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I have a doubt, kindly clarify. - Algebra - BITSAT-2

Find the domain f \left ( b \right ) = \frac{1}{\sqrt{b-5}} + b^{2} +\frac{1}{\sqrt{b+7}}

  • Option 1)

    b\epsilon \left [ -7,5 \right ]

  • Option 2)

    b\epsilon \left ( 5, \infty \right )

  • Option 3)

    b\epsilon \left ( -\infty, 7 \right )

  • Option 4)

    None of these.

 
Answers (1)
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As learnt in concept

Domain of function -

All posible values of x for f(x) to be defined is known as domain.

-

 

 \frac{1}{\sqrt {b-5}}+ b^{2}+\frac{1}{\sqrt{b+7}}

Here b-5>0

So, b>5.

and b+7>0

So,b>-7

Thus, b\in (5, \infty)

 


Option 1)

b\epsilon \left [ -7,5 \right ]

Incorrect option

Option 2)

b\epsilon \left ( 5, \infty \right )

Correct option

Option 3)

b\epsilon \left ( -\infty, 7 \right )

Incorrect option

Option 4)

None of these.

Incorrect option

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