Home
Class 12
MATHS
f(x) = (x - 1) (x - 2) (x - 3) , x in [ ...

` f(x) = (x - 1) (x - 2) (x - 3) , x in [ 0, 4 ] " find "'c' if `
LMVT can be applied .

Text Solution

Verified by Experts

`f(x) = (x - 1) (x-2) (x-3) , in [0,4]`
`therefore f(x) = x^(3) - 6x^(2) + 11x - 6`
As f(x) is a polynomial in x
(i) f(x) is continuous on [0.4]
(ii) f(x) is differentiable of LMVT are satisfied . To
verify LMVT we have to find c in (0, 4), such that
`f'(c) = (f(4)- f(0))/(4-0) ` ...(i) Now , ` f(4) = (4 - 1) (4-2) (4 - 3) = 6`
` f(0) = (0-1) (0-2) (0-3) = - 6 `
and ` f'(x) = 3x^(2) - 12x + 11`
` therefore f'(c) = 3c^(2) - 12 c + 11`
Now , ` (f (4) - f(0))/(4-0) = 3c^(2) - 12c + 11 " "` [ From (ii)]
` (6-(-6))/(4) = 3c^(2) - 12c + 11`
` 3 = 3c^(2) - 12 c + 11`
` 3c^(2) - 12c + 8 = 0 `
` c = (12pmsqrt(144 - 96))/(6)`
` c = 2 pm (2)/(3) sqrt(3)`
` therefore c = 2 pm (2)/(sqrt(3))`
Both the values of c lie between 0 and 4 .
Promotional Banner

Topper's Solved these Questions

  • MARCH 2019

    GURUKUL PUBLICATION - MAHARASHTRA PREVIOUS YEAR PAPERS|Exercise SECTION-C|8 Videos
  • MARCH 2018

    GURUKUL PUBLICATION - MAHARASHTRA PREVIOUS YEAR PAPERS|Exercise SECTION - II|20 Videos
  • OCTOBER 2014

    GURUKUL PUBLICATION - MAHARASHTRA PREVIOUS YEAR PAPERS|Exercise SECTION - II|19 Videos

Similar Questions

Explore conceptually related problems

If f(x)=((3+x)/(1+x))^(2+3x), find f'(0)

If c=(1)/(2) and f(x)=2x-x^(2) , then interval of x in which LMVT is applicable, is

Using Lagrange's theorem , find the value of c for the following functions : (i) x^(3) - 3x^(2) + 2x in the interval [0,1/2]. (ii) f(x) = 2x^(2) - 10x + 1 in the interval [2,7]. (iii) f(x) = (x-4) (x-6) in the interval [4,10]. (iv) f(x) = sqrt(x-1) in the interval [1,3]. (v) f(x) = 2x^(2) + 3x + 4 in the interval [1,2].

If f(x)=(1+x)(1+x^2)(1+x^3)(1+x^4) , then f'(0)=1

Find the value of c, of mean value theorem.when (a) f(x) = sqrt(x^(2)-4) , in the interval [2,4] (b) f(x) = 2x^(2) + 3x+ 4 in the interval [1,2] ( c) f(x) = x(x-1) in the interval [1,2].

Statement - 1 : Rolle's Theorem can be applied to the function f(x)=1+(x-2)^(4//5) in the interval [0, 4] and Statement - 2 : f(x) is continuous in [0, 4] and f(0)=f(4) .

Evaluate If f' (x) = 4x ^(3) - 3x ^(2) + 2x + k, f (0) = 1 and f (1) = 4 find f (x)