TS ICET 2019 Question Paper Shift-1 (23rd May)

$$\left(1 + x + x^2\right)^n = a_0 + a_1x + a_2x^2 + ......... + a_{2n}x^{2n}$$

Put x = 1

$$\left(1 + 1 + 1\right)^n = a_0 + a_1 + a_2 + .........+ a_{2n-1}+ a_{2n}$$

$$a_0 + a_1 + a_2 + .........+ a_{2n-1}+ a_{2n} = 3^{n}$$ ....... (1)

Put x = -1

$$\left(1 - 1 + 1\right)^n = a_0 - a_1 + a_2 - .........- a_{2n-1}+ a_{2n}$$

$$a_0 - a_1 + a_2 - .........- a_{2n-1}+ a_{2n} = 1$$ ....... (2)

From (1) and (2):

$$(a_0 + a_1 + a_2 + .........+ a_{2n-1}+ a_{2n})-(a_0 - a_1 + a_2 - .........- a_{2n-1}+ a_{2n}) = 3^{n} - 1$$

$$2a_1 + 2a_3 + 2a_5 + .........+ 2a_{2n-1} = 3^{n}-1$$

$$a_1 + a_3 + a_5 + .........+ a_{2n-1} = \frac{3^{n}-1}{2}$$

$$A = \begin{bmatrix}1 & 1 \\1 & 1 \end{bmatrix}$$

$$A^{2} = \begin{bmatrix}1 & 1 \\1 & 1 \end{bmatrix} * \begin{bmatrix}1 & 1 \\1 & 1 \end{bmatrix} = \begin{bmatrix}2 & 2 \\2 & 2 \end{bmatrix}$$

$$A^{2} = 2\begin{bmatrix}1 & 1 \\1 & 1 \end{bmatrix} = 2A$$

$$A = 2$$

$$A^{4} = \begin{bmatrix}2 & 2 \\2 & 2 \end{bmatrix} * \begin{bmatrix}2 & 2 \\2 & 2 \end{bmatrix} = \begin{bmatrix}8 & 8 \\8 & 8 \end{bmatrix}$$

$$A^{4} = (2A)^{2} = 8\begin{bmatrix}1 & 1 \\1 & 1 \end{bmatrix} = 8A$$[$$Since, A^{2} = 2A$$]

$$A^{4} = 4A^{2} = 8A = 2^{2}A^{2}$$

$$Now, (A^{4})^{504} = (2^{2})^{504}(A^{2})^{504}$$

$$A^{2016} = 2^{1008}(2^{2})^{504}$$[$$Since; A = 2$$]

$$A^{2016} = 2^{2016}$$

$$A^{2016}A^{3} = 2^{2016}A^{3}$$

$$A^{2019} = 2^{2016}A^{2}A$$

$$A^{2019} = 2^{2016}2^{2}A$$[$$Since; A = 2$$]

$$A^{2019} = 2^{2018}A$$

Sum of elements along principal diagonal of A = 3 

Sum of elements along principal diagonal of B = 4

Sum of principal diagonal elements of 3A + 5B = (3 * 3) + (5 * 4)

= 9 + 20

= 29

Since, $$cos2x = 1 - 2sin^{2}x$$

$$2sin^{2}x = 1 - cos2x$$

Similarly, we can write;

$$2sin^{2}\frac{x}{2} = 1 - cosx$$

Now, 

$$\lim_{x \rightarrow 0} \frac{1 - \cos x}{x^2} =$$

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

= $$\frac{1}{2}\lim_{x \rightarrow 0} (\frac{sin\frac{x}{2}}{\frac{x}{2}})^{2}$$

= $$\frac{1}{2}$$ * 1                                             [$$\lim_{x \rightarrow 0} \frac{sin x}{x} = 1$$]

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

$$=\frac{x^2 - \sqrt{x}}{\sqrt{x} - 1}$$

After differentiation:

$$\frac{2x - \frac{1}{2\sqrt{x}}}{\frac{1}{2\sqrt{x}}}$$

Now, 

= $$\lim_{x \rightarrow 1}\frac{2x - \frac{1}{2\sqrt{x}}}{\frac{1}{2\sqrt{x}}}$$

= $$\frac{2 - \frac{1}{2}}{\frac{1}{2}}$$

= $$\frac{\frac{3}{2}}{\frac{1}{2}}$$

= $$3$$

Let $$\angle$$A = $$30^\circ$$, $$\angle$$B = $$60^\circ$$ and $$\angle$$C = $$90^\circ$$

sin$$30^\circ$$ = $$\frac{BC}{AB}$$

$$\frac{1}{2}$$ = $$\frac{BC}{AB}$$

Let BC = x and AB = 2x

AC = $$\sqrt{AB^{2} - BC^{2}}$$

AC = $$\sqrt{3}x$$

Now, 

Required ratio = BC: AC: AB = x: $$\sqrt{3}x$$: 2x = 1: $$\sqrt{3}$$: 2

Area of rhombus = $$\frac{1}{2}$$ * d1 * d2

Where; 

d1 = First diagonal and d2 = Second diagonal

Now, 

24 = $$\frac{1}{2}$$ * 8 * d2

d2 = 6 cm

The difference between radii of both the circles = 16 - 7 = 9 cm

Since, the difference between radii of both the circles is more than the difference between their centers. This means that the smaller circle will be completely inside the bigger circle there will be no contact point. 

Since, there is no contact point of both the circle which means there will be no common tangent of both the circle.

A = (-2, 3), B = (-7, 5) and C = (3, -5)

G = $$\frac{(-2) + (-7) + 3}{3}$$,$$\frac{3 + 5 + (-5)}{3}$$ = (-2, 1)

Area of $$\triangle GAB$$ = $$\frac{1}{2}$$[(-2)(5 - 1) + (-7)(1 - 3) + (-2)(3 - 5)]

= $$\frac{1}{2}$$[-8 + 14 + 4]

= $$\frac{1}{2}$$ * 10

= 5 sq. units

Let co-ordinate of A = (x1,y1), B = (x2, y2) and C = (x3, y3)

Mid-point of AB = $$\frac{x1 + x2}{2}$$,$$\frac{y1 + y2}{2}$$ = (-4, 6)

x1 + x2 = -8 and y1 + y2 = 12 ............. (1)

Similarly, 

x2 + x3 = 4 and y2 + y3 = -6 ............. (2)

x1 + x3 = 6 and y1 + y3  = 14 ............. (3)

From (1), (2) and (3):

x1 + x2 + x3 = 1 and y1 + y2 + y3 = 10 .......... (4)

From (2) and (4):

x1 = -3 and y1 = 16

Hence, A = (-3, 16)