AP EAMCET · Maths · Application of Derivatives
The minimum value of \(2 x^2+x-1\) is :
- A \(-\frac{1}{4}\)
- B \(\frac{3}{2}\)
- C \(-\frac{9}{8}\)
- D \(\frac{9}{8}\)
Answer & Solution
Correct Answer
(C) \(-\frac{9}{8}\)
Step-by-step Solution
Detailed explanation
Let \(y=2 x^2+x-1\) \(y^{\prime}=4 x+1\) For maxima or minima, put \(y^{\prime}=0\) \(\Rightarrow \quad x=-\frac{1}{4}\) \(y^{\prime \prime}=4=+\mathrm{ve}\) \(y\) is minimum at \(x=-\frac{1}{4}\). Thus, minimum value \(=2\left(-\frac{1}{4}\right)^2+\left(-\frac{1}{4}\right)-1\)…
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