JEE Mains · Maths · STD 12 - 5. continuity and differentiation
Let \(f :[0,1] \rightarrow R\) be a twice differentiable function in \((0,1)\) such that \(f (0)=3\) and \(f (1)=5\). If the line \(y=2 x+3\) intersects the graph of \(f\) at only two distinct points in \((0,1)\), then the least number of points \(x \in(0,1)\), at which \(f ^{\prime \prime}( x )=0\), is\(......\)
- A \(3\)
- B \(4\)
- C \(2\)
- D \(5\)
Answer & Solution
Correct Answer
(C) \(2\)
Step-by-step Solution
Detailed explanation
\(f ^{\prime}( a )= f ^{\prime}( b )= f ^{\prime}( c )=2\) \(\Rightarrow f ^{\prime \prime}( x ) \text { is zero }\) for atleast \(x_{1} \in(a, b) \& x_{2} \in(b, c)\)
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