JEE Mains · Maths · STD 12 - 6. Application of derivatives
Let \(f:[-1,1] \rightarrow R\) be defined as \(f(x)=a x^{2}+b x+c\) for all \(x \in[-1,1],\) where \(a , b , c \in R\) such that \(f (-1)=2, f ^{\prime}(-1)=1\) and for \(x \in(-1,1)\) the maximum value of \(f ^{\prime \prime}( x )\) is \(\frac{1}{2} .\) If \(f ( x ) \leq \alpha\) , \(x \in[-1,1],\) then the least value of \(\alpha\) is equal to ...... .
- A \(10\)
- B \(2\)
- C \(5\)
- D \(8\)
Answer & Solution
Correct Answer
(C) \(5\)
Step-by-step Solution
Detailed explanation
\(f :[-1,1] \rightarrow R\) \(f ( x )= ax ^{2}+ bx + c\) \(f(-1)=a-b+c=2 .....(1)\) \(f ^{\prime}(-1)=-2 a + b =1.....(2)\) \(f ^{\prime \prime}( x )=2 a\) \(\Rightarrow\) Max. value of \(f ^{\prime \prime}( x )=2 a =\frac{1}{2}\)…
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