Let \( f(x) = \begin{cases} x^2 - \alpha, & \text{if } x < 1 \\ \beta x - 3, & \text{if } x \geq 1 \end{cases} \). If \( f \) is continuous at \( x = 1 \), then the value of \( \alpha + \beta \) is:
The general solution of the differential equation \( \frac{dy}{dx} = xy - 2x - 2y + 4 \) is:
An assignment of probabilities for outcomes of the sample space \( S = \{1, 2, 3, 4, 5, 6\} \) is as follows:
\[ \begin{array}{c|c c c c c c} 1 & 2 & 3 & 4 & 5 & 6 \\ \hline k & 3k & 5k & 7k & 9k & 11k \end{array} \]
If this assignment is valid, then the value of \( k \) is:
For \(1 \leq x<\infty\), let \(f(x) = \sin^{-1}\left(\frac{1}{x}\right) + \cos^{-1}\left(\frac{1}{x}\right)\). Then \(f'(x) =\)
\[ \int_0^{\frac{\pi}{4}} (\tan^3 x + \tan^5 x) \, dx \]
Let \( I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{\tan^2 x}{1+5^x} \, dx \). Then:
If a line makes angles \(\alpha\), \(\beta\), and \(\gamma\) with the positive directions of the x, y, and z-axis respectively, then \(\cos 2\alpha + \cos 2\beta + \cos 2\gamma\) equals:
Let \( f(x) = \log_e(x) \) and let \( g(x) = \frac{x - 2}{x^2 + 1} \). Then the domain of the composite function \( f \circ g \) is:
The first term and the 6th term of a G.P. are 2 and \( \frac{64}{243} \) respectively. Then the sum of first 10 terms of the G.P. is:
The value of the limit \(\lim_{x \to 0} \frac{(2 + \cos 3x) \sin^2 x}{x \tan(2x)}\) is equal to:
The value of the limit \(\lim_{t \to 0} \frac{(5-t)^2 - 25}{t}\) is equal to:
For a hyperbola, the vertices are at \( (6, 0) \) and \( (-6, 0) \). If the foci are at \( (2\sqrt{10}, 0) \) and \( -2\sqrt{10}, 0) \), then the equation of the hyperbola is:
The foci of the ellipse \(\frac{x^2}{49} + \frac{y^2}{24} = 1\) are:
The radius of the circle with centre at \((-4, 0)\) and passing through the point \((2, 8)\) is:
The centre of the hyperbola \(16x^2 - 4y^2 + 64x - 24y - 36 = 0\) is at the point:
Let \( f(x) = \frac{x^2 + 40}{7x} \), \( x \neq 0 \), \( x \in [4,5] \). The value of \( c \) in \( [4,5] \) at which \( f'(c) = -\frac{1}{7} \) is equal to: