Q.19. Let f: R R be a differentiable function such that for all x, y R and . Then the minimum val

Q: What is the correct answer for JEE Main Mathematics Q19?

The correct answer is B. Sol. Given $\frac{f(x+y)}{3} = \frac{f(x)+f(y)}{3} \Rightarrow f(x+y) = f(x)+f(y)$. This is Cauchy's functional equation. Since f is differentiable, the solution is $f(x) = cx$ for some constant c. We are given $f'(0) = 3$. $f'(x) = c$, so $f'(0) = c = 3$. Thus, $f(x) = 3x$. Now consider the function $g(x) = 3 + e^x f(x) = 3 + e^x (3x) = 3 + 3xe^x$. To find the minimum value, we find $g'(x)$ and s

Q: Q.19. Let f: R R be a differentiable function such that for all x, y R and . Then the minimum value of the function , is

The correct answer is B. Sol. Given $\frac{f(x+y)}{3} = \frac{f(x)+f(y)}{3} \Rightarrow f(x+y) = f(x)+f(y)$. This is Cauchy's functional equation. Since f is differentiable, the solution is $f(x) = cx$ for some constant c. We are given $f'(0) = 3$. $f'(x) = c$, so $f'(0) = c = 3$. Thus, $f(x) = 3x$. Now consider the function $g(x) = 3 + e^x f(x) = 3 + e^x (3x) = 3 + 3xe^x$. To find the minimum value, we find $g'(x)$ and s

Q: What are the options for this JEE Main Mathematics question?

A. 3(1-e)/e · B. 3(e-1)/e · C. 3-e/e · D. 3e

Question

Q.19. Let f: R  R be a differentiable function such that  for all x, y  R and . Then the minimum value of the function , is

FirstInTest · Accepted Answer

Sol. Given $\frac{f(x+y)}{3} = \frac{f(x)+f(y)}{3} \Rightarrow f(x+y) = f(x)+f(y)$. This is Cauchy's functional equation. Since f is differentiable, the solution is $f(x) = cx$ for some constant c. We are given $f'(0) = 3$. $f'(x) = c$, so $f'(0) = c = 3$. Thus, $f(x) = 3x$. Now consider the function $g(x) = 3 + e^x f(x) = 3 + e^x (3x) = 3 + 3xe^x$. To find the minimum value, we find $g'(x)$ and set it to 0. $g'(x) = 0 + 3(1 \cdot e^x + x \cdot e^x) = 3e^x(1+x)$. Set $g'(x) = 0 \Rightarrow 3e^x(1+x) = 0$. Since $e^x > 0$, we have $1+x=0 \Rightarrow x=-1$. To check if it's a minimum, we can use the second derivative test or observe the sign of $g'(x)$. For $x < -1$, $1+x < 0$, so $g'(x) < 0$ (g(x) is decreasing). For $x > -1$, $1+x > 0$, so $g'(x) > 0$ (g(x) is increasing). Thus, $x=-1$ is a point of local minimum. The minimum value is $g(-1) = 3 + 3(-1)e^{-1} = 3 - 3e^{-1} = 3(1 - e^{-1}) = 3(1 - \frac{1}{e}) = 3(\frac{e-1}{e})$.

Answer

3(1-e)/e

Answer

3(e-1)/e

Answer

3-e/e

Answer

3e

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