Let a curve y = f(x) pass through the point (2, (logₑ2)²) and have slope 2y/(xlog ₑx) for all positive real value of x. Then the value of f(e) is equal to ______________.

Solution

Maths sir · 4 steps

Separate y and x, then put t = ln x; the integral on the right becomes ln(ln x).

1. Separate variables. The slope is dy/dx = 2y/(x ln x). Move every y to the left and every x to the right: dy/y = (2 dx)/(x ln x).
2. Integrate both sides. For the right side put t = ln x, so dt = dx/x and the integral becomes ∫ (2 dt)/t = 2 ln t. This gives ln y = 2 ln(ln x) + C.

   y = k(ln x)²

3. Use the point. At x = 2 the curve gives y = (ln 2)², so k(ln 2)² = (ln 2)² and k = 1. The curve is f(x) = (ln x)².
4. So. f(e) = (ln e)² = 1² = 1.

The idea to keep. When you see x ln x in a denominator, think of t = ln x straight away: ∫ dx/(x ln x) = ln(ln x). Then use the given point to fix the constant.

Answer from the NTA final answer key. Solution by PrepWiser.