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JEE Main · 2023 · 13 Apr 2023 · Shift 1 · Definite Integrals

Let for

x \in R

,

S_{0} (x) = x

,

S_{k} (x) = C_{k} x + k \int_{0}^{x} S_{k - 1} (t) d t

, where

C_{0} = 1

,

C_{k} = 1 - \int_{0}^{1} S_{k - 1} (x) d x

,

k = 1, 2, 3 \dots

. Then

S_{2} (3) + 6 C_{3}

is equal to ______.

Answer: 18

Solution

With

S_{0} (x) = x

,

C_{1} = 1 - \int_{0}^{1} x d x = \frac{1}{2}

, so

S_{1} (x) = \frac{x}{2} + \int_{0}^{x} t d t = \frac{x}{2} + \frac{x ^{2}}{2}

. Then

C_{2} = 1 - \int_{0}^{1} (\frac{x}{2} + \frac{x ^{2}}{2}) d x = 1 - \frac{5}{12} = \frac{7}{12}

, and

S_{2} (x) = \frac{7 x}{12} + 2 \int_{0}^{x} (\frac{t}{2} + \frac{t ^{2}}{2}) d t = \frac{7 x}{12} + \frac{x ^{2}}{2} + \frac{x ^{3}}{3}

. Next

C_{3} = 1 - \int_{0}^{1} S_{2} (x) d x = 1 - (\frac{7}{24} + \frac{1}{6} + \frac{1}{12}) = 1 - \frac{13}{24} = \frac{11}{24}

. Therefore

S_{2} (3) = \frac{7}{4} + \frac{9}{2} + 9 = \frac{61}{4}

and

6 C_{3} = \frac{11}{4}

, so

S_{2} (3) + 6 C_{3} = \frac{61}{4} + \frac{11}{4} = 18

.

JEE Main · 2023 · 13 Apr 2023 · Shift 1 · Integer answer

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