Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 7 Applications of Differential Calculus Ex 7.4 Textbook Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.4

Question 1.

Write the Maclaurin series expansion of thef following functions:

(i) e^{x}

(ii) sin x

(iii) cos x

(iv) log (1 – x); – 1 ≤ x ≤ 1

(v) tan^{-1} (x); -1 ≤ x ≤ 1

(vi) cos² x

Solution:

(i) Let f(x) = e^{x}

f(x) = e^{x} f'(0) = e° = 1

f(x) = e^{x} f'(0) = e° = 1

f”(x) = e^{x} f”(0) = e° = 1

Maclaurin ‘s expansion is

(ii) Let f(x) = sin x

f(x) = sin x; f(0) = 0

f'(x) = cos x; f'(0) = 1

f”(x) = -sin x; f”(0) = 0

f”‘(x) = -cos x; f”'(0) = -1

f^{IV}(x) = sin x; f^{IV}(0) = 0

f^{V}(x) = cos x; f^{V}(0) = 1

f^{VI}(x) = -sin x; f^{VI}(0) = 0

f^{VII}(x) = -cos x; f^{VII}(0) = -1

Maclaurin ‘s expansion is

(iii) Let f(x) = cos x

f(x) = cos x ; f(0) = 1

f'(x) = -sin x ; f'(0) = 0

f”(x) = -cos x ; f”(0) = -1

f”'(x) = sin x ; f”'(0) = 0

f^{IV}(x) = cos x ; f^{IV}(0) = 1

f^{V}(x) = -sin x ; f^{V}(0) = 0

f^{VI}(x) = -cos x ; f^{VI}(0) = -1

Maclaurin ‘s expansion is

(iv) log (1 – x); – 1 ≤ x ≤ 1

Maclaurin ‘s expansion is

(v) tan^{-1} (x); -1 ≤ x ≤ 1

f(x) = tan^{-1} x ; f(0) = 0

f'(x) = \(\frac { 1 }{ 1+x^2 }\) f'(0) = 1

= 1 – x² + x^{4} – x^{6} + …..

f”(x) = -2x + 4x^{3} – 6x^{5} + ….. f”(0) = 0

f”'(x) = -2 + 12x² – 30x^{4} + ….. f”(0) = -2

f^{IV}(x) = 24x – 120x³ + …… f^{IV}(0) = 0

f^{V}(x) = 24 – 360x² + ….. f^{V}(0) = 24 .

f^{VI}(x) = -720x + ….. f^{VI}(0) = 0

f^{VII}(x) = -720 + … f^{VII}(0) = -720

Maclaurin ‘s expansion is

(vi) Let f(x) = cos² x

f(x) = cos² x ; f(0) = 1

f'(x) = -2 cos x sin x ; f'(0) = 0

= -sin 2 x

f”(x) = -2 cos 2x ; f”(0) = -2

f”‘(x) = 4 sin 2x ; f”‘(0) = 0

f^{IV}(x) = 8 cos 2x ; f^{IV}( 0) = 8

f^{V}(x) = -16 sin 2x ; f^{V}(0) = 0

f^{VI}(x) = -32 cos 2x ; f^{VI}(0) = -32

Maclaurin’s expansion is

Question 2.

Write down the Taylor series expansion, of the function log x about x = 1 upto three non-zero terms for x > 0.

Solution:

Let f(x) = log x

Taylor series of f(x) is

Question 3.

Expand sin x ascending powers x – \(\frac { π }{ 4 }\) upto three non-zero terms.

Solution:

Let f(x) = sin x

Taylor series of f(x) is

Question 4.

Expand the polynomial f(x) = x² – 3x + 2 in power of x – 1.

Solution:

Let f(x) = x² – 3x + 2

f(x) = x² – 3x + 2 ; f(1) = 0

f'(x) = 2x – 3 ; f'(1) = -1

f”(x) = 2 ; f”(1) = 2

Taylor series of f(x) is

f(x) = \(\sum_{n=0}^{n=\infty}\) a_{n} (x – 1)^{n}, where a_{n} = \(\frac { f^{(n)} (1)}{ n! }\)

∴ The required expansion is

x² – 3x + 2 = 0 – \(\frac { 1(x-1) }{ 1! }\) + \(\frac { 2(x-1)^2 }{ 2! }\)

= -(x – 1) + (x – 1)²

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