Q.18. Let be such that , for all and . Let be a differentiable function such that . Then is equa

Q: What is the correct answer for this exam Mathematics Q18?

The correct answer is C. Given $f(xy) = f(x)f(y)$ and $f(0) \ne 0$.This is a functional equation. If $f(x) = x^n$ or $f(x) = c$.If $f(x) = c$, then $c = c \cdot c \Rightarrow c^2=c \Rightarrow c=0$ or $c=1$.Since $f(0) \ne 0$, $c \ne 0$. So $f(x)=1$.If $f(x) = x^n$, then $(xy)^n = x^n y^n$, which is true. But $f(0) = 0^n = 0$ (for $n>0$), which contradicts $f(0) \ne 0$.So, $f(x)=1$ for all $x \in R$.Now, given $x^2g(x) =

Q: Q.18. Let be such that , for all and . Let be a differentiable function such that . Then is equal to :

The correct answer is C. Given $f(xy) = f(x)f(y)$ and $f(0) \ne 0$.This is a functional equation. If $f(x) = x^n$ or $f(x) = c$.If $f(x) = c$, then $c = c \cdot c \Rightarrow c^2=c \Rightarrow c=0$ or $c=1$.Since $f(0) \ne 0$, $c \ne 0$. So $f(x)=1$.If $f(x) = x^n$, then $(xy)^n = x^n y^n$, which is true. But $f(0) = 0^n = 0$ (for $n>0$), which contradicts $f(0) \ne 0$.So, $f(x)=1$ for all $x \in R$.Now, given $x^2g(x) =

Q: What are the options for this this exam Mathematics question?

A. $\frac{13}{8}$ · B. $\frac{11}{16}$ · C. $\frac{15}{32}$ · D. $\frac{17}{64}$

Question

Q.18. Let  be such that , for all  and . Let  be a differentiable function such that . Then  is equal to :

FirstInTest · Accepted Answer

Given $f(xy) = f(x)f(y)$ and $f(0) 
e 0$.This is a functional equation. If $f(x) = x^n$ or $f(x) = c$.If $f(x) = c$, then $c = c \cdot c \Rightarrow c^2=c \Rightarrow c=0$ or $c=1$.Since $f(0) 
e 0$, $c 
e 0$. So $f(x)=1$.If $f(x) = x^n$, then $(xy)^n = x^n y^n$, which is true. But $f(0) = 0^n = 0$ (for $n>0$), which contradicts $f(0) 
e 0$.So, $f(x)=1$ for all $x \in R$.Now, given $x^2g(x) = \int_1^x (t^2f(t) - tg(t))\,dt$.Substitute $f(t)=1$: $x^2g(x) = \int_1^x (t^2 - tg(t))\,dt$.Differentiate both sides with respect to $x$:$\frac{d}{dx}(x^2g(x)) = \frac{d}{dx}(\int_1^x (t^2 - tg(t))\,dt)$.Using product rule on LHS and Fundamental Theorem of Calculus on RHS: $2xg(x) + x^2g'(x) = x^2 - xg(x)$.Rearrange the terms: $x^2g'(x) + 3xg(x) = x^2$.Divide by $x^2$ (since $x \in [1, \infty)$, $x 
e 0$): $g'(x) + \frac{3}{x}g(x) = 1$.This is a first-order linear differential equation of the form $y' + P(x)y = Q(x)$, where $y=g(x)$, $P(x) = \frac{3}{x}$, and $Q(x)=1$.The integrating factor (IF) is $e^{\int P(x)\,dx} = e^{\int \frac{3}{x}\,dx} = e^{3\ln|x|} = e^{\ln(x^3)} = x^3$ (since $x \ge 1$).Multiply the differential equation by the IF: $x^3g'(x) + 3x^2g(x) = x^3$.The LHS is the derivative of $(x^3g(x))$: $\frac{d}{dx}(x^3g(x)) = x^3$.Integrate both sides with respect to $x$: $\int \frac{d}{dx}(x^3g(x))\,dx = \int x^3\,dx$.$x^3g(x) = \frac{x^4}{4} + C$.To find C, use the initial condition from the integral definition: $x^2g(x) = \int_1^x (t^2 - tg(t))\,dt$.When $x=1$, $1^2g(1) = \int_1^1 (t^2 - tg(t))\,dt = 0$.So $g(1)=0$.Substitute $x=1, g(1)=0$ into $x^3g(x) = \frac{x^4}{4} + C$:$(1)^3(0) = \frac{1^4}{4} + C \Rightarrow 0 = \frac{1}{4} + C \Rightarrow C = -\frac{1}{4}$.So, $x^3g(x) = \frac{x^4}{4} - \frac{1}{4}$.We need to find $g(2)$. Substitute $x=2$:$(2)^3g(2) = \frac{2^4}{4} - \frac{1}{4}$.$8g(2) = \frac{16}{4} - \frac{1}{4} = 4 - \frac{1}{4} = \frac{16-1}{4} = \frac{15}{4}$.$g(2) = \frac{15}{4 	imes 8} = \frac{15}{32}$.

Answer

$\frac{13}{8}$

Answer

$\frac{11}{16}$

Answer

$\frac{15}{32}$

Answer

$\frac{17}{64}$

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