Let for x R, S0(x) = x, Sk(x) = Ck x + k 0^x Sk-1(t) dt where C0 = 1, Ck = 1 - 0^1 Sk-1(x) dx, k = 1, 2, 3,. Then S2(3) + 6C3 is equal to:

Let for x R, S0(x) = x, Sk(x) = Ck x + k 0^x Sk-1(t) dt where C0… — Solution | JEE Mathematics

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Let for $x \in \mathbb{R}, \, S_0(x) = x, \, S_k(x) = C_k x + k \int_0^x S_{k-1}(t) \, dt$ where $C_0 = 1, \, C_k = 1 - \int_0^1 S_{k-1}(x) \, dx, \, k = 1, 2, 3, \dots$ . Then $S_2(3) + 6C_3$ is equal to:

Answer

Correct answer:18

Step-by-step solution

the solution unavailable

Working could not be extracted because the solution is unavailable. Using the provided correct answer field, the numerical value is $18$ .

Common mistakes

Using the recursion for $S_k(x)$ without first computing $C_k$ . This is wrong because each $S_k(x)$ depends explicitly on the corresponding constant $C_k$ . Always evaluate $C_k = 1 - \int_0^1 S_{k-1}(x) \, dx$ before forming $S_k(x)$ .
Confusing the variable of integration inside $\int_0^x S_{k-1}(t) \, dt$ with the outer variable $x$ . This can lead to incorrect antiderivatives. Treat $t$ as the dummy variable and substitute the upper limit only after integration.
Substituting directly into $C_3$ without computing $S_2(x)$ on the full interval $[0,1]$ . This is wrong because $C_3$ depends on $\int_0^1 S_2(x) \, dx$ , not merely on the value $S_2(3)$ . Compute the entire function first, then integrate.

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