JEE 2023 — Mathematics PYQ
JEE | Mathematics | 2023limt→0(1sin2t1+2sin2t1+⋯+nsin2t1)sin2t is equal to:
Choose the correct answer:
- A.
n2
- B.
2n(n+1)
- C.
n
(Correct Answer) - D.
n2+n
n
Explanation
Solution
Step 1: Let k=sin2t
As t→0, k→0.
Step 2: Limit ko simplify karein
L=k→0lim(1−k+2−k+3−k+⋯+n−k)k
Step 3: Logarithm lene par
lnL=k→0limkln(r=1∑nr−k)
Step 4: Limit put karein (k=0)
lnL=0×ln(10+20+⋯+n0)
lnL=0×ln(n times1+1+⋯+1)
lnL=0×ln(n)=0
Step 5: Final Result
L=e0=1
Explanation
Solution
Step 1: Let k=sin2t
As t→0, k→0.
Step 2: Limit ko simplify karein
L=k→0lim(1−k+2−k+3−k+⋯+n−k)k
Step 3: Logarithm lene par
lnL=k→0limkln(r=1∑nr−k)
Step 4: Limit put karein (k=0)
lnL=0×ln(10+20+⋯+n0)
lnL=0×ln(n times1+1+⋯+1)
lnL=0×ln(n)=0
Step 5: Final Result
L=e0=1