Let for x \in \mathbbR, 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, \ldots Then S_2(3)+6 C_3 is equal to \underline\hspace2cm .

Definite Integration PYQ — JEE Main 2023 · 13 Apr · Shift 1 · Q42 with Solution

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Correct answer: 18

We are given the recursive definitions

$S_0(x)=x,$ $S_k(x)=C_kx+k\int_0^x S_{k-1}(t)\,dt,$ with

\qquad C_k=1-\int_0^1 S_{k-1}(x)\,dx \quad (k\ge 1).$$ We need to find: $$S_2(3)+6C_3.$$ --- ## 1. Find $C_1$ Since $S_0(x)=x$, $$C_1=1-\int_0^1 x\,dx=1-\left[\frac{x^2}{2}\right]_0^1=1-\frac12=\frac12.$$ --- ## 2. Find $S_1(x)$ Using $$S_1(x)=C_1x+\int_0^x S_0(t)\,dt,$$ we get $$S_1(x)=\frac12 x+\int_0^x t\,dt =\frac12 x+\left[\frac{t^2}{2}\right]_0^x =\frac{x}{2}+\frac{x^2}{2}.$$ So, $$S_1(x)=\frac{x+x^2}{2}.$$ --- ## 3. Find $C_2$ $$C_2=1-\int_0^1 S_1(x)\,dx =1-\int_0^1 \frac{x+x^2}{2}\,dx.$$ Now, $$\int_0^1 \frac{x+x^2}{2}\,dx =\frac12\left(\int_0^1 x\,dx+\int_0^1 x^2\,dx\right) =\frac12\left(\frac12+\frac13\right) =\frac12\cdot \frac56 =\frac{5}{12}.$$ Hence, $$C_2=1-\frac{5}{12}=\frac{7}{12}.$$ --- ## 4. Find $S_2(x)$ Using $$S_2(x)=C_2x+2\int_0^x S_1(t)\,dt,$$ we get $$S_2(x)=\frac{7}{12}x+2\int_0^x \frac{t+t^2}{2}\,dt =\frac{7}{12}x+\int_0^x (t+t^2)\,dt.$$ Now, $$\int_0^x (t+t^2)\,dt=\left[\frac{t^2}{2}+\frac{t^3}{3}\right]_0^x =\frac{x^2}{2}+\frac{x^3}{3}.$$ Therefore, $$S_2(x)=\frac{7}{12}x+\frac{x^2}{2}+\frac{x^3}{3}.$$ --- ## 5. Compute $S_2(3)$ Substitute $x=3$: $$S_2(3)=\frac{7}{12}(3)+\frac{3^2}{2}+\frac{3^3}{3}.$$ That is, $$S_2(3)=\frac{21}{12}+\frac92+9 =\frac74+\frac92+9.$$ Taking denominator $4$: $$S_2(3)=\frac74+\frac{18}{4}+\frac{36}{4}=\frac{61}{4}.$$ --- ## 6. Find $C_3$ $$C_3=1-\int_0^1 S_2(x)\,dx =1-\int_0^1 \left(\frac{7}{12}x+\frac{x^2}{2}+\frac{x^3}{3}\right)dx.$$ Compute each part: $$\int_0^1 \frac{7}{12}x\,dx=\frac{7}{12}\cdot \frac12=\frac{7}{24},$$ $$\int_0^1 \frac{x^2}{2}\,dx=\frac12\cdot \frac13=\frac16,$$ $$\int_0^1 \frac{x^3}{3}\,dx=\frac13\cdot \frac14=\frac{1}{12}.$$ So, $$\int_0^1 S_2(x)\,dx=\frac{7}{24}+\frac16+\frac{1}{12}.$$ With denominator $24$: $$\frac{7}{24}+\frac{4}{24}+\frac{2}{24}=\frac{13}{24}.$$ Hence, $$C_3=1-\frac{13}{24}=\frac{11}{24}.$$ Therefore, $$6C_3=6\cdot \frac{11}{24}=\frac{11}{4}.$$ --- ## 7. Final value $$S_2(3)+6C_3=\frac{61}{4}+\frac{11}{4}=\frac{72}{4}=18.$$ So the required integer is $$\boxed{18}.$$

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View written solutionFree

Correct answer: 18

We are given the recursive definitions

$S_0(x)=x,$ $S_k(x)=C_kx+k\int_0^x S_{k-1}(t)\,dt,$ with

\qquad C_k=1-\int_0^1 S_{k-1}(x)\,dx \quad (k\ge 1).$$ We need to find: $$S_2(3)+6C_3.$$ --- ## 1. Find $C_1$ Since $S_0(x)=x$, $$C_1=1-\int_0^1 x\,dx=1-\left[\frac{x^2}{2}\right]_0^1=1-\frac12=\frac12.$$ --- ## 2. Find $S_1(x)$ Using $$S_1(x)=C_1x+\int_0^x S_0(t)\,dt,$$ we get $$S_1(x)=\frac12 x+\int_0^x t\,dt =\frac12 x+\left[\frac{t^2}{2}\right]_0^x =\frac{x}{2}+\frac{x^2}{2}.$$ So, $$S_1(x)=\frac{x+x^2}{2}.$$ --- ## 3. Find $C_2$ $$C_2=1-\int_0^1 S_1(x)\,dx =1-\int_0^1 \frac{x+x^2}{2}\,dx.$$ Now, $$\int_0^1 \frac{x+x^2}{2}\,dx =\frac12\left(\int_0^1 x\,dx+\int_0^1 x^2\,dx\right) =\frac12\left(\frac12+\frac13\right) =\frac12\cdot \frac56 =\frac{5}{12}.$$ Hence, $$C_2=1-\frac{5}{12}=\frac{7}{12}.$$ --- ## 4. Find $S_2(x)$ Using $$S_2(x)=C_2x+2\int_0^x S_1(t)\,dt,$$ we get $$S_2(x)=\frac{7}{12}x+2\int_0^x \frac{t+t^2}{2}\,dt =\frac{7}{12}x+\int_0^x (t+t^2)\,dt.$$ Now, $$\int_0^x (t+t^2)\,dt=\left[\frac{t^2}{2}+\frac{t^3}{3}\right]_0^x =\frac{x^2}{2}+\frac{x^3}{3}.$$ Therefore, $$S_2(x)=\frac{7}{12}x+\frac{x^2}{2}+\frac{x^3}{3}.$$ --- ## 5. Compute $S_2(3)$ Substitute $x=3$: $$S_2(3)=\frac{7}{12}(3)+\frac{3^2}{2}+\frac{3^3}{3}.$$ That is, $$S_2(3)=\frac{21}{12}+\frac92+9 =\frac74+\frac92+9.$$ Taking denominator $4$: $$S_2(3)=\frac74+\frac{18}{4}+\frac{36}{4}=\frac{61}{4}.$$ --- ## 6. Find $C_3$ $$C_3=1-\int_0^1 S_2(x)\,dx =1-\int_0^1 \left(\frac{7}{12}x+\frac{x^2}{2}+\frac{x^3}{3}\right)dx.$$ Compute each part: $$\int_0^1 \frac{7}{12}x\,dx=\frac{7}{12}\cdot \frac12=\frac{7}{24},$$ $$\int_0^1 \frac{x^2}{2}\,dx=\frac12\cdot \frac13=\frac16,$$ $$\int_0^1 \frac{x^3}{3}\,dx=\frac13\cdot \frac14=\frac{1}{12}.$$ So, $$\int_0^1 S_2(x)\,dx=\frac{7}{24}+\frac16+\frac{1}{12}.$$ With denominator $24$: $$\frac{7}{24}+\frac{4}{24}+\frac{2}{24}=\frac{13}{24}.$$ Hence, $$C_3=1-\frac{13}{24}=\frac{11}{24}.$$ Therefore, $$6C_3=6\cdot \frac{11}{24}=\frac{11}{4}.$$ --- ## 7. Final value $$S_2(3)+6C_3=\frac{61}{4}+\frac{11}{4}=\frac{72}{4}=18.$$ So the required integer is $$\boxed{18}.$$

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