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Application of Integration question

2015 · Shift 1 · Q31
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  5. /2015 · Shift 1 · Q31

Application of Integration question

2015 · Shift 1 · Q31

JEE AdvancedMathematicsApplication of IntegrationNumerical+4 / −1
Let F(x)=∫xx2+π62cos⁡2t(dt)F\left( x \right) = \int\limits_x^{{x^2} + {\pi \over 6}} {2{{\cos }^2}t\left( {dt} \right)}F(x)=x∫x2+6π​​2cos2t(dt) for all x∈Rx \in Rx∈R and f:[0,12]→[0,∞]f:\left[ {0,{1 \over 2}} \right] \to \left[ {0,\infty } \right]f:[0,21​]→[0,∞] be a continuous function. For a∈[0,12], a \in \left[ {0,{1 \over 2}} \right],\,a∈[0,21​], F′(a)+2F'(a)+2F′(a)+2 is the area of the region bounded by x=0,y=0,y=f(x)x=0, y=0, y=f(x)x=0,y=0,y=f(x) and x=a,x=a,x=a, then f(0)f(0)f(0) is
Numerical answer
View written solutionFree

Correct answer: 3

  1. Differentiate F(x)F(x)F(x) using Leibniz rule

Given

F(x)=∫xx2+π/62cos⁡2t dtF(x)=\int_x^{x^2+\pi/6} 2\cos^2 t\,dtF(x)=∫xx2+π/6​2cos2tdt

Let g(t)=2cos⁡2tg(t)=2\cos^2 tg(t)=2cos2t Then by Leibniz rule,

F′(x)=g(x2+π/6)⋅ddx(x2+π6)−g(x)⋅ddx(x)F'(x)=g(x^2+\pi/6)\cdot \frac{d}{dx}\left(x^2+\frac\pi6\right)-g(x)\cdot \frac{d}{dx}(x)F′(x)=g(x2+π/6)⋅dxd​(x2+6π​)−g(x)⋅dxd​(x)

So,

F′(x)=2cos⁡2(x2+π6)(2x)−2cos⁡2xF'(x)=2\cos^2\left(x^2+\frac\pi6\right)(2x)-2\cos^2 xF′(x)=2cos2(x2+6π​)(2x)−2cos2x

Hence,

F′(x)=4xcos⁡2(x2+π6)−2cos⁡2xF'(x)=4x\cos^2\left(x^2+\frac\pi6\right)-2\cos^2 xF′(x)=4xcos2(x2+6π​)−2cos2x
  1. Use the area condition

The area bounded by x=0x=0x=0, y=0y=0y=0, y=f(x)y=f(x)y=f(x) and x=ax=ax=a is

∫0af(x) dx\int_0^a f(x)\,dx∫0a​f(x)dx

Given that for every a∈[0,1/2]a\in[0,1/2]a∈[0,1/2],

F′(a)+2=∫0af(x) dxF'(a)+2=\int_0^a f(x)\,dxF′(a)+2=∫0a​f(x)dx

We need f(0)f(0)f(0).

  1. Differentiate both sides with respect to aaa

By Fundamental Theorem of Calculus,

dda(∫0af(x) dx)=f(a)\frac{d}{da}\left(\int_0^a f(x)\,dx\right)=f(a)dad​(∫0a​f(x)dx)=f(a)

Therefore,

f(a)=F′′(a)f(a)=F''(a)f(a)=F′′(a)

So,

f(0)=F′′(0)f(0)=F''(0)f(0)=F′′(0)
  1. Compute F′′(0)F''(0)F′′(0)

We have

F′(x)=4xcos⁡2(x2+π6)−2cos⁡2xF'(x)=4x\cos^2\left(x^2+\frac\pi6\right)-2\cos^2 xF′(x)=4xcos2(x2+6π​)−2cos2x

Differentiate term by term.

For the first term,

ddx[4xcos⁡2(x2+π6)]=4cos⁡2(x2+π6)+4x⋅ddx[cos⁡2(x2+π6)]\frac{d}{dx}\left[4x\cos^2\left(x^2+\frac\pi6\right)\right] =4\cos^2\left(x^2+\frac\pi6\right)+4x\cdot \frac{d}{dx}\left[\cos^2\left(x^2+\frac\pi6\right)\right]dxd​[4xcos2(x2+6π​)]=4cos2(x2+6π​)+4x⋅dxd​[cos2(x2+6π​)]

At x=0x=0x=0, the second part vanishes because of the factor xxx. Thus this contributes

4cos⁡2(π6)=4⋅(32)2=4⋅34=34\cos^2\left(\frac\pi6\right)=4\cdot \left(\frac{\sqrt3}{2}\right)^2=4\cdot \frac34=34cos2(6π​)=4⋅(23​​)2=4⋅43​=3

For the second term,

ddx[−2cos⁡2x]=−2⋅(−sin⁡2x)=2sin⁡2x\frac{d}{dx}[-2\cos^2 x]= -2\cdot (-\sin 2x)=2\sin 2xdxd​[−2cos2x]=−2⋅(−sin2x)=2sin2x

At x=0x=0x=0,

2sin⁡0=02\sin 0=02sin0=0

Therefore,

F′′(0)=3+0=3F''(0)=3+0=3F′′(0)=3+0=3

Hence,

f(0)=3f(0)=3f(0)=3
  1. Comparison with stored answer

Derived answer is 333, which matches the stored correct answer.

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