JEE MAIN - Mathematics (2019 - 8th April Evening Slot - No. 4)
Let $$f(x) = \int\limits_0^x {g(t)dt} $$ where g is a non-zero even
function. If ƒ(x + 5) = g(x), then $$ \int\limits_0^x {f(t)dt} $$ equals-
5$$\int\limits_{x + 5}^5 {g(t)dt} $$
$$\int\limits_{x + 5}^5 {g(t)dt} $$
$$\int\limits_{5}^{x+5} {g(t)dt} $$
2$$\int\limits_{5}^{x+5} {g(t)dt} $$
Explanation
$$f(x) = \int\limits_0^x {g(t)dt} $$
$$f\left( { - x} \right) = \int\limits_0^{ - x} {g\left( t \right)} dt$$
Put t = -v
= $$ - \int\limits_0^x {g\left( { - v} \right)} dv$$
= $$ - \int\limits_0^x {g\left( v \right)} d(v)$$ [ as g(v) is an even function.]
= - f(x)
$$ \Rightarrow $$ f(-x) = -f(x)
$$ \therefore $$ f(x) is an odd function.
Given ƒ(x + 5) = g(x)
$$ \therefore $$ g(- x) = ƒ(- x + 5)
$$ \Rightarrow $$ g(x) = - f(x - 5) [as g(x) is even and f(x) is an odd function]
Replacing x by x + 5, we get
f(x) = - g(x + 5) ......(1)
Now
$$ \int\limits_0^x {f(t)dt} $$
= $$ - \int\limits_0^x {g\left( {t + 5} \right)} dt$$
= $$ - \int\limits_5^{x + 5} {g\left( t \right)} dt$$
= $$\int\limits_{x + 5}^5 {g(t)dt} $$
$$f\left( { - x} \right) = \int\limits_0^{ - x} {g\left( t \right)} dt$$
Put t = -v
= $$ - \int\limits_0^x {g\left( { - v} \right)} dv$$
= $$ - \int\limits_0^x {g\left( v \right)} d(v)$$ [ as g(v) is an even function.]
= - f(x)
$$ \Rightarrow $$ f(-x) = -f(x)
$$ \therefore $$ f(x) is an odd function.
Given ƒ(x + 5) = g(x)
$$ \therefore $$ g(- x) = ƒ(- x + 5)
$$ \Rightarrow $$ g(x) = - f(x - 5) [as g(x) is even and f(x) is an odd function]
Replacing x by x + 5, we get
f(x) = - g(x + 5) ......(1)
Now
$$ \int\limits_0^x {f(t)dt} $$
= $$ - \int\limits_0^x {g\left( {t + 5} \right)} dt$$
= $$ - \int\limits_5^{x + 5} {g\left( t \right)} dt$$
= $$\int\limits_{x + 5}^5 {g(t)dt} $$
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