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Definite Integration question

2021 · Shift 2 · Q22
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  5. /2021 · Shift 2 · Q22

Definite Integration question

2021 · Shift 2 · Q22

JEE AdvancedMathematicsDefinite IntegrationMultiple correct+4 / −2
Let f:[−π2,π2]→Rf:\left[ { - {\pi \over 2},{\pi \over 2}} \right] \to Rf:[−2π​,2π​]→R be a continuous function such that f(0)=1f(0) = 1f(0)=1 and ∫0π3f(t)dt=0\int_0^{{\pi \over 3}} {f(t)dt = 0}∫03π​​f(t)dt=0. Then which of the following statements is(are) TRUE?
  1. A
    The equation f(x)−3cos⁡3x=0f(x) - 3\cos 3x = 0f(x)−3cos3x=0 has at least one solution in (0,π3)\left( {0,{\pi \over 3}} \right)(0,3π​)
  2. B
    The equation f(x)−3sin⁡3x=−6πf(x) - 3\sin 3x = - {6 \over \pi }f(x)−3sin3x=−π6​ has at least one solution in (0,π3)\left( {0,{\pi \over 3}} \right)(0,3π​)
  3. C
    lim⁡x→0x∫0xf(t)dt1−ex2=−1\mathop {\lim }\limits_{x \to 0} {{x\int_0^x {f(t)dt} } \over {1 - {e^{{x^2}}}}} = - 1x→0lim​1−ex2x∫0x​f(t)dt​=−1
  4. D
    lim⁡x→0sin⁡x∫0xf(t)dtx2=−1\mathop {\lim }\limits_{x \to 0} {{\sin x\int_0^x {f(t)dt} } \over {{x^2}}} = - 1x→0lim​x2sinx∫0x​f(t)dt​=−1
View written solutionFree

Correct answer: A, B, C

  1. Given data

We have a continuous function f:[−π2,π2]→R,f:\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\to \mathbb R,f:[−2π​,2π​]→R, with f(0)=1,∫0π/3f(t) dt=0.f(0)=1, \qquad \int_0^{\pi/3} f(t)\,dt=0.f(0)=1,∫0π/3​f(t)dt=0.

We must check each option.

  1. Option A: f(x)−3cos⁡3x=0f(x)-3\cos 3x=0f(x)−3cos3x=0 has at least one solution in (0,π3)\left(0,\frac{\pi}{3}\right)(0,3π​).

Define g(x)=f(x)−3cos⁡3x.g(x)=f(x)-3\cos 3x.g(x)=f(x)−3cos3x. Since fff is continuous, ggg is continuous on [0,π3]\left[0,\frac{\pi}{3}\right][0,3π​].

Now, g(0)=f(0)−3cos⁡0=1−3=−2<0.g(0)=f(0)-3\cos 0=1-3=-2<0.g(0)=f(0)−3cos0=1−3=−2<0. Also, g(π3)=f(π3)−3cos⁡π=f(π3)+3.g\left(\frac{\pi}{3}\right)=f\left(\frac{\pi}{3}\right)-3\cos \pi=f\left(\frac{\pi}{3}\right)+3.g(3π​)=f(3π​)−3cosπ=f(3π​)+3. This alone does not give sign change, so use the integral condition.

Assume, for contradiction, that g(x)<0∀x∈(0,π3).g(x)<0 \quad \forall x\in \left(0,\frac{\pi}{3}\right).g(x)<0∀x∈(0,3π​). Then f(x)<3cos⁡3x∀x∈(0,π3).f(x)<3\cos 3x \quad \forall x\in \left(0,\frac{\pi}{3}\right).f(x)<3cos3x∀x∈(0,3π​). Integrating from 000 to π3\frac{\pi}{3}3π​, ∫0π/3f(t) dt<∫0π/33cos⁡3t dt.\int_0^{\pi/3} f(t)\,dt < \int_0^{\pi/3} 3\cos 3t\,dt.∫0π/3​f(t)dt<∫0π/3​3cos3tdt. But ∫0π/33cos⁡3t dt=[sin⁡3t]0π/3=sin⁡π−sin⁡0=0.\int_0^{\pi/3} 3\cos 3t\,dt = \left[\sin 3t\right]_0^{\pi/3}=\sin \pi - \sin 0=0.∫0π/3​3cos3tdt=[sin3t]0π/3​=sinπ−sin0=0. So we get ∫0π/3f(t) dt<0,\int_0^{\pi/3} f(t)\,dt<0,∫0π/3​f(t)dt<0, contradicting the given value 000.

Hence ggg cannot be negative throughout the interval. Since g(0)=−2<0g(0)=-2<0g(0)=−2<0 and ggg is continuous, either:

  • there exists some x∈(0,π/3)x\in (0,\pi/3)x∈(0,π/3) with g(x)=0g(x)=0g(x)=0, or
  • ggg becomes positive somewhere, and then by IVT there is a zero in (0,π/3)(0,\pi/3)(0,π/3).

Therefore, Option A is true.

  1. Option B: f(x)−3sin⁡3x=−6πf(x)-3\sin 3x=-\dfrac{6}{\pi}f(x)−3sin3x=−π6​ has at least one solution in (0,π3)\left(0,\frac{\pi}{3}\right)(0,3π​).

Let h(x)=f(x)−3sin⁡3x+6π.h(x)=f(x)-3\sin 3x+\frac{6}{\pi}.h(x)=f(x)−3sin3x+π6​. We need to show h(x)=0h(x)=0h(x)=0 for some x∈(0,π/3)x\in (0,\pi/3)x∈(0,π/3).

Compute the average value of hhh on [0,π3]\left[0,\frac{\pi}{3}\right][0,3π​]:

=\int_0^{\pi/3} f(t)\,dt -\int_0^{\pi/3} 3\sin 3t\,dt +\int_0^{\pi/3} \frac{6}{\pi}\,dt.$$ Now, $$\int_0^{\pi/3} f(t)\,dt=0.$$ Also, $$\int_0^{\pi/3} 3\sin 3t\,dt =\left[-\cos 3t\right]_0^{\pi/3} =-\cos \pi + \cos 0=1+1=2.$$ And $$\int_0^{\pi/3} \frac{6}{\pi}\,dt=\frac{6}{\pi}\cdot \frac{\pi}{3}=2.$$ Therefore, $$\int_0^{\pi/3} h(t)\,dt=0-2+2=0.$$ Since $h$ is continuous on $\left[0,\frac{\pi}{3}\right]$, if $h$ were always positive or always negative on $(0,\pi/3)$, its integral could not be $0$. Hence either: - $h(x)=0$ for some $x\in (0,\pi/3)$ directly, or - it changes sign, and by continuity has a zero in $(0,\pi/3)$. Thus there exists $x\in (0,\pi/3)$ such that $$f(x)-3\sin 3x+\frac{6}{\pi}=0,$$ i.e. $$f(x)-3\sin 3x=-\frac{6}{\pi}.$$ Therefore, **Option B is true**. --- 4. **Option C**: $$\lim_{x\to 0}\frac{x\int_0^x f(t)\,dt}{1-e^{x^2}}=-1.$$ Let $$I(x)=\int_0^x f(t)\,dt.$$ Since $f$ is continuous and $f(0)=1$, we know $$\lim_{x\to 0}\frac{I(x)}{x}=f(0)=1.$$ So near $x=0$, $$I(x)\sim x.$$ Also, $$e^{x^2}=1+x^2+o(x^2),$$ so $$1-e^{x^2}=-(x^2+o(x^2))\sim -x^2.$$ Hence $$\frac{xI(x)}{1-e^{x^2}}\sim \frac{x\cdot x}{-x^2}=-1.$$ Therefore, $$\lim_{x\to 0}\frac{x\int_0^x f(t)\,dt}{1-e^{x^2}}=-1.$$ So **Option C is true**. --- 5. **Option D**: $$\lim_{x\to 0}\frac{\sin x\int_0^x f(t)\,dt}{x^2}=-1.$$ Again using $$\int_0^x f(t)\,dt \sim x$$ as $x\to 0$, and $$\sin x \sim x,$$ we get $$\frac{\sin x\int_0^x f(t)\,dt}{x^2}\sim \frac{x\cdot x}{x^2}=1.$$ So the limit should be $$1,$$ not $-1$. Therefore, **Option D is false**. --- 6. **Final conclusion** The true statements are: $$\boxed{A,\ B,\ C}$$ --- 7. **Comparison with stored answer** Stored correct answer: **A, B, C** Our derived answer matches exactly.
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