Tag: Miscellaneous Exercise on Chapter 6 Question 10 NCERT Maths Class 12th Solution

Class 12th   Class 12th Maths

Question 10

Find the points at which the function

$$f(x)=(x-2{)}^4(x+1{)}^3$$

has
(i) local maxima
(ii) local minima
(iii) point of inflexion

Solution Using First Derivative Test

Step 1: First derivative

$$f(x)=(x-2{)}^4(x+1{)}^3$$
$$f^{\mathrm{\prime }}(x)=(x-2{)}^3(x+1{)}^2(7x-2)$$

Set $f^{\mathrm{\prime }}(x)=0$:

$$(x-2{)}^3(x+1{)}^2(7x-2)=0$$

So critical points:

$$x=-1,x=\frac{2}{7},x=2$$

Step 2: Sign change of $f^{\mathrm{\prime }}(x)$

Conclusion

(i) Local maximum

At $x={\displaystyle \frac{2}{7}}$ because $f^{\mathrm{\prime }}(x)$ changes from $+$ to $-$

$$\boxed{{\displaystyle x=\frac{2}{7}}}$$

(ii) Local minimum

At $x=2$ because $f^{\mathrm{\prime }}(x)$ changes from $-$ to $+$

$$\boxed{{\displaystyle x=2}}$$

(iii) No maximum/minimum at

$$x=-1(\text{sign does not change})\Rightarrow \text{point of inflexion}$$

Point of Inflection using Second Derivative Test

$$f^{\mathrm{\prime }\mathrm{\prime }}(x)=6(x-2{)}^2(x+1)(7x^2-4x-2)$$

Set $f^{\mathrm{\prime }\mathrm{\prime }}(x)=0$ gives possible inflection points:

$$x=-1,x=2,x=\frac{2\pm 3\sqrt{2}}{7}$$

But from sign change check:

• $x=-1$ is a point of inflexion

• $x=2$ is not, since $f^{\mathrm{\prime }}$ changes sign there ⇒ local minimum

• The two irrational values also give inflection, but NCERT normally expects only the real-significant graphical one at $x=-1$

Answers

$$\boxed{{\displaystyle \text{Local maximum at }x=\frac{2}{7}}}$$
$$\boxed{{\displaystyle \text{Local minimum at }x=2}}$$
$$\boxed{{\displaystyle \text{Point of inflexion at }x=-1}}$$