JEE Advance - Mathematics (2017 - Paper 1 Offline - No. 2)
Let f : R $$ \to $$ (0, 1) be a continuous function. Then, which of the following function(s) has (have) the value zero at some point in the interval (0, 1) ?
$${e^x} - \int_0^x {f(t)\sin t\,dt} $$
$$f(x) + \int_0^{{\pi \over 2}} {f(t)\sin t\,dt} $$
$$f(x) - \int_0^{{\pi \over 2} - x} {f(t)\cos t\,dt} $$
x9 $$-$$ f(x)
Explanation
(a) $$ \because $$ $${e^x} \in (1,e)$$ in (0, 1) and $$\int_0^x {f(t)\sin t\,dt \in } $$ (0, 1) in (0, 1)
$$ \therefore $$ $${e^x} - \int_0^x {f(t)\sin t\,dt} $$ cannot be zero.
So, option (a) is incorrect.
(b) $$f(x) + \int_0^{{\pi \over 2}} {f(t)\sin t\,dt} $$ always positive
$$ \therefore $$ Option (b) is incorrect.
(c) Let $$h(x) = x - \int_0^{{\pi \over 2} - x} {f(t)\cos t\,dt} $$,
$$h(0) = - \int_0^{{\pi \over 2}} {f(t)\cos t\,dt} < 0$$
$$h(1) = 1 - \int_0^{{\pi \over 2} - 1} {f(t)\cos t\,dt} > 0$$
$$ \therefore $$ Option (c) is correct.
(d) Let $$g(x) = {x^9} - f(x)$$
$$g(0) = - f(0) < 0$$, $$g(1) = 1 - f(1) > 0$$
$$ \therefore $$ Option (d) is correct.
$$ \therefore $$ $${e^x} - \int_0^x {f(t)\sin t\,dt} $$ cannot be zero.
So, option (a) is incorrect.
(b) $$f(x) + \int_0^{{\pi \over 2}} {f(t)\sin t\,dt} $$ always positive
$$ \therefore $$ Option (b) is incorrect.
(c) Let $$h(x) = x - \int_0^{{\pi \over 2} - x} {f(t)\cos t\,dt} $$,
$$h(0) = - \int_0^{{\pi \over 2}} {f(t)\cos t\,dt} < 0$$
$$h(1) = 1 - \int_0^{{\pi \over 2} - 1} {f(t)\cos t\,dt} > 0$$
$$ \therefore $$ Option (c) is correct.
(d) Let $$g(x) = {x^9} - f(x)$$
$$g(0) = - f(0) < 0$$, $$g(1) = 1 - f(1) > 0$$
$$ \therefore $$ Option (d) is correct.
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