MHT CET · Maths · Application of Derivatives
The equation \(x^3+x-1=0\) has
- A no real root.
- B exactly two real roots.
- C exactly one real root.
- D all three real roots.
Answer & Solution
Correct Answer
(C) exactly one real root.
Step-by-step Solution
Detailed explanation
Given equation \(x^3+x-1=0\)
Let \(\mathrm{f}(x)=x^3+x-1\)
\(\begin{array}{ll}
\therefore \quad & \mathrm{f}^{\prime}(x)=3 x^2+1 \\
& \Rightarrow \mathrm{f}^{\prime}(x)>0 \forall x \in \mathrm{R}
\end{array}\)
\(\Rightarrow \mathrm{f}(x) \text { is increasing }\)
\(\therefore \quad\) for \(x_2>x_1, \mathrm{f}\left(x_2\right)>\mathrm{f}\left(x_1\right)\)
Now, \(\mathrm{f}(0)=-1\) and \(\mathrm{f}(1)=1\)
\(\therefore \quad \mathrm{f}(x)=0\) for some \(x \in(0,1)\)
\(\therefore \quad\) Equation has one real root.
Let \(\mathrm{f}(x)=x^3+x-1\)
\(\begin{array}{ll}
\therefore \quad & \mathrm{f}^{\prime}(x)=3 x^2+1 \\
& \Rightarrow \mathrm{f}^{\prime}(x)>0 \forall x \in \mathrm{R}
\end{array}\)
\(\Rightarrow \mathrm{f}(x) \text { is increasing }\)
\(\therefore \quad\) for \(x_2>x_1, \mathrm{f}\left(x_2\right)>\mathrm{f}\left(x_1\right)\)
Now, \(\mathrm{f}(0)=-1\) and \(\mathrm{f}(1)=1\)
\(\therefore \quad \mathrm{f}(x)=0\) for some \(x \in(0,1)\)
\(\therefore \quad\) Equation has one real root.
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