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QJEEM21Mac223: JEE Main 2021 Mathematics - Class 12 JEEM21Mac223 Solution: Let f(x) and g(x) be two functions satisfying f(x 2 ) + g(4 x)…

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 ∫−44f(x)2dx\int\limits_{ - 4}^4 {f{{(x)}^2}dx} is

Step-by-Step Verified Solution

I=2∫04f(x2)dxI = 2\int\limits_0^4 {f({x^2})dx} ............(1)

⇒I=2∫04f((4−x)2)dx\Rightarrow I = 2\int\limits_0^4 {f({{(4 - x)}^2})dx} ..............(2)

Adding equation (1) & (2)

2I=2∫04[f(x)2+f(4−x)2] dx2I = 2\int\limits_0^4 {\left[ {f{{(x)}^2} + f{{(4 - x)}^2}} \right]} \,dx ............(3)

Now using f(x2)+g(4−x)=4x3f{(x^2)} + g(4 - x) = 4{x^3} ............. (4)

x→4−xx \to 4 - x

f((4−x)2)+g(x)=4(4−x)3f({(4 - x)^2}) + g(x) = 4{(4 - x)^3} ..............(5)

Adding equation (4) & (5)

f(x2)+f(4−x2)+g(x)+g(4−x)=4(x3+(4−x)3]f({x^2}) + f(4 - {x^2}) + g(x) + g(4 - x) = 4({x^3} + {(4 - x)^3}]

⇒f(x2)+f(4−x2)=4(x3+(4−x)3]\Rightarrow f({x^2}) + f(4 - {x^2}) = 4({x^3} + {(4 - x)^3}]

Now, I=4∫04(x3+(4−x)3)dx=512I = 4\int\limits_0^4 {\left( {{x^3} + {{(4 - x)}^3}} \right)dx = 512}

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