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Ex 13.2, 5 - Chapter 13 Class 11 Limits and Derivatives (Term 1 and Term 2)
Last updated at March 30, 2023 by Teachoo
Ex 13.2
Ex 13.2, 1
Ex 13.2, 2
Ex 13.2, 3
Ex 13.2, 4 (i) Important
Ex 13.2, 4 (ii)
Ex 13.2, 4 (iii) Important
Ex 13.2, 4 (iv)
Ex 13.2, 5 You are here
Ex 13.2, 6
Ex 13.2, 7 (i) Important
Ex 13.2, 7 (ii)
Ex 13.2, 7 (iii) Important
Ex 13.2, 8
Ex 13.2, 9 (i)
Ex 13.2, 9 (ii) Important
Ex 13.2, 9 (iii)
Ex 13.2, 9 (iv) Important
Ex 13.2, 9 (v)
Ex 13.2, 9 (vi)
Ex 13.2, 10 Important
Ex 13.2, 11 (i)
Ex 13.2, 11 (ii) Important
Ex 13.2, 11 (iii) Important
Ex 13.2, 11 (iv)
Ex 13.2, 11 (v) Important
Ex 13.2, 11 (vi)
Ex 13.2, 11 (vii) Important
Transcript
Ex 13.2, 5 For the function f(x) = x100100 + x99100 +….+ x22 + x + 1. Prove that f’(1) = 100 f’(0) We have f (x) = 𝑥100100 + 𝑥9999 + …… + 𝑥22 + x + 1 f’ (x) = 1100 x100 + 199 x99 + …… + 12 x2 + x1 + 1′ f’ (x) = 1100 × 100x100 – 1 + 199 × 99x99 – 1 + … + 12 × 2x2 – 1 + 1.x1-1 + 0 = 100100 x99 + 9999 x98 + …+ 22 x1 + x0 + 0 = x99 + x98 + …..+ x + 1 + 0 = x99 + x98 + … + x + 1 Hence , f’ (x) = x99 + x98 + … + x + 1 We need to prove f’(1) = 100 f’(0) Hence R.H.S = L.H.S Hence proved
