UPSC NDA 2 2019 Mathematics Solved Paper

Concept:The slope of the curve $y = f(x)$ at a point is given by $f'(x)$.The equation of a tangent line with slope $m$ passing through $(x_0, y_0)$ is:$y - y_0 = m (x - x_0)$.Explanation:Step 1: The curve $y = m e^{mx}$ (with $m > 0$) meets the $y$-axis at point $P$.On the $y$-axis, $x = 0$. Substitute $x = 0$ into the curve:$y = m e^{m \cdot 0} = m e^0 = m \cdot 1 = m$.Thus $P = (0, m)$.Step 2: Find the slope of the curve at $P$.Differentiate $y = m e^{mx}$ with respect to $x$:$\frac{dy}{dx} = m \cdot m e^{mx} = m^2 e^{mx}$.At $P$, $x = 0$, so slope $= m^2 e^{0} = m^2 \cdot 1 = m^2$.Step 3: Write the tangent equation using the point-slope form.Slope $= m^2$, point $(0, m)$:$y - m = m^2 (x - 0)$$\Rightarrow y - m = m^2 x$$\Rightarrow y = m^2 x + m$.Answer:$y = m^{2}x + m$ (Option D).