If the function f(x)=x−1c−x2+1 does not take any value in the internal [−1,−13] , then the largest integral value that c can attain is equal to

Lety=f(x)=x−1c−x2+1 Takey=−t , wheret∈[13,1] , ∴−t=x−1c−x2+1 ⇒x2−c−1=x−1t⇒x2−1tx+1t−c−1=0 As −t∈[−1,(−13)] , hence the above must not posses real solution ∴ (1t)2−4(1t−c−1)<0⇒1t2−4t+4<−4c ⇒c<−14(1t−2)2 Now, 13≤t≤l⇒1≤1t−2≤(−1)⇒(−14≤14(1t−2)2≤0) Hence, c∈(−∞,−14] .

If the function f(x)=x−1c−x2+1 does not take any value in the internal [−1,−13] , then the largest integral value that c can attain is equal to

Lety=f(x)=x−1c−x2+1 Takey=−t , wheret∈[13,1] , ∴−t=x−1c−x2+1 ⇒x2−c−1=x−1t⇒x2−1tx+1t−c−1=0 As −t∈[−1,(−13)] , hence the above must not posses real solution ∴ (1t)2−4(1t−c−1)<0⇒1t2−4t+4<−4c ⇒c<−14(1t−2)2 Now, 13≤t≤l⇒1≤1t−2≤(−1)⇒(−14≤14(1t−2)2≤0) Hence, c∈(−∞,−14] .

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If the function $f (x) = \frac{x - 1}{c - x^{2} + 1}$ does not take any value in the internal $[- 1, - \frac{1}{3}]$ , then the largest integral value that c can attain is equal to

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Detailed Solution

Let $y = f (x) = \frac{x - 1}{c - x^{2} + 1}$

Take $y = - t$ , where $t \in [\frac{1}{3}, 1]$ ,

$∴ - t = \frac{x - 1}{c - x^{2} + 1}$

$\Rightarrow x^{2} - c - 1 = \frac{x - 1}{t} \Rightarrow x^{2} - \frac{1}{t} x + \frac{1}{t} - c - 1 = 0$

As $- t \in [- 1, (\frac{- 1}{3})]$ , hence the above must not posses real solution

$∴ {(\frac{1}{t})}^{2} - 4 (\frac{1}{t} - c - 1) < 0 \Rightarrow \frac{1}{t^{2}} - \frac{4}{t} + 4 < - 4 c$

$\Rightarrow c < - \frac{1}{4} {(\frac{1}{t} - 2)}^{2}$

Now, $\frac{1}{3} \leq t \leq l \Rightarrow 1 \leq \frac{1}{t} - 2 \leq (- 1) \Rightarrow (- \frac{1}{4} \leq \frac{1}{4} {(\frac{1}{t} - 2)}^{2} \leq 0)$