TS ICET 20th August 2021 Shift-1

Find the term independent of $$x$$, where $$x \neq 0$$, in the expansion of $$\left(\frac{3x^{2}}{2} -\frac{1}{3x}\right)^{15}$$

If $$A = \begin{bmatrix}2 & 2 & 1\\1& 0 & 2\\2 & 1 & 2\end{bmatrix}$$ is the inverse of $$B = \begin{bmatrix} -2 & a & 4\\ 2 & 2 & -3\\ 1 & 2 & -2\end{bmatrix}$$ then a = 

If $$A = \begin{bmatrix}1 & 2 & 3\\ 0& 1 & 2\\0 & 0 & 1\end{bmatrix}, B = \begin{bmatrix}x\\ y\\ z\end{bmatrix}$$ and $$AB = \begin{bmatrix}6\\3\\ 1\end{bmatrix}$$ then $$x + y + z =$$

$$\lim_{x \rightarrow 0}\frac{2 \sin x - \sin 2x}{x^{3}} =$$

If $$y = \frac{x^{3} - 3}{7x^{2} + 2}$$, then $$x = 2$$, $$\frac{dy}{dx} =$$

In the adjacent figure $$\angle A=90^{\circ}$$ and $$AD \perp BC$$. If $$AC=10\sqrt{3}$$ and $$AD = 5$$ then $$DC =$$

In the quadrilateral ABCD, AB = AD. The bisectors of $$\angle$$ BAC and $$\angle$$ CAD meet the sides BC and CD at points E and F respectively. If CE = 6 cm, BE = 4 cm and BD = 15 cm, then the length of EF (in cm) is

A, B, C are three points on the circumference of a circle. If AB = 6 cm, AC = 8 cm and $$\angle BAC =90^{\circ}$$ then the radius of the circle (in cm) is

The perimeter of the triangle ABC with the vertices $$A(3, 2); B(-3, 2); C(0, 2 - 3 \sqrt{3})$$

The fourth vertex D of a parallelogram ABCD whose three vertices are A(-2. 3), B(6, 7) and C(8, 3) is