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) ex
(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) = ex
f(x) = ex f'(0) = e° = 1
f(x) = ex f'(0) = e° = 1
f”(x) = ex 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
fIV(x) = sin x; fIV(0) = 0
fV(x) = cos x; fV(0) = 1
fVI(x) = -sin x; fVI(0) = 0
fVII(x) = -cos x; fVII(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
fIV(x) = cos x ; fIV(0) = 1
fV(x) = -sin x ; fV(0) = 0
fVI(x) = -cos x ; fVI(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 + x4 – x6 + …..
f”(x) = -2x + 4x3 – 6x5 + ….. f”(0) = 0
f”'(x) = -2 + 12x² – 30x4 + ….. f”(0) = -2
fIV(x) = 24x – 120x³ + …… fIV(0) = 0
fV(x) = 24 – 360x² + ….. fV(0) = 24 .
fVI(x) = -720x + ….. fVI(0) = 0
fVII(x) = -720 + … fVII(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
fIV(x) = 8 cos 2x ; fIV( 0) = 8
fV(x) = -16 sin 2x ; fV(0) = 0
fVI(x) = -32 cos 2x ; fVI(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) = $$\sum_{n=0}^{n=\infty}$$ an (x – 1)n, where an = $$\frac { f^{(n)} (1)}{ n! }$$
∴ The required expansion is
x² – 3x + 2 = 0 – $$\frac { 1(x-1) }{ 1! }$$ + $$\frac { 2(x-1)^2 }{ 2! }$$
= -(x – 1) + (x – 1)²