Previous Year Papers

Download Solved Question Papers Free for Offline Practice and view Solutions Online.

Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.

Continuity and Differentiability

Short Answer Type

591. Verify Roll's Theorem for the function :f(x) = x (x - 3)² in 0 ≤ x ≤ 3

83 Views

592. Verify Roll's Theorem for the function :f(x) = (x² - 1) (x - 2) in [-1, 2]

Let f (x) = (x² - 1) (x - 2) = x³ - 2 x² - x + 2
It is a polynomial in x
(a)    Since every polynomial in x is a continuous function for every value of x.
∴ f(x) is continuous in [-1, 2].
(b)    f'(x) = 3 x² - 4 x - 1, which exists in (-1, 2)
∴ f is derivable in (-1, 2).
(c)    f(-1) = (1 - 1) (-1 -2) = 0
f(2) = (4 - 1) (2 - 1) (2 - 2) = 0
∴ f(-1) = f(2)
∴ f satisfies all the conditions of Rolle's Theorem
∴ there must exist at least one value c ofx such that f'(c) = 0 where - 1 < c < 2.
Now f'(c) = 0 gives us 3 c² - 4 c - 1 = 0
$or space straight c equals fraction numerator 4 plus-or-minus square root of 16 plus 12 end root over denominator 6 end fraction equals fraction numerator 4 plus-or-minus 2 square root of 7 over denominator 6 end fraction equals fraction numerator 2 plus-or-minus square root of 7 over denominator 3 end fraction Now space minus 1 less than fraction numerator 2 minus square root of 7 over denominator 3 end fraction less than fraction numerator 2 plus square root of 7 over denominator 3 end fraction less than 2 therefore space straight c equals fraction numerator 2 minus square root of 7 over denominator 3 end fraction space and space fraction numerator 2 plus square root of 7 over denominator 3 end fraction space both space lie space in space left parenthesis negative 1 comma space 2 right parenthesis$
∴ Rolle's Theorem is verified.