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JEE Mains · Maths · STD 12 - 7.2 definite integral

माना \(\mathrm{x} \in \mathbb{R}\) के लिए \(\mathrm{S}_0(\mathrm{x})=\mathrm{x}\), \(\mathrm{S}_{\mathrm{k}}(\mathrm{x})=\mathrm{C}_{\mathrm{k}} \mathrm{x}+\mathrm{k} \int_0^{\mathrm{x}} \mathrm{S}_{\mathrm{k}-1}(\mathrm{t}) \mathrm{dt}\), हैं, जहाँ \(\mathrm{C}_0=1, \mathrm{C}_{\mathrm{k}}=1-\int_0^{\mathrm{l}} \mathrm{S}_{\mathrm{k}-1}(\mathrm{x}) \mathrm{dx}, \mathrm{k}=1,2,3 \ldots\) हैं।  तो \(\mathrm{S}_2(3)+6 \mathrm{C}_3\) बराबर है

  1. A \(17\)
  2. B \(16\)
  3. C \(18\)
  4. D \(11\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(18\)

Step-by-step Solution

Detailed explanation

\(S _{ k }( x )=C_k x + k \int_0^x S_{k-1}(t) d t\) Put \(k=2\) and \(x=3\) \(S _2(3)= C _2(3)+2 \int_0^3 S _1( t ) dt\) Also, \(S_1(x)=C_1(x)+\int_0^x S_0(t) d t\) \(=C_1 x+\frac{x^2}{2}\) \(S_2(3)=3 C_2+2 \int_0^3\left(C_1 t+\frac{t^2}{2}\right) d t\) \(=3 C_2+9 C_1+9\) Also,…
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