TS EAMCET 9-Sep-2020 Shift 1 Solved Paper

(A) We know that,
−1≤cos‌x≤1
⇒0≤cos2x≤1
⇒1≤1+cos2x≤2
∴[1+cos2x]={1,2}
So, range of sec−1[1+cos2x]={sec−11,sec−12}
(A)→(V)
(B) We have,
f(x+‌

1

x

)=x2+‌

1

x2

f(x+‌

1

x

)=(x+‌

1

x

)2−2
Put y=x+‌

1

x

∴‌‌f(y)=y2−2
Now, AM≥GM
Case I When x>0,‌

x+ 1 x

2

≥√x⋅‌

1

x

⇒‌‌x+‌

1

x

≥2
Case II When∴x+‌

1

x

∈R
R−(−2,2)
So, domain of f(x) is R−(−2,2)
(B) → (IV)
(C) We have
f(x+y)=f(x)+f(y)
Put x=y=0, we get
f(0)=f(0)+f(0)
⇒‌‌f(0)=0
Now, put y=−x, we get
f(x−x)=f(x)+f(−x)
⇒‌‌f(0)=f(x)+f(−x)
⇒f(x)+f(−x)=0[∵f(0)=0]
⇒f(−x)=−f(x)
So, f(x) is a odd function.
(C)→I
D. We have
sin−1x−cos−1x+sin−1(1−x)=0
⇒‌‌sin−1(1−x)=cos−1x−sin−1x
⇒‌‌sin−1(1−x)=‌

π

2

−2sin−1x
⇒‌‌1−x=sin(‌

π

2

−2sin−1x)
⇒‌‌1−x=cos(2sin−1x)
⇒‌‌1−x=cos(sin−12x√1−x2)
⇒‌‌1−x=cos‌cos−1(1−2x2)
⇒1−x=1−2x2
⇒2x2−x=0⇒x(2x−1)=0
⇒x=0,

1

2

∴x∈{0,‌

1

2

}
‌ (D) ‌→&‌ II ‌