AP ICET 2014

The middle term will be the 21st term whose coefficient will be $$^{^{40}C}20=\frac{40!}{20!20!}=\frac{21\times\ 22\times\ 23.....\times\ 40}{20!}$$

AB=A and BA=B

Hence A and B are identity matrices

The square of an identity matrix is the matrix itself

Hence $$A^2=A\ and\ B^2=B$$

So $$A^2+B^2=A+B$$

Converting x degrees to radian. $$180^o=\pi^c\ \ or\ x^o=\frac{\pi}{180}x^c\ \ $$

$$Now\ \lim\ x\longrightarrow\ 0\ \frac{\tan x}{x}=1\ $$

Hence $$Now\ \lim\ x\longrightarrow\ 0\ \frac{\tan x}{x}\left(\frac{180}{\pi\ }\ \right)=\frac{180}{\pi\ }$$

Given, $$x=\sqrt{\ x+y}$$

$$\Rightarrow$$  $$x^2=\ x+y$$

$$\Rightarrow$$  $$y=x^2-x$$

$$\Rightarrow$$  $$\frac{dy}{dx}=2x-1$$

Hence, the correct answer is Option D

From mid-point theorem of triangles,we can say that DE||CA and DE=1/2(CA)=6

Similarly EF||AB and EF=1/2(AB)=4

And FD||BC and FD=1/2(BC)=7.5

EF+FD+DE=6+4+7.5=17.5

Quadrilateral formed after joining the mid-points of another quadrilateral is always a parallelogram.

$$\therefore\ $$Quadrilateral PQRS is a parallelogram.

Hence, the correct answer is Option B

$$\frac{AC}{CB}=2\ or\ AC=2CB$$

Hence C divides AB in the ratio of 2:1

Using the proportionality theorem of coordinate geometry

$$x=\frac{2\times\ \left(-5\right)+1\times\ 3}{3}=-\frac{7}{3}$$

$$x=\frac{2\times\ \left(4\right)+1\times\ \left(-5\right)}{3}=1$$

Hence the coordinates are (7/3,1)

D is the midpoint of BC. Hence coordinates of D is (3.5,4.5)

Now applying distance formula , AD is $$\sqrt{\ \left(0.5\right)^2+\left(2.5\right)^2}=\sqrt{\ 0.25+6.25}=\sqrt{\ 6.5}$$

$$2AD^2=2\times\ 6.5=13$$