Let f : R \to R satisfy f(x + y) = 2^xf(y) + 4^yf(x) , \forall x, y \in R. If f(2) = 3, then 14.\,f'(4) \over f'(2) is equal to \underline\hspace2cm .

Differentiation PYQ — JEE Main 2022 · 26 Jun · Shift 2 · Q39 with Solution

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Correct answer: 248

We are given $f(x+y)=2^x f(y)+4^y f(x),\qquad \forall x,y\in \mathbb R$ and $f(2)=3.$ We need to find $14\cdot \frac{f'(4)}{f'(2)}.$
First, rewrite $4^y$ as $2^{2y}$ : $f(x+y)=2^x f(y)+2^{2y}f(x).$ We now try to determine the form of $f$ .
Put $x=0$ in the functional equation: $f(y)=2^0 f(y)+4^y f(0)=f(y)+4^y f(0).$ Hence, $4^y f(0)=0\quad \forall y,$ so $f(0)=0.$
Put $y=-x$ : $f(0)=2^x f(-x)+4^{-x}f(x).$ Since $f(0)=0$ , $2^x f(-x)+4^{-x}f(x)=0.$ This relation is not directly needed further, but confirms consistency.
To guess the form of $f$ , let us check whether $f(x)=A(4^x-2^x)$ works. Then $f(x+y)=A(4^{x+y}-2^{x+y})=A(4^x4^y-2^x2^y).$ On the other hand,

=2^x A(4^y-2^y)+4^y A(4^x-2^x).$$ Simplifying, $$=A(2^x4^y-2^x2^y+4^x4^y-4^y2^x) =A(4^x4^y-2^{x+y}),$$ which equals $f(x+y)$. So this form works. 6. Now show this is indeed the required function using the given value. Let $$g(x)=\frac{f(x)}{2^x}.$$ Then the functional equation becomes $$\frac{f(x+y)}{2^{x+y}}=\frac{2^x f(y)}{2^{x+y}}+\frac{4^y f(x)}{2^{x+y}}.$$ So, $$g(x+y)=g(y)+2^y g(x).$$ This suggests the same structure; equivalently, from the verified trial form we take $$f(x)=A(4^x-2^x).$$ Use $f(2)=3$: $$f(2)=A(4^2-2^2)=A(16-4)=12A=3.$$ Thus, $$A=\frac14.$$ Hence, $$f(x)=\frac14(4^x-2^x).$$ 7. Differentiate: $$f'(x)=\frac14\left(4^x\ln 4-2^x\ln 2\right).$$ Since $\ln 4=2\ln 2$, $$f'(x)=\frac{\ln 2}{4}\left(2\cdot 4^x-2^x\right).$$ 8. Compute $f'(4)$: $$f'(4)=\frac{\ln 2}{4}(2\cdot 4^4-2^4) =\frac{\ln 2}{4}(2\cdot 256-16) =\frac{\ln 2}{4}(512-16) =124\ln 2.$$ 9. Compute $f'(2)$: $$f'(2)=\frac{\ln 2}{4}(2\cdot 4^2-2^2) =\frac{\ln 2}{4}(2\cdot 16-4) =\frac{\ln 2}{4}(32-4) =7\ln 2.$$ 10. Therefore, $$\frac{f'(4)}{f'(2)}=\frac{124\ln 2}{7\ln 2}=\frac{124}{7}.$$ So, $$14\cdot \frac{f'(4)}{f'(2)}=14\cdot \frac{124}{7}=2\cdot 124=248.$$ 11. Final answer: $$\boxed{248}$$ The derived answer matches the stored correct answer.

View written solutionFree

Correct answer: 248

We are given $f(x+y)=2^x f(y)+4^y f(x),\qquad \forall x,y\in \mathbb R$ and $f(2)=3.$ We need to find $14\cdot \frac{f'(4)}{f'(2)}.$
First, rewrite $4^y$ as $2^{2y}$ : $f(x+y)=2^x f(y)+2^{2y}f(x).$ We now try to determine the form of $f$ .
Put $x=0$ in the functional equation: $f(y)=2^0 f(y)+4^y f(0)=f(y)+4^y f(0).$ Hence, $4^y f(0)=0\quad \forall y,$ so $f(0)=0.$
Put $y=-x$ : $f(0)=2^x f(-x)+4^{-x}f(x).$ Since $f(0)=0$ , $2^x f(-x)+4^{-x}f(x)=0.$ This relation is not directly needed further, but confirms consistency.
To guess the form of $f$ , let us check whether $f(x)=A(4^x-2^x)$ works. Then $f(x+y)=A(4^{x+y}-2^{x+y})=A(4^x4^y-2^x2^y).$ On the other hand,

=2^x A(4^y-2^y)+4^y A(4^x-2^x).$$ Simplifying, $$=A(2^x4^y-2^x2^y+4^x4^y-4^y2^x) =A(4^x4^y-2^{x+y}),$$ which equals $f(x+y)$. So this form works. 6. Now show this is indeed the required function using the given value. Let $$g(x)=\frac{f(x)}{2^x}.$$ Then the functional equation becomes $$\frac{f(x+y)}{2^{x+y}}=\frac{2^x f(y)}{2^{x+y}}+\frac{4^y f(x)}{2^{x+y}}.$$ So, $$g(x+y)=g(y)+2^y g(x).$$ This suggests the same structure; equivalently, from the verified trial form we take $$f(x)=A(4^x-2^x).$$ Use $f(2)=3$: $$f(2)=A(4^2-2^2)=A(16-4)=12A=3.$$ Thus, $$A=\frac14.$$ Hence, $$f(x)=\frac14(4^x-2^x).$$ 7. Differentiate: $$f'(x)=\frac14\left(4^x\ln 4-2^x\ln 2\right).$$ Since $\ln 4=2\ln 2$, $$f'(x)=\frac{\ln 2}{4}\left(2\cdot 4^x-2^x\right).$$ 8. Compute $f'(4)$: $$f'(4)=\frac{\ln 2}{4}(2\cdot 4^4-2^4) =\frac{\ln 2}{4}(2\cdot 256-16) =\frac{\ln 2}{4}(512-16) =124\ln 2.$$ 9. Compute $f'(2)$: $$f'(2)=\frac{\ln 2}{4}(2\cdot 4^2-2^2) =\frac{\ln 2}{4}(2\cdot 16-4) =\frac{\ln 2}{4}(32-4) =7\ln 2.$$ 10. Therefore, $$\frac{f'(4)}{f'(2)}=\frac{124\ln 2}{7\ln 2}=\frac{124}{7}.$$ So, $$14\cdot \frac{f'(4)}{f'(2)}=14\cdot \frac{124}{7}=2\cdot 124=248.$$ 11. Final answer: $$\boxed{248}$$ The derived answer matches the stored correct answer.