JEE Mains · Maths · STD 11 - 4.2 Quadratic equations and inequations
The real number \(k\) for which the equation, \(2{x^2} + 3x + k = 0\) has two distinct real roots in \([0, 1]\)
- A lies between \(1 \) and \( 2 \)
- B lies between \(2 \) and \( 3 \)
- C lies between \(-1 \) and \( 0 \)
- D does not exist
Answer & Solution
Correct Answer
(D) does not exist
Step-by-step Solution
Detailed explanation
\(f(x)=2 x^{3}+3 x+k\) \(f^{\prime}(x)=6 x^{2}+3>0 \quad \forall x \in R\) \(\Rightarrow f(x)\) is strictly increasing function \(\Rightarrow f(x)=0\) has only one real root, so two roots are not possible
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