Join Our JEE Preparation Group
Prep with like-minded aspirants; Get access to free daily tests and study material.
Sign in
Please select an account to continue using cracku.in
↓ →
Join Our JEE Preparation Group
Prep with like-minded aspirants; Get access to free daily tests and study material.
If $$f(x) = \frac{x^4 + x^2 + 1}{x^2 - x + 1}$$, then the minimum value of $$f(x)$$ is
$$x^4 + x^2 + 1 = (x^2 + x + 1)(x^2 - x + 1)$$ and $$x^2 - x + 1 > 0$$ for all $$x$$, so $$f(x) = x^2 + x + 1 = \left(x + \frac{1}{2}\right)^2 + \frac{3}{4}$$. The minimum value is $$\frac{3}{4}$$, attained at $$x = -\frac{1}{2}$$.
Was this solution helpful?
Predict your JEE Main percentile, rank & performance in seconds
Educational materials for JEE preparation
Ask our AI anything
AI can make mistakes. Please verify important information.
AI can make mistakes. Please verify important information.