JEE 2023 — Mathematics PYQ
JEE | Mathematics | 2023Let for x∈R, S0(x)=x, Sk(x)=Ckx+k∫0πSk−1(t)dt where C0=1, Ck=1−∫01Sk−1(x)dx,k=1,2,3,…. Then, S2(3)+6C3 is equal to ________.
Choose the correct answer:
- A.
18
(Correct Answer) - B.
7
- C.
8
- D.
9
18
Explanation
Given that:
Sk(x)=Ckx+k∫0xSk−1(t)dt
Putting k=2 and x=3:
Putting k=1:
-
C1=1−∫01S0(x)dx=1−[2x2]01=21
-
Diye gaye calculation ke mutabiq C2=1−∫01S1(x)dx=0
Step 2: S2(3) ki value
Image ke formula Sk(x)=Ckx+k∫0xSk−1(t)dt ke hisaab se:
Step 3: C3 nikalna
Step 4: Final Value
Value =S2(3)+6C3
Value =(3C2+9C1+9)+6C3
Value =0+29+9+6(43)=29+9+29=9+9=18
Answer: 18
Explanation
Given that:
Sk(x)=Ckx+k∫0xSk−1(t)dt
Putting k=2 and x=3:
Putting k=1:
-
C1=1−∫01S0(x)dx=1−[2x2]01=21
-
Diye gaye calculation ke mutabiq C2=1−∫01S1(x)dx=0
Step 2: S2(3) ki value
Image ke formula Sk(x)=Ckx+k∫0xSk−1(t)dt ke hisaab se:
Step 3: C3 nikalna
Step 4: Final Value
Value =S2(3)+6C3
Value =(3C2+9C1+9)+6C3
Value =0+29+9+6(43)=29+9+29=9+9=18
Answer: 18

