Practicing Success
Practicing Success
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If x lies in [0, 1], then minimum value of x2 + x + 1 is |
$\frac{3}{4}$ 1 3 None of these |
1 |
$f(x)=x^2+x+1 \Rightarrow f'(x) 2 x+1>0$ for $x \in[0, 1]$ ∴ f(x) is monotonically increasing on [0, 1] ∴ Min. value of f(x) is f(0) = 1 |
