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www.akhiai. www.akhiai. MEAN VALUE THEOREMS Rolle'sTheorem : If a function f : [a, b] →R is such that i) It is continuous on [a, b]

ii) It is derivable on (a, b) and iii) f(a) = f(b) then there exists at least one (,)ca b∈ such that ()0f c′=.

Lagrange's mean -value theorem or first mean - value theorem :

If a function f : [a, b] →R is such that i) It is continuous on [a, b].

ii)It is derivable on (a, b) then there exists at least one (,)ca b∈ such that( ) ( )()f b f afcb a−′=− VSAQ’S 1.Verify Rolle’s theorem for the function x2 – 1 on [–1, 1].

Sol. Let f(x) = x2 – 1 f is continuous on [–1, 1] and f is differentiable on (–1, 1) since f(–1) = f(1) = 0 and ∴ By Rolle’s theoremc(1,1)∃ ∈ − Such that f′(c) = 0

f′(x) = 2x = 0 ∴ = f′(c) = 0 2c = 0 ⇒ c = 0 The point c = 0 ∈ (–1, 1) Then Rolle’s theorem is verified.

2. sin x – sin 2x on [0, π].

Sol. Let f(x) = sin x – sin 2x f is continuous on [0, π] f is differentiable on (0, π) since f(0) = f(π) = 0 and ∴ By Rolle’s theoremc(0,)∃ ∈ π www.akhiai.

www.akhiai. Such that f′(c) = 0 f′(x) = cos x – 2 cos 2x f′(c) = 0 ⇒cos c – 2 cos 2x = 0 ⇒cos c – 2(2cos2 c – 1) = 0 cos c – 4 cos2 c + 2 = 0

4cos2 c – cos c – 2 = 0 11 1 32 1 33cosc881 33c cos8−± + ±= = ±∴= 3. log(x2 + 2) – log 3 on [–1, 1].

Sol. Let f(x) = log (x2 + 2) – log 3 f is continuous on [–1, 1] and f is differentiable on (–1, 1) since f(–1) = f(1) = 0 ∴ By Rolle’s theoremc(1,1)∃ ∈ − Such that f′(c) = 0 f′(x) = 21x2+(2x)

f′(c) = 22cc2+ = 0 2c = 0 ⇒ c = 0 The point c = 0 ∈ (–1, 1)4. It is given that Rolle’s theorem holds for the function f(x) = x3 + bx2 + ax on [1, 3] with1c 2t3= + . Find the values of a and b.

Sol. Given f(x) = x3 + bx2 + ax www.akhiai. www.akhiai. 222

222f (x) 3x 2bx af'(x)0 3c 2bc a 02b 4b 12ac6b b 3ac3

1 b b 3a23 3b b 3a 12 and3 3 3′= + +∴=⇔+ + =−±−⇔=−±−=

−±−+ =− −= = 2b 6 and b 3a 336 3 3a 33 3a a 11⇔=−=⇒−=⇒=⇒= Hence a = 11, b = –6.

5. Find a point on the graph of the curve y = x3, where the tangent is parallel to the chord joining (1, 1) and (3, 27).

Sol. Given Points (1, 1) and (3, 27).

Slope of chord = 27 1133 1−=−Given y = x3 222 2dy3xdx

Slope 3x1313 3x x313 39x33=⇒== ⇒=⇒= =

3339 39 39 13 39y x327 9 = = = = ∴ The point on the curve is 39 13 39,3 9 6. Find ‘c’, so that f(b)f(a)f (c)ba−′=− in the following case. f(x) = x2 – 3x – 1, a = –11/7,

b = 13/7.

Sol. ( )13 169 3(13)fb f 17 49 7 = =− − www.akhiai. www.akhiai. 169 273 49 15349 49− − −= 11 121 3(11)f(a)f 17 49 7121 231 49 303

49 49f (x) 2x 3f (c) 2c 3− − = =− − +−= =′=−

′=− Given f(b)f(a)f (c)ba−′=− 153 303 45649 49 492c 313 11 2477 7456 7 192c 349 24 719 2 12c 3 c77 7− −−

−= =+− −−= × =−= + =⇒=7. Verify the Rolle’s theorem for the function (x2 – 1)(x – 2) on [–1, 2]. Find the point in

the interval where the derivate vanishes.

Sol. Let f(x) = (x2 – 1)(x – 2) = x3 – 2x2 – x + 2 f is continuous on [–1, 2] since f(–1) = f(2) = 0 and f is differentiable on (–1, 2) ∴ By Rolle’s theoremc(1,2)∃ ∈ −

Let f′(c) = 0 f′(x) = 3x2 – 4x – 1 3c2 – 4c – 1 = 0 416 12 4 28c6 6± + ±= = 27c3±⇒=

8. Verify the conditions of the Lagrange’s mean value theorem for the following function. In each case find a point ‘c’ in the interval as stated by the theorem.

sin x – sin 2x on [0, π].

Sol. Let f(x) = sin x – sin 2x www.akhiai. www.akhiai. f is continuous on [0, π] and f is differentiable on (0,π) Given f(x) = sinx – sin 2x

f′(x) = cos x – 2 cos 2x By Lagrange’s mean value than c(0,)∃ ∈ π such there 2 22

1f ( ) f (0)f (c) cos c 2cos 2c 00cos c 2(2cos c 1) 0 cos c 4cos c 2 01 1 32 1 334cos c cos c 2 0 cos c881 33c cos8−π −′=⇒−=π−⇒− −=⇒−+ =

± + ±⇒− −=⇒= = ±⇒=