Tag: Miscellaneous Exercise on Chapter 6 Question 13 NCERT Class 12th Maths Solution

Class 12th   Class 12th Maths

Question 13.
Let $f$ be a function defined on $[a,b]$ such that $f^{\mathrm{\prime }}(x)>0$ for all $x\in (a,b)$.
Prove that $f$ is an increasing function on $(a,b)$.

Proof

Given:

$$f^{\mathrm{\prime }}(x)>0\text{for all }x\in (a,b)$$

We need to prove:

$$x_1<x_2\Rightarrow f(x_1)<f(x_2)$$

which means $f$ is increasing on $(a,b)$.

Using the Mean Value Theorem (MVT)

Since $f$ is differentiable on $(a,b)$ and continuous on $[a,b]$, by Mean Value Theorem, for any two points $x_1,x_2\in (a,b)$ with $x_1<x_2$, there exists some $c$ in $(x_1,x_2)$ such that:

$$f^{\mathrm{\prime }}(c)=\frac{f(x_2)-f(x_1)}{x_2-x_1}$$

Given $f^{\mathrm{\prime }}(x)>0$ for all $x$, we have:

$$f^{\mathrm{\prime }}(c)>0$$

Therefore:

$$\frac{f(x_2)-f(x_1)}{x_2-x_1}>0$$

Since $x_2-x_1>0$, multiplying both sides gives:

$$f(x_2)-f(x_1)>0$$
$$\Rightarrow f(x_2)>f(x_1)$$

So, whenever $x_1<x_2$, $f(x_1)<f(x_2)$, meaning:

$$\boxed{{\displaystyle f\text{ is an increasing function on }(a,b)}}$$

Conclusion

Since $f^{\mathrm{\prime }}(x)>0$ for every point in $(a,b)$, the slope of the tangent line to the graph of $f$ is always positive, so the function always rises as x increases. Hence, $f$ is increasing on $(a,b)$.