JEE Main 2024 · 5 April · morning · Single correct · +4 / −1
If 1/(√1+√2)+1/(√2+√3)+…+1/(√99+√100)=m and 1/(1 · 2)+1/(2 · 3)+…+1/(99 · 100)=n, then the point (m, n) lies on the line
- (a) 11(x-1)-100 y=0
- (b) 11 x-100 y=0
- (c) 11(x-1)-100(y-2)=0
- (d) 11(x-2)-100(y-1)=0
Answer and solution
Answer (b) 11 x-100 y=0
Solution
Maths sir · 4 steps
Rationalise the first sum, split the second into partial fractions, and both telescope.
- Rationalise each term. Multiply the top and bottom of 1/(√k+√(k+1)) by √(k+1)-√k. The denominator becomes (k+1)-k = 1, so each term is just √(k+1)-√k.
m = (√2-√1) + (√3-√2) + … + (√100-√99) = √100-√1 = 9
- Split the second sum. Write 1/k(k+1) = 1/k-1/(k+1). Adding from k=1 to 99, the middle terms cancel and n = 1-1/100 = 99/100.
- Test the point. Put (9, 99/100) into option (b): 11 × 9 - 100 × 99/100 = 99 - 99 = 0. In (a) the left side is 88-99=-11, in (c) it is 189 and in (d) it is 78, so none of these is zero.
- So. The point (m, n) = (9, 99/100) lies on 11x-100y=0, which is option (b).
The idea to keep. When a term has √k+√(k+1) in the denominator, rationalise it. When it has k(k+1), use partial fractions. Then only the first and last pieces survive.
Answer from the NTA final answer key. Solution by PrepWiser.