CBSE Class 11 Mathematics Logarithms Notes Set 01

Logarithms

Definition

“The logarithm of a given number to a given base is the index of the power to which the base must be raised in order to equal the given number.”
If a > 0 and a ≠ 1, then the logarithm of a positive number N is defined as the index x of that power of ‘a’ which equals N, i.e. \( \log_a N = x \) iff \( a^x = N \Rightarrow a^{\log_a N} = N \), a > 0, a ≠ 1 and N > 0.
It is also known as the fundamental logarithmic identity.
The function defined by \( f(x) = \log_a x \), a > 0, a ≠ 1 is called the logarithmic function. Its domain is (0, ∞) and range is R. a is called the base of the logarithmic function.
When the base is ‘e’, the logarithmic function is called the natural or Napierian logarithmic function, and when the base is 10, it is called the common logarithmic function.
Note: (i) The logarithm of a number is unique, i.e. no number can have two different logarithms to a given base.
(ii) \( \log_e a = \log_e 10 \cdot \log_{10} a \) or \( \log_{10} a = \frac{\log_e a}{\log_e 10} = 0.434 \log_e a \)

Characteristic and Mantissa

(1) The integral part of a logarithm is called the characteristic and the fractional part is called the mantissa.
\( \log_{10} N = \) integer (characteristic) + fraction, +ve (mantissa)
(2) The mantissa part of the log of a number is always kept positive.
(3) If the characteristic of \( \log_{10} N \) is n, then the number of digits in N is (n + 1).
(4) If the characteristic of \( \log_{10} N \) is (−n), then there exist (n − 1) zeros after the decimal point of N.

Example 1. For \( y = \log_a x \) to be defined ‘a’ must be [IIT 1990]
(a) Any positive real number
(b) Any number
(c) ≥ e
(d) Any positive real number ≠ 1
Answer:
Ans. (d) Any positive real number ≠ 1
It is obvious from the definition: the base of a logarithm must be positive and different from 1.

Example 2. Logarithm of \( 32\sqrt[5]{4} \) to the base \( 2\sqrt{2} \) is
(a) 3.6
(b) 5
(c) 5.6
(d) None of these
Answer:
Ans. (a) 3.6
Let x be the required logarithm; then by definition \( (2\sqrt{2})^x = 32\sqrt[5]{4} \)
\( (2 \cdot 2^{1/2})^x = 2^5 \cdot 2^{2/5} \); ∴ \( 2^{3x/2} = 2^{5 + \frac{2}{5}} \)
Here, by equating the indices, \( \frac{3}{2}x = \frac{27}{5} \), ∴ \( x = \frac{18}{5} = 3.6 \)

Properties of Logarithms

Let m and n be arbitrary positive numbers such that a > 0, a ≠ 1, b > 0, b ≠ 1; then
(1) \( \log_a a = 1,\ \log_a 1 = 0 \)
(2) \( \log_a b \cdot \log_b a = 1 = \log_a a = \log_b b \Rightarrow \log_a b = \frac{1}{\log_b a} \)
(3) \( \log_c a = \log_b a \cdot \log_c b \) or \( \log_c a = \frac{\log_b a}{\log_b c} \)
(4) \( \log_a (mn) = \log_a m + \log_a n \)
(5) \( \log_a \left(\frac{m}{n}\right) = \log_a m - \log_a n \)
(6) \( \log_a m^n = n \log_a m \)
(7) \( a^{\log_a m} = m \)
(8) \( \log_a \left(\frac{1}{n}\right) = -\log_a n \)
(9) \( \log_{a^\beta} n = \frac{1}{\beta}\log_a n \)
(10) \( \log_{a^\beta} n^\alpha = \frac{\alpha}{\beta}\log_a n,\ (\beta \neq 0) \)
(11) \( a^{\log_c b} = b^{\log_c a} \), (a, b, c > 0 and c ≠ 1)

Example 3. The number \( \log_2 7 \) is [IIT 1990]
(a) An integer
(b) A rational number
(c) An irrational number
(d) A prime number
Answer:
Ans. (c) An irrational number
Suppose, if possible, \( \log_2 7 \) is rational, say p/q where p and q are integers, prime to each other.
Then \( \frac{p}{q} = \log_2 7 \Rightarrow 7 = 2^{p/q} \Rightarrow 2^p = 7^q \),
which is false since L.H.S. is even and R.H.S. is odd. Obviously \( \log_2 7 \) is not an integer and hence not a prime number.

Example 4. If \( \log_7 2 = m \), then \( \log_{49} 28 \) is equal to [Roorkee 1999]
(a) 2(1 + 2m)
(b) \( \frac{1 + 2m}{2} \)
(c) \( \frac{2}{1 + 2m} \)
(d) 1 + m
Answer:
Ans. (b) \( \frac{1 + 2m}{2} \)
\( \log_{49} 28 = \frac{\log 28}{\log 49} = \frac{\log 7 + \log 4}{2\log 7} = \frac{1}{2} + \frac{1}{2}\log_7 4 = \frac{1}{2} + \log_7 2 = \frac{1}{2} + m = \frac{1 + 2m}{2} \)

Example 5. If \( \log_e \left(\frac{a + b}{2}\right) = \frac{1}{2}(\log_e a + \log_e b) \), then the relation between a and b will be [UPSEAT 2000]
(a) a = b
(b) \( a = \frac{b}{2} \)
(c) 2a = b
(d) \( a = \frac{b}{3} \)
Answer:
Ans. (a) a = b
\( \log_e \left(\frac{a + b}{2}\right) = \frac{1}{2}(\log_e a + \log_e b) = \frac{1}{2}\log_e (ab) = \log_e \sqrt{ab} \)
\( \Rightarrow \frac{a + b}{2} = \sqrt{ab} \Rightarrow a + b = 2\sqrt{ab} \Rightarrow \left(\sqrt{a} - \sqrt{b}\right)^2 = 0 \Rightarrow \sqrt{a} = \sqrt{b} \Rightarrow a = b \)

Example 6. If \( \log_{10} 3 = 0.477 \), the number of digits in \( 3^{40} \) is [IIT 1992]
(a) 18
(b) 19
(c) 20
(d) 21
Answer:
Ans. (c) 20
Let \( y = 3^{40} \). Taking log on both sides, \( \log y = 40 \log 3 = 40 \times 0.477 = 19.08 \)
The characteristic is 19, ∴ number of digits in y = 19 + 1 = 20.

Example 7. Which is the correct order for a given number α (α > 1) in increasing order [Roorkee 2000]
(a) \( \log_2 \alpha, \log_3 \alpha, \log_e \alpha, \log_{10} \alpha \)
(b) \( \log_{10} \alpha, \log_3 \alpha, \log_e \alpha, \log_2 \alpha \)
(c) \( \log_{10} \alpha, \log_e \alpha, \log_2 \alpha, \log_3 \alpha \)
(d) \( \log_3 \alpha, \log_e \alpha, \log_2 \alpha, \log_{10} \alpha \)
Answer:
Ans. (b) \( \log_{10} \alpha, \log_3 \alpha, \log_e \alpha, \log_2 \alpha \)
\( \log_b \alpha = \frac{\log \alpha}{\log b} \); for α > 1 the numerator is positive, so a larger base gives a smaller logarithm.
Since 10, 3, e, 2 are in decreasing order,
\( \log_{10} \alpha, \log_3 \alpha, \log_e \alpha, \log_2 \alpha \) are in increasing order.

Logarithmic Inequalities

(1) If a > 1, p > 1 ⇒ \( \log_a p > 0 \)
(2) If 0 < a < 1, p > 1 ⇒ \( \log_a p < 0 \)
(3) If a > 1, 0 < p < 1 ⇒ \( \log_a p < 0 \)
(4) If p > a > 1 ⇒ \( \log_a p > 1 \)
(5) If a > p > 1 ⇒ \( 0 < \log_a p < 1 \)
(6) If 0 < a < p < 1 ⇒ \( 0 < \log_a p < 1 \)
(7) If 0 < p < a < 1 ⇒ \( \log_a p > 1 \)
(8) If \( \log_m a > b \Rightarrow \begin{cases} a > m^b, & \text{if } m > 1 \\ a < m^b, & \text{if } 0 < m < 1 \end{cases} \)
(9) If \( \log_m a < b \Rightarrow \begin{cases} a < m^b, & \text{if } m > 1 \\ a > m^b, & \text{if } 0 < m < 1 \end{cases} \)
(10) \( \log_p a > \log_p b \Rightarrow a > b \) if the base p is positive and > 1, or a < b if the base p is positive and < 1, i.e. 0 < p < 1.
In other words, if the base is greater than 1 then the inequality remains the same, and if the base is positive but less than 1 then the sign of inequality is reversed.

Example 8. If \( x = \log_3 5,\ y = \log_{17} 25 \), which one of the following is correct [W.B. JEE 1993]
(a) x < y
(b) x = y
(c) x > y
(d) None of these
Answer:
Ans. (c) x > y
\( y = \log_{17} 25 = 2\log_{17} 5 \), ∴ \( \frac{1}{y} = \frac{1}{2}\log_5 17 \)
\( \frac{1}{x} = \log_5 3 = \frac{1}{2}\log_5 9 \)
Clearly \( \frac{1}{y} > \frac{1}{x} \) (as 17 > 9), ∴ x > y

Example 9. If \( \log_{0.3}(x - 1) < \log_{0.09}(x - 1) \), then x lies in the interval
(a) (2, ∞)
(b) (−2, −1)
(c) (1, 2)
(d) None of these
Answer:
Ans. (a) (2, ∞)
\( \log_{0.3}(x - 1) < \log_{(0.3)^2}(x - 1) = \frac{1}{2}\log_{0.3}(x - 1) \)
∴ \( \frac{1}{2}\log_{0.3}(x - 1) < 0 \), or \( \log_{0.3}(x - 1) < 0 = \log_{0.3} 1 \)
As the base is less than 1, the inequality is reversed: (x − 1) > 1, or x > 2 ⇒ x lies in (2, ∞).

Indices and Surds

Definition of Indices

If a is any non-zero real or imaginary number and m is a positive integer, then \( a^m = a \cdot a \cdot a \cdots a \) (m times). Here a is called the base and m the index, power or exponent.

Laws of Indices

(1) \( a^0 = 1,\ (a \neq 0) \)
(2) \( a^{-m} = \frac{1}{a^m},\ (a \neq 0) \)
(3) \( a^{m + n} = a^m \cdot a^n \), where m and n are rational numbers
(4) \( a^{m - n} = \frac{a^m}{a^n} \), where m and n are rational numbers, a ≠ 0
(5) \( (a^m)^n = a^{mn} \)
(6) \( a^{p/q} = \sqrt[q]{a^p} \)
(7) If x = y, then \( a^x = a^y \), but the converse may not be true. For example: \( (1)^6 = (1)^8 \), but 6 ≠ 8.
(i) If a ≠ ±1 or 0, then x = y
(ii) If a = 1, then x, y may be any real numbers
(iii) If a = −1, then x, y may be both even or both odd
(iv) If a = 0, then x, y may be any non-zero real numbers
But if we have to solve equations like \( [f(x)]^{\phi(x)} = [f(x)]^{\psi(x)} \), then we have to solve: (a) f(x) = 1   (b) f(x) = −1   (c) f(x) = 0   (d) φ(x) = ψ(x). Verification should be done in cases (b) and (c).
(8) \( a^m \cdot b^m = (ab)^m \) is not always true. In the real domain, \( \sqrt{a}\sqrt{b} = \sqrt{ab} \) only when a ≥ 0, b ≥ 0. In the complex domain, \( \sqrt{a} \cdot \sqrt{b} = \sqrt{ab} \) if at least one of a and b is positive.
(9) If \( a^x = b^x \), then consider the following cases: (i) if a ≠ ±b, then x = 0; (ii) if a = b ≠ 0, then x may have any real value; (iii) if a = −b, then x is even.
If we have to solve an equation of the form \( [f(x)]^{\phi(x)} = [g(x)]^{\phi(x)} \), i.e. same index, different bases, then we have to solve (a) f(x) = g(x), (b) f(x) = −g(x), (c) φ(x) = 0. Verification should be done in cases (b) and (c).

Example 10. For x ≠ 0, \( \left(\frac{x^l}{x^m}\right)^{l^2 + lm + m^2} \left(\frac{x^m}{x^n}\right)^{m^2 + mn + n^2} \left(\frac{x^n}{x^l}\right)^{n^2 + nl + l^2} = \)
(a) 1
(b) x
(c) Does not exist
(d) None of these
Answer:
Ans. (a) 1
\( = (x^{l - m})^{l^2 + lm + m^2} (x^{m - n})^{m^2 + mn + n^2} (x^{n - l})^{n^2 + nl + l^2} \)
\( = x^{l^3 - m^3} \cdot x^{m^3 - n^3} \cdot x^{n^3 - l^3} = x^{l^3 - m^3 + m^3 - n^3 + n^3 - l^3} = x^0 = 1 \)

Example 11. If \( 2^x = 4^y = 8^z \) and xyz = 288, then \( \frac{1}{2x} + \frac{1}{4y} + \frac{1}{8z} = \)
(a) 11/48
(b) 11/24
(c) 11/8
(d) 11/96
Answer:
Ans. (d) 11/96
\( 2^x = 2^{2y} = 2^{3z} \), i.e. x = 2y = 3z = k (say). Then \( xyz = \frac{k^3}{6} = 288 \), so k = 12.
∴ x = 12, y = 6, z = 4. Therefore, \( \frac{1}{2x} + \frac{1}{4y} + \frac{1}{8z} = \frac{1}{24} + \frac{1}{24} + \frac{1}{32} = \frac{11}{96} \)

Example 12. \( \frac{2 \cdot 3^{n + 1} + 7 \cdot 3^{n - 1}}{3^{n + 2} - 2(1/3)^{1 - n}} = \)
(a) 1
(b) 3
(c) −1
(d) 0
Answer:
Ans. (a) 1
\( \frac{2 \cdot 3^{n + 1} + 7 \cdot 3^{n - 1}}{3^{n + 2} - 2\left(\frac{1}{3}\right)^{1 - n}} = \frac{2 \cdot 3^{n - 1} \cdot 3^2 + 7 \cdot 3^{n - 1}}{3^{n - 1} \cdot 3^3 - 2 \cdot 3^{n - 1}} = \frac{3^{n - 1}[18 + 7]}{3^{n - 1}[27 - 2]} = 1 \)

Example 13. If \( \left(\frac{2}{3}\right)^{x + 2} = \left(\frac{3}{2}\right)^{2 - 2x} \), then x = [UPSEAT 1999]
(a) 1
(b) 3
(c) 4
(d) 0
Answer:
Ans. (c) 4
\( \left(\frac{2}{3}\right)^{x + 2} = \left(\frac{3}{2}\right)^{2 - 2x} \Rightarrow \left(\frac{2}{3}\right)^{x + 2} = \left(\frac{2}{3}\right)^{2x - 2} \)
Clearly x + 2 = 2x − 2 ⇒ x = 4

Example 14. The equation \( 4^{(x^2 + 2)} - 9 \cdot 2^{(x^2 + 2)} + 8 = 0 \) has the solution
(a) x = 1
(b) x = −1
(c) \( x = \sqrt{2} \)
(d) \( x = -\sqrt{2} \)
Answer:
Ans. (a, b) x = 1; x = −1
\( 4^{(x^2 + 2)} - 9 \cdot 2^{(x^2 + 2)} + 8 = 0 \Rightarrow \left(2^{(x^2 + 2)}\right)^2 - 9 \cdot 2^{(x^2 + 2)} + 8 = 0 \)
Put \( 2^{(x^2 + 2)} = y \). Then \( y^2 - 9y + 8 = 0 \), which gives y = 8, y = 1.
When y = 8: \( 2^{x^2 + 2} = 2^3 \Rightarrow x^2 + 2 = 3 \Rightarrow x^2 = 1 \Rightarrow x = 1, -1 \)
When y = 1: \( 2^{x^2 + 2} = 2^0 \Rightarrow x^2 + 2 = 0 \Rightarrow x^2 = -2 \), which is not possible.

Definition of Surds

Any root of a number which cannot be exactly found is called a surd.
Let a be a rational number and n a positive integer. If the n^th root of a, i.e. \( a^{1/n} \), is irrational, then it is called a surd of order n.
Order of a surd is indicated by the number denoting the root.
For example \( \sqrt{7}, \sqrt[3]{9}, (11)^{3/5}, \sqrt[n]{3} \) are surds of second, third, fifth and n^th order respectively.
A second order surd is often called a quadratic surd, and a surd of third order is called a cubic surd.
Note: If a is not rational, \( \sqrt[n]{a} \) is not a surd. For example, \( \sqrt{5 + \sqrt{7}} \) is not a surd as \( 5 + \sqrt{7} \) is not a rational number.

Types of Surds

(1) Simple surd: A surd consisting of a single term. For example \( 2\sqrt{3}, 6\sqrt{5}, \sqrt{5} \) etc.
(2) Pure and mixed surds: A surd consisting wholly of an irrational number is called a pure surd. Example: \( \sqrt{5}, \sqrt[3]{7} \). A surd consisting of the product of a rational number and an irrational number is called a mixed surd. Example: \( 5\sqrt{3} \).
(3) Compound surds: An expression consisting of the sum or difference of two or more surds. Example: \( \sqrt{5} + \sqrt{2},\ 2 - \sqrt{3} + 3\sqrt{5} \) etc.
(4) Similar surds: If the surds are different multiples of the same surd, they are called similar surds. Example: \( \sqrt{45}, \sqrt{80} \) are similar surds because they are equal to \( 3\sqrt{5} \) and \( 4\sqrt{5} \) respectively.
(5) Binomial surds: A compound surd consisting of two surds is called a binomial surd. Example: \( \sqrt{5} - \sqrt{2},\ 3 + \sqrt[3]{2} \) etc.
(6) Binomial quadratic surds: Binomial surds consisting of pure (or simple) surds of order two, i.e. surds of the form \( a\sqrt{b} \pm c\sqrt{d} \) or \( a \pm b\sqrt{c} \), are called binomial quadratic surds.
Two binomial quadratic surds which differ only in the sign which connects their terms are said to be conjugate or complementary to each other. The product of a binomial quadratic surd and its conjugate is always rational.
For example: the conjugate of the surd \( 2\sqrt{7} + 5\sqrt{3} \) is the surd \( 2\sqrt{7} - 5\sqrt{3} \).

Properties of Quadratic Surds

(1) The square root of a rational number cannot be expressed as the sum or difference of a rational number and a quadratic surd.
(2) If two quadratic surds cannot be reduced to others which have the same irrational part, their product is irrational.
(3) One quadratic surd cannot be equal to the sum or difference of two others not having the same irrational part.
(4) If \( a + \sqrt{b} = c + \sqrt{d} \), where a and c are rational and \( \sqrt{b}, \sqrt{d} \) are irrational, then a = c and b = d.

Example 15. The greatest number among \( \sqrt[3]{9}, \sqrt[4]{11}, \sqrt[6]{17} \) is
(a) \( \sqrt[3]{9} \)
(b) \( \sqrt[4]{11} \)
(c) \( \sqrt[6]{17} \)
(d) Cannot be determined
Answer:
Ans. (a) \( \sqrt[3]{9} \)
The L.C.M. of 3, 4, 6 is 12.
∴ \( \sqrt[3]{9} = 9^{1/3} = (9^4)^{1/12} = (6561)^{1/12} \), \( \sqrt[4]{11} = (11^3)^{1/12} = (1331)^{1/12} \), \( \sqrt[6]{17} = (17^2)^{1/12} = (289)^{1/12} \)
Hence \( \sqrt[3]{9} \) is the greatest number.

Example 16. The value of \( \frac{15}{\sqrt{10} + \sqrt{20} + \sqrt{40} - \sqrt{5} - \sqrt{80}} \) is
(a) \( \sqrt{5}(5 + \sqrt{2}) \)
(b) \( \sqrt{5}(2 + \sqrt{2}) \)
(c) \( \sqrt{5}(1 + \sqrt{2}) \)
(d) \( \sqrt{5}(3 + \sqrt{2}) \)
Answer:
Ans. (c) \( \sqrt{5}(1 + \sqrt{2}) \)
Given fraction \( = \frac{15}{\sqrt{10} + 2\sqrt{5} + 2\sqrt{10} - \sqrt{5} - 4\sqrt{5}} = \frac{15}{3\sqrt{10} - 3\sqrt{5}} = \frac{5}{\sqrt{10} - \sqrt{5}} \)
\( = \frac{5}{\sqrt{10} - \sqrt{5}} \cdot \frac{\sqrt{10} + \sqrt{5}}{\sqrt{10} + \sqrt{5}} = \sqrt{10} + \sqrt{5} = \sqrt{5}(\sqrt{2} + 1) \)

Example 17. If \( x = \sqrt[3]{\sqrt{2} + 1} - \sqrt[3]{\sqrt{2} - 1} \), then \( x^3 + 3x = \)
(a) 2
(b) 6
(c) 6x
(d) None of these
Answer:
Ans. (a) 2
\( x = (\sqrt{2} + 1)^{1/3} - (\sqrt{2} - 1)^{1/3} \)
\( x^3 = (\sqrt{2} + 1) - (\sqrt{2} - 1) - 3(\sqrt{2} + 1)^{1/3}(\sqrt{2} - 1)^{1/3}\left[\sqrt[3]{\sqrt{2} + 1} - \sqrt[3]{\sqrt{2} - 1}\right] \)
\( x^3 = 2 - 3(2 - 1)^{1/3}x \Rightarrow x^3 + 3x = 2 \)

Rationalisation Factors

If two surds are such that their product is rational, then each one of them is called a rationalising factor of the other.
Thus each of \( 2\sqrt{3} \) and \( \sqrt{3} \) is a rationalising factor of the other. Similarly \( \sqrt{3} + \sqrt{2} \) and \( \sqrt{3} - \sqrt{2} \) are rationalising factors of each other, as \( (\sqrt{3} + \sqrt{2})(\sqrt{3} - \sqrt{2}) = 1 \), which is rational.
To find the factor which will rationalise any given binomial surd:
Case I: Suppose the given surd is \( \sqrt[p]{a} - \sqrt[q]{b} \). Suppose \( a^{1/p} = x,\ b^{1/q} = y \) and let n be the L.C.M. of p and q. Then \( x^n \) and \( y^n \) are both rational.
Now \( x^n - y^n \) is divisible by x − y for all values of n, and \( x^n - y^n = (x - y)(x^{n-1} + x^{n-2}y + x^{n-3}y^2 + \ldots + y^{n-1}) \).
Thus the rationalising factor is \( x^{n-1} + x^{n-2}y + x^{n-3}y^2 + \ldots + y^{n-1} \) and the rational product is \( x^n - y^n \).
Case II: Let the given surd be \( \sqrt[p]{a} + \sqrt[q]{b} \). Let x, y, n have the same meaning as in Case I.
(1) If n is even, then \( x^n - y^n \) is divisible by x + y and \( x^n - y^n = (x + y)(x^{n-1} - x^{n-2}y + x^{n-3}y^2 - \ldots - y^{n-1}) \). Thus the rationalising factor is \( x^{n-1} - x^{n-2}y + x^{n-3}y^2 - \ldots - y^{n-1} \) and the rational product is \( x^n - y^n \).
(2) If n is odd, \( x^n + y^n \) is divisible by x + y, and \( x^n + y^n = (x + y)(x^{n-1} - x^{n-2}y + \ldots - xy^{n-2} + y^{n-1}) \). Thus the rationalising factor is \( x^{n-1} - x^{n-2}y + \ldots - xy^{n-2} + y^{n-1} \) and the rational product is \( x^n + y^n \).

Example 18. The rationalising factor of \( a^{1/3} + a^{-1/3} \) is
(a) \( a^{1/3} - a^{-1/3} \)
(b) \( a^{2/3} + a^{-2/3} \)
(c) \( a^{2/3} - a^{-2/3} \)
(d) \( a^{2/3} + a^{-2/3} - 1 \)
Answer:
Ans. (d) \( a^{2/3} + a^{-2/3} - 1 \)
Let \( x = a^{1/3},\ y = a^{-1/3} \); then \( a = x^3,\ a^{-1} = y^3 \)
\( x^3 + y^3 = (x + y)(x^2 - xy + y^2) \)
So the rationalising factor is \( (x^2 - xy + y^2) \). Put the values of x and y (xy = 1):
Thus the required rationalising factor is \( a^{2/3} + a^{-2/3} - 1 \).

Square Roots of a + √b and a + √b + √c + √d where √b, √c, √d are Surds

Let \( \sqrt{a + \sqrt{b}} = \sqrt{x} + \sqrt{y} \), where x, y > 0 are rational numbers.
Then squaring both sides we have \( a + \sqrt{b} = x + y + 2\sqrt{x}\sqrt{y} \Rightarrow a = x + y,\ \sqrt{b} = 2\sqrt{xy} \Rightarrow b = 4xy \)
So, \( (x - y)^2 = (x + y)^2 - 4xy = a^2 - b \). After solving we can find x and y.
Similarly the square root of \( a - \sqrt{b} \) can be found by taking \( \sqrt{a - \sqrt{b}} = \sqrt{x} - \sqrt{y} \), x > y.
To find the square root of \( a + \sqrt{b} + \sqrt{c} + \sqrt{d} \): Let \( \sqrt{a + \sqrt{b} + \sqrt{c} + \sqrt{d}} = \sqrt{x} + \sqrt{y} + \sqrt{z} \) (x, y, z > 0) and take \( \sqrt{a + \sqrt{b} - \sqrt{c} - \sqrt{d}} = \sqrt{x} + \sqrt{y} - \sqrt{z} \). Then by squaring and equating, we get equations in x, y, z. On solving these equations, we can find the required square roots.
Note: (i) If \( a^2 - b \) is not a perfect square, the square root of \( a + \sqrt{b} \) is complicated, i.e. we cannot find the value of \( \sqrt{a + \sqrt{b}} \) in the form of a compound surd.
(ii) If \( \sqrt{a + \sqrt{b}} = \sqrt{x} + \sqrt{y} \), x > y, then \( \sqrt{a - \sqrt{b}} = \sqrt{x} - \sqrt{y} \)
(iii) \( \sqrt{a + \sqrt{b}} = \sqrt{\frac{a + \sqrt{a^2 - b}}{2}} + \sqrt{\frac{a - \sqrt{a^2 - b}}{2}} \)
(iv) \( \sqrt{a - \sqrt{b}} = \sqrt{\frac{a + \sqrt{a^2 - b}}{2}} - \sqrt{\frac{a - \sqrt{a^2 - b}}{2}} \)
(v) If a is a rational number and \( \sqrt{b}, \sqrt{c}, \sqrt{d} \) are surds such that \( \sqrt{a + \sqrt{b} + \sqrt{c} + \sqrt{d}} = \sqrt{x} + \sqrt{y} + \sqrt{z} \) with b = 4xy, c = 4yz, d = 4zx, then \( x = \sqrt{\frac{bd}{4c}},\ y = \sqrt{\frac{bc}{4d}},\ z = \sqrt{\frac{cd}{4b}} \) and a = x + y + z. For example, \( \sqrt{10 + \sqrt{24} + \sqrt{40} + \sqrt{60}} = \sqrt{2} + \sqrt{3} + \sqrt{5} \). When some of the surds carry a minus sign, the same working is done with the corresponding terms \( \sqrt{x}, \sqrt{y}, \sqrt{z} \) taken with suitable signs.

Example 19. \( \sqrt{3 + \sqrt{5}} \) is equal to
(a) \( \sqrt{5} + 1 \)
(b) \( \sqrt{3} + \sqrt{2} \)
(c) \( \frac{\sqrt{5} + 1}{\sqrt{2}} \)
(d) \( \frac{1}{2}(\sqrt{5} + 1) \)
Answer:
Ans. (c) \( \frac{\sqrt{5} + 1}{\sqrt{2}} \)
Let \( \sqrt{3 + \sqrt{5}} = \sqrt{x} + \sqrt{y} \Rightarrow 3 + \sqrt{5} = x + y + 2\sqrt{xy} \)
Obviously x + y = 3 and 4xy = 5. So \( (x - y)^2 = 9 - 5 = 4 \), or x − y = 2.
After solving, \( x = \frac{5}{2},\ y = \frac{1}{2} \). Hence \( \sqrt{3 + \sqrt{5}} = \sqrt{\frac{5}{2}} + \sqrt{\frac{1}{2}} = \frac{\sqrt{5} + 1}{\sqrt{2}} \)

Example 20. \( \sqrt{10 - \sqrt{24} - \sqrt{40} + \sqrt{60}} = \)
(a) \( \sqrt{5} + \sqrt{3} + \sqrt{2} \)
(b) \( \sqrt{5} + \sqrt{3} - \sqrt{2} \)
(c) \( \sqrt{5} - \sqrt{3} + \sqrt{2} \)
(d) \( \sqrt{2} + \sqrt{3} - \sqrt{5} \)
Answer:
Ans. (b) \( \sqrt{5} + \sqrt{3} - \sqrt{2} \)
Let \( 10 - \sqrt{24} - \sqrt{40} + \sqrt{60} = (\sqrt{a} - \sqrt{b} + \sqrt{c})^2 = a + b + c - 2\sqrt{ab} - 2\sqrt{bc} + 2\sqrt{ca} \), a, b, c > 0.
Then a + b + c = 10, ab = 6, bc = 10, ca = 15.
\( a^2b^2c^2 = 900 \Rightarrow abc = 30 \). So a = 3, b = 2, c = 5.
Therefore, \( \sqrt{10 - \sqrt{24} - \sqrt{40} + \sqrt{60}} = \sqrt{3} + \sqrt{5} - \sqrt{2} \) (taking the positive square root).

Example 21. \( \sqrt[4]{17 + 12\sqrt{2}} = \)
(a) \( \sqrt{2} + 1 \)
(b) \( 2^{1/4}(\sqrt{2} + 1) \)
(c) \( 2\sqrt{2} + 1 \)
(d) None of these
Answer:
Ans. (a) \( \sqrt{2} + 1 \)
\( \sqrt{17 + 12\sqrt{2}} = \sqrt{3^2 + (2\sqrt{2})^2 + 2 \cdot 3 \cdot 2\sqrt{2}} = 3 + 2\sqrt{2} \)
∴ \( \sqrt[4]{17 + 12\sqrt{2}} = \sqrt{3 + 2\sqrt{2}} = \sqrt{2} + 1 \)

Cube Root of a Binomial Quadratic Surd

If \( (a + \sqrt{b})^{1/3} = x + \sqrt{y} \), then \( (a - \sqrt{b})^{1/3} = x - \sqrt{y} \), where a is a rational number and \( \sqrt{b} \) is a surd.
The procedure of finding \( (a + \sqrt{b})^{1/3} \) is illustrated with the help of an example:
Taking \( (37 - 30\sqrt{3})^{1/3} = x - \sqrt{y} \), we get on cubing both sides, \( 37 - 30\sqrt{3} = x^3 + 3xy - (3x^2 + y)\sqrt{y} \)
∴ \( x^3 + 3xy = 37 \) and \( (3x^2 + y)\sqrt{y} = 30\sqrt{3} = 15\sqrt{12} \)
As \( \sqrt{3} \) cannot be reduced, let us assume y = 3; we get \( 3x^2 + y = 3x^2 + 3 = 30 \), ∴ x = 3, which does not satisfy \( x^3 + 3xy = 37 \).
Again taking y = 12, we get \( 3x^2 + 12 = 15 \), ∴ x = 1.
x = 1, y = 12 satisfy \( x^3 + 3xy = 37 \).
∴ \( \sqrt[3]{37 - 30\sqrt{3}} = 1 - \sqrt{12} = 1 - 2\sqrt{3} \)

Example 22. \( \sqrt[3]{61 - 46\sqrt{5}} = \)
(a) \( 1 - 2\sqrt{5} \)
(b) \( 1 - \sqrt{5} \)
(c) \( 2 - \sqrt{5} \)
(d) None of these
Answer:
Ans. (a) \( 1 - 2\sqrt{5} \)
Let \( \sqrt[3]{61 - 46\sqrt{5}} = a - \sqrt{b} \Rightarrow 61 - 46\sqrt{5} = (a - \sqrt{b})^3 = a^3 + 3ab - (3a^2 + b)\sqrt{b} \)
\( \Rightarrow 61 = a^3 + 3ab,\ 46\sqrt{5} = (3a^2 + b)\sqrt{b} \Rightarrow 61 = (a^2 + 3b)a,\ 23\sqrt{20} = (3a^2 + b)\sqrt{b} \)
So a = 1, b = 20. Therefore \( \sqrt[3]{61 - 46\sqrt{5}} = 1 - \sqrt{20} = 1 - 2\sqrt{5} \)

Equations Involving Surds

While solving equations involving surds, we usually have to square; on squaring, the domain of the equation extends and we may get some extraneous solutions, so we must verify the solutions and neglect those which do not satisfy the equation.
Note that from ax = bx, to conclude a = b is not correct. The correct procedure is x(a − b) = 0, i.e. x = 0 or a = b. Here, verification is necessary.

Example 23. The equation \( \sqrt{x + 1} - \sqrt{x - 1} = \sqrt{4x - 1} \), x ∈ R has
(a) One solution
(b) Two solutions
(c) Four solutions
(d) No solution
Answer:
Ans. (d) No solution
Given \( \sqrt{x + 1} - \sqrt{x - 1} = \sqrt{4x - 1} \) .....(i)
Squaring both sides, we get \( -2\sqrt{x^2 - 1} = 2x - 1 \)
Squaring again, we get \( x = \frac{5}{4} \), which does not satisfy equation (i) (L.H.S. = 1, R.H.S. = 2).
Hence, there is no solution of the given equation.

Partial Fractions

Definition

An expression of the form \( \frac{f(x)}{g(x)} \), where f(x) and g(x) are polynomials in x, is called a rational fraction.
(1) Proper rational functions: Functions of the form \( \frac{f(x)}{g(x)} \), where f(x) and g(x) are polynomials and g(x) ≠ 0, are called rational functions of x. If the degree of f(x) is less than the degree of g(x), then \( \frac{f(x)}{g(x)} \) is called a proper rational function. Example: \( \frac{x + 2}{x^2 + 2x + 4} \) is a proper rational function.
(2) Improper rational functions: If the degree of f(x) is greater than or equal to the degree of g(x), then \( \frac{f(x)}{g(x)} \) is called an improper rational function. For example: \( \frac{x^3}{(x - 1)(x - 2)} \) is an improper rational function.
(3) Partial fractions: Any proper rational function can be broken up into a group of different rational fractions, each having a simple factor of the denominator of the original rational function. Each such fraction is called a partial fraction.
If by some process we can break a given rational function \( \frac{f(x)}{g(x)} \) into different fractions whose denominators are the factors of g(x), then the process of obtaining them is called the resolution or decomposition of \( \frac{f(x)}{g(x)} \) into its partial fractions.

Different Cases of Partial Fractions

(1) When the denominator consists of non-repeated linear factors: To each linear factor (x − a) occurring once in the denominator of a proper fraction, there corresponds a single partial fraction of the form \( \frac{A}{x - a} \), where A is a constant to be determined.
If \( g(x) = (x - a_1)(x - a_2)(x - a_3) \cdots (x - a_n) \), then we assume that \( \frac{f(x)}{g(x)} = \frac{A_1}{x - a_1} + \frac{A_2}{x - a_2} + \ldots + \frac{A_n}{x - a_n} \)
where \( A_1, A_2, \ldots, A_n \) are constants, which can be determined by equating the numerator of L.H.S. to the numerator of R.H.S. (after L.C.M.) and substituting \( x = a_1, a_2, \ldots, a_n \).
Note: (i) The remainder when a polynomial f(x) is divided by (x − a) is f(a). e.g. the remainder of x² + 3x − 7 when divided by x − 2 is (2)² + 3(2) − 7 = 3.
(ii) \( \frac{px + q}{(x - a)(x - b)} = \frac{pa + q}{(x - a)(a - b)} + \frac{pb + q}{(b - a)(x - b)} \)

Example 24. The remainder obtained when the polynomial \( x^{64} + x^{27} + 1 \) is divided by (x + 1) is
(a) 1
(b) −1
(c) 2
(d) −2
Answer:
Ans. (a) 1
Remainder of \( x^{64} + x^{27} + 1 \) when divided by x + 1 is \( (-1)^{64} + (-1)^{27} + 1 = 1 - 1 + 1 = 1 \)

Example 25. If \( \frac{2x + 3}{(x + 1)(x - 3)} = \frac{a}{x + 1} + \frac{b}{x - 3} \), then a + b = [MNR 1993]
(a) 1
(b) 2
(c) \( \frac{9}{4} \)
(d) \( -\frac{1}{4} \)
Answer:
Ans. (b) 2
2x + 3 = a(x − 3) + b(x + 1)
Put x = −1: 2(−1) + 3 = a(−1 − 3) ⇒ 1 = −4a ⇒ \( a = -\frac{1}{4} \)
Now put x = 3: 2(3) + 3 = b(3 + 1) ⇒ 9 = 4b ⇒ \( b = \frac{9}{4} \)
Therefore, \( a + b = -\frac{1}{4} + \frac{9}{4} = 2 \)

Example 26. If \( \frac{3x + a}{x^2 - 3x + 2} = \frac{A}{x - 2} - \frac{10}{x - 1} \), then
(a) a = 7
(b) a = −7
(c) A = −13
(d) A = 13
Answer:
Ans. (a, d) a = 7; A = 13
\( \frac{3x + a}{x^2 - 3x + 2} = \frac{A}{x - 2} - \frac{10}{x - 1} \Rightarrow 3x + a = A(x - 1) - 10(x - 2) \)
⇒ 3 = A − 10, a = −A + 20 (on equating the coefficients of x and the constant term)
⇒ A = 13, a = 7.

(2) When the denominator consists of linear factors, some repeated: To each linear factor (x − a) occurring r times in the denominator of a proper rational function, there corresponds a sum of r partial fractions.
Let \( g(x) = (x - a)^k (x - a_1)(x - a_2) \cdots (x - a_r) \). Then we assume that
\( \frac{f(x)}{g(x)} = \frac{A_1}{x - a} + \frac{A_2}{(x - a)^2} + \ldots + \frac{A_k}{(x - a)^k} + \frac{B_1}{x - a_1} + \ldots + \frac{B_r}{x - a_r} \)
where \( A_1, A_2, \ldots, A_k \) are constants. To determine the values of the constants, adopt the procedure as above.

Example 27. If \( \frac{3x + 4}{(x + 1)^2(x - 1)} = \frac{A}{x - 1} + \frac{B}{x + 1} + \frac{C}{(x + 1)^2} \), then A = [EAMCET 1994]
(a) \( -\frac{1}{2} \)
(b) \( \frac{15}{4} \)
(c) \( \frac{7}{4} \)
(d) \( -\frac{1}{4} \)
Answer:
Ans. (c) \( \frac{7}{4} \)
We have \( \frac{3x + 4}{(x + 1)^2(x - 1)} = \frac{A}{x - 1} + \frac{B}{x + 1} + \frac{C}{(x + 1)^2} \)
\( \Rightarrow 3x + 4 = A(x + 1)^2 + B(x + 1)(x - 1) + C(x - 1) \)
Putting x = 1, we get \( 7 = A(2)^2 \Rightarrow A = \frac{7}{4} \)

Example 28. The partial fractions of \( \frac{x^2}{(x - 1)^3(x - 2)} \) are [IIT 1992]
(a) \( \frac{-1}{(x - 1)^3} + \frac{3}{(x - 1)^2} - \frac{4}{x - 1} + \frac{4}{x - 2} \)
(b) \( \frac{-1}{(x - 1)^3} - \frac{3}{(x - 1)^2} + \frac{4}{x - 1} + \frac{4}{x - 2} \)
(c) \( \frac{-1}{(x - 1)^3} + \frac{-3}{(x - 1)^2} + \frac{-4}{x - 1} + \frac{4}{x - 2} \)
(d) None of these
Answer:
Ans. (c) \( \frac{-1}{(x - 1)^3} + \frac{-3}{(x - 1)^2} + \frac{-4}{x - 1} + \frac{4}{x - 2} \)
Put the repeated factor x − 1 = y ⇒ x = y + 1. ∴ \( \frac{x^2}{(x - 1)^3(x - 2)} = \frac{(1 + y)^2}{y^3(y - 1)} = \frac{1 + 2y + y^2}{y^3(-1 + y)} \)
Dividing the numerator \( 1 + 2y + y^2 \) by −1 + y till \( y^3 \) appears as a factor, we get \( \frac{1 + 2y + y^2}{-1 + y} = (-1 - 3y - 4y^2) + \frac{4y^3}{-1 + y} \)
Given expression \( = \frac{-1}{y^3} - \frac{3}{y^2} - \frac{4}{y} + \frac{4}{-1 + y} = \frac{-1}{(x - 1)^3} + \frac{-3}{(x - 1)^2} + \frac{-4}{x - 1} + \frac{4}{x - 2} \)

(3) When the denominator consists of non-repeated quadratic factors: To each irreducible non-repeated quadratic factor \( ax^2 + bx + c \), there corresponds a partial fraction of the form \( \frac{Ax + B}{ax^2 + bx + c} \), where A and B are constants to be determined.
Example: \( \frac{4x^2 + 2x + 3}{(x^2 + 4x + 9)(x - 2)(x + 3)} = \frac{Ax + B}{x^2 + 4x + 9} + \frac{C}{x - 2} + \frac{D}{x + 3} \)
Note: (i) \( \frac{px + q}{x^2(x - a)} = \frac{-q}{ax^2} - \frac{pa + q}{a^2 x} + \frac{pa + q}{a^2(x - a)} \)
(ii) \( \frac{px + q}{x(x - a)^2} = \frac{q}{a^2 x} - \frac{q}{a^2(x - a)} + \frac{pa + q}{a(x - a)^2} \)
(iii) \( \frac{px + q}{x(x^2 + a^2)} = \frac{q}{a^2 x} + \frac{pa^2 - qx}{a^2(x^2 + a^2)} \)

Example 29. The partial fractions of \( \frac{3x - 1}{(1 - x + x^2)(2 + x)} \) are [MNR 1995]
(a) \( \frac{x}{x^2 - x + 1} + \frac{1}{x + 2} \)
(b) \( \frac{1}{x^2 - x + 1} + \frac{x}{x + 2} \)
(c) \( \frac{x}{x^2 - x + 1} - \frac{1}{x + 2} \)
(d) \( \frac{-1}{x^2 - x + 1} + \frac{x}{x + 2} \)
Answer:
Ans. (c) \( \frac{x}{x^2 - x + 1} - \frac{1}{x + 2} \)
\( \frac{3x - 1}{(1 - x + x^2)(2 + x)} = \frac{Ax + B}{x^2 - x + 1} + \frac{C}{x + 2} \Rightarrow (3x - 1) = (Ax + B)(x + 2) + C(x^2 - x + 1) \)
Comparing the coefficients of like terms, we get A + C = 0, 2A + B − C = 3, 2B + C = −1 ⇒ A = 1, B = 0, C = −1
∴ \( \frac{3x - 1}{(1 - x + x^2)(2 + x)} = \frac{x}{x^2 - x + 1} - \frac{1}{x + 2} \)

Example 30. If \( \frac{(x + 1)^2}{x^3 + x} = \frac{A}{x} + \frac{Bx + C}{x^2 + 1} \), then \( \sin^{-1}\left(\frac{A}{C}\right) = \) [EAMCET 1997, 98]
(a) \( \frac{\pi}{6} \)
(b) \( \frac{\pi}{4} \)
(c) \( \frac{\pi}{3} \)
(d) \( \frac{\pi}{2} \)
Answer:
Ans. (a) \( \frac{\pi}{6} \)
\( \frac{(x + 1)^2}{x^3 + x} = \frac{A}{x} + \frac{Bx + C}{x^2 + 1} \Rightarrow (x + 1)^2 = A(x^2 + 1) + (Bx + C)x \)
⇒ A + B = 1, C = 2, A = 1 ⇒ B = 0
Therefore \( \sin^{-1}\left(\frac{A}{C}\right) = \sin^{-1}\left(\frac{1}{2}\right) = 30° = \frac{\pi}{6} \)

(4) When the denominator consists of repeated quadratic factors: To each irreducible quadratic factor \( ax^2 + bx + c \) occurring r times in the denominator of a proper rational fraction, there corresponds a sum of r partial fractions of the form
\( \frac{A_1 x + B_1}{ax^2 + bx + c} + \frac{A_2 x + B_2}{(ax^2 + bx + c)^2} + \ldots + \frac{A_r x + B_r}{(ax^2 + bx + c)^r} \)
where the A’s and B’s are constants to be determined.

Example 31. If \( \frac{x}{(x - 1)(x^2 + 1)^2} = \frac{1}{4}\left[\frac{1}{x - 1} - \frac{x + 1}{x^2 + 1}\right] + y \), then y =
(a) \( \frac{1 - x}{2(x^2 + 1)^2} \)
(b) \( \frac{1 - x}{3(x^2 + 1)} \)
(c) \( \frac{1 + x}{2(x^2 - 1)^2} \)
(d) None of these
Answer:
Ans. (a) \( \frac{1 - x}{2(x^2 + 1)^2} \)
Let \( y = \frac{Ax + B}{(x^2 + 1)^2} \). Then \( \frac{x}{(x - 1)(x^2 + 1)^2} = \frac{1}{4}\left[\frac{1}{x - 1} - \frac{x + 1}{x^2 + 1}\right] + \frac{Ax + B}{(x^2 + 1)^2} \)
\( \Rightarrow 4x = (x^2 + 1)^2 - (x + 1)(x - 1)(x^2 + 1) + 4(Ax + B)(x - 1) \)
⇒ 4A + 2 = 0, 4B − 4A = 4 ⇒ \( A = -\frac{1}{2},\ B = \frac{1}{2} \)
∴ \( y = \frac{Ax + B}{(x^2 + 1)^2} = \frac{1}{2} \cdot \frac{1 - x}{(x^2 + 1)^2} \)

Partial Fractions of Improper Rational Functions

If the degree of f(x) is greater than or equal to the degree of g(x), then \( \frac{f(x)}{g(x)} \) is called an improper rational function, and every such rational function can be transformed into a proper rational function by dividing the numerator by the denominator.
We divide the numerator by the denominator until a remainder is obtained which is of lower degree than the denominator, i.e. \( \frac{f(x)}{g(x)} = Q(x) + \frac{R(x)}{g(x)} \), where degree of R(x) < degree of g(x).
For example, \( \frac{x^3}{x^2 - 5x + 6} \) is an improper rational function and can be expressed as \( (x + 5) + \frac{19x - 30}{x^2 - 5x + 6} \), which is the sum of a polynomial (x + 5) and a proper rational function \( \frac{19x - 30}{x^2 - 5x + 6} \).

Example 32. If \( \frac{x^3 - 6x^2 + 10x - 2}{x^2 - 5x + 6} = f(x) + \frac{A}{x - 2} + \frac{B}{x - 3} \), then f(x) =
(a) x − 1
(b) x + 1
(c) x
(d) None of these
Answer:
Ans. (a) x − 1
Dividing \( x^3 - 6x^2 + 10x - 2 \) by \( x^2 - 5x + 6 \): the first term of the quotient is x, and \( x^3 - 6x^2 + 10x - 2 - x(x^2 - 5x + 6) = -x^2 + 4x - 2 \)
The next term of the quotient is −1, and \( -x^2 + 4x - 2 - (-1)(x^2 - 5x + 6) = -x + 4 \), the remainder.
So \( \frac{x^3 - 6x^2 + 10x - 2}{x^2 - 5x + 6} = (x - 1) + \frac{-x + 4}{x^2 - 5x + 6} \). ∴ f(x) = x − 1.

General Method of Finding the Constants

(1) Express the given fraction in its partial fractions in accordance with the rules written above.
(2) Then multiply both sides by the denominator of the given fraction to get an identity which holds for all values of x.
(3) Equate the coefficients of like powers of x in the resulting identity and solve the equations so obtained simultaneously to find the various constants. Sometimes, as a short method, we substitute particular values of the variable x in the identity obtained after clearing of fractions to find some or all the constants. For non-repeated linear factors, the values of x used are those for which the denominators of the corresponding partial fractions become zero.
Note: If the given fraction is improper, then before finding partial fractions, the given fraction must be expressed as the sum of a polynomial and a proper fraction by division.
Important Tip: Sometimes a suitable substitution transforms the given function into a rational fraction which can be broken into partial fractions.

Example 33. The coefficient of \( x^n \) in the expression \( \frac{5x + 6}{(2 + x)(1 - x)} \) when expanded in ascending powers of x is
(a) \( \frac{-2}{3} \cdot \frac{(-1)^n}{2^n} + \frac{11}{3} \)
(b) \( \frac{2}{3} + \frac{(-1)^n}{2^n} - \frac{11}{3} \)
(c) \( -\frac{2}{3} + \frac{(-1)^n}{3} - \frac{11}{2^n} \)
(d) None of these
Answer:
Ans. (a) \( \frac{-2}{3} \cdot \frac{(-1)^n}{2^n} + \frac{11}{3} \)
\( \frac{5x + 6}{(2 + x)(1 - x)} = \frac{-4/3}{2 + x} + \frac{11/3}{1 - x} \)
Rewriting the denominators, \( = \frac{-4/3}{2\left(1 + \frac{x}{2}\right)} + \frac{11/3}{1 - x} = \frac{-2}{3}\left(1 + \frac{x}{2}\right)^{-1} + \frac{11}{3}(1 - x)^{-1} \)
\( = \frac{-2}{3}\left[1 - \frac{x}{2} + \frac{x^2}{4} - \frac{x^3}{8} + \ldots + (-1)^n \frac{x^n}{2^n} + \ldots\right] + \frac{11}{3}[1 + x + x^2 + \ldots + x^n + \ldots] \)
The coefficient of \( x^n \) in the given expression is \( \frac{-2}{3}(-1)^n \frac{1}{2^n} + \frac{11}{3} \).