Samacheer Kalvi Class 12 Maths Solutions Chapter 7 Applications of Differential Calculus Exercise 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); \( - 1 \leq x \leq 1 \)
(v) \( \tan^{-1} (x) \); \( -1 \leq x \leq 1 \)
Answer:
(i) For \( f(x) = e^x \):
First, find the derivatives and their values at \( x = 0 \):
\( f(x) = e^x \implies f(0) = e^0 = 1 \)
\( f'(x) = e^x \implies f'(0) = e^0 = 1 \)
\( f''(x) = e^x \implies f''(0) = e^0 = 1 \)
\( f'''(x) = e^x \implies f'''(0) = e^0 = 1 \)
The Maclaurin series expansion is given by the formula:
\[ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n \] This means we sum up terms where each term uses a higher derivative and power of x.
\[ f(x) = f(0) + \frac{f'(0)}{1!} x + \frac{f''(0)}{2!} x^2 + \dots + \frac{f^{(n)}(0)}{n!} x^n + \dots \] Substitute the values of the derivatives at \( x=0 \):
\[ e^x = 1 + \frac{1}{1!} x + \frac{1}{2!} x^2 + \frac{1}{3!} x^3 + \dots \]
(ii) For \( f(x) = \sin x \):
Calculate the derivatives and their values at \( x = 0 \):
\( f(x) = \sin x \implies f(0) = 0 \)
\( f'(x) = \cos x \implies f'(0) = 1 \)
\( f''(x) = -\sin x \implies f''(0) = 0 \)
\( f'''(x) = -\cos x \implies f'''(0) = -1 \)
\( f^{(iv)}(x) = \sin x \implies f^{(iv)}(0) = 0 \)
\( f^{(v)}(x) = \cos x \implies f^{(v)}(0) = 1 \)
\( f^{(vi)}(x) = -\sin x \implies f^{(vi)}(0) = 0 \)
\( f^{(vii)}(x) = -\cos x \implies f^{(vii)}(0) = -1 \)
Using the Maclaurin series expansion formula:
\[ \sin x = f(0) + \frac{f'(0)}{1!} x + \frac{f''(0)}{2!} x^2 + \frac{f'''(0)}{3!} x^3 + \dots \] Substitute the values:
\[ \sin x = 0 + \frac{1}{1!} x + \frac{0}{2!} x^2 + \frac{-1}{3!} x^3 + \frac{0}{4!} x^4 + \frac{1}{5!} x^5 + \dots \] So, the series is:
\[ \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \dots \]
(iii) For \( f(x) = \cos x \):
Find the derivatives and their values at \( x = 0 \):
\( f(x) = \cos x \implies f(0) = 1 \)
\( f'(x) = -\sin x \implies f'(0) = 0 \)
\( f''(x) = -\cos x \implies f''(0) = -1 \)
\( f'''(x) = \sin x \implies f'''(0) = 0 \)
\( f^{(iv)}(x) = \cos x \implies f^{(iv)}(0) = 1 \)
\( f^{(v)}(x) = -\sin x \implies f^{(v)}(0) = 0 \)
\( f^{(vi)}(x) = -\cos x \implies f^{(vi)}(0) = -1 \)
Using the Maclaurin series expansion formula:
\[ \cos x = f(0) + \frac{f'(0)}{1!} x + \frac{f''(0)}{2!} x^2 + \frac{f'''(0)}{3!} x^3 + \dots \] Substitute the values:
\[ \cos x = 1 + \frac{0}{1!} x + \frac{-1}{2!} x^2 + \frac{0}{3!} x^3 + \frac{1}{4!} x^4 + \frac{0}{5!} x^5 + \frac{-1}{6!} x^6 + \dots \] So, the series is:
\[ \cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \dots \]
(iv) For \( f(x) = \log (1-x) \), where \( -1 \leq x \leq 1 \):
Calculate the derivatives and their values at \( x = 0 \):
\( f(x) = \log (1-x) \implies f(0) = \log(1) = 0 \)
\( f'(x) = -\frac{1}{1-x} \implies f'(0) = -1 \)
\( f''(x) = -\frac{1}{(1-x)^2} \implies f''(0) = -1 \)
\( f'''(x) = -\frac{2}{(1-x)^3} \implies f'''(0) = -2 \)
\( f^{(iv)}(x) = -\frac{6}{(1-x)^4} \implies f^{(iv)}(0) = -6 \)
Using the Maclaurin series expansion formula:
\[ \log (1-x) = f(0) + \frac{f'(0)}{1!} x + \frac{f''(0)}{2!} x^2 + \frac{f'''(0)}{3!} x^3 + \frac{f^{(iv)}(0)}{4!} x^4 + \dots \] Substitute the values:
\[ \log (1-x) = 0 + \frac{-1}{1!} x + \frac{-1}{2!} x^2 + \frac{-2}{3!} x^3 + \frac{-6}{4!} x^4 + \dots \] This simplifies to:
\[ \log (1-x) = -x - \frac{x^2}{2} - \frac{x^3}{3} - \frac{x^4}{4} - \dots \]
(v) For \( f(x) = \tan^{-1} (x) \), where \( -1 \leq x \leq 1 \):
Find the derivatives and their values at \( x = 0 \):
\( f(x) = \tan^{-1} (x) \implies f(0) = 0 \)
\( f'(x) = \frac{1}{1+x^2} = 1 - x^2 + x^4 - x^6 + \dots \implies f'(0) = 1 \)
\( f''(x) = -2x + 4x^3 - 6x^5 + \dots \implies f''(0) = 0 \)
\( f'''(x) = -2 + 12x^2 - 30x^4 + \dots \implies f'''(0) = -2 \)
\( f^{(iv)}(x) = 24x - 120x^3 + \dots \implies f^{(iv)}(0) = 0 \)
\( f^{(v)}(x) = 24 - 360x^2 + \dots \implies f^{(v)}(0) = 24 \)
\( f^{(vi)}(x) = -720x + \dots \implies f^{(vi)}(0) = 0 \)
\( f^{(vii)}(x) = -720 + \dots \implies f^{(vii)}(0) = -720 \)
Using the Maclaurin series expansion formula:
\[ \tan^{-1} x = f(0) + \frac{f'(0)}{1!} x + \frac{f''(0)}{2!} x^2 + \frac{f'''(0)}{3!} x^3 + \frac{f^{(iv)}(0)}{4!} x^4 + \dots \] Substitute the values:
\[ \tan^{-1} x = 0 + \frac{1}{1!} x + \frac{0}{2!} x^2 + \frac{-2}{3!} x^3 + \frac{0}{4!} x^4 + \frac{24}{5!} x^5 + \frac{0}{6!} x^6 + \frac{-720}{7!} x^7 + \dots \] Simplifying gives:
\[ \tan^{-1} x = x - \frac{2}{6} x^3 + \frac{24}{120} x^5 - \frac{720}{5040} x^7 + \dots \] \[ \tan^{-1} x = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \dots \]In simple words: The Maclaurin series helps us write a function as a very long sum of simple power terms. We find the function's value and its derivatives at x=0, and then put these values into a special formula. This way, we can approximate the function using polynomials.

🎯 Exam Tip: Remember to calculate enough derivatives to identify the pattern of non-zero terms. Pay close attention to the signs and factorials in the denominator.

Question 2. Write down the Taylor series expansion, of the function log x about x = 1 three non-zero terms for x > 0.
Answer:
Let \( f(x) = \log x \). We need to find the Taylor series expansion about \( x = 1 \).
First, calculate the derivatives of \( f(x) \) and their values at \( x = 1 \):
\( f(x) = \log x \implies f(1) = \log 1 = 0 \)
\( f'(x) = \frac{1}{x} \implies f'(1) = \frac{1}{1} = 1 \)
\( f''(x) = -\frac{1}{x^2} \implies f''(1) = -\frac{1}{1^2} = -1 \)
\( f'''(x) = \frac{2}{x^3} \implies f'''(1) = \frac{2}{1^3} = 2 \)
\( f^{(iv)}(x) = -\frac{6}{x^4} \implies f^{(iv)}(1) = -\frac{6}{1^4} = -6 \)
The Taylor series expansion of \( f(x) \) about \( x = a \) is given by:
\[ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x-a)^n \] For our problem, \( a = 1 \). So, the formula becomes:
\[ \log x = f(1) + \frac{f'(1)}{1!} (x-1) + \frac{f''(1)}{2!} (x-1)^2 + \frac{f'''(1)}{3!} (x-1)^3 + \frac{f^{(iv)}(1)}{4!} (x-1)^4 + \dots \] Now, substitute the values of the derivatives at \( x=1 \):
\[ \log x = 0 + \frac{1}{1!} (x-1) + \frac{-1}{2!} (x-1)^2 + \frac{2}{3!} (x-1)^3 + \frac{-6}{4!} (x-1)^4 + \dots \] Simplify the terms:
\[ \log x = (x-1) - \frac{1}{2} (x-1)^2 + \frac{2}{6} (x-1)^3 - \frac{6}{24} (x-1)^4 + \dots \] \[ \log x = (x-1) - \frac{(x-1)^2}{2} + \frac{(x-1)^3}{3} - \frac{(x-1)^4}{4} + \dots \] The first three non-zero terms are \( (x-1) - \frac{(x-1)^2}{2} + \frac{(x-1)^3}{3} \).In simple words: A Taylor series lets us write a function like log x as a sum of simpler terms, but instead of around zero (Maclaurin), we build it around a different point, like x=1. We find the function's value and its derivatives at that point, then use a formula to create the series. This series helps us approximate the function near the point we chose.

🎯 Exam Tip: When asked for "about x=a", remember that the terms will be \( (x-a)^n \). Also, carefully calculate derivatives and their values at 'a' to avoid errors.

Question 3. Expand sin x ascending powers x \( - \frac { \pi }{ 4 } \) upto three non-zero terms.
Answer:
Let \( f(x) = \sin x \). We need to expand it in ascending powers of \( x - \frac{\pi}{4} \), which means finding the Taylor series expansion about \( a = \frac{\pi}{4} \).
First, find the derivatives of \( f(x) \) and their values at \( x = \frac{\pi}{4} \):
\( f(x) = \sin x \implies f\left(\frac{\pi}{4}\right) = \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}} \)
\( f'(x) = \cos x \implies f'\left(\frac{\pi}{4}\right) = \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}} \)
\( f''(x) = -\sin x \implies f''\left(\frac{\pi}{4}\right) = -\sin \frac{\pi}{4} = -\frac{1}{\sqrt{2}} \)
\( f'''(x) = -\cos x \implies f'''\left(\frac{\pi}{4}\right) = -\cos \frac{\pi}{4} = -\frac{1}{\sqrt{2}} \)
The Taylor series expansion of \( f(x) \) about \( x = a \) is given by:
\[ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x-a)^n \] For this problem, \( a = \frac{\pi}{4} \). So, the formula becomes:
\[ \sin x = f\left(\frac{\pi}{4}\right) + \frac{f'\left(\frac{\pi}{4}\right)}{1!} \left(x-\frac{\pi}{4}\right) + \frac{f''\left(\frac{\pi}{4}\right)}{2!} \left(x-\frac{\pi}{4}\right)^2 + \frac{f'''\left(\frac{\pi}{4}\right)}{3!} \left(x-\frac{\pi}{4}\right)^3 + \dots \] Substitute the calculated values:
\[ \sin x = \frac{1}{\sqrt{2}} + \frac{1/\sqrt{2}}{1!} \left(x-\frac{\pi}{4}\right) + \frac{-1/\sqrt{2}}{2!} \left(x-\frac{\pi}{4}\right)^2 + \frac{-1/\sqrt{2}}{3!} \left(x-\frac{\pi}{4}\right)^3 + \dots \] To express this neatly, we can factor out \( \frac{1}{\sqrt{2}} \):
\[ \sin x = \frac{1}{\sqrt{2}} \left[ 1 + \left(x-\frac{\pi}{4}\right) - \frac{1}{2!} \left(x-\frac{\pi}{4}\right)^2 - \frac{1}{3!} \left(x-\frac{\pi}{4}\right)^3 + \dots \right] \] The first three non-zero terms are \( \frac{1}{\sqrt{2}} \left[ 1 + \left(x-\frac{\pi}{4}\right) - \frac{1}{2!} \left(x-\frac{\pi}{4}\right)^2 \right] \).In simple words: This question asks us to write the sine function as a series, but centered around the angle \( \frac{\pi}{4} \) instead of 0. We find the value of sine and its repeated derivatives at \( \frac{\pi}{4} \). Then we plug these into the Taylor series formula. This gives us a polynomial that can closely estimate the sine function's value near \( \frac{\pi}{4} \).

🎯 Exam Tip: When expanding about a non-zero point like \( \frac{\pi}{4} \), ensure all derivatives are evaluated at that specific point, not at 0. Remember the values of trigonometric functions for common angles.

Question 4. Expand the polynomial f(x) = x² – 3x + 2 in power of x – 1.
Answer:
Let \( f(x) = x^2 - 3x + 2 \). We need to expand this polynomial in powers of \( x - 1 \), which means finding its Taylor series about \( x = 1 \).
First, find the derivatives of \( f(x) \) and their values at \( x = 1 \):
\( f(x) = x^2 - 3x + 2 \implies f(1) = 1^2 - 3(1) + 2 = 1 - 3 + 2 = 0 \)
\( f'(x) = 2x - 3 \implies f'(1) = 2(1) - 3 = 2 - 3 = -1 \)
\( f''(x) = 2 \implies f''(1) = 2 \)
\( f'''(x) = 0 \implies f'''(1) = 0 \)
All higher derivatives will also be zero, as it's a polynomial.
The Taylor series expansion of \( f(x) \) about \( x = a \) is:
\[ f(x) = f(a) + \frac{f'(a)}{1!} (x-a) + \frac{f''(a)}{2!} (x-a)^2 + \frac{f'''(a)}{3!} (x-a)^3 + \dots \] Here, \( a = 1 \). Substitute the values we found:
\[ f(x) = 0 + \frac{-1}{1!} (x-1) + \frac{2}{2!} (x-1)^2 + \frac{0}{3!} (x-1)^3 + \dots \] Simplify the terms:
\[ f(x) = -(x-1) + \frac{2}{2} (x-1)^2 \] \[ f(x) = -(x-1) + (x-1)^2 \] This is the required expansion.In simple words: We want to write the given polynomial in a new form where everything is based on \( (x-1) \) instead of just \( x \). We do this by finding the value of the polynomial and its derivatives at \( x=1 \), then using the Taylor series formula. Since it's a simple polynomial, the series will be short and exact, not an infinite sum.

🎯 Exam Tip: For polynomials, the Taylor series will always be a finite sum. Ensure you calculate all non-zero derivatives and evaluate them correctly at the expansion point, then simplify the factorials.