Skip to main content
Back to Differential Equations
JEE Main 2021
Differential Equations
Differential Equations
Hard

Question

Let f(x)f(x) be a real differentiable function such that f(0)=1f(0)=1 and f(x+y)=f(x)f(y)+f(x)f(y)f(x+y)=f(x) f^{\prime}(y)+f^{\prime}(x) f(y) for all x,yRx, y \in \mathbf{R}. Then \sum_\limits{n=1}^{100} \log _e f(n) is equal to :

Options

Solution

Key Concepts and Formulas

  • Functional Equation: An equation where the unknown is a function. Solving often involves substitution and differentiation.
  • Differential Equations: Equations involving derivatives. Separable equations are of the form dydx=g(x)h(y)\frac{dy}{dx} = g(x)h(y), solved by separating variables and integrating.
  • Logarithm Properties: loge(ab)=bloge(a)\log_e(a^b) = b \log_e(a).
  • Arithmetic Series: k=1nk=n(n+1)2\sum_{k=1}^n k = \frac{n(n+1)}{2}.

Step-by-Step Solution

Step 1: Analyze the Given Information

We are given the functional equation f(x+y)=f(x)f(y)+f(x)f(y)()f(x+y) = f(x)f'(y) + f'(x)f(y) \quad (*) and the initial condition f(0)=1f(0) = 1. We need to find n=1100logef(n)\sum_{n=1}^{100} \log_e f(n).

Step 2: Find f(0)f'(0)

To find f(0)f'(0), substitute x=0x=0 and y=0y=0 into the functional equation ()(*): f(0+0)=f(0)f(0)+f(0)f(0)f(0+0) = f(0)f'(0) + f'(0)f(0) f(0)=2f(0)f(0)f(0) = 2f(0)f'(0) Since f(0)=1f(0) = 1, we have 1=2(1)f(0)1 = 2(1)f'(0) f(0)=12f'(0) = \frac{1}{2}

Step 3: Derive a Differential Equation

Substitute y=0y=0 into the functional equation ()(*): f(x+0)=f(x)f(0)+f(x)f(0)f(x+0) = f(x)f'(0) + f'(x)f(0) f(x)=f(x)f(0)+f(x)f(0)f(x) = f(x)f'(0) + f'(x)f(0) Substitute f(0)=1f(0) = 1 and f(0)=12f'(0) = \frac{1}{2}: f(x)=f(x)(12)+f(x)(1)f(x) = f(x)\left(\frac{1}{2}\right) + f'(x)(1) f(x)=12f(x)+f(x)f(x) = \frac{1}{2}f(x) + f'(x) f(x)=12f(x)f'(x) = \frac{1}{2}f(x)

Step 4: Solve the Differential Equation

We have the differential equation dfdx=12f(x)\frac{df}{dx} = \frac{1}{2}f(x) Separate the variables: dff(x)=12dx\frac{df}{f(x)} = \frac{1}{2}dx Integrate both sides: dff(x)=12dx\int \frac{df}{f(x)} = \int \frac{1}{2}dx lnf(x)=12x+C\ln|f(x)| = \frac{1}{2}x + C Exponentiate both sides: f(x)=e12x+C=e12xeC|f(x)| = e^{\frac{1}{2}x + C} = e^{\frac{1}{2}x}e^C f(x)=Ae12xf(x) = Ae^{\frac{1}{2}x} where A=±eCA = \pm e^C is a constant.

Step 5: Determine the Constant A

Use the initial condition f(0)=1f(0) = 1: f(0)=Ae12(0)=Ae0=Af(0) = Ae^{\frac{1}{2}(0)} = Ae^0 = A Since f(0)=1f(0) = 1, we have A=1A = 1. Therefore, f(x)=ex2f(x) = e^{\frac{x}{2}}

Step 6: Evaluate the Summation

We need to find S=n=1100logef(n)=n=1100loge(en2)S = \sum_{n=1}^{100} \log_e f(n) = \sum_{n=1}^{100} \log_e \left(e^{\frac{n}{2}}\right) Using the property loge(ek)=k\log_e(e^k) = k, we have S=n=1100n2=12n=1100nS = \sum_{n=1}^{100} \frac{n}{2} = \frac{1}{2} \sum_{n=1}^{100} n Using the formula for the sum of the first nn natural numbers, k=1nk=n(n+1)2\sum_{k=1}^n k = \frac{n(n+1)}{2}, we get S=12(100(100+1)2)=12(100(101)2)=12(101002)=101004=2525S = \frac{1}{2} \left(\frac{100(100+1)}{2}\right) = \frac{1}{2} \left(\frac{100(101)}{2}\right) = \frac{1}{2} \left(\frac{10100}{2}\right) = \frac{10100}{4} = 2525

Common Mistakes & Tips

  • Remember to use the initial condition to find the constant of integration after solving the differential equation.
  • Don't forget the properties of logarithms when simplifying the summation.
  • Double-check your calculations, especially when dealing with fractions and summations.

Summary

We used the given functional equation and initial condition to derive a differential equation for f(x)f(x). Solving this differential equation and applying the initial condition, we found f(x)=ex/2f(x) = e^{x/2}. Then, we evaluated the sum n=1100logef(n)\sum_{n=1}^{100} \log_e f(n), which simplifies to 12n=1100n\frac{1}{2} \sum_{n=1}^{100} n. Using the arithmetic series formula, we found the sum to be 2525.

The final answer is \boxed{2525}. This corresponds to option (C).

Practice More Differential Equations Questions

View All Questions