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Question: If \[f:R \to S\] defined by \[f\left( x \right) = \sin x - \sqrt 3 \cos x + 1\] is onto, then the in...

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

Explanation

Solution

Hint : Basically we have to find the range of the function f(x)=sin⁡x−3cos⁡x+1f\left( x \right) = \sin x - \sqrt 3 \cos x + 1. We know sin⁡x\sin x and cos⁡x\cos x both have a range [−1,1]\left[ { - 1,1} \right] but for the same value of xx 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⁡(x−a)=sin⁡x×cos⁡a−sin⁡a×cos⁡x\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 3\sqrt 3 is not a value of Sine function of known angle but we know 32\dfrac{{\sqrt 3 }}{2}is equal to sin⁡ 60∘\sin \,{60^ \circ }
So to reduce the trigonometric part of the function we will take 22common from that part
sin⁡x−3cos⁡x\sin x - \sqrt 3 \cos x
= 2(sin⁡x×12−32cos⁡x)= \,2\left( {\sin x \times \dfrac{1}{2} - \dfrac{{\sqrt 3 }}{2}\cos x} \right)
After putting the value of 12\dfrac{1}{2}and 32\dfrac{{\sqrt 3 }}{2} in terms of Cosine and Sine function respectively we get
=2 (sin⁡x×cos⁡ 60∘−sin⁡ 60∘×cos⁡x)= 2\,\left( {\sin x \times \cos \,{{60}^ \circ } - \sin \,{{60}^ \circ } \times \cos x} \right)
=2 sin⁡(x− 60∘)= 2\,\sin \left( {x - \,{{60}^ \circ }} \right)
Using formula we get the last line .
So our function become
f(x)=sin⁡x−3cos⁡x+1f\left( x \right) = \sin x - \sqrt 3 \cos x + 1
=2 sin⁡(x− 60∘)+1= 2\,\sin \left( {x - \,{{60}^ \circ }} \right) + 1
Value of sin⁡(x− 60∘)\sin \left( {x - \,{{60}^ \circ }} \right) vary between [−1,1]\left[ { - 1,1} \right]
So range of 2 sin⁡(x− 60∘)2\,\sin \left( {x - \,{{60}^ \circ }} \right) is [−2,2]\left[ { - 2,2} \right]
So the range of the function 2 sin⁡(x− 60∘)+12\,\sin \left( {x - \,{{60}^ \circ }} \right) + 1 is [−2+1 ,2+1]\left[ { - 2 + 1\,,2 + 1} \right]
Which is [−1,3]\left[ { - 1,3} \right].
This is our answer .
Option 44is 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⁡(x+a)=cos⁡x×cos⁡ a−sin⁡x×sin⁡a\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+ca\,\cos x + b\,\sin x + crange vary between c−a2+b2c - \sqrt {{a^2} + {b^2}} and c+a2+b2c + \sqrt {{a^2} + {b^2}}
Putting the value of a,b & ca,b\,\& \,c for this question we will end up with the same answer .
Whatever method students follow will give the correct unique answer .