Application of Derivatives — MATHS STD 12 Science — Question
Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSApplication of Derivatives2 Marks
Question
Find the maximum and minimum value $,f(x) = (2x - 1)^2 + 3$
✓
Answer
It is given that $f(x) = (2x - 1)^2 + 3$
Now, we can see that $(2x - 1)^2 \ge 0$ for every $x \in R$
$\Rightarrow f(x) = (2x - 1)^2 + 3 \ge 3$ for every $x \in R$
The minimum value of $f$ is attained when $2x - 1 = 0$
$2x - 1 = 0$
$\Rightarrow \mathrm{x}=\frac{1}{2}$
Then, Minimum value of $f=f\left(\frac{1}{2}\right)=\left(2 \cdot \frac{1}{2}-1\right)^{2}+3=3$
Now, $f\ ' (x) = 4x - 2 = 0,$
$\Rightarrow x = \frac12$ is the only critical point which is a minimum.
Therefore, function $f$ does not have a maximum value.
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.