Verified Solution • 2025–26 Session moderate

QJEEM23Ma8411: JEE Main 2023 Mathematics - Class 12 JEEM23Ma8411 Solution: Let for , where Then is equal to ____________.

Let for x∈R,S0(x)=x,Sk(x)=Ckx+k∫0xSk−1(t)dtx \in \mathbb{R}, S_{0}(x)=x, S_{k}(x)=C_{k} x+k \int_{0}^{x} S_{k-1}(t) d t , where

C0=1,Ck=1−∫01Sk−1(x)dx,k=1,2,3,…C_{0}=1, C_{k}=1-\int_{0}^{1} S_{k-1}(x) d x, k=1,2,3, \ldots Then S2(3)+6C3S_{2}(3)+6 C_{3} is equal to ____________.

Step-by-Step Verified Solution

Given,

Sk(x)=Ckx+k∫0xSk−1(t)dtS_k(x)=C_k x+k \int_0^x S_{k-1}(t) d t

Put k=2\mathrm{k}=2 and x=3\mathrm{x}=3

S2(3)=C2(3)+2∫03 S1(t)dt\mathrm{S}_2(3)=\mathrm{C}_2(3)+2 \int_0^3 \mathrm{~S}_1(\mathrm{t}) \mathrm{dt} .........(i)

Also,

amp;S1(x)=C1(x)+∫0xS0(t)dtamp;=C1x+x22amp;S2(3)=3C2+2∫03(C1t+t22)dtamp;=3C2+9C1+9\begin{aligned}& 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\end{aligned}

Also,

amp;C1=1−∫01 S0(x)dx=12amp;C2=1−∫01 S1(x)dx=0amp;C3=1−∫01 S2(x)dxamp;=1−∫01(C2x+C1x2+x33)dxamp;=34\begin{aligned}& \mathrm{C}_1=1-\int_0^1 \mathrm{~S}_0(\mathrm{x}) \mathrm{dx}=\frac{1}{2} \\\\& \mathrm{C}_2=1-\int_0^1 \mathrm{~S}_1(\mathrm{x}) \mathrm{dx}=0 \\\\& \mathrm{C}_3=1-\int_0^1 \mathrm{~S}_2(\mathrm{x}) \mathrm{dx} \\\\& =1-\int_0^1\left(\mathrm{C}_2 \mathrm{x}+\mathrm{C}_1 \mathrm{x}^2+\frac{\mathrm{x}^3}{3}\right) \mathrm{dx} \\\\& =\frac{3}{4}\end{aligned}

amp;S2(x)=C2x+2∫0xS1(t)dtamp;=C2x+C1x2+x33amp;=S2(3)+6C3=6C3+3C2+9C1+9amp;=18\begin{aligned}& S_2(x)=C_2 x+2 \int_0^x S_1(t) d t \\\\&=C_2 x+C_1 x^2+\frac{x^3}{3} \\\\& = S_2(3)+6 C_3=6 C_3+3 C_2+9 C_1+9 \\\\&=18\end{aligned}

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring