Let f(x)= cos -1( cos 2)+ sin -1( sin 1) x-x2. If operatornamesgn(f(x)) is maximum for true set of values of x ∈(a, b), then (a+b) is equal to

Question

Let f(x)= cos -1( cos 2)+ sin -1( sin 1) x-x2. If operatornamesgn(f(x)) is maximum for true set of values of x ∈(a, b), then (a+b) is equal to (A) 0 (B) 1 (C) -1 (D) 3

Tardigrade Admin · Accepted Answer

f ( x )=2+ x - x 2=-( x 2- x -2)=-( x -2)( x +1) Θ sgn ( f ( x ))=1 ⇒ f ( x )>0 ∴ x ∈(-1,2) ∴ text sum =1

Q. Let $f (x) = cos^{- 1} (cos 2) + sin^{- 1} (sin 1) x - x^{2}$ . If $sgn (f (x))$ is maximum for true set of values of $x \in (a, b)$ , then $(a + b)$ is equal to

0

1

-1

3

$f (x) = 2 + x - x^{2} = - (x^{2} - x - 2) = - (x - 2) (x + 1)$
$Θ s g n (f (x)) = 1$
$\Rightarrow f (x) > 0$
$∴ x \in (- 1, 2)$
$∴ sum = 1$

Q. Let f(x)=cos−1(cos2)+sin−1(sin1)x−x2. If sgn(f(x)) is maximum for true set of values of x∈(a,b), then (a+b) is equal to

0

1

-1

3

f(x)=2+x−x2=−(x2−x−2)=−(x−2)(x+1) Θsgn(f(x))=1 ⇒f(x)>0 ∴x∈(−1,2) ∴ sum =1

Q. Let $f (x) = cos^{- 1} (cos 2) + sin^{- 1} (sin 1) x - x^{2}$ . If $sgn (f (x))$ is maximum for true set of values of $x \in (a, b)$ , then $(a + b)$ is equal to

$f (x) = 2 + x - x^{2} = - (x^{2} - x - 2) = - (x - 2) (x + 1)$
$Θ s g n (f (x)) = 1$
$\Rightarrow f (x) > 0$
$∴ x \in (- 1, 2)$
$∴ sum = 1$