JEE Advance - Mathematics (2021 - Paper 2 Online - No. 3)
Let $$f:\left[ { - {\pi \over 2},{\pi \over 2}} \right] \to R$$ be a continuous function such that $$f(0) = 1$$ and $$\int_0^{{\pi \over 3}} {f(t)dt = 0} $$. Then which of the following statements is(are) TRUE?
The equation $$f(x) - 3\cos 3x = 0$$ has at least one solution in $$\left( {0,{\pi \over 3}} \right)$$
The equation $$f(x) - 3\sin 3x = - {6 \over \pi }$$ has at least one solution in $$\left( {0,{\pi \over 3}} \right)$$
$$\mathop {\lim }\limits_{x \to 0} {{x\int_0^x {f(t)dt} } \over {1 - {e^{{x^2}}}}} = - 1$$
$$\mathop {\lim }\limits_{x \to 0} {{\sin x\int_0^x {f(t)dt} } \over {{x^2}}} = - 1$$
Explanation
Given, f(0) = 1 and $$\int_0^{\pi /3} {f(t)\,dt = 0} $$
For option (a)
Consider a function
$$g(x) = \int_0^\pi {f(t)\,dt - \sin 3x} $$
g(x) is continuous and differentiable function and g(0) = g($$\pi$$/3) = 0
$$\therefore$$ By Rolle's theorem, g'(x) = 0 has at least one solution in (0, $$\pi$$/3).
i.e. g'(x) = f(x) $$\times$$ 1 $$-$$ 3cos 3x = 0 for some $$x \in \left( {0,{\pi \over 3}} \right)$$
For option (b)
Consider the function
$$\phi (x) = \int_0^x {f(t)dt + \cos 3x + {6 \over \pi }x} $$
$$\phi$$(x) is continuous and differentiable function as well as $$\phi$$(0) = $$\phi$$($$\pi$$/3) = 1
Hence, by Rolle's theorem, $$\phi$$'(x) = 0 has at least one solution in (0, $$\pi$$ / 3).
i.e. $$\phi$$'(x) = f(x) $$\times$$ 1 $$-$$ 3sin 3x + $${6 \over \pi } = 0$$ for some $$x \in \left( {0,{\pi \over 3}} \right)$$.
For option (c)
Let $$L = \mathop {\lim }\limits_{x \to 0} {{x\int_0^x {f(t)dt} } \over {1 - {e^{{x^2}}}}}$$ (form $${0 \over 0}$$)
$$ \Rightarrow L = \mathop {\lim }\limits_{x \to 0} {{xf(x) + \int_0^x {f(t)dt} } \over { - 2x{e^{{x^2}}}}}$$ (Using L-Hospital Rule)
Again, using L'-Hospital Rule ($$\therefore$$ form $${0 \over 0}$$)
$$L = \mathop {\lim }\limits_{x \to 0} {{xf'(x) + f(x) + f(x)} \over { - 4{x^2}{e^{{x^2}}} - 2{e^{{x^2}}}}} = {{0 + 2f(0)} \over { - 0 - 2}} = - 1$$ ($$\because$$ f(0) = 1)
For option (d)
Let $$P = \mathop {\lim }\limits_{x \to 0} {{\sin x.\int_0^x {f(t)dt} } \over {{x^2}}}$$ (form $${0 \over 0}$$)
Applying L-Hospital Rule,
$$P = \mathop {\lim }\limits_{x \to 0} {{\sin x.f(x) + \cos x.\int_0^x {f(t)dt} } \over {2x}}$$ (form $${0 \over 0}$$)
Again using L-Hospital Rule,
$$P = \mathop {\lim }\limits_{x \to 0} {{[\cos x.f(x) + \sin x.f'(x) + \cos x.f(x) - \sin x\int_0^x {f(t)dt]} } \over 2}$$
$$ \Rightarrow P = {{1 \times f(0) + 0 \times f'(0) + 1 \times f(0) - 0 \times 0} \over 2}$$
$$ \Rightarrow P = {{1 + 0 + 1 - 0} \over 2} = 1$$
For option (a)
Consider a function
$$g(x) = \int_0^\pi {f(t)\,dt - \sin 3x} $$
g(x) is continuous and differentiable function and g(0) = g($$\pi$$/3) = 0
$$\therefore$$ By Rolle's theorem, g'(x) = 0 has at least one solution in (0, $$\pi$$/3).
i.e. g'(x) = f(x) $$\times$$ 1 $$-$$ 3cos 3x = 0 for some $$x \in \left( {0,{\pi \over 3}} \right)$$
For option (b)
Consider the function
$$\phi (x) = \int_0^x {f(t)dt + \cos 3x + {6 \over \pi }x} $$
$$\phi$$(x) is continuous and differentiable function as well as $$\phi$$(0) = $$\phi$$($$\pi$$/3) = 1
Hence, by Rolle's theorem, $$\phi$$'(x) = 0 has at least one solution in (0, $$\pi$$ / 3).
i.e. $$\phi$$'(x) = f(x) $$\times$$ 1 $$-$$ 3sin 3x + $${6 \over \pi } = 0$$ for some $$x \in \left( {0,{\pi \over 3}} \right)$$.
For option (c)
Let $$L = \mathop {\lim }\limits_{x \to 0} {{x\int_0^x {f(t)dt} } \over {1 - {e^{{x^2}}}}}$$ (form $${0 \over 0}$$)
$$ \Rightarrow L = \mathop {\lim }\limits_{x \to 0} {{xf(x) + \int_0^x {f(t)dt} } \over { - 2x{e^{{x^2}}}}}$$ (Using L-Hospital Rule)
Again, using L'-Hospital Rule ($$\therefore$$ form $${0 \over 0}$$)
$$L = \mathop {\lim }\limits_{x \to 0} {{xf'(x) + f(x) + f(x)} \over { - 4{x^2}{e^{{x^2}}} - 2{e^{{x^2}}}}} = {{0 + 2f(0)} \over { - 0 - 2}} = - 1$$ ($$\because$$ f(0) = 1)
For option (d)
Let $$P = \mathop {\lim }\limits_{x \to 0} {{\sin x.\int_0^x {f(t)dt} } \over {{x^2}}}$$ (form $${0 \over 0}$$)
Applying L-Hospital Rule,
$$P = \mathop {\lim }\limits_{x \to 0} {{\sin x.f(x) + \cos x.\int_0^x {f(t)dt} } \over {2x}}$$ (form $${0 \over 0}$$)
Again using L-Hospital Rule,
$$P = \mathop {\lim }\limits_{x \to 0} {{[\cos x.f(x) + \sin x.f'(x) + \cos x.f(x) - \sin x\int_0^x {f(t)dt]} } \over 2}$$
$$ \Rightarrow P = {{1 \times f(0) + 0 \times f'(0) + 1 \times f(0) - 0 \times 0} \over 2}$$
$$ \Rightarrow P = {{1 + 0 + 1 - 0} \over 2} = 1$$
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