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Question: Let $F(x) = \int_{\sin x}^{\cos x} e^{(1+\arcsin t)^2} dt$ on $[0, \frac{\pi}{2}]$ then...

Let F(x)=∫sin⁡xcos⁡xe(1+arcsin⁡t)2dtF(x) = \int_{\sin x}^{\cos x} e^{(1+\arcsin t)^2} dt on [0,π2][0, \frac{\pi}{2}] then

A

F''(c) = 0 for all c in (0, \frac{\pi}{2})

B

F''(c) = 0 for some c in (0, \frac{\pi}{2})

C

F'(c) != 0 for all c in (0, \frac{\pi}{2})

D

F'(c) = 0 for some c in (0, \frac{\pi}{2})

Answer

(B), (C)

Explanation

Solution

Let f(t)=e(1+arcsin⁡t)2f(t) = e^{(1+\arcsin t)^2}. Then F(x)=∫sin⁡xcos⁡xf(t)dtF(x) = \int_{\sin x}^{\cos x} f(t) dt. By the Fundamental Theorem of Calculus and the chain rule, F′(x)=f(cos⁡x)⋅(−sin⁡x)−f(sin⁡x)⋅(cos⁡x)F'(x) = f(\cos x) \cdot (-\sin x) - f(\sin x) \cdot (\cos x). For x∈[0,π2]x \in [0, \frac{\pi}{2}], we have arcsin⁡(cos⁡x)=π2−x\arcsin(\cos x) = \frac{\pi}{2} - x and arcsin⁡(sin⁡x)=x\arcsin(\sin x) = x. So, F′(x)=−e(1+π2−x)2sin⁡x−e(1+x)2cos⁡xF'(x) = -e^{(1+\frac{\pi}{2}-x)^2} \sin x - e^{(1+x)^2} \cos x. For x∈(0,π2)x \in (0, \frac{\pi}{2}), sin⁡x>0\sin x > 0, cos⁡x>0\cos x > 0, and the exponential terms are positive. Thus, −e(1+π2−x)2sin⁡x<0-e^{(1+\frac{\pi}{2}-x)^2} \sin x < 0 and −e(1+x)2cos⁡x<0-e^{(1+x)^2} \cos x < 0. Therefore, F′(x)<0F'(x) < 0 for all x∈(0,π2)x \in (0, \frac{\pi}{2}). This implies F′(c)≠0F'(c) \neq 0 for some c∈(0,π2)c \in (0, \frac{\pi}{2}), making option (C) true and option (D) false.

To analyze F′′(x)F''(x), we differentiate F′(x)F'(x): F′′(x)=ddx(−e(1+π2−x)2sin⁡x)−ddx(e(1+x)2cos⁡x)F''(x) = \frac{d}{dx}(-e^{(1+\frac{\pi}{2}-x)^2} \sin x) - \frac{d}{dx}(e^{(1+x)^2} \cos x) Using the product rule and chain rule: ddx(e(1+π2−x)2sin⁡x)=e(1+π2−x)2⋅2(1+π2−x)(−1)sin⁡x+e(1+π2−x)2cos⁡x\frac{d}{dx}(e^{(1+\frac{\pi}{2}-x)^2} \sin x) = e^{(1+\frac{\pi}{2}-x)^2} \cdot 2(1+\frac{\pi}{2}-x)(-1) \sin x + e^{(1+\frac{\pi}{2}-x)^2} \cos x =−2(1+π2−x)e(1+π2−x)2sin⁡x+e(1+π2−x)2cos⁡x= -2(1+\frac{\pi}{2}-x)e^{(1+\frac{\pi}{2}-x)^2} \sin x + e^{(1+\frac{\pi}{2}-x)^2} \cos x

ddx(e(1+x)2cos⁡x)=e(1+x)2⋅2(1+x)cos⁡x+e(1+x)2(−sin⁡x)\frac{d}{dx}(e^{(1+x)^2} \cos x) = e^{(1+x)^2} \cdot 2(1+x) \cos x + e^{(1+x)^2} (-\sin x) =2(1+x)e(1+x)2cos⁡x−e(1+x)2sin⁡x= 2(1+x)e^{(1+x)^2} \cos x - e^{(1+x)^2} \sin x

F′′(x)=−[−2(1+π2−x)e(1+π2−x)2sin⁡x+e(1+π2−x)2cos⁡x]−[2(1+x)e(1+x)2cos⁡x−e(1+x)2sin⁡x]F''(x) = -[-2(1+\frac{\pi}{2}-x)e^{(1+\frac{\pi}{2}-x)^2} \sin x + e^{(1+\frac{\pi}{2}-x)^2} \cos x] - [2(1+x)e^{(1+x)^2} \cos x - e^{(1+x)^2} \sin x] F′′(x)=2(1+π2−x)e(1+π2−x)2sin⁡x−e(1+π2−x)2cos⁡x−2(1+x)e(1+x)2cos⁡x+e(1+x)2sin⁡xF''(x) = 2(1+\frac{\pi}{2}-x)e^{(1+\frac{\pi}{2}-x)^2} \sin x - e^{(1+\frac{\pi}{2}-x)^2} \cos x - 2(1+x)e^{(1+x)^2} \cos x + e^{(1+x)^2} \sin x

Consider the limits as x→0+x \to 0^+ and x→π2−x \to \frac{\pi}{2}^-. As x→0+x \to 0^+: F′′(x)≈2(1+π2)e(1+π2)2⋅x−e(1+π2)2⋅1−2(1)e12⋅1+e12⋅0F''(x) \approx 2(1+\frac{\pi}{2})e^{(1+\frac{\pi}{2})^2} \cdot x - e^{(1+\frac{\pi}{2})^2} \cdot 1 - 2(1)e^{1^2} \cdot 1 + e^{1^2} \cdot 0 F′′(x)≈0−e(1+π2)2−2e=−e(1+π2)2−2eF''(x) \approx 0 - e^{(1+\frac{\pi}{2})^2} - 2e = -e^{(1+\frac{\pi}{2})^2} - 2e. This is negative.

As x→π2−x \to \frac{\pi}{2}^- (let x=π2−ϵx = \frac{\pi}{2} - \epsilon for small ϵ>0\epsilon > 0): 1+π2−x=1+ϵ1+\frac{\pi}{2}-x = 1+\epsilon sin⁡x=sin⁡(π2−ϵ)=cos⁡ϵ≈1\sin x = \sin(\frac{\pi}{2}-\epsilon) = \cos \epsilon \approx 1 cos⁡x=cos⁡(π2−ϵ)=sin⁡ϵ≈ϵ\cos x = \cos(\frac{\pi}{2}-\epsilon) = \sin \epsilon \approx \epsilon 1+x=1+π2−ϵ1+x = 1+\frac{\pi}{2}-\epsilon

F′′(x)≈2(1+ϵ)e(1+ϵ)2⋅1−e(1+ϵ)2⋅ϵ−2(1+π2−ϵ)e(1+π2−ϵ)2⋅ϵ+e(1+π2−ϵ)2⋅1F''(x) \approx 2(1+\epsilon)e^{(1+\epsilon)^2} \cdot 1 - e^{(1+\epsilon)^2} \cdot \epsilon - 2(1+\frac{\pi}{2}-\epsilon)e^{(1+\frac{\pi}{2}-\epsilon)^2} \cdot \epsilon + e^{(1+\frac{\pi}{2}-\epsilon)^2} \cdot 1 As ϵ→0+\epsilon \to 0^+, F′′(x)≈2(1)e12⋅1−e12⋅0−2(1+π2)e(1+π2)2⋅0+e(1+π2)2⋅1F''(x) \approx 2(1)e^{1^2} \cdot 1 - e^{1^2} \cdot 0 - 2(1+\frac{\pi}{2})e^{(1+\frac{\pi}{2})^2} \cdot 0 + e^{(1+\frac{\pi}{2})^2} \cdot 1 F′′(x)≈2e+e(1+π2)2F''(x) \approx 2e + e^{(1+\frac{\pi}{2})^2}. This is positive.

Since F′′(x)F''(x) is continuous on (0,π2)(0, \frac{\pi}{2}) and changes sign from negative to positive, by the Intermediate Value Theorem, there exists some c∈(0,π2)c \in (0, \frac{\pi}{2}) such that F′′(c)=0F''(c) = 0. Thus, option (B) is true. Option (A) is false because F′′(x)F''(x) is not identically zero. The correct options are (B) and (C).