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

let the boys are standing in the triangle $$\triangleABC$$  here D is the point in between AB 

so the height is CD = 80 m and $$\angleCAD$$ = 45 and   $$\angleCBD$$ = 60  

so we can say that 

tan45 = CD\AD = 80\AD $$\Rightarrow$$ AD = 80 

tan60 = CD\BD = 80\BD $$\Rightarrow$$ BD = 80\$$\sqrt{3}$$

so that the total ditance is AD + BD = 80 + 80\$$\sqrt{3}$$

$$\frac{80}{3}(3+\sqrt{3})$$   answer 

we have given that 

$$3x^2 - 5x - 8 < 0$$,

now factorize  the equation we get 

$$3x^2 - 8x  + 3x  - 8 < 0$$

 $$3x^2 + 3x -  8x - 8 < 0$$ 

so we get 

(x + 1) (3x - 8) < 0

x = -1 , 8\3  answer 

we have given that the root of the polynomonal equation 

$$\frac{3}{2}, \frac{2}{3}, \pm$$ \sqrt{3}$$ is mean that 

(x - 3\2 = 0 )   ( x - 2\3 = 0 )    ( x$$ \pm $$ $$\sqrt{3}$$ = 0 ) 

so we have (3x - 2) (2x - 3) ($$ x^{2}$$ - 3)  = 0  is the equation so the 

(3x - 2) (2x - 3) ($$ x^{2}$$ - 3) = 0    answer 

we have given 

$$\left(x^2 - 3x + 2) $$ =0                             equation 1 

and $$ left(x^3 - 6x^2 + Ax + B) $$ = 0           equation 2 

now we have to factorize  equation 1 we get the equation 

(x - 1) (x - 2) = 0 

x = 1, 2 

put the value of x in equation 2 we get 

2a +  b = 16                                                       equation 3

a + b = -5                                                          equation 4 

on solving eq 3 and eq 4 we get   

a = 11

b = 6 

then a^2 + b^2 = 157 answer 

we have given that 

 $$x - 3$$ = 0  so we get x = 3 put the value of x in the following equation 

 $$x^4 - 2x^3 + 3x^2 - mx + 3$$ = 0

3^4 - 2$$\times{3^3}$$ + 3$$\times{3^2}$$ - m$$\times{3}$$ + 3 = 0 

m$$\times{3}$$ = 57 

m = 19 answer 

we have given that 

$$\frac{4}{x - 3} + \frac{6}{y - 4} = 5$$  equation 1

$$\frac{5}{x - 3} - \frac{3}{y - 4} = 1$$  equation 2 

now    2$$\times{equation}$$  -  equation 1 

 2{$$\frac{5}{x - 3} - \frac{3}{y - 4} = 1$$}  -

 $$\frac{4}{x - 3} + \frac{6}{y - 4}  = 5$$

then we get x= 5 put the value of x in equation 1 

$$\frac{4}{5 - 3} + \frac{6}{y - 4} = 5$$   we get y = 6 

then x + y = 11 answer 

we have given 

 x + y = 30,    and     x, y(x < y)

then we have 

y - x  > 0   

y  > x 

then y should be positive integer and the  x have the following condition 

x > 0

x = 0 

x < 0  

so we have three pair of satisfying the condition of (x,y) in equation x + y = 30     answer 

Let a and d are first term and common difference of an A.P.

$$n^{th}$$ term will be  $$a_{n}$$ = a + (n-1)d then we have 

$$a_{7}$$ = a + (7-1)d     

$$a_{7}$$ = a + 6d             equation 1

$$a_{11}$$ = a + (11-1)d    

$$a_{11}$$ = a + 10d           equation 2 

now eq2 - eq1 

we get  4d = 16  then  d = 4    put the value in eq 1 

we get  a  = 31 

so the 15^(th) term will be 

$$a_{15}$$ = 31 + (15-1)4 

$$a_{15}$$ = 63  answer 

let the first term is a and common ratio is r then 

$$T_{6}$$= {a}$$\times{r}^{5}$$ =24           equation 1                                                            

$$T_{13}$$= {a}$$\times{r}^{12}$$ =3\16     equation 2

equation 1 \equation 2

24\{3\16} = 1\r^7

128  = 1\r^7 

2 = 1\r 

r = 1\2 

put the value of r in equation 1 

a$$\times\frac{1}{2}^{5}$$ =24 

so we get a= 768  so the series is 

768, 384,192,96,48,24,12,6,3,3\2,3\2^2,.............. so the 25^{th} term is 

 
 $$\frac{3}{2^{16}}$$   answer

we know that general term of expansion (a+b)^n is 

$$ T_{r+1}$$ = $$nC_{r}$${a}^{n-r}b^n       x$$\geq0$$

 here in expression  $$\left(2x^2 + \frac{1}{x^2}\right)$$  we have given   n=1    a=2x^2     b=1\x^2

$$ T_{r+1}$$ = $$1C_{r}$${2x^2}^{1-r}{1\x^2}^1

                   =   $$1C_{r}$${2}^{1-r}{x^2}^{1-r}{1\x^2}^1

                   =    $$1C_{r}$${2}^{1-r}{x}^{2-2r}{x}^-2r

                  =     $$1C_{r}$${2}^{1-r}{x}^-4r

so  from  that   n-r =0  and  r=4      so     n=$$5^{th}$$ term      answer