x+1 + ^ -1 1 x^2+3 x+3 + ^ -1 1 x^2+5 x+7 + .to n terms, then d y d x =

Options

  1. A. 1 x^2+n^2 - 1 x^2+1
  2. B. 1 (x+n)^2+1 - 1 x^2+1
  3. C. 1 x^2+(n+1)^2 - 1 x^2+1
  4. D. None of these

Answer

B. 1 (x+n)^2+1 - 1 x^2+1

Step-by-step solution

Given, y= ^ -1 1 x^2+x+1 + ^ -1 1 x^2+3 x+3 + ^ -1 1 x^2+5 x+7 + to n terms = ^ -1 \ 1 1+x(x+1) \ + ^ -1 \ 1 1+(x+1)(x+2) \ + ^ -1 \ 1 1+(x+2)(x+3) \ + + ^ -1 \ 1 1+(x+(n-1))(x+n) \ = ^ -1 \ (x+1)-x 1+(x+1) x \ + ^ -1 \ (x+2)-(x+1) 1+(x+2)(x+1) \ + ^ -1 \ (x+3)-(x+2) 1+(x+3)(x+2) \ + .+ ^ -1 ( (x+n)-(x+n-1) 1+(x+n)(x+n-1) ) y= \ ^ -1 (x+1)- ^ -1 (x) \ + \ ^ -1 (x+2)- ^ -1 (x+1) \ + \ ^ -1 (x+3)- ^ -1 (x+2) \ + . .+ \ ^ -1 (x+n)- ^ -1 [x+(n-1)] \ So, y= ^ -1 (x+n)- ^ -1 (x) On differentiating both sides w.r.t. x, we get d y d x = 1 1+(x+n)^2 - 1 1+x^2

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