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If \frac{6a^{2}-5a-3}{a^{2}-2a+6} \leq 4   the least and the highest value of 4a^{2} are

  • Option 1)

    0 and 81

  • Option 2)

    9 and 81

  • Option 3)

    36 and 81

  • Option 4)

    None of these

 

Answers (1)

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}

 

 \frac{6a^{2}-5a-3}{a^{2}-2a+6}-4\leq 0\\*\\*\frac{2a^{2}+3a-27}{a^{2}-2a+6}\leq 0\\\*\\*\frac{2a^{2}+9a-6a-27}{a^{2}-2a+6}\leq 0\\*\\*\frac{\left ( a-3 \right )(2a+9)}{a^{2}-2a+6}\leq 0\ \\*\\*a\in \left ( \frac{-9}{2},3 \right )

Thus maximum value of

a^{2}=\frac{81}{4}\\*\\*= > 4a^{2}=81

 


Option 1)

0 and 81

Correct

Option 2)

9 and 81

Incorrect

Option 3)

36 and 81

Incorrect

Option 4)

None of these

Incorrect

Posted by

prateek

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