NIMCET 2025 — Mathematics PYQ
NIMCET | Mathematics | 2025Find the value of dxd(∫sinx22sinxet2dt) at x=π
Choose the correct answer:
- A.
-1
- B.
1
- C.
-2
(Correct Answer) - D.
2
-2
Explanation
Formula (Leibniz Rule):
dxd(∫LowerUpperf(t)dt)=f(Upper)⋅dxd(Upper)−f(Lower)⋅dxd(Lower)
Step-by-Step Calculation:
Step 1: Upper limit ka derivative nikalo
Upper=2sinx
dxd(2sinx)=2cosx
Step 2: Lower limit ka derivative nikalo
Lower=sinx2
dxd(sinx2)=cos(x2)⋅2x=2xcosx2
Step 3: Formula me values put karo
dxdI(x)=[e(2sinx)2⋅2cosx]−[e(sinx2)2⋅2xcosx2]
Step 4: x=π put karo
Hum jaante hain ki:
sinπ=0
cosπ=−1
Ab bas value rakh do:
Value=[e(2⋅0)2⋅2(−1)]−[e(sinπ2)2⋅2πcos(π2)]
Value=[e0⋅(−2)]−[Second Term]
Value=1⋅(−2)−[Second Term]
Value=−2−2πcos(π2)esin2(π2)
Explanation
Formula (Leibniz Rule):
dxd(∫LowerUpperf(t)dt)=f(Upper)⋅dxd(Upper)−f(Lower)⋅dxd(Lower)
Step-by-Step Calculation:
Step 1: Upper limit ka derivative nikalo
Upper=2sinx
dxd(2sinx)=2cosx
Step 2: Lower limit ka derivative nikalo
Lower=sinx2
dxd(sinx2)=cos(x2)⋅2x=2xcosx2
Step 3: Formula me values put karo
dxdI(x)=[e(2sinx)2⋅2cosx]−[e(sinx2)2⋅2xcosx2]
Step 4: x=π put karo
Hum jaante hain ki:
sinπ=0
cosπ=−1
Ab bas value rakh do:
Value=[e(2⋅0)2⋅2(−1)]−[e(sinπ2)2⋅2πcos(π2)]
Value=[e0⋅(−2)]−[Second Term]
Value=1⋅(−2)−[Second Term]
Value=−2−2πcos(π2)esin2(π2)

