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] .