Let \( I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{\tan^2 x}{1+5^x} \, dx \). Then:
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:
Two circles \(C_1\) and \(C_2\) have radii 18 and 12 units, respectively. If an arc of length \( \ell \) of \(C_1\) subtends an angle 80° at the centre, then the angle subtended by an arc of same length \( \ell \) of \(C_2\) at the centre is:
If \( 2 \) is a solution of the inequality \( \frac{x-a}{a-2x}<-3 \), then \( a \) must lie in the interval:
Let \[ A = \begin{pmatrix} 2 & -1 & 1 \\ -1 & 0 & 2 \\ 1 & -2 & -1 \end{pmatrix} \] and let \( B = \frac{1}{|A|} A \). Then the value of \( |B| \) is equal to:
If \( f'(x) = 4x\cos^2(x) \sin\left(\frac{x}{4}\right) \), then \( \lim_{x \to 0} \frac{f(\pi + x) - f(\pi)}{x} \) is equal to:
The integral \(\int e^x \sqrt{e^x} \, dx\) equals: