Lecture notes on Advanced Engineering mathematics

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Engineering Mathematics – I Dr. V. Lokesha 2011 Engineering Mathematics – I (10 MAT11) LECTURE NOTES (FOR I SEMESTER B E OF VTU) VTU-EDUSAT Programme-15 Dr. V. Lokesha Professor and Head DEPARTMENT OF MATHEMATICS ACHARYA INSTITUTE OF TECNOLOGY Soldevanahalli, Bangalore – 90 10 MAT11 1 Engineering Mathematics – I Dr. V. Lokesha 2011 UNIT - I DIFFERENTIAL CALCULUS – I Introduction: The mathematical study of change like motion, growth or decay is calculus. The Rate of change of given function is derivative or differential. The concept of derivative is essential in day to day life. Also applicable in Engineering, Science, Economics, Medicine etc. Successive Differentiation: Let y = f (x) (1) be a real valued function. dy 1 The first order derivative of y denoted by or y’ or y or∆ 1 dx 2 d y 2 The Second order derivative of y denoted by or y’’ or y or ∆ 2 2 dx Similarly differentiating the function (1) n-times, successively, n d y th n n the n order derivative of y exists denoted by or y or y or ∆ n n dx nd The process of finding 2 and higher order derivatives is known as Successive Differentiation. th n derivative of some standard functions: ax 1. y = e ax Sol : y = a e 1 2 ax y = a e 2 Differentiating Successively n ax y = a e n n ax n ax ie. D e = a e n x x For, a =1 D e = e 10 MAT11 3 Engineering Mathematics – I Dr. V. Lokesha 2011 10 MAT11 4 Engineering Mathematics – I Dr. V. Lokesha 2011 10 MAT11 5 Engineering Mathematics – I Dr. V. Lokesha 2011 10 MAT11 6 Engineering Mathematics – I Dr. V. Lokesha 2011 10 MAT11 7 Engineering Mathematics – I Dr. V. Lokesha 2011 Leibnitz’s Theorem : th It provides a useful formula for computing the n derivative of a product of two functions. th th Statement : If u and v are any two functions of x with u and v as their n derivative. Then the n n n derivative of uv is n n n (uv) = u v + C u v + C u v + …+ C u v +u v n 0 n 1 1 n-1 2 2 n-2 n-1 n-1 1 n 0 Note : We can interchange u & v (uv) = (vu) n n, n n n C = n , C = n(n-1) /2 , C = n(n-1)(n-2) /3 … 1 2 3 th ax 1. Find the n derivations of e cos(bx + c) ax ax Solution: y = e – b sin (bx +c) + a e cos (b x + c), by product rule. 1 ax .i.e, y = e a cos() bx + c − b sin(bx + c) 1 Let us put a = r cos θ , and b = r sin θ . 2 2 2 ∴ a + b = r and tan θ = b / a 2 2 -1 .ie., r = a + b and θ = tan (b/a) ax Now, y = e r cos θcos(bx + c) − r sin θsin(bx + c) 1 ax Ie., y = r e cos () θ + bx + c 1 where we have used the formula cos A cos B – sin A sin B = cos (A + B) Differentiating again and simplifying as before, 2 ax y = r e cos () 2θ + bx + c . 2 3 ax Similarly y = r e cos () 3θ + bx + c . 3 ……………………………………… n ax () Thus y = r e cos nθ + bx + c n 2 2 -1 Where r = a + b and θ = tan (b/a). n ax Thus D e cos (b x + c) 2 2 n ax −1 = ( a + b ) e cosn tan() b / a + bx + c 10 MAT11 8 Engineering Mathematics – I Dr. V. Lokesha 2011 2 th 2. Find the n derivative of log 4x + 8x +3 2 2 ½ Solution : Let y = log 4x + 8x + 3 = log (4x + 8x +3) 1 2 n ie., log (4x + 8x +3) ∵ log x = n log x y = 2 1 y = log (2x + 3) (2x+1), by factorization. 2 1 ∵ y = log (2x + 3) + log (2x + 1) 2 n−1 n−1 n n ⎧ ⎫ 1 () −1 (n −1)2 (−1) (n −1)2 y = + Now n ⎨ ⎬ n n 2 () 2x + 3 () 2x +1 ⎩ ⎭ ⎧ ⎫ 1 1 n-1 n-1 Ie., y = 2 (-1) (n-1) + n ⎨ ⎬ n n () 2x + 3 () 2x +1 ⎩ ⎭ th 3 5 3. Find the n derivative of log (1-2x) (8x+1) 10 3 5 Solution : Let y = log 1-2x) (8x+1) 10 It is important to note that we have to convert the logarithm to the base e by the property: log x e log x = 10 log 10 e 1 3 5 Thus y = log () 1− 2x(8x +1) e log 10 e 1 Ie., y = 3log() 1− 2x + 5log(8x +1) log 10 e n−1 n n−1 n ⎧ ⎫ 1 () −1 (n −1)(− 2) (−1) (n −1)8 ∴y = 3. + 5 ⎨ ⎬ n n n log 10 () 1− 2x () 8x +1 e ⎩ ⎭ n−1 n n n ⎧ ⎫ () −1 (n −1)2 3(−1) 5(4) y = + Ie., n ⎨ ⎬ n n log 10 () 1− 2x () 8x +1 e ⎩ ⎭ 10 MAT11 9 Engineering Mathematics – I Dr. V. Lokesha 2011 th 2x 2 4. Find the n derivative of e cos x sin x 1+ cos 2x ⎡ ⎤ 2x 2 2x Solution : let y = e cos x sin x = e sin x ⎢ ⎥ 2 ⎣ ⎦ 2x e ie., y = (sin x + sin x cos 2x) 2 2x e 1 ⎧ ⎫ = sin x + sin 3x + sin() − x ⎨ ⎬ 2 2 ⎩ ⎭ 2x e = () 2sin x + sin 3x − sin x∵ sin (-x) = -sin x 4 2x e ∴y = (sin x + sin 3x) 4 1 n 2x n 2x Now y = D() e sin x + D (e sin 3x) n 4 n n 1 2x −1 2x −1 Thus y = () 5 e sin n tan () 1 2 + x + ( 13) e sinn tan() 3 2 + 3x n 4 2x n n e −1 −1 ∴y = () 5 sin n tan() 1 2 + x + ( 13) sin n tan() 3 2 + 3x n 4 th 2x 3 5. Find the n derivative of e cos x 1 2x 3 2x Solution : Let y=e cos x = e . (3 cos x + cos 3x) 4 1 2x 2x Ie., y = (3 e cos x + e cos 3x) 4 1 n 2x n 2x 3D (e cos x) + D (e cos 3x) ∴ y = n 4 n n 1 2 x −1 2 x −1 y = () 3 5 e cos n tan () 1 2 + x + ( 13) e cosn tan() 3 2 + 3x n 4 2x n n e −1 −1 Thus y = 3() 5 cos n tan() 1 2 + x + ( 13) cos n tan() 3 2 + 3x n 4 10 MAT11 10 Engineering Mathematics – I Dr. V. Lokesha 2011 2 x th 6. Find the n derivative of () 2x +1(2x + 3) 2 x Solution : is an improper fraction because; the degree of the y = () 2x +1(2x + 3) numerator being 2 is equal to the degree of the denominator. Hence we must divide and rewrite the fraction. 2 2 x 1 4x y = = . for convenience. 2 2 4x + 8x + 3 4 4x + 8x + 3 1 2 4x 2 4x + 8x + 3 2 4x +8x +3 −8x − 3 1 − 8x − 3 ⎡ ⎤ ∴ y = 1+ ⎢ 2 ⎥ 4 4x + 8x + 3 ⎣ ⎦ 1 1 8x + 3 ⎡ ⎤ Ie., y = − 2 ⎢ ⎥ 4 4 4x + 8x + 3 ⎣ ⎦ The algebraic fraction involved is a proper fraction. 1 8x + 3 ⎡ ⎤ n Now y = 0 − D ⋅ n 2 ⎢ ⎥ 4 4x + 8x + 3 ⎣ ⎦ 8x + 3 A B Let = + () 2x +1(2x + 3) 2x +1 2x + 3 Multiplying by (2x + 1) (2x + 3) we have, 8x + 3 = A (2x + 3) + B (2x + 1) ................(1) By setting 2x + 1 = 0, 2x + 3 = 0 we get x = -1/2, x = -3/2. Put x = -1/2 in (1): -1 -1 + A (2) ⇒ A = -1/2 Put x = -3/2 in (1): -9 = B (-2) ⇒ B = 9/2 1 ⎧ 1 1 9 1 ⎫ ⎡ ⎤ ⎡ ⎤ n n ∴ y = − − D + D ⎨ ⎬ n ⎢ ⎥ ⎢ ⎥ 4 2 2x +1 2 2x + 3 ⎣ ⎦ ⎣ ⎦ ⎩ ⎭ 10 MAT11 11 Engineering Mathematics – I Dr. V. Lokesha 2011 n n n n ⎧ ⎫ 1 () −1 n2 (−1) n2 = − () −1 ⋅ + 9⋅ ⎨ ⎬ n+1 n+1 8 () 2x +1 () 2x + 3 ⎩ ⎭ n+1 n ⎧ ⎫ () −1 n2 1 9 ie., y = + ⎨ ⎬ n n+1 n+1 8 (2x +1) () 2x + 3 ⎩ ⎭ 4 x th 7. Find the n derivative of (x +1)(x + 2) 4 x Solution : y = is an improper fraction. (x +1)(x + 2) (deg of nr. = 4 deg. of dr. = 2) 4 2 On dividing x by x + 3 x + 2, We get ⎡ −15x −14 ⎤ 2 y = ( x – 3x + 7 ) + 2 ⎢ ⎥ x + 3x + 2 ⎣ ⎦ 15x −14 ⎡ ⎤ n 2 n ∴ y = D (x -3x+7)-D n 2 ⎢ ⎥ x + 3x + 2 ⎣ ⎦ 2 2 2 But D = ( x – 3x + 7 ) = 2x – 3, D ( x – 3x + 7 ) = 2 3 2 n 2 D ( x – 3x + 7 ) = 0......... D ( x – 3x + 7 ) = 0 if n 2 ⎡ ⎤ 15x +14 n Hence y = -D n ⎢ ⎥ (x +1)(x + 2) ⎣ ⎦ 15x +14 A B n Now, let D = + 2 (x +1) (x + 2) x + 3x + 2 = 15x+ 14 = A(x+2) + B(x+ 1 ) Put x = - 1 ; - 1 = A ( 1 ) or A = - 1 Put x = - 2 ; - 16 = B ( - 1 ) or B = 16 ⎧ 1 1 ⎫ ⎡ ⎤ ⎡ ⎤ n n Y = − D +16D n ⎨ ⎬ ⎢ ⎥ ⎢ ⎥ x +1 x + 2 ⎣ ⎦ ⎣ ⎦ ⎩ ⎭ 10 MAT11 12 Engineering Mathematics – I Dr. V. Lokesha 2011 nn nn (1−− )nn 1 ( 1) 1 =−16 nn ++ 11 (1xx ++ ) ( 2) ⎧⎫ 116 n yn =−(1) 2 − n ⎨⎬ n nn ++ 11 (1xx ++ ) ( 2) ⎩⎭ 8. Show that n n d log x (−1) n 1 1 1 ⎛ ⎞ ⎧ ⎫ = log x −1− − − ⎜ ⎟ ⎨ ⎬ n n+1 x 2 3 n dx x ⎝ ⎠ ⎩ ⎭ log x 1 1 Solution : Let y = = log x. and let u = log x, v = x x x We have Leibnitz theorem, (uv) = uv + n u v + n u v + .... + u v …… (1) n n C 1 n−1 C 2 n−2 n 1 2 n−1 (−1) (n −1) Now, u = log x ∴u = n n x n 1 (−1) n v = ∴v = n n+1 x x Using these in (1) by taking appropriate values for n we get, n n−1 ⎛ log x ⎞ (−1) n 1 (−1) (n −1) D = = log x.. + n . n ⎜ ⎟ n+1 n x x x x ⎝ ⎠ n−2 n(n −1) 1 (−1) (n − 2) ⎛ ⎞ + − ⎜ ⎟ 2 n−1 1. 2 x x ⎝ ⎠ n−1 (−1) () n −1 1 + ......+ . n x x n n−1 (−1) n (−1) n Ie.. = log x + n+1 n+1 x x n−2 n−1 (−1) n (−1) (n −1) − +....+ n+1 n+1 2x x n−2 −2 −1 (−1) n⎡ (−1) (−1) (n −1)⎤ −1 − log x(−1) − +....+ ⎢ ⎥ n+1 1 x 2 n ⎣ ⎦ 1 1 −2 -1 Note : (-1) = = −1;(−1) = =1 2 −1 (−1) 10 MAT11 13 Engineering Mathematics – I Dr. V. Lokesha 2011 (n −1) (n −1) 1 Also = = n n(n −1) n n n d ⎡log x ⎤ (−1) n⎡ 1 1 1 ⎤ ∴ = log x −1− − ... − n n+1 ⎢ ⎥ ⎢ ⎥ dx x x 2 3 n ⎣ ⎦ ⎣ ⎦ n n 9. If y = D (x logx) n Prove that y = n y +(n-1) and hence deduce that n n-1 1 1 1 ⎛ ⎞ y = n log x +1+ + +....+ n ⎜ ⎟ 2 3 n ⎝ ⎠ n n n-1 n Solution : y = D (x log x) = D D (x log x n 1 ⎧ ⎫ n n−1 n-1 x . + nx log x ⎨ ⎬ = D x ⎩ ⎭ n-1 n-1 n-1 n-1 = D (x ) + nD (x log x ∴ y = (n-1) +ny . This proves the first part. n n-1 Now Putting the values for n = 1, 2, 3...we get y = 0 + 1 y = 1 + log x = 1 (log x + 1 ) 1 0 y = l + 2y = l+2 (l + log x) 2 1 1 ⎛ ⎞ ie., y = 21og x + 3 = 2(log x + 3/2) = 2 log x +1+ 2 ⎜ ⎟ 2 ⎝ ⎠ y = 2 + 3y = 2 + 3(2 log x + 3) 3 2 1 1 ⎛ ⎞ ie., y = 61og x+ll = 6 (log x + ll/6) = 3 log x +1+ + 3 ⎜ ⎟ 2 3 ⎝ ⎠ ………………………………………………………………………….. 1 1 1 ⎛ ⎞ y = n log x +1+ + + ...+ ⎜ ⎟ n 2 3 n ⎝ ⎠ 10. If y = a cos (log x) + b sin ( log x), show that 2 x y + xy + y = 0. Then apply Leibnitz theorem to differentiate this result n times. 2 1 or If y = a cos (log x) + b sin (log x ), show that 2 2 x y + (2n+l)xy +(n +1)y = 0. July-03 n + 2 n + l n 10 MAT11 14 Engineering Mathematics – I Dr. V. Lokesha 2011 Solution : y = a cos (log x) + b sin (log x) Differentiate w.r.t x 1 1 ∴ y = -a sin (log x) + b cos (log x). 1 x x (we avoid quotient rule to find y ) . 2 = xy = - a sin (log x) + b cos (log x) 1 Differentiating again w.r.t x we have, 1 xy + 1 y = - a cos (log x) + b sin ( log x) 2 1 x 2 or x y + xy = - a cos (log x) + b sin (log x) = -y 2 1 2 ∴ x y +xy +y = 0 2 1 Now we have to differentiate this result n times. n 2 n n ie., D (x y ) + D (xy ) + D (y) = 0 2 1 We have to employ Leibnitz theoreom for the first two terms. Hence we have, ⎧ ⎫ n(n −1) 2 n n−1 n−2 2) x . D (y ) + n. 2x.D (y ) + . 2 .D (y ) ⎨ 2 2 ⎬ 1. 2 ⎩ ⎭ n n−1 x. D (y ) + n. 1 . D ( y )+ y = 0 1 1 n 2 ie., x y + 2n x y + n (n – 1)y + xy +ny +y = 0 n + 2 n + 1 n n+1 n n 2 2 ie., x y + 2n x y + n y - ny + xy +ny +y = 0 n + 2 n + 1 n n n+1 n n 2 2 ie., x y + (2n+l)xy + (n +l)y = 0 n + 2 n+l n -1 n 11. If cos (y/b ) = log (x/n) , then show that 2 2 x y + (2n+l) xy + 2n y = 0 n + 2 n+l n -1 m Solution :By data, cos (y/b) = n log (x/n) ∴log(a ) = m log a y = = cos n log (x/n ) b or y = b . cos n log (x/n) Differentiating w.r.t x we get, 10 MAT11 15 Engineering Mathematics – I Dr. V. Lokesha 2011 1 1 ⋅ n ⋅ ⋅ y = -b sin n log (x/n) 1 () x / n n or xy = - n b sin n log (x/n ) 1 Differentiating w.r.t x again we get, 1 1 xy + 1. y = - n . b cos n log (x/n ) n . 2 1 (x / n) n 2 2 or x (xy +y ) = n b cos n log (x/n) =-n y, by using (1). 2 1 2 2 or x y +xy + n y = 0 2 l Differentiating each term n times we have, 2 n 2 n D(x y ) + D (xy ) + n D (y) = 0 2 1 Applying Leibnitz theorem to the product terms we have, ⎧ n(n −1) ⎫ 2 x y + n. 2x. y + . 2 . y ⎨ ⎬ n+2 n+1 n 1. 2 ⎩ ⎭ 2 + xy + n. 1 .y + n y = 0 n+1 n n 2 2 2 ie x y + 2 x y + n y + xy + ny + n y =0 n+2 n+1 n n+1 n n 2 2 or x y + (2n + l) xy + 2n y = 0 n+2 n+1 n 2 12. If y = sin( log (x + 2 x + 1)), or Feb-03 -1 If sin y = 2 log (x + 1), show that 2 2 (x+l) y + (2n+1)(x+1)y + (n + 4)y = 0 n + 2 n+l n 2 Solution : By data y = sin log (x + 2 x + 1 ) 1 2 ∴ y = cos log (x + 2 x + 1) 2x + 2 1 2 (x +1) 1 2 ie., y1 = cos log (x + 2 x + 1) 2 (x + 1) 2 x + 2x +1 2 2cos log(x + 2 x + 1 ) ie., y = 1 (x +1) 2 or (x + 1) y1 = 2 cos log (x + 2 x + 1 ) Differentiating w.r.t x again we get 10 MAT11 16 Engineering Mathematics – I Dr. V. Lokesha 2011 1 2 . 2(x +1) (x+1)y + 1 y = -2 sin log (x + 2x + 1) 2 1 2 (x +1) 2 or (x + 1) y + (x+1) y = -4y 2 1 2 or (x+l ) y + (x+l) y + 4y = 0 , 2 1 Differentiating each term n times we have, n 2 n n D (x + 1) y +D (x+ 1)y + D y = 0 2 1 Applying Leibnitz theorem to the product terms we have, ⎧ n(n −1) ⎫ 2 (x +1) y + n. 2(x +1). y + .2 .y ⎨ ⎬ n+2 n+1 n 1. 2 ⎩ ⎭ + (x+l) y +n. 1 .y + 4yn = 0 n + 1 n 2 ie., (x+l) y + 2n (x+1)y n + 2 n+1 2 + n y -ny + (x+l)y + ny + 4y = 0 n n n+l n n 2 2 ie., (x+l) y + (2n + l) (x + l) y + (n + 4)y = 0 n + 2 n+ 1 n 2 13. If = log (x + 1+ x ) prove that 2 2 (1 + x ) y + (2n + 1) xy + n y = 0 n+2 n+1 n 2 By data, y = log (x + 1+ x ) ⎧ ⎫ 1 1 ⎪ ⎪ ∴y = 1+ . 2x 1 ⎨ ⎬ 2 2 (x + 1+ x ) ⎪ 2 1+ x ⎪ ⎩ ⎭ 2 1 1+ x + x 1 Ie., y = 1 2 2 2 (x + 1+ x ) 1+ x 1+ x 2 or 1+ x y = 1 1 Differentiating w.r.t.x again we get 1 2 1+ x y + .2 x.y = 0 2 1 2 2 1+ x ) 2 or (1+x )y + xy = 0 2 1 n 2 n Now D (l+x )y + D xy = 0 2 1 Applying Leibnitz theorem to each term we get, 10 MAT11 17 Engineering Mathematics – I Dr. V. Lokesha 2011 ⎧ ⎫ n(n −1) 2 (1+ x )y + n. 2x .y + .2 .y ⎨ ⎬ n+2 n+1 n 1.2 ⎩ ⎭ + x . y +n .1 y = 0 n + 1 n 2 2 Ie., (1 + x ) y + 2 n x y + n y – ny + xy + ny = 0 n +2 n + 1 n n n+l n 2 2 or (l+x )y + (2n + l)xy +n y = 0 n + 2 n+1 n 14. If x = sin t and y = cos mt, prove that 2 2 2 (l-x )y -(2n+1)xy + (m -n )y = 0. Feb-04 n + 2 n+l n Solution : By data x = sin t and y = cos mt -1 x = sin t = t = sin x and y = cos mt becomes -1 y = cos m sin x) Differentiating w.r.t.x we get m -1 y = - sin (m sin x) 1 2 1− x 2 -1 or 1+ x y = - m sin (m sin x) 1 Differentiating again w.r.f .x we get, 1 m 2 −1 1− x y + (−2x) y = −m cos(msin x). 2 1 2 2 2 1− x 1− x 2 2 or (1 -x )y -xy = -m y 2 l 2 2 or (1 -x )y –xy +m y = 0 2 1 2 2 2 Thus (1-x )y -(2n+1)xy +(m -n )y =0 n+2 n+1 n 15. If x = tan ( log y), find the value of 2 (l+x )y + (2nx-l) y +n(n-1)y July-04 n+1 n n-1 -1 tan-1 Solution : By data x = tan(log y) = tan x = log y or y = e x Since the desired relation involves y y and y we can find y and differentiate n times the result associated with y and y. n+1, n n-1 1 1 −1 −1 1 tan x tan x Consider y = e ⋅ ∴ y. = e ⋅ 2 1+ x 2 or (1 +x )y = y 1 Differentiating n times we have 10 MAT11 18 Engineering Mathematics – I Dr. V. Lokesha 2011 n 2 n D (l+x )y =D y 1 Anplying Leibnitz theorem onto L.H.S, we have, 2 n n-1 (l+x )D (y ) + n .2x .D (y ) 1 1 n(n −1) n−2 + .2 .D (y ) = y 1 n 1.2 2 Ie., (1+x )y +2n x y + n (n-1) y -y =0 n+1 n n-1 n 2 Or (l+x )y + (2nx-l)y + n(n-l)y = 0 n + 1 n n-1 10 MAT11 19 Engineering Mathematics – I Dr. V. Lokesha 2011 Continuity & Differentiability Some Fundamental Definitions A function f (x) is defined in the interval I, then it is said to be continuous at a point x = a lim f (x) = f (a) if x→a fx() +h −f(a) lim = f '(a) exists a∈I A function f (x) is said to be differentiable at x = a if h→0 h − x −1 ≤ x ≤ 0 ⎧ x = Ex : Consider a function f (x) is defined in the interval -1,1 by f (x) = ⎨ x 0 ≤ x ≤ 1 ⎩ It is continuous at x = 0 But not differentiable at x = 0 Note : If a function f (x) is differentiable then it is continuous, but converse need not be true. Geometrically : (1) If f (x) is Continuous at x =a means, f (x) has no breaks or jumps at the point x = a −− 1 1≤ x≤ 0 ⎧ fx() = ⎨ Ex : x 0 x ≤ 1 ⎩ Is discontinuous at x=0 (2) If f (x) is differentiable at x = a means, the graph of f (x) has a unique tangent at the point or graph is smooth at x = a 1. Give the definitions of Continuity & Differentiability: Solution: A function f (x) is said to be continuous at x = a, if corresponding to an arbitrary positive number ε, however small, their exists another positive number δ such that. ⏐f (x) – f (a)⏐ ε, where ⏐x - a⏐ δ It is clear from the above definition that a function f (x) is continuous at a point ‘a’. If (i) it exists at x = a (ii) Lt f (x) = f (a) x→a i.e, limiting value of the function at x = a is to the value of the function at x = a 10 MAT11 20 Engineering Mathematics – I Dr. V. Lokesha 2011 Differentiability: A function f (x) is said to be differentiable in the interval (a, b), if it is differentiable at every point in the interval. 1 In Case a,b the function should posses derivatives at every point and at the end points a & b i.e., Rf 1 (a) and Lf (a) exists. 2. State Rolle’s Theorem with Geometric Interpretation. Statement: Let f (x) be a function is defined on a,b & it satisfies the following Conditions. (i) f (x) is continuous in a,b (ii) f (x) is differentiable in (a,b) (iii) f (a) = f (b) 1 Then there exists at least a point C∈ (a,b), Here a b such that f ( c ) = 0 Proof: Geometrical Interpretation of Rolle’s Theorem: Y y = f (x) P R A B A B f(a) Q S →f(a) → f(b) O x = a x = c x = b x =a c c c c x = b 1 2 3 4 Let us consider the graph of the function y = f (x) in xy – plane. A (a,.f(a)) and B (b, f( b ) ) be the two points in the curve f (x) and a, b are the corresponding end points of A & B respectively. Now, explained the conditions of Rolle’s theorem as follows. (i) f (x) is continuous function in a,b, Because from figure without breaks or jumps in between A & B on y = f (x). (ii) f (x) is a differentiable in (a,b), that means let us joining the points A & B, we get a line AB. ∴ Slope of the line AB = 0 then ∃ a point C at P and also the tangent at P (or Q or R or S) is Parallel to x –axis. ∴ Slope of the tangent at P (or Q or R or S) to be Zero even the curve y = f (x) decreases or increases, i.e., f (x) is Constant. 10 MAT11 21

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