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For the following questions answer them individually
If the fourth term in the binomial expansion of $$\left(2^x. x^2 + \frac{1}{x^2}\right)^6$$ is 160, then x =
The middle term in the expansion of $$\left(x^2 - \frac{1}{x}\right)^8$$ is
$$A = \begin{bmatrix}1 & 2 & 2 \\2 & 1 & 2 \\2 & 2 & 1 \end{bmatrix} \Rightarrow A^2 - 4A =$$
If det A = 5 for a $$3 \times 3$$ matrix A, then det(5A) =
$$\lim_{x \rightarrow 3}\frac{x - 3}{\sqrt{x - 2} - \sqrt{4 - x}} =$$
$$\lim_{x \rightarrow \frac{\pi}{2}}\frac{1 - \sin^3 x}{\cos^2 x} =$$
In $$\triangle ABC, \angle BAC = 90^\circ, AD \perp BC, AD = 6$$ cm and BD = 4 cm. Then DC =
The vertices A, B, C, D in that order of a quadrilateral lie on a circle. Then $$\angle ABC + \angle ADC =$$
If a square is inscribed in a circle of radius $$\sqrt{\frac{2015}{2}}$$ then area of the square is
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