Verify Rolle's Theorem for the function f(x) = x (x2-4) in the

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.
Advertisement

 Multiple Choice QuestionsShort Answer Type

581. Verify Rolle's Theorem for the function x2 + 2 in [-2, 2]
75 Views

582. Verify Rolle's Theorem for the function x2 + 2 x - 8 in [-4, 2]
83 Views

583. Verify Rolle's Theorem for the function f(x) = x(x - 1)2 in [0, 1]. 
92 Views

584. Verily space Rolle apostrophe straight s space theorem space for space the space function
straight f left parenthesis straight x right parenthesis equals straight x cubed over 3 minus fraction numerator 5 straight x squared over denominator 3 end fraction plus 2 straight x comma space straight x element of left square bracket 0 comma 3 right square bracket
84 Views

Advertisement
585. Verily space Rolle apostrophe straight s space theorem space for space the space function space straight f left parenthesis straight x right parenthesis equals 8 straight x minus straight x squared space in space left square bracket 0 comma space 8 right square bracket
96 Views

586. Verify Rolle's Theorem for the function :f(x) = x2 in the interval [- 1, 1]
78 Views

587. Verify Rolle's Theorem for the function : f(x) = x2 - 1 in the interval [- 1, 1]
85 Views

588. Verify Rolle's Theorem for the function :f(x)= x2 - 4 x + 3 in the interval 1 ≤ x ≤ 3.
100 Views

Advertisement
589. Verify Rolle's Theorem for the function : x2 - 5 x + 4 on [1, 4]
84 Views

Advertisement

590.
Verify Rolle's Theorem for the function f(x) = x (x2-4) in the interval [-2, 2].


 Here f(x) = x (x2-4) = x3-4 x
It is a polynomial in x
(i) Since every polynomial in x is a continuous function for every value of x
∴ f(x) is continuous in [-2, 2]
(ii) f'(x) = 3 x2-4, which exists in (-2, 2)
∴ f(x) is derivable in (-2, 2)
(iii) f(-2) = 0, f(2) = 0
∴ f(-2) = f(2)
∴ f(x) satisfies all the conditions of Rolle's Theorem.
∴ there must exist at least one real value of c such that f'(c) = 0 where - 2 < c < 2
Now space straight f apostrophe left parenthesis straight c right parenthesis equals 0 space given space us space 3 straight c squared minus 4 equals 0
therefore space straight c squared equals 4 over 3 space space space rightwards double arrow space straight c equals plus-or-minus fraction numerator 2 over denominator square root of 3 end fraction equals plus-or-minus fraction numerator 2 square root of 3 over denominator 3 end fraction equals plus-or-minus fraction numerator 2 left parenthesis 1.7 right parenthesis over denominator 3 end fraction equals plus-or-minus fraction numerator 3.4 over denominator 3 end fraction equals plus-or-minus space 1.13 element of left parenthesis negative 2 comma space 2 right parenthesis
∴ Rolle's Theorem is verified.

100 Views

Advertisement
Advertisement