Q62 / 150 · MHT CET 2023 (09 May Shift 2) · Mathematics

The equation $x^3+x-1=0$ has

  • A
    no real root.
  • B
    exactly two real roots.
  • Correct answer
    C
    exactly one real root.
  • D
    all three real roots.

Solution

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.

Hence, the correct answer is option: 'C'.

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