Question

Let $f(x) = 1 + \int_{0}^{1}(xe^y+ye^x)f(y)dy$ where $x$ and $y$ are independent variables.

If complete solution set of 'x' for which function $h(x) = f(x) + 3x$ is strictly increasing is $(-\infty, k)$ then $\left[\frac{4}{3}e^k\right]$ equals to : (where [.] denotes greatest integer function):

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (C)

3

Answer 2

The given equation is $f(x) = 1 + \int_{0}^{1}(xe^y+ye^x)f(y)dy$. We can split the integral into two parts: $f(x) = 1 + \int_{0}^{1}xe^y f(y)dy + \int_{0}^{1}ye^x f(y)dy$ Since $x$ is independent of the integration variable $y$, we can take $x$ and $e^x$ out of the integrals: $f(x) = 1 + x \int_{0}^{1}e^y f(y)dy + e^x \int_{0}^{1}y f(y)dy$.

Let $A = \int_{0}^{1}e^y f(y)dy$ and $B = \int_{0}^{1}y f(y)dy$. A and B are constants. Then $f(x) = 1 + Ax + Be^x$.

Now we substitute this form of $f(y)$ back into the definitions of A and B. For A: $A = \int_{0}^{1}e^y (1 + Ay + Be^y)dy = \int_{0}^{1}(e^y + Aye^y + Be^{2y})dy$ $A = \int_{0}^{1}e^y dy + A\int_{0}^{1}ye^y dy + B\int_{0}^{1}e^{2y} dy$. $\int_{0}^{1}e^y dy = [e^y]_0^1 = e - 1$. $\int_{0}^{1}ye^y dy$: Use integration by parts $\int u dv = uv - \int v du$ with $u=y, dv=e^y dy$. Then $du=dy, v=e^y$. $\int_{0}^{1}ye^y dy = [ye^y]_0^1 - \int_{0}^{1}e^y dy = (1 \cdot e^1 - 0 \cdot e^0) - [e^y]_0^1 = e - (e - 1) = 1$. $\int_{0}^{1}e^{2y} dy = [\frac{1}{2}e^{2y}]_0^1 = \frac{1}{2}(e^2 - e^0) = \frac{e^2 - 1}{2}$. Substituting these values into the equation for A: $A = (e - 1) + A(1) + B\left(\frac{e^2 - 1}{2}\right)$ $A = e - 1 + A + B\frac{(e-1)(e+1)}{2}$. $0 = e - 1 + B\frac{(e-1)(e+1)}{2}$. Since $e \neq 1$, $e-1 \neq 0$. We can divide by $(e-1)$: $0 = 1 + B\frac{e+1}{2}$. $B\frac{e+1}{2} = -1 \implies B = -\frac{2}{e+1}$.

For B: $B = \int_{0}^{1}y (1 + Ay + Be^y)dy = \int_{0}^{1}(y + Ay^2 + Bye^y)dy$ $B = \int_{0}^{1}y dy + A\int_{0}^{1}y^2 dy + B\int_{0}^{1}ye^y dy$. $\int_{0}^{1}y dy = [\frac{y^2}{2}]_0^1 = \frac{1}{2} - 0 = \frac{1}{2}$. $\int_{0}^{1}y^2 dy = [\frac{y^3}{3}]_0^1 = \frac{1}{3} - 0 = \frac{1}{3}$. $\int_{0}^{1}ye^y dy = 1$ (calculated above). Substituting these values into the equation for B: $B = \frac{1}{2} + A\left(\frac{1}{3}\right) + B(1)$. $B = \frac{1}{2} + \frac{A}{3} + B$. $0 = \frac{1}{2} + \frac{A}{3}$. $\frac{A}{3} = -\frac{1}{2} \implies A = -\frac{3}{2}$.

So, the function $f(x)$ is $f(x) = 1 + Ax + Be^x = 1 - \frac{3}{2}x - \frac{2}{e+1}e^x$.

Now consider the function $h(x) = f(x) + 3x$. $h(x) = \left(1 - \frac{3}{2}x - \frac{2}{e+1}e^x\right) + 3x = 1 + \left(3 - \frac{3}{2}\right)x - \frac{2}{e+1}e^x$ $h(x) = 1 + \frac{3}{2}x - \frac{2}{e+1}e^x$.

For $h(x)$ to be strictly increasing, its derivative $h'(x)$ must be positive. $h'(x) = \frac{d}{dx}\left(1 + \frac{3}{2}x - \frac{2}{e+1}e^x\right) = 0 + \frac{3}{2} - \frac{2}{e+1}e^x$. $h'(x) > 0 \implies \frac{3}{2} - \frac{2}{e+1}e^x > 0$. $\frac{3}{2} > \frac{2}{e+1}e^x$. Multiply both sides by $\frac{e+1}{2}$: $\frac{3}{2} \cdot \frac{e+1}{2} > e^x$ $\frac{3(e+1)}{4} > e^x$.

To solve for $x$, take the natural logarithm of both sides: $\ln\left(\frac{3(e+1)}{4}\right) > \ln(e^x)$ $\ln\left(\frac{3(e+1)}{4}\right) > x$.

The complete solution set of $x$ for which $h(x)$ is strictly increasing is $(-\infty, \ln\left(\frac{3(e+1)}{4}\right))$. This set is given as $(-\infty, k)$, so $k = \ln\left(\frac{3(e+1)}{4}\right)$.

We need to calculate $\left[\frac{4}{3}e^k\right]$. First, find $e^k$: $e^k = e^{\ln\left(\frac{3(e+1)}{4}\right)} = \frac{3(e+1)}{4}$.

Now, calculate $\frac{4}{3}e^k$: $\frac{4}{3}e^k = \frac{4}{3} \cdot \frac{3(e+1)}{4} = e+1$.

Finally, we need to find the greatest integer of $e+1$, i.e., $[e+1]$. The value of $e$ is approximately $2.71828$. So, $e+1$ is approximately $2.71828 + 1 = 3.71828$. The greatest integer function $[x]$ gives the largest integer less than or equal to $x$. $[e+1] = [3.71828] = 3$.

The value of $\left[\frac{4}{3}e^k\right]$ is 3.