Question

Consider the function f(x)=x3−px2+qx defined on [1,3]. If f satisfies the hypothesis of Rolle's theorem such that c=32, then the value of 4p+q+16 is

Solution

f(x)=x3−px2+qx

f satisfies the hypothesis of Rolle's theorem.

⇒f(1)=f(3)

⇒1−p+q=27−9p+3q

⇒8p−2q=26

⇒4p−q=13 ⋯(1)

Also, f′(c)=0

⇒3c2−2pc+q=0

⇒3(32)2−2p(32)+q=0

⇒27−12p+4q=0 ⋯(2)

Solving (1) and (2), we get

p=254, q=12

∴4p+q+16=25+12+16=53

f satisfies the hypothesis of Rolle's theorem.

⇒f(1)=f(3)

⇒1−p+q=27−9p+3q

⇒8p−2q=26

⇒4p−q=13 ⋯(1)

Also, f′(c)=0

⇒3c2−2pc+q=0

⇒3(32)2−2p(32)+q=0

⇒27−12p+4q=0 ⋯(2)

Solving (1) and (2), we get

p=254, q=12

∴4p+q+16=25+12+16=53

Suggest corrections

0 Upvotes

Similar questions

View More...

People also searched for

View More...

- About Us
- Contact Us
- Investors
- Careers
- BYJU'S in Media
- Students Stories - The Learning Tree
- Faces of BYJU'S – Life at BYJU'S
- Social Initiative - Education for All
- BYJU'S APP
- FAQ
- Support

© 2021, BYJU'S. All rights reserved.