If the function f ( x ) = c − x 2 + 1 x − 1 does not take any value in the interval [ − 1 , − 3 1 ] , then the largest integral value that c can attain is equal to k, find 8k

Question

If the function f ( x ) = c − x 2 + 1 x − 1 ​ does not take any value in the interval [ − 1 , − 3 1 ​ ] , then the largest integral value that c can attain is equal to k, find 8k

TestHub · Accepted Answer

Let y = f ( x ) = c − x 2 + 1 x − 1 ​ Take y = − t , where t ∈ [ 3 1 ​ , 1 ] ​ ∴ − t = c − x 2 + 1 x − 1 ​ ​ ⇒ x 2 − c − 1 = t x − 1 ​ ⇒ x 2 − t 1 ​ x + t 1 ​ − c − 1 = 0 ​ As − t ∈ [ − 1 , − 3 1 ​ ] , hence the above must not possess real solution ∴ ( t 1 ​ ) 2 − 4 ( t 1 ​ − c − 1 ) < 0 ⇒ t 2 1 ​ − t 4 ​ + 4 ≤ − 4 c

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