Question

If \[f:R \to S\] defined by \[f\left( x \right) = \sin x - \sqrt 3 \cos x + 1\] is onto, then the interval of \[S\] is
A.\[\left[ {0,3} \right]\]
B.\[\left[ { - 1,1} \right]\]
C.\[\left[ {0,1} \right]\]
D.\[\left[ { - 1,3} \right]\]

Answer 1

Hint: Basically we have to find the range of the function \[f\left( x \right) = \sin x - \sqrt 3 \cos x + 1\]. We know \[\sin x\] and \[\cos x\] both have a range \[\left[ { - 1,1} \right]\] but for the same value of \[x\] they do not attain the same value . So to know the exact range of the function we have to reduce the trigonometric part to a single part ; either in terms of Sine or Cosine .
FORMULA USED:
\[\sin \left( {x - a} \right) = \sin x \times \cos a - \sin a \times \cos x\]

Complete step-by-step answer:
To know the exact range of the function we have to reduce the trigonometric part in a single term . Here we will reduce it in terms of the Sine function .
We can see \[\sqrt 3 \] is not a value of Sine function of known angle but we know \[\dfrac{{\sqrt 3 }}{2}\]is equal to \[\sin \,{60^ \circ }\]
So to reduce the trigonometric part of the function we will take \[2\]common from that part
\[\sin x - \sqrt 3 \cos x\]
\[ = \,2\left( {\sin x \times \dfrac{1}{2} - \dfrac{{\sqrt 3 }}{2}\cos x} \right)\]
After putting the value of \[\dfrac{1}{2}\]and \[\dfrac{{\sqrt 3 }}{2}\] in terms of Cosine and Sine function respectively we get
\[ = 2\,\left( {\sin x \times \cos \,{{60}^ \circ } - \sin \,{{60}^ \circ } \times \cos x} \right)\]
\[ = 2\,\sin \left( {x - \,{{60}^ \circ }} \right)\]
Using formula we get the last line .
So our function become
\[f\left( x \right) = \sin x - \sqrt 3 \cos x + 1\]
 \[ = 2\,\sin \left( {x - \,{{60}^ \circ }} \right) + 1\]
Value of \[\sin \left( {x - \,{{60}^ \circ }} \right)\] vary between \[\left[ { - 1,1} \right]\]
So range of \[2\,\sin \left( {x - \,{{60}^ \circ }} \right)\] is \[\left[ { - 2,2} \right]\]
So the range of the function \[2\,\sin \left( {x - \,{{60}^ \circ }} \right) + 1\] is \[\left[ { - 2 + 1\,,2 + 1} \right]\]
Which is \[\left[ { - 1,3} \right]\].
This is our answer .
Option \[4\]is the answer .
So, the correct answer is “Option 4”.

Note: we have to reduce the trigonometric part .
 This problem can also be solved by reducing in terms of Cosine function using the formula \[\cos \left( {x + a} \right) = \cos x \times \cos \,a - \sin x \times \sin a\] . In this method also we will get the same answer .
 For very quick solution we can use following method :
For the function like \[a\,\cos x + b\,\sin x + c\]range vary between \[c - \sqrt {{a^2} + {b^2}} \] and \[c + \sqrt {{a^2} + {b^2}} \]
Putting the value of \[a,b\,\& \,c\] for this question we will end up with the same answer .
Whatever method students follow will give the correct unique answer .