JEE Main202429 Jan 2024Morning ShiftMathematicsLimitsActual
lim x → π 2 1 x - π 2 2 ∫ x 3 π 2 3 cos 1 t 3 d t is equal to
Options
- A3 π 8
- B3 π 2 4
- C3 π 2 8
- D3 π 4
Correct answer
C. 3 π 2 8
Step-by-step solution
Let, y = lim x → π 2 ∫ x 3 π 2 3 cos t 1 3 d t x - π 2 2 ⇒ y = lim h → 0 ∫ π 2 - h 3 π 2 3 cos t 1 3 d t π 2 - h - π 2 2 ⇒ y = lim h → 0 ∫ π 2 - h 3 π 2 3 cos t 1 3 d t h 2 Applying L-Hospital's rule and Newton Leibnitz Theorem we get, ⇒ y = lim h → 0 0 + cos π 2 - h × 3 × π 2 - h 2 2 h ⇒ y = lim h → 0 sin h × 3 × π 2 - h 2 2 h We know that, lim h → 0 sin h h = 1 ⇒ y = lim h → 0 3 × π 2 - h 2 2 ⇒ y = 3 π 2 8