TS ICET 2019 Question Paper Shift-1 (24th May)

$$(1 - 2x + 3x^2)^9$$

substitute  x=1

$$(1 - 2\times1 + 3\times1^2)^9$$

$$(1 - 2 + 3)^9$$

$$(2)^9$$

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

by help of $$$$$$A = \begin{bmatrix}a1 & a2 & a3 \\b1 & b2 & b3\\c1 & c2 & c3 \end{bmatrix} \

$$$$$$A = \begin{bmatrix}a2*b3+b2*a3 & a1*c3+c1*a3 & a1*b2+a2*b1 \\\end{bmatrix} \

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

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

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

A^{2019} = 3^{2018}\begin{bmatrix}1 & 1 & 1 \\1 & 1 & 1 \\1 & 1 & 1 \end{bmatrix} \

A^{2019} = 3^{2018}\begin{bmatrix}A \end{bmatrix} \

$$y = x \sin x$$, then at $$x = \frac{\pi}{2

$$y = x \sin x$$, then at $$x = \frac{180}{2}

\frac{dsin90}{dx}

1

In a $$\ \triangle\ $$ ABC,

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

$$\angle\ $$A=$$\angle\ $$B=$$\angle\ $$C=60$$^{\circ\ }$$

since, bisector of $$\angle\ $$B and$$\angle\ $$C meet at D

$$\therefore\ $$  $$\angle\ $$DBC=$$\angle\ $$DCB=30$$^{\circ\ }$$

In$$\triangle\ $$DBC,

$$\angle\ $$DBC+$$\angle\ $$DCB+$$\angle\ $$BDC=180$$^{\circ\ }$$

30$$^{\circ\ }$$+30$$^{\circ\ }$$+$$^{ }\angle\ $$BDC=180$$^{\circ\ }$$

$$\angle\ $$BDC=120$$^{\circ\ }$$

Let ABCD be rhombus with AC and BD it's diagonals bisecting each other at O ,where

AC=9cm,AO=OC=4.5cm

BD=12cm, BO=OD=6 cm 

In $$\triangle\ $$AOB,

$$\angle\ $$AOB=90$$^{\circ\ }$$(we know diagonals of rhombus bisect each other at 90$$^{\circ\ }$$)

Apply Pythagoras theorem,

$$AO^{ 2}$$+$$OB^{2 }$$=$$AB^{ 2}$$

$$4.5^{ 2}$$+$$6^{ 2}$$=$$AB^{ 2}$$

20.25$$\ +\ $$36=$$AB^{2 }$$

$$AB^{ 2}$$=56.25

AB=7.5

Perimeter of rhombus=4$$\times\ $$7.5=30

Using tangent secant theorem,

$$PT^{2 }$$=PA$$\times\ $$PB

5$$\times\ $$5=PA$$\times\ $$4

PA=$$\ \frac{25}{ 4}$$=6.25

Consider a traingle ABC,Let A(1,2)=$$\ (x_1,y_1)$$

B(2,3)=$$\ (x_2,y_2)$$

C(x,4)=$$\ (x_3,y_3)$$  

Area of traingle=

$$\ \frac{1 }{ 2}$$ [$$\ x_1*(y_2$$-$$y_3)$$+$$\ x_2*(y_3$$-$$y_1) $$+$$\ x_3*(y_1$$-$$y_2) $$]

40=$$\ \frac{1 }{2 }$$[1*(3$$\ -\ $$4)+2*(4$$\ -\ $$2)+x*(2$$\ -\ $$3)]

80=$$\ -\ $$1$$\ +\ $$4$$\ +\ $$x($$\ -\ $$1)

x=$$\ -\ $$77

$$\mid x-3 \mid$$ = $$\mid -77-3\mid$$=80

Area of quadrilateral ABCD= Area of $$\triangle\ $$DAB+ Area of $$\triangle\ $$BCD

Area of $$\triangle\ $$DAB=$$\ \frac{1 }{ 2}$$*[$$\ -\ $$3($$\ -\ $$1$$\ -\ $$3)+2(3$$\ -\ $$($$\ -\ $$2))+4($$\ -\ $$2$$\ -\ $$($$\ -\ $$1))]

=$$\ \frac{1}{ 2}$$[12$$\ \ +\ $$10$$\ -\ $$4]

=9sq.units

Area of $$\ \triangle\ $$BCD= $$\ \frac{1 }{ 2}$$*[4(2$$\ -\ $$($$\ -\ $$2))+($$\ -\ $$1)($$\ -\ $$2$$\ -\ $$3)+($$\ -\ $$3)(3$$\ -\ $$2)]

=$$\ \frac{1 }{2 }$$*[16$$\ +\ $$5$$\ -\ $$3]=9sq.units

Area of quadrilateralABCD=9$$\ +\ $$9=18sq.units