AP ICET 2nd May 2018 Shift-2

The power of x in the first term of expansion would be 18
The power of x in the second term would be 16(of the first term in original bracket) - 1 (of the second term in original bracket) = 15
The power of x in the third term would be 14 - 2 = 12
The power of x in the fourth term would be 12 - 3 = 9
The power of x in the fifth term would be 10 - 4 = 6
The power of x in the sixth term would be 8 - 5 = 3
The power of x in the seventh term would be 6-6 = 0

The seventh term would be independent of x

The value of seventh term would be (9C6)*$$\left(\frac{3}{2}\right)^3\cdot\left(\frac{1}{3}\right)^6$$ = 7/18

Upon successive multiplication, obtain the matrix $$A^5$$
Its terms are:
$$A_{11}=6140$$, $$A_{12}=8097$$, $$A_{21}=10796$$, $$A_{22}=14237$$

Cross multiplication and subtraction of terms results in determinant = -32

As can be seen from the options, $$A^{-1}$$ is going to be a power of A itself. 
We can continuing multiplying A successively to its own powers till we arrive at an identity matrix. 
If the identity matrix is arrived at $$A^n$$ , then we can say that $$A^{n-1\ }=\ A^{-1}$$

Starting with the multiplication, we can see that a 2x2 identity matrix is arrived at at $$A^4$$
Hence, $$A^3$$ is the inverse of matrix A. 

Multiplying numerator and denominator by $$\sqrt{\ 1+x}+\sqrt{\ 1-x}$$, we get:

$$\lim\ x\longrightarrow\ 0\ $$ $$\frac{\left(1+x\right)-\left(1-x\right)}{x\cdot\left(\sqrt{\ 1+x}+\sqrt{\ 1-x}\right)}$$
$$=\frac{2}{\left(\sqrt{\ 1+x}+\sqrt{\ 1-x}\right)}$$

Putting x = 0 now, we get 

Result = 2/2 = 1

Given,  $$y = x \sin x$$

$$\Rightarrow$$  $$\frac{dy}{dx}=x\cdot\left(\frac{d}{dx}\sin\ x\right)+\sin\ x\cdot\left(\frac{d}{dx}x\right)=x\cdot\cos\ x+\sin\ x$$

Let the length of equal sides of right angled isosceles triangle = a

In the triangle,

a$$^2$$ + a$$^2$$ = 8$$^2$$

$$\Rightarrow$$  2a$$^2$$ = 64

$$\Rightarrow$$  a$$^2$$ = 32

$$\Rightarrow$$  a = $$4\sqrt{2}$$ cm

$$\therefore\ $$Area of the right angled isosceles triangle = $$\frac{1}{2}\times a\times a$$

$$=\frac{1}{2}\times4\sqrt{2}\times4\sqrt{2}$$

$$=$$ 16 cm$$^2$$

We can draw the figure as:

We can conclude that AD will be 280 units in length and hence, perimeter = 620

In $$\triangle$$ABC,

$$\angle$$A + $$\angle$$B + $$\angle$$C = $$180^\circ$$

$$\angle$$A + $$60^\circ$$ + $$70^\circ$$ = $$180^\circ$$

$$\angle$$A = $$50^\circ$$

Angle subtended by the arc at the centre is twice the angle subtended by the arc at any point circle.

$$\Rightarrow$$ Angle subtended by the arc BC at centre O is twice the angle subtended by arc BC at point A.

$$\Rightarrow$$  $$\angle$$BOC = 2$$\angle$$A

$$\Rightarrow$$  $$\angle$$BOC = $$100^\circ$$

The midpoint of BC to which the median will connect from A : (3.5, 4.5)

Length of median = Distance between A(4,2) and (3.5, 4.5) = $$\sqrt{\left(3.5-4\right)^2+\left(4.5-2\right)^2}$$ = $$\sqrt{6.5}=\frac{\sqrt{\ 26}}{2}$$