JEE MAIN - Mathematics (2021 - 18th March Morning Shift - No. 17)
Let f(x) and g(x) be two functions satisfying f(x2) + g(4 $$-$$ x) = 4x3 and g(4 $$-$$ x) + g(x) = 0, then the value of $$\int\limits_{ - 4}^4 {f{{(x)}^2}dx} $$ is
Answer
512
Explanation
$$I = 2\int\limits_0^4 {f({x^2})dx} $$ ............(1)
$$ \Rightarrow I = 2\int\limits_0^4 {f({{(4 - x)}^2})dx} $$ ..............(2)
Adding equation (1) & (2)
$$2I = 2\int\limits_0^4 {\left[ {f{{(x)}^2} + f{{(4 - x)}^2}} \right]} \,dx$$ ............(3)
Now using $$f{(x^2)} + g(4 - x) = 4{x^3}$$ ............. (4)
$$x \to 4 - x$$
$$f({(4 - x)^2}) + g(x) = 4{(4 - x)^3}$$ ..............(5)
Adding equation (4) & (5)
$$f({x^2}) + f(4 - {x^2}) + g(x) + g(4 - x) = 4({x^3} + {(4 - x)^3}]$$
$$ \Rightarrow f({x^2}) + f(4 - {x^2}) = 4({x^3} + {(4 - x)^3}]$$
Now, $$I = 4\int\limits_0^4 {\left( {{x^3} + {{(4 - x)}^3}} \right)dx = 512} $$
$$ \Rightarrow I = 2\int\limits_0^4 {f({{(4 - x)}^2})dx} $$ ..............(2)
Adding equation (1) & (2)
$$2I = 2\int\limits_0^4 {\left[ {f{{(x)}^2} + f{{(4 - x)}^2}} \right]} \,dx$$ ............(3)
Now using $$f{(x^2)} + g(4 - x) = 4{x^3}$$ ............. (4)
$$x \to 4 - x$$
$$f({(4 - x)^2}) + g(x) = 4{(4 - x)^3}$$ ..............(5)
Adding equation (4) & (5)
$$f({x^2}) + f(4 - {x^2}) + g(x) + g(4 - x) = 4({x^3} + {(4 - x)^3}]$$
$$ \Rightarrow f({x^2}) + f(4 - {x^2}) = 4({x^3} + {(4 - x)^3}]$$
Now, $$I = 4\int\limits_0^4 {\left( {{x^3} + {{(4 - x)}^3}} \right)dx = 512} $$
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