Step 1:
The equation of the curve is given by:
\[ \left( \frac{x}{a} \right)^n + \left( \frac{y}{b} \right)^n = 2 \] We are asked to find the value of \( n \) such that this curve touches the straight line \( \frac{x}{a} + \frac{y}{b} = 2 \) at the point \( (a, b) \).
Step 2:
Substituting the point \( (a, b) \) into the equation of the curve: \[ \left( \frac{a}{a} \right)^n + \left( \frac{b}{b} \right)^n = 2 \] This simplifies to: \[ 1^n + 1^n = 2 \] which holds true for any \( n \).
Step 3:
To ensure the curve touches the straight line at the point \( (a, b) \), the curve must be in a form that can represent a degenerate conic section (like an ellipse or hyperbola) that meets the straight line exactly at one point. This is achieved when \( n = 2 \), which corresponds to the equation of an ellipse, and the curve touches the line.
Step 4:
Therefore, the correct value of \( n \) is \( n = 2 \).
If the probability density function of a random variable is given by $$ f(x) = \begin{cases} 12x^2(1 - x), & 0 \leq x \leq 1 \\ 0, & \text{elsewhere} \end{cases} $$ then the mean and variance are respectively:
The area of a parallelogram whose diagonals are given by $ \vec{u} + \vec{v} $ and $ \vec{v} + \vec{w} $, where:
$ \vec{u} = 2\hat{i} - 3\hat{j} + \hat{k}, \quad \vec{v} = -\hat{i} + \hat{k}, \quad \vec{w} = 2\hat{j} - \hat{k} $ is:
The direction ratios of the normal to the plane passing through the points
$ (1, 2, -3), \quad (1, -2, 1) \quad \text{and parallel to the line} \quad \frac{x - 2}{2} = \frac{y + 1}{3} = \frac{z}{4} \text{ is:} $
At 15 atm pressure, $ \text{NH}_3(g) $ is being heated in a closed container from 27°C to 347°C and as a result, it partially dissociates following the equation: $ 2\text{NH}_3(g) \rightleftharpoons \text{N}_2(g) + 3\text{H}_2(g) $ If the volume of the container remains constant and pressure increases to 50 atm, then calculate the percentage dissociation of $ \text{NH}_3(g) $
If equilibrium constant for the equation $ A_2 + B_2 \rightleftharpoons 2AB \quad \text{is} \, K_p, $ then find the equilibrium constant for the equation $ AB \rightleftharpoons \frac{1}{2} A_2 + \frac{1}{2} B_2. $
Consider the following reaction: $ \text{CO}(g) + \frac{1}{2} \text{O}_2(g) \rightarrow \text{CO}_2(g) $ At 27°C, the standard entropy change of the process becomes -0.094 kJ/mol·K. Moreover, standard free energies for the formation of $ \text{CO}_2(g) $ and $ \text{CO}(g) $ are -394.4 and -137.2 kJ/mol, respectively. Predict the nature of the above chemical reaction.