Question

For f(x)=x + 1/x, x ∈ [1,3] the value of c for mean-value theorem and for f(x)=x²-4x+3 for roll's theorem are.......,Select Proper option from the given options.
(a) √3,1
(b) 2,1
(c) √3,2
(d) 2,√3

Answer 1

if f(x) is defined and continuous on the interval [a,b] and differentiable on (a,b), then there is at least one number c in the interval (a,b) (that is a< c < b) such that $\bf{f'(c)=\frac{f(b)-f(a)}{b-a}}$

here given, f(x) = x + 1/x , x belongs to [1, 3]
let's take a number in [1, 3] such that
f'(c) = [f(3) - f(1)]/(3 - 1)
1 - 1/c² = [3 + 1/3 - 1 - 1/1 ]/2
(c² - 1)/c² = 2/3
3c² - 3 = 2c²
c² = 3
c = ±√3 , but in [1,3] , c = √3

The special case, when f'(c) = 0, is known as Rolle's Theorem. 
means, f'(c) = 0
2c - 4 = 0 [ f'(x) = 2x - 4]
c = 2

hence, option (c) is correct.

Answer 2

Dear Student,

Solution:

1) Mean value theorem:

f'(c) =[f(b)-f(a)]/[b-a]

For f(x)=x + 1/x, x ∈ [1,3]

f'(x) =
$1 - \frac{1}{ {x}^{2} }$
f(b) = f(3) = 3+1/3 = 10/3

f(a) = f(1) = 1+1= 2

f'(c) =[f(b)-f(a)]/[b-a]

$1 - \frac{1}{ {c}^{2} } = \frac{ \frac{10}{3} - 2}{3 - 1} \\ \\ 1 - \frac{1}{ {c}^{2} } = \frac{4}{6} = \frac{2}{3} \\ \frac{ {c}^{2} - 1 }{ {c}^{2} } = \frac{2}{3} \\ \\ 3 {c}^{2} - 3 = 2 {c}^{2} \\ {c}^{2} = 3 \\ c = + - \sqrt{3} \\ \\ c = \sqrt{3}$
discarded c= -√3

2) Rolle's theorem:

f(x)=x²-4x+3

f'(c) =0

f'(x) = 2x-4

f'(c) = 2c-4= 0

c-2 =0

c= 2

Option C is correct ( √3,2)

Hope it helps you.