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Question

# If f(x)=∣∣ ∣ ∣∣cosxex22x cos2x2x2secxsinx+x312x+tanx∣∣ ∣ ∣∣, then the value of π/2∫−π/2(x2+1)[f(x)+f′′(x)] dx

A
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B
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C
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D
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Solution

## The correct option is D 0As, f(x)=∣∣ ∣ ∣∣cosxex22x cos2x2x2secxsinx+x312x+tanx∣∣ ∣ ∣∣⇒f(−x)=∣∣ ∣ ∣ ∣∣cosxex2−2x cos2x2x2secx−(sinx+x3)12−(x+tanx)∣∣ ∣ ∣ ∣∣⇒f(−x)=−f(x) So, f(x) is an odd function. Differentiating w.r.t x ⇒−f′(−x)=−f′(x)⇒f′(−x)=f′(x) Again Differentiating ⇒−f′′(−x)=f′′(x)⇒f′′(−x)=−f′′(x) So, f′′(x) is an odd function Thus, f(x)+f′′(x) is odd function Let, ϕ(x)=(x2+1)⋅[f(x)+f′′(x)]⇒ϕ(−x)=−ϕ(x) Therefore, ϕ(x) is an odd function. ∴π/2∫−π/2ϕ(x)dx=0

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