By A. Ganesh (Author), Balasubramanian G. (Author)

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2! 3! n! (3) This is called the Maclaurin’s series for the function f (x). If f (x) = y and f ′ (x), f ′′ (x), .......... are denoted by y1, y2, ...... the Maclaurin’s series can also be written in the form: y = y (0) + xy1 (0) + x2 xn y2 (0) + ...... y (0) + ...... 2! n! n WORKED OUT EXAMPLES 1. Verify Rolle’s theorem for the function f (x) = x2 – 4x + 8 in the interval [1, 3]. Solution f (x) = x2 – 4x + 8 is continuous in [1, 3] and f ′ (x) = 2x – 4 exist for all values in (1, 3) ∴ f (1) = 1 – 4 + 8 = 5; f (3) = 32 – 4 (3) + 8 = 5 ∴ f (1) = f (3) Hence all the three conditions of the theorem are satisfied.

T. (4) x = 1 2y4 (1) + 3 – 3 = 0 y4 (1) = 0 Substituting these values in the expansion, we get tan–1 x = π 1 + 4 2 R| U S|b x – 1g – b x –21g + b x –61g |V| T W 2 3 13. By using Maclaurin’s theorem expand log sec x up to the term containing x6. Solution Let y (x) = log sec x, y (0) = log sec 0 = 0 We have y1 = sec x tan x = tan x sec x y1 (0) = 0 y2 = sec2 x y2 (0) = 1 To find the higher order derivatives Consider y2 = 1 + tan2 x y2 = 1 + y12 This gives, Hence, y3 = 2y1 y2 y3 (0) = 2y1 (0) y2 (0) = 0 y4 = 2 [ y1 y3 + y22] so that and y4 (0) = 2 [ y1 (0) y3 (0) + y22 (0)] = 2 [0·1 + 12] = 2 This yields y5 = 2 [ y1 y4 + y2 y3 + 2y2 y3] = 2 [ y1 y4 + 3y2 y3] y5 (0) = 2 [ y1 (0) y4 (0) + 3y2 (0) y3 (0)] = 2 [0·2 + 3·1·0] = 0 y6 = 2 [ y1 y5 + y2 y4 + 3 {y2 y4 + y32}] 45 DIFFERENTIAL CALCULUS—I = 2 [ y1 y5 + 4y2 y4 + 3y32] This yields y6 (0) = 2 [ y1 (0) y5 (0) + 4y2 (0) y4 (0) + 3y32 (0)] = 2 [0·0 + 4·1·2 + 3·0] = 16 Therefore by Maclaurin’s expansion, we have y = y (0) + xy1 (0) + log sec x = 0 + x · 0 + = x2 y (0) + ........

The given f (x) is continuous in [a, b] since 0 < a < b in LM x + ab OP MN ba + bg x PQ 2 f (x) = log The given function can be written in the form f (x) = log (x2 + ab) – log (a + b) – log x f ′ (x) = 2x 1 –0– x x + ab 2 d 2 x 2 – x 2 + ab = dx 2 i + ab x i= x 2 – ab dx 2 f ′ (x) exists in (a, b) L a + ab OP = log 1 = 0 f (a) = log M MN ba + bg a PQ L b + ab OP = log 1 = 0 f (b) = log M MN ba + bg b PQ 2 Also 2 ∴ f (a) = f (b) = 0 Hence all the conditions of the theorem are satisfied. , ∴ that c = ± ab c = + ab ∈ (a, b) and we know ab is the geometric mean of a and b.