TS EAMCET 4 May 2018 Shift 1 Solved Paper

Consider the equation.
x2−(λa−5)x+(λ−6)=0
since,
A=

π

2

Therefore,
B+C=

π

2

And,
r=4R‌sin‌

A

2

‌sin‌

B

2

‌sin‌

C

2

r

R

=4 (

1

√2

)‌sin‌

B

2

‌sin(

π

4

−

B

2

)

5

2

=2√2‌sin‌

B

2

[

1

√2

‌cos‌

B

2

−

1

√2

‌sin‌

B

2

]

5

4

=sin‌

B

2

‌cos‌

B

2

−sin2

B

2

……. (I)
And,
r1=4R‌sin‌

A

2

‌cos‌

B

2

‌cos‌

C

2

λ

2

=2√2‌cos‌

B

2

[

1

√2

‌cos‌

B

2

+

1

√2

‌sin‌

B

2

]

λ

4

=cos2

B

2

+cos‌

B

2

‌sin‌ ‌

B

2

……. (II)
From equation (I) and (II)

λ

4

−

5

4

=1
λ−5=4
λ=9
Therefore, the quadratic equation becomes
x2−(9−5)x+(9−6)=0
x2−4x+3=0
(x−1)(x−3)=0
So, the roots of the equation are 1,3.