Q56 / 170 · Manipal (MET) 2018 · Mathematics
The minimum value of $x^2+\frac{1}{1+x^2}$ is, at
- Correct answerA$x=0$
- B$x=1$
- C$x=4$
- D$x=3$
Solution
Let $f(x)=x^2+\frac{1}{1+x^2}$
On differentiating w.r.t. $x$, we get $f^{\prime}(x)=2 x-\frac{1}{\left(1+x^2\right)^2} 2 x$ For a minimum, put $f^{\prime}(x)=0$ $\Rightarrow \quad x=0$
So, the function has minimum value at $x=0$.
On differentiating w.r.t. $x$, we get $f^{\prime}(x)=2 x-\frac{1}{\left(1+x^2\right)^2} 2 x$ For a minimum, put $f^{\prime}(x)=0$ $\Rightarrow \quad x=0$
So, the function has minimum value at $x=0$.
Hence, the correct answer is option: 'A'.