Application of Derivatives — Class 12 Maths Solution

ncert exercise SA NCERT Ex. 6.2, Q.11,Page 206
Question

Prove that the function $f$ given by $f\left( x \right) = {x^2} - x + 1$ is neither strictly increasing nor strictly decreasing on $( - 1,1)$ .

Step-by-step Solution

We have, $f(x) = {x^2} - x + 1\forall x \in ( - 1,1)$ …(i)

Differentiating (i) w.r.t. $x$, we get $f(x) = 2x - 1$

For increasing, $f'(x) > 0 \Rightarrow 2x - 1 > 0 \Rightarrow x > \cfrac{1}{2}$

For decreasing, $f'(x) < 0 \Rightarrow 2x - 1 < 0 \Rightarrow x < \cfrac{1}{2}$
$\Rightarrow f'(x) < 0$ for all $x \in \left( { - 1,\;\cfrac{1}{2}} \right)$

and $f'(x) > 0$ for all $x \in \left( {\cfrac{1}{2},1} \right)$

Hence, $f$ is neither increasing nor decreasing on $( - 1,1)$

NCERT & Exemplar solution for CBSE Class 12 Mathematics, Application of Derivatives. Curated by Sachin Sharma. Free for all students.