RD Sharma - Mathematics (Volume 1)

Book: RD Sharma - Mathematics (Volume 1)

Chapter: 17. Increasing and Decreasing Functions

Subject: Maths - Class 12th

Q. No. 19 of Exercise 17.2

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19

Show that the function x2 – x + 1 is neither increasing nor decreasing on (0, 1).

Given:- Function f(x) = x2 – x + 1


Theorem:- Let f be a differentiable real function defined on an open interval (a,b).


(i) If f’(x) > 0 for all , then f(x) is increasing on (a, b)


(ii) If f’(x) < 0 for all , then f(x) is decreasing on (a, b)


Algorithm:-


(i) Obtain the function and put it equal to f(x)


(ii) Find f’(x)


(iii) Put f’(x) > 0 and solve this inequation.


For the value of x obtained in (ii) f(x) is increasing and for remaining points in its domain it is decreasing.


Here we have,


f(x) = x2 – x + 1


⇒


⇒ f’(x) = 2x – 1


Taking different region from (0, 1)


a) let


⇒ 2x – 1 < 0


⇒ f’(x) < 0


Thus f(x) is decreasing in


b) let


⇒ 2x – 1 > 0


⇒ f’(x) > 0


Thus f(x) is increasing in


Therefore, from above condition we find that


⇒ f(x) is decreasing in and increasing in


Hence, condition for f(x) neither increasing nor decreasing in (0, 1)


Chapter Exercises

Exercise 17.1
Exercise 17.2
MCQ

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