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If f:R\rightarrow S, defined by f\left ( x \right )= \sin x-\sqrt{3}\cos x+1 is onto, then the interval  of S is

  • Option 1)

    \left [ 0,1 \right ]

  • Option 2)

    \left [-1,1 \right ]

  • Option 3)

    \left [ 0,3 \right ]

  • Option 4)

    \left [ -1,3 \right ]

 

Answers (2)

best_answer

As we learnt in

Range -

The range of the relation R is the set of all second elements of the ordered pairs in a relation R.

- wherein

eg. R={(a,b),(c,d)}. Then Range is {b,d}

 

 f(x)=sinx-\sqrt{3}\;cos x + 1

    =2\left(\frac{1}{2}sinx-\frac{\sqrt{3}}{2}cos x \right )+1

    =2\left(cos 60^{o}. sin y-sin 60^{o}cos x\right )+1

    =2\left(sin x. cos\frac{\pi}{3}-cos x .sin\frac{\pi}{3}\right )+1

    =2sin\left(x-\frac{\pi}{3} \right )+1

Now =sin\left(x-\frac{\pi}{3} \right )\leq 1

\therefore        -2 + 1,        2 + 1

            [- 1, 3]

Correct option is 4.

 


Option 1)

\left [ 0,1 \right ]

This is an incorrect option.

Option 2)

\left [-1,1 \right ]

This is an incorrect option.

Option 3)

\left [ 0,3 \right ]

This is an incorrect option.

Option 4)

\left [ -1,3 \right ]

This is the correct option.

Posted by

Aadil

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