Question

Let f(x) be a differentiable function satisfying $x(f(x)+1) = x^2 + \frac{x}{e} + \int_{0}^{x^2} f(\frac{t}{x}) dt \forall x \in R_0$.

Let $f'(1) = 1$. If curves $y = f(x)$, $y = \ln x^2$ intersects at $(x_n,y_n)$ for some $n = 1,2,3,4,...$ where $y_1 < y_2 < y_3 < y_4,...$, then value of $[y_1] + [y_2] + n$ is ([.] denotes greatest integer function)

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (B)

2

Answer 2

We are given that for all non‐zero real numbers x

$$x(f(x)+1)= x^2+\frac{x}{e}+\int_{0}^{x^2} f(\frac{t}{x})\,dt.$$

Step 1. Simplify the Integral

Let

$$u=\frac{t}{x} \quad \Rightarrow \quad t=xu, \quad dt=x\,du.$$

When $t=0$, $u=0$ and when $t=x^2$, $u=x$. Hence,

$$\int_{0}^{x^2} f(\frac{t}{x})\,dt = \int_{0}^{x} f(u)\,x\,du = x\int_0^x f(u)\,du.$$

Thus the equation becomes (for $x\ne 0$ ):

$$x(f(x)+1)= x^2+\frac{x}{e}+x\int_0^x f(u)\,du.$$

Divide by $x$:

$$f(x)+1=x+\frac{1}{e}+\int_0^x f(u)\,du.$$

Step 2. Differentiate Both Sides

Differentiate with respect to $x$:

• LHS derivative: $f'(x)$.
• RHS derivative: $\frac{d}{dx}[x+\tfrac{1}{e}]=1$ and $\frac{d}{dx}\int_0^x f(u)\,du = f(x)$ by the Fundamental Theorem of Calculus.

Thus,

$$f'(x)=1+f(x).$$

Step 3. Solve the Differential Equation

This is a first–order linear ODE. Its integrating factor is $e^{-x}$. Multiplying:

$$e^{-x} f'(x) - e^{-x} f(x) = e^{-x} \quad \Longrightarrow \quad \frac{d}{dx}(e^{-x}f(x))=e^{-x}.$$

Integrate both sides:

$$e^{-x}f(x)= -e^{-x}+C \quad \Longrightarrow \quad f(x)= -1+Ce^{x}.$$

Step 4. Use Condition $f'(1)=1$ to Determine $C$

Differentiate $f(x)= -1+Ce^x$ to get

$$f'(x)= Ce^x.$$

At $x=1$:

$$f'(1)= Ce=1 \quad \Longrightarrow \quad C=\frac{1}{e}.$$

Thus,

$$f(x)= -1+\frac{1}{e}e^x = e^{x-1}-1.$$

Step 5. Find the Intersection with $y=\ln x^2$

We need to solve

$$f(x)=\ln x^2.$$

That is,

$$e^{x-1}-1=2\ln|x|.$$

Notice that, because $\ln x^2=2\ln|x|$ is defined for $x\neq 0$, intersections can occur for positive and negative $x$. A brief analysis shows:

1. For $x>0$:

   • At $x=1$: $$f(1)= e^{0}-1=0,\quad \ln1^2=0.$$ So one intersection is $(1,0)$.
   • There is another intersection for some $x \in (1.5, 2)$ where numerically one finds $f(x)\approx 1.12$ (since $e^{1.75-1}-1\approx1.12$ and $2\ln1.75\approx1.12$).

2. For $x<0$:
   Let $u=|x|$ (with $u>0$). Then

   $$f(-u)= e^{-u-1}-1 \quad \text{and} \quad \ln((-u)^2)=2\ln u.$$

   A numerical check shows there is a solution for some $u\in(0,1)$ approximately yielding $f(-u)\approx -0.81$.

Thus, there are three intersections:

• Intersection 1 (with lowest y–value): $y_1\approx -0.81$ (at some $x<0$).
• Intersection 2: $y_2=0$ (at $x=1$).
• Intersection 3: $y_3\approx 1.12$ (at $x\approx1.75$).

Let the total number of intersection points be $n = 3$.

Step 6. Compute the Required Expression

We are to find:

$$[y_1] + [y_2] + n,$$

where $[\,\cdot\,]$ is the greatest integer function.

• $y_1\approx -0.81$ so $[y_1] = -1$.
• $y_2=0$ so $[y_2] = 0$.
• $n=3$.

Thus,

$$[y_1]+[y_2]+n = (-1) + 0 + 3 = 2.$$