If f:A arrow B defined by f(x)=sinx-cosx+3√2 is an invertible function, then the correct statement can be

Question

If f:A arrow B defined by f(x)=sinx-cosx+3√2 is an invertible function, then the correct statement can be (A) A=[(π /4) , (5 π /4)],B=[3 √2 , 4 √2] (B) A=[(- π /4) , (5 π /4)],B=[2 √2 , 4 √2] (C) A=[(- π /4) , (3 π /4)],B=[√2 , 4 √2] (D) A=[(- π /4) , (3 π /4)],B=[2 √2 , 4 √2]

Tardigrade Admin · Accepted Answer

f(x)=√2((1/√2) sin x - (1/√2) cos â¡ x)+3√2=√2sin(x - (π /4))+3√2 Clearly, for options A and B f(x) is not injective and for the option C f(x) is not surjective Hence, f(x) is bijective only for option D

Q. If $f : A \to B$ defined by $f (x) = s in x - cos x + 32$ is an invertible function, then the correct statement can be

$A = [\frac{π}{4}, \frac{5 π}{4}], B = [32, 42]$

$A = [\frac{- π}{4}, \frac{5 π}{4}], B = [22, 42]$

$A = [\frac{- π}{4}, \frac{3 π}{4}], B = [2, 42]$

$A = [\frac{- π}{4}, \frac{3 π}{4}], B = [22, 42]$

$f (x) = 2 (\frac{1}{2} s in x - \frac{1}{2} cos x) + 32 = 2 s in (x - \frac{π}{4}) + 32$
Clearly, for options $A$ and $B$ $f (x)$ is not injective
and for the option $C$ $f (x)$ is not surjective
Hence, $f (x)$ is bijective only for option $D$

Q. If f:A→B defined by f(x)=sinx−cosx+32​ is an invertible function, then the correct statement can be

A=[4π​,45π​],B=[32​,42​]

A=[4−π​,45π​],B=[22​,42​]

A=[4−π​,43π​],B=[2​,42​]

A=[4−π​,43π​],B=[22​,42​]