Class 11

Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 14 Mathematical Reasoning Ex 14.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Maths Chapter 14 Mathematical Reasoning Ex 14.4

Question 1.
Rewrite the following statement with “if-then” in five different ways conveying the same meaning.
If a natural number is odd, then its square is also odd.
Answer.
(i) A natural number is odd implies that its square is odd.
(ii) A natural number is odd only if its square is odd.
(iii) If the square of a natural number is not odd, then the natural number is also not odd.
(iv) For a natural number to be odd it is necessary that its square is odd.
(v) For a square of a natural number to be odd, it is sufficient that the number is odd.

Question 2.
Write the contrapositive and converse of the following statements.
(i) If x is a prime number, then x is odd.
(ii) It the two lines are parallel, then they do not intersect in the same plane.
(iii) Something is cold implies that it has low temperature.
(iv) You cannot comprehend geometry if you do not know how to reason deductively.
(v) x is an even number implies that x is divisible by 4.
Answer.
(i) Contrapositive statement: If a number x is not odd, then xis not a prime number.
Converse statement: If x is odd, then it is a prime number.

(ii) Contrapositive statement: If two straight lines intersect in a plane than the lines are not parallel.
Converse statement: If two lines do not intersect in the sample plane, then the two lines are parallel.

(iii) Contrapositive statement : If the temperature of something is not low, then it is not cold.
Converse statement : If something has low temperature, then it is cold.

(iv) Contrapositive statement: If you know how to reason deductively, then you comprehend geometry.

(v) Converse statement: If you do not know how to reason deductively, then you can not comprehend geometry.
The given statement can be written as :
“If x is an even number, then x is divisible by 4”.
Contrapositive statement: If x is not divisible by 4, then x is not an even number.
Converse statement: If x is divisible by 4 then x is an even number.

Question 3.
Write each of the following statement in the form “if-then”.
(i) You get a job implies that your credentials are good.
(ii) The Banana trees will bloom if it stays warm for a month.
(iii) A quadrilateral is a parallelogram if its diagonals bisect each other.
(iv) To get A+ in the class, it is necessary that you do the exercises of the book.
Answer.
(i) If you get a job, then your credential are good.
(ii) If the banana trees,stays warm for a month, then it will bloom.
(iii) If the diagonals of a quadrilateral bisect each other, then it is a parallelogram.
(iv) If you get A+ in the class, then you do all exercises in the book.

Question 4.
Given statements in (a) and (b). Identify the statements given below as contrapositive or converse of each other.
(a) If you live in Delhi, then you have winter clothes.
(i) If you do not have winter clothes, then you do not live in Delhi.
(ii) If you have winter clothes, then you live in Delhi.

(b) If a quadrilateral is a parallelogram, then its diagonals bisect each other.
(i) If the diagonals of a quadrilateral do not bisect each other, then the quadrilateral is not a parallelogram.
(ii) If the diagonals of a quadrilateral bisect each other, then it is a parallelogram.
Answer.
(a) (i) Contrapositive statement
(ii) Converse statement.

(b) (i) Contrapositive statement
(ii) Converse statement.

Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 14 Mathematical Reasoning Ex 14.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Maths Chapter 14 Mathematical Reasoning Ex 14.3

Question 1.
For each of the following compound statements first identify the connecting words and then break it into component statements.
(i) All rational numbers are real and all real numbers are not complex.
(ii) Square of an integer is positive or negative.
(iii) The sand heats up quickly in the Sun and does not cool down fast at night.
(iv) x = 2 and x = 3 are the roots of the equation 3x² – x – 10 = 0.
Answer.
(i) Here, the connecting word is ‘and’.
The component statements are as follows.
p : All rational numbers are real.
q : All real numbers are not complex.

(ii) Here, the connecting word is ‘or’.
The component statements are as follows,
p : Square of an integer is positive.
q : Square of an integer is negative.

(iii) Here, the connecting word is ‘and’.
The component statements are as follows.
p : The sand heats up quickly in the sun.
q : The sand does not cool down fast at night.

(iv) Here, the connecting word is ‘and’.
The component statements are as follows.
p : x = 2 is a root of the equation 3x² – x -10 = 0
q : x = 3 is a root of the equation 3x² – x -10 = 0

Question 2.
Identify the quantifier in the following statements and write the negation of the statements.
(i) There exists a number which is equal to its square.
(ii) For every real numbers, x is less than x + 1.
(iii) There exists a capital for every state in India.
Answer.
(i) Quantifier : There exists.
p : There exists a number which is equal to its square
not p : There does not exist a number which is equal to its square.

(ii) Quantifier : For every
p : For every real number x, x is less than x + 1
~p : For every real number x, x is not less than x + 1

(iii) Quantifier : There exists
p : There exists a capital for every state of India.
~ p : There does not exist a capital for every state of India.

Question 3.
Check whether the following pair of statements are negation of each other. Give reasons for the answer.
(i) x + y = y + x is true for every real numbers x andy.
(ii) There exists real number x and y for which x + y = y + x.
Answer.
Let p: x + y = y + x is true for every real numbers x and y.
q : There exists real numbers x and y for which x + y = y + x.
Now ~ p : There exists real numbers x and y for which x + y ≠ y + x. Thus ~ p ≠ q.

Question 4.
State whether the “Or” used in the following statements is exclusive “or” inclusive. Give reasons for your answer.
(i) Sun rises or Moon sets.
(ii) To apply for a driving licence, you should have a ration card or a passport.
(iii) All integers are positive or negative.
Answer.
(i) Here, “or” is exclusive because it is not possible for the Sun to rise and the Moon to set together.
(ii) Here, “or” is inclusive since a person can have both a ration card and a passport to apply for a driving licence.
(iii) Here, “or” is exclusive because all integers cannot be both positive and negative.

Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 14 Mathematical Reasoning Ex 14.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Maths Chapter 14 Mathematical Reasoning Ex 14.2

Question 1.
Write the negation of the following statements:
(i) Chennai is the capital of Tamil Nadu.
(ii) √2 is not a complex number.
(iii) All triangles are not equilateral triangle.
(iv) The number 2 is greater than 7.
(v) Every natural number is an integer.
Answer.
The negation of the given statements is as follows:
(i) Chennai is not the capital of Tamil Nadu.
(ii) √2 is a complex number.
(iii) All triangles are equilateral triangles.
(iv) The number 2 is not greater than 7.
(v) Every natural number is not an integer.

Question 2.
Are the following pairs of statements negations of each other?
(i) The number x is not a rational number.
The number x is not an irrational number.
(ii) The number x is a rational number.
The number x is an irrational number.
Answer.
(i) The negation of the first statement:
The number x is “a rational number”.
Which is the same as the second statement.
This is because when a number is statement not rational, it is rational.
Therefore, given statements are negation of each other.

(ii) The negation of the first statement:
The number x is an irrational number.
The second statement, which is the same as the second statement.
Therefore they are negation of each other.

Question 3.
Find the component statements of the following compound statements and check whether they are true or false.
(i) Number 3 is prime or it is odd.
(ii) All integers are positive or negative.
(iii) 100 is divisible by 3,11 and 5.
Answer.
(i) The component statements are as follows.
p : Number 3 is prime.
q : Number 3 is odd.
Both the statements are true.

(ii) The component statements are as follows.
p : All integers are positive.
q : All integers are negative.
Both the statements are false.

(iii) The component statements are as follows.
p : 100 is divisible by 3.
q : 100 is divisible by 11.
r : 100 is divisible by 5.
Here, the statements, p and q, are false and statement r is true.

Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 14 Mathematical Reasoning Ex 14.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Maths Chapter 14 Mathematical Reasoning Ex 14.1

Question 1.
Which of the following sentences are statements? Give reasons for your answer.
(i) There are 35 days in a month.
(ii) Mathematics is difficult.
(iii) The sum of 5 and 7 is greater than 10.
(iv) The square of a number is an even number.
(v) The side of a quadrilateral have equal length.
(vi) Answer this question.
(vii) The product of (-1) and 8 is 8.
(viii) The sum of all interior angles of a triangle is 180°.
(ix) Today is a windy day.
(x) All real numbers are complex numbers.
Answer.
(i) This sentence is incorrect because the maximum number of days in a month is 31. Hence, it is a statement.
(ii) This sentence is subjective in the sense that for some people, mathematics can be easy and for some others, it can be difficult. Hence, it is not a statement.
(iii) The sum of 5 and 7 is 12, which is greater than 10. Therefore, this sentence is always correct. Hence, it is a statement.
(iv) This sentence is sometimes correct and sometimes incorrect. For example, the square of 2 is an even number. However, the square of 3 is an odd number. Hence, it is not a statement.
(v) This sentence is sometimes correct and sometimes incorrect. For example, squares and rhombus have sides of equal lengths. However, trapezium and rectangles have sides of unequal lengths. Hence, it is not a statement.
(vi) It is an order. Therefore, it is not a statement.
(vii) The product of (- 1) and 8 is (- 8). Therefore, the given sentence is incorrect. Hence, it is a statement.
(viii) This sentence is correct and hence, it is a statement.
(ix) The day that is being referred to is not evident from the sentence. Hence, it is not a statement.
(x) All real numbers can be expressed as a + ib. Therefore, the given sentence is always correct. Hence, it is a statement.

Question 2.
Give three examples of sentences which are not statements. Give reasons for the answers.
Answer.
The three examples of sentences, which are not statements, are as follows:
(i) He is a doctor.
It is not evident from the sentence as to whom ‘he’ is referred to. Therefore, it is not a statement.
(ii) Geometry is difficult.
This is not a statement because for some people, geometry can be easy and for some others, it can be difficult.
(iii) Where is she going?
This is a question, which also contains ‘she’, and it is not evident as to who ‘she’ is. Hence, it is not a statement.

Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 13 Limits and Derivatives Miscellaneous Exercise Questions and Answers.

PSEB Solutions for Class 11 Maths Chapter 13 Limits and Derivatives Miscellaneous Exercise

Question 1.
Find the derivative of the following functions from first principle:
(i) – x
(ii) (- x)^{– 1}
(iii) sin (x + 1)
(iv) cos (x – \(\frac{\pi}{8}\))
Answer.
(i) Let f(x) = – x. Accordingly, f(x + h) = – (x + h)
By first prinicple,
f’(x) = \(\lim -{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\)

= \(\lim -{h \rightarrow 0} \frac{-(x+h)-(-x)}{h}\)

= \(\lim -{h \rightarrow 0} \frac{-x-h+x}{h}=\lim -{h \rightarrow 0} \frac{-h}{h}\)

= \(\lim -{h \rightarrow 0}\) (- 1) = – 1.

(ii) Let f(x) = (- x)^{– 1} = \(=\frac{1}{-x}=\frac{-1}{x}\)
Accordingly, f(x + h) = =\frac{1}{-x}=\frac{-1}{x}
By first prinicple,

(iii) Let f(x) = sin (x +1).
Accordingly, f(x + h) = sin (x + h + 1)
By first prinicple,
f’(x) = \(\lim -{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\)

= \(\lim -{h \rightarrow 0} \frac{1}{h}\) [sin (x + h + 1) – sin (x + 1)]

= \(\lim -{h \rightarrow 0} \frac{1}{h}\left[2 \cos \left(\frac{x+h+1+x+1}{2}\right) \sin \left(\frac{x+h+1-x-1}{2}\right)\right]\)

(iv) Let f(x) = cos (x – \(\frac{\pi}{8}\))
By using first principle of derivative,
We have,
f'(x) = \(\lim -{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\)

Question 2.
Find the derivative of the following functions (it is to be understood that a, b, e, d, p, q, r and s are fixed non-zero constants and m and n are integers): (x + a)
Answer.
Let f(x) = x + a.
Accordingly, f(x + h) = x + h + a
By first prinicple,
f’(x) = \(\lim -{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\)

= \(\lim -{h \rightarrow 0} \frac{x+h+a-x-a}{h}=\lim -{h \rightarrow 0}\left(\frac{h}{h}\right)\)

= \(\lim -{h \rightarrow 0}\) (1) = 1.

Question 3.
(Px + q) (\(\frac{r}{x}\) + s).
Answer.
Let f(x) = (Px + q) (\(\frac{r}{x}\) + s)
By Leibnitz product rule,
f’(x) = (px + q) (\(\frac{r}{x}\) + s) + (\(\frac{r}{x}\) + s) (px + q)’
= (px + q)(rx^{– 1} + s)’ + (\(\frac{r}{x}\) + s) (p)
= (px + q) (- rx^{– 2}) + (\(\frac{r}{x}\) + s) p
= (px + q) \(\left(\frac{-r}{x^{2}}\right)\) + (\(\frac{r}{x}\) + s) p
= \(\frac{-p r}{x}-\frac{q r}{x^{2}}+\frac{p r}{x}\) + ps
= ps – \(\frac{q r}{x^{2}}\).

Question 4.
(ax + b) (cx + d)².
Answer.
Let f(x) = (ax + b) (cx + d)²
By Leibnitz product rule,
f(x) = (ax + b) \(\frac{d}{d x}\) (cx + d)² + (cx + d)² \(\frac{d}{d x}\) (ax + b)
= (ax + b) \(\frac{d}{d x}\) (c²x² + 2cdx + d²) + (cx + d)² \(\frac{d}{d x}\) (ax + b)
= (ax + b) [\(\frac{d}{d x}\) (c²x²) + \(\frac{d}{d x}\) (2cdx) + \(\frac{d}{d x}\) d²] + (cx + d)² [\(\frac{d}{d x}\) ax + \(\frac{d}{d x}\) b]
= (ax + b) (2c²x + 2cd) + (cx + d²) a
= 2c (ax + b) (cx+ d) + a (cx + d)².

Question 5.
\(\frac{a x+b}{c x+d}\)
Answer.

Question 6.
\(\frac{1+\frac{1}{x}}{1-\frac{1}{x}}\)
Answer.

Question 7.
\(\frac{1}{a x^{2}+b x+c}\)
Answer.
Let f(x) = \(\frac{1}{a x^{2}+b x+c}\)
By quotient rule,
f'(x) = \(\frac{\left(a x^{2}+b x+c\right) \frac{d}{d x}(1)-\frac{d}{d x}\left(a x^{2}+b x+c\right)}{\left(a x^{2}+b x+c\right)^{2}}\)

= \(\frac{\left(a x^{2}+b x+c\right)(0)-(2 a x+b)}{\left(a x^{2}+b x+c\right)^{2}}\)

= \(\frac{-(2 a x+b)}{\left(a x^{2}+b x+c\right)^{2}}\)

Question 8.
\(\frac{a x+b}{p x^{2}+q x+r}\)
Answer.

Question 9.
\(\frac{p x^{2}+q x+r}{a x+b}\)
Answer.

Question 10.
\(\frac{a}{x^{4}}-\frac{b}{x^{2}}\) + cos x
Answer.

Question 11.
4√x – 2
Answer.
Let f(x) = 4√x – 2
f'(x) = \(\frac{d}{d x}\) (4√x – 2)
= \(\frac{d}{d x}\) (4√x) – \(\frac{d}{d x}\) (2)
= 4 \(\frac{d}{d x}\) (x^{\(\frac{1}{2}\)}) – 0
= 4 \(\left(\frac{1}{2} x^{\frac{1}{2}-1}\right)\)
= \(\left(2 x^{-\frac{1}{2}}\right)=\frac{2}{\sqrt{x}}\)

Question 12.
(ax + b)ⁿ
Answer.
Let f(x) = (ax + b)ⁿ
Accordingly, f(x + h) = {a(x + h) + b}ⁿ
= (ax + ah + b)ⁿ
By first principle,
f'(x) = \(\lim -{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\)

= \(\lim -{h \rightarrow 0} \frac{(a x+a h+b)^{n}-(a x+b)^{n}}{h}\)

Question 13.
(ax + b)ⁿ (cx + d)^m.
Answer.
Let f(x) = (ax + b)ⁿ (cx + d)^m
By Leibnitz product rule,

= \(\frac{m c(c x+d)^{m}}{(c x+d)}\)

= mc (cx + d)^{m – 1}
\(\frac{d}{d x}\) (cx + d)^m = mc (cx + d)^{m – 1} …………….(ii)
Similarly, (ax + b)ⁿ = na(ax + b)^{n – 1} ………….(iii)
Therefore, from eqs. (i), (ii) and (iii), we obtain
f'(x) = (ax + b)ⁿ {mc (cx + d)^{m – 1}} + (cx + d)^m {na(ax + b)^{n – 1}}
= (ax + b)^{n – 1} (a + d)^{m – 1} [mc (ax + b) + na (cx + d)].

Question 14.
sin (x + a).
Answer.
Let f(x) = sin (x + a), f(x + h) = sin (x + h + a)
By first principle,

Question 15.
cosec x cot x
Answer.
Let f(x) = cosec x cot x
By Leibnitz product rule,
f’(x) = cosec x(cot x’ + cot x (cosec x)’ ……………(i)
Let f₁(x) = cot x,
Accordingly, f₁(x + h) = cot(x + h)
By first prinicple,
f₁‘(x) = \(\lim -{h \rightarrow 0} \frac{f-{1}(x+h)-f-{1}(x)}{h}\)

= \(\lim -{h \rightarrow 0} \frac{\cot (x+h)-\cot x}{h}\)

= \(\lim -{h \rightarrow 0} \frac{1}{h}\left(\frac{\cos (x+h)}{\sin (x+h)}-\frac{\cos x}{\sin x}\right)\)

Question 16.
\(\frac{\cos x}{1+\sin x}\)
Answer.

Question 17.
\(\frac{\sin x+\cos x}{\sin x-\cos x}\)
Answer.

Question 18.
\(\frac{\sec x-1}{\sec x+1}\)
Answer.
Let f(x) = \(\frac{\sec x-1}{\sec x+1}\)

Question 19.
sinⁿ x.
Ans.
Let y = sinⁿ x
Accordingly, for n = 1, y = sin x.
∴ \(\frac{d y}{d x}\) = cos x, i.e., \(\frac{d}{d x}\) sin x = cos x
For n = 2, y = sin² x
∴ \(\frac{d y}{d x}\) = \(\frac{d}{d x}\) (sin x sin x)
= (sin x)’ sin x + sin x (sin x)’ [by Leibnitz product rule]
= cos x sin x + sin x cos x
= 2 sin x cos x ………………..(i)
For n = 3, y = sin³ x
∴ \(\frac{d y}{d x}\) = \(\frac{d}{d x}\) (sin x sin² x)
= (sin x)’ sin² x + sin x (sin² x)’ [by Leibnitz product rule]
= cos x sin² x + sin x(2 sin x cos x) [using eq. (i)]
= cos x sin² x + 2 sin² x cos x
= 3 sin² x cos x
We assert that \(\frac{d}{d x}\) (sinⁿ x) = n sin^{(n – 1)} x cos x
Let our assertion be true for n = k.
ie., \(\frac{d}{d x}\) (sin^k x) = k sin^{(k – 1)} x cos x …………….(ii)
Consider, (sin^{k + 1} x) = \(\frac{d}{d x}\) (sin x sin^k x)
= (sin x)’ sin^k x + sin x (sin^k x)’ [by Leibnitz product rule]
= cos x sin^k x + sin x (k sin^{(k – 1)} x cos x) [using eq. (ii)]
= cos x sin^k x + k sin^k x cos x
= (k + 1) sin^k x cos x
Thus, our assertion is true for n = k + 1.
Hence, by mathematical induction, \(\frac{d}{d x}\) (sinⁿ x) = n sin^{(n – 1)} x cos x.

Question 20.
\(\frac{a+b \sin x}{c+d \cos x}\)
Answer.

Question 21.
\(\frac{\sin (x+a)}{\cos x}\)
Answer.

Question 22.
x⁴ (5 sin x – 3 cos x)
Answer.
Let f(x) = x⁴ (5 sin x – 3 cos x)
By product rule,
f’(x) = x⁴ \(\frac{d}{d x}\) (5 sin x – 3 cos x) + (5 sin x – 3 cos x) \(\frac{d}{d x}\) (x⁴)
= x⁴ [5 \(\frac{d}{d x}\) (sin x) – 3 (cos x)] + (5 sin x – 3 cos x) (4x³)
= x⁴[5 cos x – 3(- sin x)] + (5 sin x – 3 cos x) (4x³)
= x³ [ 5x cos x + 3x sin x +20 sin x – 12 cos x].

Question 23.
(x² + 1) cos x.
Answer.
Let f(x) = (x² + 1) cos x
By product rule,
f’(x) = (x² + 1) \(\frac{d}{d x}\) (cos x) + cos x \(\frac{d}{d x}\) (x² + 1)
= (x² + 1) (- sin x) + cos x (2x)
= – x² sin x – sin x + 2x cos x.

Question 24.
(ax² + sin x) (p + q cos x).
Answer.
Let f(x) = (ax² + sin x) (p + q cos x)
By product rule,
f’(x) = (ax² + sin x) (p + q cos x) + (p + q cos x) (ax² + sin x)
= (ax² + sin x)(- q sin x) + (p + q cos x)(2ax + cos x)
= – q sin x (ax² + sin x) + (p + q cos x) (2ax + cos x).

Question 25.
(x + cos x) (x – tan x).
Answer.
Let y = (x + cos x) (x – tan x)
On differentiating both sides w.r.t. x, we get
ciy d d
= (x + cos x) . \(\frac{d}{d x}\) (x – tan x) + (x – tan x) (x + cos x)
[∵ \(\frac{d}{d x}\) (u . v) = u \(\frac{d v}{d x}\) + v \(\frac{d u}{d x}\)]
= (x + cos x) (1 – sec² x) + (x – tan x) (1 – sin x)
= (x + cos x) (- tan 2x) + (x – tan x) (1 – sin x)

Question 26.
\(\frac{4 x+5 \sin x}{3 x+7 \cos x}\)
Answer

Question 27.
\(\frac{x^{2} \cos \left(\frac{\pi}{4}\right)}{\sin x}\)
Answer.
Let f(x) = \(\frac{x^{2} \cos \left(\frac{\pi}{4}\right)}{\sin x}\)

By quotient rule, f'(x) = \(\cos \frac{\pi}{4} \cdot\left[\frac{\sin x \frac{d}{d x}\left(x^{2}\right)-x^{2} \frac{d}{d x}(\sin x)}{\sin ^{2} x}\right]\)

= \(\cos \frac{\pi}{4} \cdot\left[\frac{\sin x \cdot 2 x-x^{2} \cos x}{\sin ^{2} x}\right]\)

= \(\frac{x \cos \frac{\pi}{4}[2 \sin x-x \cos x]}{\sin ^{2} x}\)

Question 28.
\(\frac{x}{1+\tan x}\)
Answer.

Question 29.
(x + secx)(x-tanx).
Answer.
Let f(x) = (x + sec x) (x – tan x)
By product rule,
f’(x) = (x + sec x) \(\frac{d}{d x}\) (x – tan x) + (x – tan x) \(\frac{d}{d x}\) (x + sec x)
= (x + sec x) [\(\frac{d}{d x}\) (x) – \(\frac{d}{d x}\) tan x] + (x – tan x) [\(\frac{d}{d x}\) (x) + \(\frac{d}{d x}\) sec x]
= (x + sec x) [1 – \(\frac{d}{d x}\) tan x] + (x – tan x) [1 + \(\frac{d}{d x}\) sec x] ……………..(i)
Let f₁(x) = tan x, f₂(x) = sec x
Accordingly, f₁ (x + h) = tan (x + h) and f₂ (x + h) = sec (x + h)

From eqs. (i), (ii) and (iii), we obtain
f'(x) = (x + sec x) (1 – sec² x) + (x – tan x)(1 + sec x tan x).

Question 30.
\(\frac{x}{\sin ^{n} x}\)
Answer.

Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 9 Sequences and Series Ex 9.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.1

Question 1.
Write the first five terms of the sequence whose nth term is a_n = n (n + 2).
Answer.
a_n = n(n +2)
Substituting n = 1, 2, 3, 4 and 5, we obtain
a₁ = 1 (1 + 2) = 3,
a₂ = 2 (2 + 2) = 8,
a₃ = 3 (3 + 2) = 15,
a₄ = 4 (4 + 2) = 24,
a₅ = 5 (5 + 2) = 35
Therefore, the required terms are 3, 8, 15, 24 and 35.

Question 2.
Write the first five terms of the sequence whose nth term is a_n = \(\frac{n}{n+1}\).
Answer.
a_n = \(\frac{n}{n+1}\)
Sustituting n = 1, 2, 3, 4, 5, we otain
a_n = \(\frac{1}{1+1}=\frac{1}{2}\)

a_n = \(\frac{2}{2+1}=\frac{2}{3}\)

a_n = \(\frac{3}{3+1}=\frac{3}{4}\)

a_n = \(\frac{4}{4+1}=\frac{4}{5}\)

a_n = \(\frac{5}{5+1}=\frac{5}{6}\)

Therefore, the required terms are \(\frac{1}{2}\), \(\frac{2}{3}\) , \(\frac{3}{4}\) , \(\frac{4}{5}\) and \(\frac{5}{6}\) .

Question 3.
Write the first five terms of the sequence whose nth term is a_n = 2ⁿ.
Answer.
a_n = 2ⁿ
Substituting n = 1, 2, 3, 4, 5, we obtain
a₁ = 2¹ = 2
a₂ = 2² = 4
a₃ = 2³ = 8
a₄ = 2⁴ = 16
a₅ = 2⁵ = 32
Therefore, the required terms are 2, 4, 8, 16 and 32.

Question 4.
Write the first five terms of the sequence whose nth term is a_n = \(\frac{2 n-3}{6}\).
Answer.
Substituting n = 1,2, 3, 4, 5, we obtain

Question 5.
WrIte the first five terms of the sequence whose nth term is
a_n = (- 1)^{n – 1} 5^{n + 1}.
Ans.
Substituting n = 1,2, 3, 4, 5, we obtain
a₁ = (- 1)^{1 – 1} 5^{1 + 1} = 5² = 25,
a₂ = (- 1)^{2 – 1} 5^{2 + 1} = – 5³ = -125,
a₃ = (- 1)^{3 – 1} 5^{3 + 1} = 5⁴ = 625,
a₄ = (- 1)^{4 – 1} 5^{4 + 1} = 5⁵ = – 3125,
a₅ = (- 1)^{5 – 1} 5^{5 + 1} = 5⁶ = 15625
Therefore, the required terms are 25, – 125, 625, – 3125, and 15625.

Question 6.
Write the first five terms of the sequence whose nth term is a_n = \(n \frac{n^{2}+5}{4}\).
Answer.
Substituting n = 1, 2, 3, 4, 5, we obtain

Question 7.
Find the indicated terms in the following sequence whose nth term is a_n =4n – 3; a₁₇, a₂₄.
Answer.
We have a_n = 4n -3
On putting n =17, we get
a₁₇ = 4 × 17 – 3
= 68 – 3 = 65
On putting n = 24, we get
a₂₄ = 4 × 24 – 3
= 96 – 3 = 93.

Question 8.
Find the indicate term in the following sequence whose nth term is a_n = \(\frac{n^{2}}{2^{n}}\); a₇.
Answer.
Substituting n = 7, we obtain
a₇ = \(\frac{7^{2}}{2^{7}}=\frac{49}{128}\).

Question 9.
Find the indicated term in the following sequence whose nt1 term is a_n = (- 1)^{n – 1} n³; a₉.
Answer.
Substituting n = 9, we obtain
a₉ = (- 1)^{9 – 1} (9)³ = 729.

Question 10.
Find the indicated term in the following sequence whose nth term is a_n = \(\frac{n(n-2)}{n+3}\); a₂₀.
Answer.
Substituting n = 20, we obtain
a₂₀ = \(\frac{20(20-2)}{20+3}=\frac{20(18)}{23}=\frac{360}{23}\).

Question 11.
Write the first five terms of the following sequence and obtain the corresponding series:
a₁ = 3, a_n = 3a_{n – 1} + 2 for all n > 1.
Answer.
a₁ = 3, a_n = 3 a_{n – 1} + 2 for all n > 1
=> a₂ = 3a_{2 – 1} + 2
= 3a₁ + 2
= 3(3) + 2 = 11

a₃ = 3a_{3 – 1} + 2
= 3a₂ + 2
= 3(11) + 2 = 35

a₄ = 3a_{4 – 1} + 2
= 3a₃ + 2
= 3(35) + 2 = 107

a₅ = 3a_{5 – 1} + 2
= 3a₄ + 2
= 3(107) + 2 = 323.
Hence, the first five terms of the sequence are 3, 11, 35, 107 and 323.
The corresponding series is 3 + 11 + 35 + 107 + 323 + …

Question 12.
Write the first five terms of the following sequence and obtain the corresponding series:
a₁ = – 1, a_n = \(\frac{\boldsymbol{a}_{n-1}}{n}\), n > 2
Answer.

Question 13.
Write the first five terms of the following sequence and obtain the corresponding series:
a₁ = a₂ = 2, a_n = a_{n – 1} – 1, n > 2.
Answer.
a₁ = a₂ = 2,
a_n = a_{n – 1}, n > 2
=> a₃ = a₂ – 1 = 2 – 1 = 1,
a₄ = a₃ – 1 = 1 – 1 = 0
a₅ = a₄ – 1 = 0 – 1 = – 1.
Hence, the first five terms of the sequence are 2, 2, 1, 0 and – 1.
The corresponding series is 2 + 2 +1 + 0+ (- 1) + ……….

Question 14.
The Fibonacci sequence is defined by 1 = a₁ = a₂ and a_n = a_{n – 1} + a_{n – 2} n > 2. Find \(\frac{\boldsymbol{a}_{n+1}}{\boldsymbol{a}_{n}}\), for n = 1, 2, 3, 4, 5.
Answer.
1 = a₁ = a₂
a_n = a_{n – 1} + a_{n – 2}, n > 2
a₃ = a₂ + a₁
= 1 + 1 = 2,
a₄ = a₃ + a₂
= 2 + 1 = 3,
a₅ = a₄ + a₃
= 3 + 2 = 5,
a₆ = a₅ + a₃
= 5 + 3 = 8

For n = 1,
\(\frac{a_{n+1}}{a_{n}}=\frac{a_{2}}{a_{1}}=\frac{1}{1}\) = 1

For n = 2,
\(\frac{a_{n+1}}{a_{n}}=\frac{a_{3}}{a_{2}}=\frac{2}{1}\) = 2

For n = 3,
\(\frac{a_{n+1}}{a_{n}}=\frac{a_{4}}{a_{3}}=\frac{3}{2}\)

For n = 4,
\(\frac{a_{n+1}}{a_{n}}=\frac{a_{5}}{a_{4}}=\frac{5}{3}\)

For n = 5,
\(\frac{a_{n+1}}{a_{n}}=\frac{a_{6}}{a_{5}}=\frac{8}{5}\)

Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 8 Binomial Theorem Miscellaneous Exercise Questions and Answers.

PSEB Solutions for Class 11 Maths Chapter 8 Binomial Theorem Miscellaneous Exercise

Question 1.
Find a, b and n in the expansion of (a + b)ⁿ if the first three terms of the expansion are 729, 7290 and 30375, respectively.
Answer.
It is known that (r + 1)th term, (T_{r + 1}), in the binomial expansion of (a + b)ⁿ is given by T_{r + 1} = \({ }^{n} C_{r}\) a^{n – r r= a}
The first three terms of the expansion are given as 729, 7290, and 30375 respectively.
Therefore, we obtain
T₁ = \({ }^{2} C_{0} a^{n-0} b^{0}\)
= aⁿ = 729 ……………..(i)

T₂ = \({ }^{n} C_{1} a^{n-1} b^{1}\)
= n a^{n – 1} b = 7290 …………….(ii)

T₃ = \({ }^{n} C_{2} a^{n-2} b^{2}\)
= \(\frac{n(n-1)}{2} a^{n-2} b^{2}\) = 30375 ……………….(iii)

Dividing eq. (ii) by eq. (i), we obtain \(\frac{n a^{n-1} b}{a^{n}}=\frac{7290}{729}\)
⇒ \(\frac{n b}{a}\) = 10 ……………….(iv)

Dividing eq. (iii) by eq. (ii), we obtain

Question 2.
Find a if the coefficient of x² and x³ in the expansion of (3 + ax)⁹ are equal.
Answer.
It is known that (r + 1)th term, (T_{r + 1}), in the binomial expansion of (a + b)ⁿ is given by T_{r + 1} = \({ }^{n} C_{r} a^{n-r} b^{r}\)

Assuming that x² occurs in the (r + 1)th term in the expansion of (3 + ax)⁹, we obtain
T_{r + 1} = \({ }^{9} C_{r}(3)^{9-r}(a x)^{r}\)
= \({ }^{9} C_{r}(3){ }^{9-r} a^{r} x^{r}\)

Comparing the indices of x in x² and in T_{r + 1} we obtain r = 2
Thus, the coefficient of x² is
\({ }^{9} C_{2}(3)^{9-2} a^{2}=\frac{9 !}{2 ! 7 !}(3)^{7} a^{2}=36(3)^{7} a^{2}\)

Assuming that x³ occurs in the (k + 1)th term in the expansion of (3 + ax)⁹, we obtain
T_{k + 1} = \({ }^{9} C_{k}(3)^{9-k}(a x)^{k}={ }^{9} C_{k}(3)^{9-k} a^{k} x^{k}\)

Comparing the indices of x in x³ in T_{r + 1}, we obtain k = 3
Thus, the coefficient of x³ is
\({ }^{9} C_{3}(3)^{9-3} a^{3}=\frac{9 !}{3 ! 6 !}(3)^{6} a^{3}=84(3)^{6} a^{3}\)

It is given that the coefficients of x² and x³ are same.
84 (3)⁶ a³ = 36 (3)⁷ a²
⇒ 84a = 36 × 3
⇒ a = \(\frac{36 \times 3}{84}=\frac{108}{84}\)
⇒ a = \(\frac{9}{7}\)

Thus, the required value of a is \(\frac{9}{7}\).

Question 3.
Find the coefficient of x5 in the product (1 + 2x)⁶(1 – x)⁷ using binomial theorem.
Answer.
Using Binomial Theorem, the expressions, (1 + 2x)⁶ and (1 – x)⁷, can be expanded as

The complete multiplication of the two brackets is not required to be carried out.
Only those terms, which involve x⁵, are required.
The terms containing x⁵ are
1 (- 21x⁵) + (12x) (35x⁴) + (60x²) (- 35x³) + (160x³) (21x²) + (240x⁴) (- 7x) + (192 x⁵)(1)
= 171 x⁵
Thus, the coefficient of x⁵ in the given product.
= – 21 + 420 – 2100 + 3360 -1 680 +192 = 171.

Question 4.
If a and b are distinct integers, prove that a – b is a factor of aⁿ – bⁿ, whenever n is a positive integer.
[Hint : write aⁿ = (a – b + b)ⁿ and expand]
Answer.
In order to prove that (a – b) is a factor of (aⁿ – bⁿ), it has to be proved that
aⁿ – bⁿ = k(a – b), where k is some natural number
It can be written that, a = a – b + b
∴ aⁿ = (a – b + b)ⁿ = [(a – b) + b]ⁿ
= \({ }^{n} C_{0}(a-b)^{n}+{ }^{n} C_{1}(a-b){ }^{n-1} b+\ldots+{ }^{n} C_{n-1}(a-b) b^{n-1}+{ }^{n} C_{n} b^{n}\)

= \((a-b)^{n}+{ }^{n} C_{1}(a-b)^{n-1} b+\ldots+{ }^{n} C_{n-1}(a-b) b^{n-1}+b^{n}\)

\(a^{n}-b^{n}=(a-b)\left[(a-b)^{n-1}+{ }^{n} C_{1}(a-b)^{n-2} b+\ldots+{ }^{n} C_{n-1} b^{n-1}\right]\)

aⁿ – bⁿ = k (a -b)

where k = \(\left[(a-b)^{n-1}+{ }^{n} C_{1}(a-b)^{n-2} b+\ldots+{ }^{n} C_{n-1} b^{n-1}\right]\) is a natural number
This shows that (a – b) is a factor of (aⁿ – bⁿ), where n is a positive integer.

Question 5.
Evaluate (√3 + √2)⁶ – (√3 – √2)⁶.
Answer.
Firstly, the expression (a + b)⁶ – (a – b)⁶ is simplified by using Binomial Theorem.
This can be done as
(a + b)⁶ = \({ }^{6} C_{0}\) a⁶ + \({ }^{6} C_{1}\) a⁵ b + \({ }^{6} C_{2}\) a⁴ b² + \({ }^{6} C_{3}\) a³b³ + \({ }^{6} C_{4}\) a²b⁴ + \({ }^{6} C_{5}\) ab⁵ + \({ }^{6} C_{6}\) b⁶

= a⁶ + 6 a⁵b + 15 a⁴b² + 20 a³b³ + 15 a²b⁴ + 6 ab⁵ + b⁶ (a – b)⁶

= \({ }^{6} C_{0}\) a⁶ – \({ }^{6} C_{1}\) a⁵b + \({ }^{6} C_{2}\) a⁴b² – \({ }^{6} C_{3}\) a³b³ + \({ }^{6} C_{4}\) a²b⁴ – \({ }^{6} C_{5}\) a¹b⁵ + \({ }^{6} C_{6}\) b⁶

= a⁶ – 6 a⁵b + 15 a⁴b² – 20 a³b³ + 15 a²b⁴ – 6ab⁵ + b⁶

.-. (a + b)⁶ – (a – b)⁶ = 2[6a⁵b + 20a³b³ + 6ab⁵]

Putting a = √3 and b = √2, we obtain (√3 + √2)⁶ – (√3 – √2)⁶
= 2[6(√3)⁵ (√2) + 20 (√3)³ (√2)³ + 6 (√3) (√2)⁵]
= 2[54 √6 +120 √6 + 24 √6]
= 2 × 198 √6 = 396√6.

Question 6.
Find the value of (a² + \(\sqrt{a^{2}-1}\))⁴ + (a² – \(\sqrt{a^{2}-1}\))⁴.
Answer.
Firstly, the expression (x + y)⁴ + (x – y)⁴ is simplified by using Binomial Theorem.
This can be done as
(x + y)⁴ = \({ }^{4} C_{0}\) x⁴ + \({ }^{4} C_{1}\) x³y + \({ }^{4} C_{2}\) x²y² + \({ }^{4} C_{3}\) xy³ + \({ }^{4} C_{4}\) y⁴
= x⁴ + 4x³y + 6x²y² + 4xy³ + y⁴

(x – y)⁴ = \({ }^{4} C_{0}\) x⁴ – \({ }^{4} C_{1}\) x³y + \({ }^{4} C_{2}\) x²y² – \({ }^{4} C_{3}\) xy³ + \({ }^{4} C_{4}\) y⁴
= x⁴ – 4x³y + 6x²y² – 4xy³ + y⁴

∴ (x + y)⁴ + (x – y)⁴ = 2(x⁴ + 6x²y² + y⁴).

Putting x = a² and y = \(\sqrt{a^{2}-1}\), we obtain
(a² + \(\sqrt{a^{2}-1}\))⁴ + (a² – \(\sqrt{a^{2}-1}\))⁴
= 2 [(a²)⁴ + 6(a²)² (Ja2 -l)² + \(\left(\sqrt{a^{2}-1}\right)\)⁴]
= 2 [a⁸ + 6a⁴ (a² – 1) + (a² – 1)²]
= 2 [a⁸ + 6a⁶ – 6a⁴ + a⁴ – 2a² + 1]
= 2 [a⁸ + 6a⁶ – 5a⁴ – 2a² + 1]
= 2a⁸ + 12a⁶ – 10a⁴ – 4a² + 2.

Question 7.
Find an approximation of (0.99)⁵ using the first three terms of its expansion.
Answer.
We have, (0.99)⁵ = (1 – 0.01)⁵
= 1 – \({ }^{5} C_{1}\) × (0.01) + \({ }^{5} C_{2}\) × (0.01)² – ……….
= 1 – 0.05 + 10 × 0.0001 – ………..
= 1.001 – 0.05 = 0.951.

Question 8.
Find n, if the ratio of the fifth term from the beginning to the fifth term from the end in the expression of \(\left(\sqrt[4]{2}+\frac{1}{\sqrt[4]{3}}\right)^{n}\) is √6 : 1.
Answer.

2n – 16 = 4
⇒ 2n = 20
⇒ n = \(\frac{20}{2}\)
⇒ n = 10.

Question 9.
Expand using Binomial Theorem \(\left(1+\frac{x}{2}-\frac{2}{x}\right)^{4}\), x ≠ 0.
Answer.
Using Binomial Theorem, the given expression \(\left(1+\frac{x}{2}-\frac{2}{x}\right)^{4}\) can be expanded as \(\left[\left(1+\frac{x}{2}\right)-\frac{2}{x}\right]^{4}\),

= \(1+2 x+\frac{3}{2} x^{2}+\frac{x^{3}}{2}+\frac{x^{4}}{16}-\frac{8}{x}-12-6 x\) – \(x^{2}+\frac{8}{x^{2}}+\frac{24}{x}+6-\frac{32}{x^{3}}+\frac{16}{x^{4}}\)

= \(\frac{16}{x}+\frac{8}{x^{2}}-\frac{32}{x^{3}}+\frac{16}{x^{4}}-4 x+\frac{x^{2}}{2}+\frac{x^{3}}{2}+\frac{x^{4}}{16}-5\).

Question 10.
Find the expansion of (3x² – 2ax + 3a²)³ using binomial theorem.
Answer.
Using Binomial Theorem, the given expression (3x² – 2ax + 3a²)³ can be expanded as [(3x² – 2ax) + 3a²]³

= \({ }^{3} C_{0}\) (3x² – 2ax)3 + \({ }^{3} C_{1}\) (3x² – 2ax)² (3a²) + \({ }^{3} C_{2}\) (3x² – 2ax) (3a²)² + \({ }^{3} C_{3}\) (3a²)³

= (3x² – 2ax)³ + 3 (9x⁴ – 12ax³ + 4a²x²) (3a²) + 3 (3x² – 2ax) (9a⁴) + 27a⁶

= (3x² – 2ax)³ + 81a²x⁴ – 108a³x³ + 36a⁴x² + 81a⁴x² – 54a⁵x + 27a⁶

= (3x² -2ax)³ + 81 a²x⁴ – 108a³x³ + 117a⁴x² – 54a⁵x + 27a⁶ …………….(i)

Again by using Binomial Theorem, we obtain (3x² – 2ax)³
= \({ }^{3} C_{0}\) (3x²)³ – \({ }^{3} C_{1}\) (3x²)² (2ax) + \({ }^{3} C_{2}\) (3x²) (2ax)² – \({ }^{3} C_{3}\)(2ax)³
= 27x⁶ – 3(9x⁴) (2ax) + 3(3x²) (4a²x²) – 8a³x³
= 27x⁶ – 54ax⁵ + 36a²x⁴ – 8a³x³ ………….(ii)

From eqs. (i) and (ii), we obtain (3x² – 2ax + 3a²)³
= 27x⁶ – 54ax⁵ + 36a² x⁴ – 8a³x³ + 81a²x⁴ – 108a³x³ + 117a⁴x² – 54a⁵x + 27a⁶
= 27x⁶ – 54ax⁵ + 117a²x⁴ – 116a³x³ + 117a⁴x² – 54a⁵x + 27a⁶.

Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 8 Binomial Theorem Ex 8.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Maths Chapter 8 Binomial Theorem Ex 8.2

Question 1.
Find the coefficient of x⁵ in (x + 3)⁸
Answer.
It is known that (r + 1)th term, (T_{r + 1}), in the binomial expansion of (a + b)ⁿ is given by T_{r + 1} = \({ }^{n} C_{r}\) a^{n – r} b^r
Assuming that x⁵ occurs in the (r + 1)th term of the expansion (x + 3)⁸, we obtain
T_{r + 1} = \({ }^{8} C_{r}\) (x)^{8 – r} (3)^r
Comparing the indices of x in x⁵ and in T_{r + 1}, we obtain r = 3.
Thus, the coefficient of x⁵ is \({ }^{8} C_{3}\) (3)³
= \(\frac{8 !}{3 ! 5 !} \times 3^{3}\)

= \(\frac{8 \cdot 7 \cdot 6 \cdot 5 !}{3 \cdot 2.5 !} \cdot 3^{3}\) = 1512.

Question 2.
Find the coefficient of a⁵ b⁷ in (a – 2b)¹².
Answer.
It is known that (r + 1)th term, (T_{r + 1}), in the binomial expansion of (a + b)ⁿ is given by T_{r + 1} = \(\) a^{n – r} b^r
Assuming that a⁵b⁷ occurs in the (r + 1)th term of the expansion (a – 2b)¹²,we obtain
T_{r + 1} = \({ }^{12} C_{r}\) (a)^{12 – r} (- 2b)^r
= \({ }^{12} C_{r}\) (- 2)^r (a)^{12 – r} (b)^r
Comparing the indices of a and b in a⁵b⁷ and in T_{r + 1}, we obtain r = 7
Thus, the coefficient of a⁵b⁷ is
12C7 (- 2)⁷ = \(\frac{12 !}{7 ! 5 !} \cdot 2^{7}\)
= \(-\frac{12.11 .10 .9 .8 .7 !}{5.4 .3 .2 .7 !} \cdot 2^{7}\)
= – (792) (128)
= – 101376

Question 3.
Write the general term ¡n the expansion of (x² – y)⁶.
Answer.
General term in the expansion of (a + b)ⁿ is given by
T_{r + 1} = \({ }^{n} C_{r}\) a^{n – r} b^r
T_{r + 1} = \({ }^{6} C_{r}\) (x²)^{6 – r} (y)^r
= \({ }^{6} C_{r}\) x^{12 – r} (- y)^r

Question 4.
Write the general term in the expansion of (x² – 2y)¹², x ≠ 0.
Answer.
It is known that the general term T_{r + 1} {which is the (r + 1)th term} in the binomial expansion of (a + b)ⁿ is given by T_{r + 1} = \({ }^{n} C_{r}\) a^{n – r} b^r
Thus, the general term in the expansion of (x² – yx)¹² is
T_{r + 1} = \({ }^{12} C_{r}\) (x²)^{12 – r} (- yx)^r
= (- 1)^r \({ }^{12} C_{r}\) x24-2r . y^r . x^r
= (- 1)^r \({ }^{12} C_{r}\) x^{24 – r} . y^r

Question 5.
Find the 4th term in the expansion of (x – 2y)¹².
Answer.
It is known that (r + 1)th term, (T_{r + 1}), in the binomial expansion of (a + b)ⁿ is given by T_{r + 1} = \({ }^{n} C_{r}\) a^{n – r} b^r
Thus, the 4th term in the expansion of (x – 2y)¹² is
T₄ = T_{3 + 1}
= \({ }^{12} C_{3}[/latex (x)^{12 – 3} (- 2y)³
= (- 1)³ . [latex]\frac{12 !}{3 ! 9 !}\) . x⁹ . (2)³ . y³
= – \(\frac{12 \cdot 11 \cdot 10}{3.2}\) . x⁹ . (2)³ . y³
= – 1760 x⁹ . x⁹ . y³

Question 6.
Find the 13th term in the expansion of (9x – \(\frac{1}{3 \sqrt{x}}\))¹⁸, x ≠ 0.
Answer.

Question 7.
Find the middle terms in the expansions of \(\left(3-\frac{x^{3}}{6}\right)^{7}\).
Answer.
It is known that in the expansion of (a + b)ⁿ, if n is odd, then there are two middle terms, namely \(\left(\frac{n+1}{2}\right)^{t h}\) term and \(\left(\frac{n+1}{2}+1\right)^{t h}\) term.

Therefore, the middle terms in the expansion of \(\left(3-\frac{x^{3}}{6}\right)^{7}\) are \(\left(\frac{7+1}{2}\right)^{t h}\)

Thus, the middle terms in the expansion of \(\left(3-\frac{x^{3}}{6}\right)^{7}\) are – \(\frac{105}{8}\) x⁹ and \(\frac{35}{48}\) x¹².

Question 8.
Find the middle terms in the expansions of \(\left(\frac{x}{3}+9 y\right)^{10}\).
Answer.
It is known that in the expansion (a + b)ⁿ, if n is even, then the middle term is \(\left(\frac{n}{2}+1\right)^{t h}\) term.

Therefore, the middle term in the expansion of \(\left(\frac{x}{3}+9 y\right)^{10}\) is 61236 x⁵ y⁵.

Question 9.
In the expansion of (1 + a)^{m + n}, prove that coefficients of a^m and aⁿ are equal.
Answer.
It is known that (r + 1)th term, (T_{r + 1}), in the binomial expansion of (a + b)ⁿ is given by
T_{r + 1} = \(\) a^{n – r} b^r.

Question 10.
The coefficients of the (r – 1)th, rth and (r + 1)th terms in the expansion of (x + 1)ⁿ are in the ratio 1 : 3 : 5. Find n and r.
Answer.

Question 11.
Prove that the coefficient of xⁿ in the expansion of (1 + x)²ⁿ is twice the coefficient of xⁿ in the expansion of (1 + x)^{2n – 1}.
Answer.

From eqs. (i) and (ii), it is observed that \(\frac{1}{2}{ }^{2 n} C_{n}={ }^{2 n-1} C_{n}\).

\({ }^{2 n} C_{n}=2\left({ }^{2 n-1} C_{n}\right)\)

Therefore, the coefficient of xⁿ in the expansion of (1 + x)²ⁿ is twice the coefficient of xⁿ in the expansion of (1 + x)^{2n – 1}.
Hence proved.

Question 12.
Find a positive value of m for which the coefficient of xⁿ in the expansion (1 + x)^m is 6.
Answer.
It is known that (r + 1)th term, (T_{r + 1}), in the binomial expansion of (a + b)ⁿ is given by T_{r + 1} = \(\) a^{n – r}b^r.

Assuming that x² occurs in the (r + 1)th term of the expansion (1 + x)^m, we obtain
\(T_{r+1}={ }^{m} C_{r}(1)^{m-r}(x)^{r}={ }^{m} C_{r}(x)^{r}\)

Comparing the indices of x in x² and in T_{r + 1}, we obtain r = 2
Therefore, the coefficient of x₂ is \({ }^{m} C_{2}\)

It is given that the coefficient of x² in the expansion (1 + x)^m is 6.

\({ }^{m} C_{2}\) = 6

⇒ \(\frac{m !}{2 !(m-2) !}\) = 6

⇒ \(\frac{m(m-1)(m-2) !}{2 \times(m-2) !}\) = 6
⇒ m (m – 1) = 12
⇒ m² – m – 12 = 0
⇒ m² – 4m + 3m – 12 = 0
⇒ m(m – 4) + 3(m – 4) = 0
⇒ (m – 4) (m + 3) = 0
⇒ (m – 4) = 0 or (m + 3) = 0
⇒ m = 4 or m = – 3.
Thus, the positive value of m, for which the coefficient of x² in the expansion (1 + x)^m is 6, is 4.

Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 8 Binomial Theorem Ex 8.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Maths Chapter 8 Binomial Theorem Ex 8.1

Question 1.
Expand the expression (1 – 2x)⁵.
Answer.
Here, (1 – 2x)⁵ = [1 + (- 2x)]⁵
= \({ }^{5} C_{0}\) + \({ }^{5} C_{1}\) (-2X) + \({ }^{5} C_{2}\) (-2x)2 + \({ }^{5} C_{3}\) (-2x)3 + \({ }^{5} C_{4}\) (-2x)4 + \({ }^{5} C_{5}\) (-2x)s
= 1 + 5 (- 2x) + 10 (4x²) + 10 (- 8x³) + 5(16x⁴) + 1 (- 32x⁵)
= 1 – 10x + 40x² – 80x³ + 80x⁴ – 32x⁵
which is the required expansion.

Question 2.
Expand the expression \(\left(\frac{2}{x}-\frac{x}{2}\right)^{5}\).
Answer.
By using Binomial Theorem, the expression \(\left(\frac{2}{x}-\frac{x}{2}\right)^{5}\) can be expanded as

Question 3.
Expand the expression (2x – 3)⁶.
Answer.
By using Binomial Theorem, the expression (2x – 3)⁶ can be expanded as
(2x – 3) = \({ }^{6} C_{0}\) (2x)⁶ – \({ }^{6} C_{1}\) (2x)⁵ (3) + \({ }^{6} C_{2}\) (2x)⁴ (3)² – \({ }^{6} C_{3}\) (2x)³ (3)³ + \({ }^{6} C_{4}\) (2x)² (3)⁴ – \({ }^{6} C_{5}\) (2x) (3)⁵ + \({ }^{6} C_{6}\) (3)⁶
= 64x⁶ – 6(32x⁵) (3) + (15) (16x⁴) (9) – 20(8x³) (27) + 15(4x²) (81) – 6 (2x) (243) + 729
= 64x⁶ – 576x⁵ + 2160x⁴ – 4320x³ + 4860x² – 2916x + 729

Question 4.
Expand the expression \(\left(\frac{x}{3}+\frac{1}{x}\right)^{5}\).
Answer.
Using Binomial Theorem,

\((a+b)^{n}={ }^{n} C_{0} a^{n}+{ }^{n} C_{1} a^{n-1} b+{ }^{n} C_{2} a^{n-2} b^{2}+\ldots+{ }^{n} C_{n} b^{n}\)

Question 5.
Expand: \(\left(x+\frac{1}{x}\right)^{6}\).
Answer.

Question 6.
Using Binomial Theorem, evaluate (96)³.
Answer.
96 can be expressed as the sum or difference of two numbers whose powers are easier to calculate and then, Binomial Theorem can be applied.
It can be written that, 96 = 100 – 4
∴ (96)³ = (100 – 4)³
= \({ }^{3} C_{0}\) (100)³ – \({ }^{3} C_{1}\) (100)² (4) + \({ }^{3} C_{2}\) (100) (4)² – 3C\({ }^{3} C_{0}\) (4)³
= (100)³ – 3(100)² (4) + 3(100) (4)² – (4)³
= 1000000 – 120000 + 4800 – 64 = 884736.

Question 7.
Using Binomial Theorem, evaluate (102)⁵.
Answer.
102 can be expressed as the sum or difference of two numbers whose powers are easier to calculate and then, Binomial Theorem can be applied.
It can be written that, 102 = 100 + 2
∴ (102)⁵ = (100 + 2)⁵

= \({ }^{5} C_{0}\) (100)⁵ + \({ }^{5} C_{1}\) (100)⁴ (2) + \({ }^{5} C_{2}\) (100)³ (2)² + \({ }^{5} C_{3}\) (100)² (2)³ + \({ }^{5} C_{4}\) (100) (2)⁴ + \({ }^{5} C_{5}\) (2)⁵

= (100)⁵ + 5(100)⁴ (2) + 10 (100)³ (2)² + 10 (100)² (2)³ + 5(100) (2)⁴ + (2)⁵
= 10000000000 + 1000000000 + 40000000 + 800000 + 8000 + 32 = 11040808032

Question 8.
Using Binomial Theorem, evaluate (101)⁴.
Answer.
101 can be expressed as the sum or difference of two numbers whose powers are easier to calculate and then,
Binomial Theorem can be applied.
If can be written that, 101 = 100 + 1
∴ (101)⁴ =(100 + 1)⁴
= \({ }^{4} C_{0}\) (100)⁴ + \({ }^{4} C_{1}\)(100)³ (1) + \({ }^{4} C_{2}\) (100)² (1)² + \({ }^{4} C_{3}\) (100)(1)³ + \({ }^{4} C_{4}\) (1)⁴.

= (100)⁴ + 4(100)³ + 6(100)² + 4(100) + (1)⁴
= 100000000 + 4000000 + 60000 + 400 +1 =104060401.

Question 9.
Using Binomial Theorem, evaluate (99)⁵.
Answer.
99 can be written as the sum or difference of two numbers whose powers are easier to calculate and then, Binomial Theorem can be applied.
It can be written that, 99 = 100 – 1
(99)⁵ =(100 – 1)⁵
= \({ }^{5} C_{0}\) (1oo)⁵ – \({ }^{5} C_{1}\) (100)⁴ (1) + \({ }^{5} C_{2}\) (100)³ (1)² – \({ }^{5} C_{3}\) (100)² (1)³ + \({ }^{5} C_{4}\) (100) (1)⁴ – \({ }^{5} C_{5}\) (1)⁵

= (100)⁵ – 5(100)⁴ + 10(100)³ – 10 (100)² + 5 (100) – 1
= 10000000000 – 500000000 + 10000000 – 100000 + 500 – 1
= 10010000500 – 500100001 = 9509900499.

Question 10.
Using Binomial Theorem, indicate which number is larger (1.1)10000 or 1000.
Answer.
By splitting 1.1 and then applying Binomial Theorem, the first few terms of (1.1)¹⁰⁰⁰⁰ can be obtained as (1.1)¹⁰⁰⁰⁰ = (1 + 0.1)¹⁰⁰⁰⁰
= \({ }^{10000} C_{0}\)+ \({ }^{10000} C_{1}\) (1.1) + Other positive terms
= 1 + 10000 × 1.1 + Other positive terms
= 1 + 11000 + Other positive terms > 1000
Hence, (1.1)¹⁰⁰⁰⁰ > 1000.

Question 11.
Find (a + b)⁴ – (a – b)⁴.
Hence, evaluate (√3 + √2)⁴ – (√3 – √2)⁴.
Answer.
Using Binomial Theorem, the expression, (a + b)⁴ and (a – b)⁴, can be expanded as
(a + b)⁴ = \({ }^{4} C_{0}\) a⁴ + \({ }^{4} C_{1}\)a³b + \({ }^{4} C_{2}\) a²b² + \({ }^{4} C_{3}\) ab³ + \({ }^{4} C_{4}\) b⁴

(a – b)⁴ = \({ }^{4} C_{0}\) a⁴ – \({ }^{4} C_{1}\)a³b + \({ }^{4} C_{2}\) a²b² – \({ }^{4} C_{3}\) ab³ + \({ }^{4} C_{4}\) b⁴

∴ (a + b)⁴ – (a + b)⁴ =
= \({ }^{4} C_{0}\) a⁴ + \({ }^{4} C_{1}\) a³b + \({ }^{4} C_{2}\) a²b² + \({ }^{4} C_{3}\) ab³ + \({ }^{4} C_{4}\) b⁴ – [ \({ }^{4} C_{0}\) a⁴ – \({ }^{4} C_{1}\) a³b + \({ }^{4} C_{2}\) a²b² – \({ }^{4} C_{3}\) ab³ + \({ }^{4} C_{4}\) b⁴]

= 2 (\({ }^{4} C_{1}\) a³b + \({ }^{4} C_{1}\) ab³)
= 2(4a³b + 4ab³)
= 8ab(a² + b²)
By putting a = √3 and b = √2 , we obtain ,
(√3 + √2)⁴ – (√3 – √2)⁴ = 8(√3) (√2) {(√3)² + ((√2)²)
= 8(√6) {3 + 2} = 40√6.

Question 12.
Find (x + 1)⁶ + (x – 1)⁶. Hence or otherwise evaluate (√2 + 1)⁶ + (√2 – 1)⁶.
Answer.
We have,

(x + 1)⁶ = x⁶ – 6x⁵ + 15x⁴ + 20x³ + 15x² + 6x + 1 ……………..(i)
Similarly, (x – 1)⁶ = x⁶ – 6x⁵ + 15x⁴ – 20x³ + 15x² – 6x + 1 ………….(ii)
Now, adding eqs. (i) and (ii), we get
(x + 1)⁶ + (x – 1)⁶ = 2 [x⁶ + 15x⁴ + 15x² + 1]
Now, putting x = V2, we get
(√2 + 1)⁶ + (√2 – 1)⁶ = 2[(V2)⁶ +15(V2)⁴ + 15(V2)² + 1]
= 2 [2³ + 15 × 2² + 15 × 2 + 1]
= 2(8 + 15 × 4 + 30 + 1]
= 2 [8 + 60 + 30 + 1]
= 2 [99] = 198.

Question 13.
Show that 9^{N + 1} – 8n – 9 is divisible by 64, whenever n is a positive integer.
Answer.
In order to show that 9^{n + 1} – 8n – 9 is divisible by 64, it has to be proved that,
9^{n + 1} – 8n – 9 = 64k, where k is some natural number.
By Binomial Theorem,

⇒ 9^{n + 1} – 8n – 9 = 64k, where k
= \({ }^{n+1} C_{2}+{ }^{n+1} C_{3} \times 8+\ldots+{ }^{n+1} C_{n+1}(8)^{n-1}\) is a natural number.
Thus, 9^{n + 1} – 8n – 9,is divisible by 64, wherever n is a positive integer.

Question 14.
Prove that \(\sum_{r=0}^{n} \mathbf{3}^{r}{ }^{n} C_{r}\) = 4ⁿ.
Answer.
By Binomial Theorem, \(\sum_{r=0}^{n}{ }^{n} C_{r} a^{n-r} b^{r}\) = (a + b)ⁿ
By putting b = 3^and a = 1 in the above equation, we obtain
\(\sum_{r=0}^{n}{ }^{n} C_{r}(1)^{n-r}(3)^{r}\) = (1 + 3)ⁿ
⇒ \(\sum_{r=0}^{n} 3^{r}{ }^{n} C_{r}\) = 4ⁿ
Hence, proved.

Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 7 Permutations and Combinations Miscellaneous Exercise Questions and Answers.

PSEB Solutions for Class 11 Maths Chapter 7 Permutations and Combinations Miscellaneous Exercise

Question 1.
How many words, with or without meaning, each of 2 vowels and 3 consonants can be formed from the letters of the word DAUGHTER?
Answer.
There are 8 letters in the word ‘DAUGHTER’ including 3 vowels and 5 consonants.
We have to select 2 vowels out of 3 vowels and 3 consonants out of 5 consonants.
∴ Number of ways of selection = \({ }^{3} C_{2} \times{ }^{5} C_{3}\) = 3 × 10 = 30
Now, each word contains 5 letters which can be arranged among themselves in 5! ways.
So, total number of words = 5 ! x 30 = 120 × 30 = 3600

Question 2.
How many words, with or without meaning, can be formed using all the letters of the word EQUATION at a time so that the vowels and consonants occur together?
Answer.
The word EQUATION consists of 5 vowels and 3 consonants.
∴ 5 vowels can be arranged in 5! = 120 ways.
3 consonants can be arranged in 3! = 6 ways
The two blockof vowels and consonants can be arranged in 2! = 2 ways.
∴ The no. of world which can be formed with 1 letters of the word EQUATION so that vowels and consonants occur together = 120 × 6 × 2 = 1440.

Question 3.
A committee of 7 has to be formed from 9 boys and 4 girls. In how many ways can this be done when the committee consists of:
(i) exactly 3 girls?
(ii) atleast 3 girls?
(iii) atmost 3 girls?
Answer.
A committee of 7 has to be formed from 9 boys and 4 girls.
(i) Since exactly 3 girls are to be there in every committee, each committee must consist of (7 – 3) = 4 boys only.
Thus, in this case, required number of ways = \({ }^{4} C_{3} \times{ }^{9} C_{4}\)
= \(\frac{4 !}{3 ! 1 !} \times \frac{9 !}{4 ! 5 !}\)
= \(4 \times \frac{9 \times 8 \times 7 \times 6 \times 5 !}{4 \times 3 \times 2 \times 1 \times 5 !}\) = 504.

(ii) Since atleast 3 girls are to be there in every committee, the committee can consist of
(a) 3 girls and 4 boys or
(b) 4 girls and 3 boys
3 girls and 4 boys can be selected in \({ }^{4} C_{3} \times{ }^{9} C_{4}\) ways.
4 girls and 3 boys can be selected in \({ }^{4} \mathrm{C}_{4} \times{ }^{9} \mathrm{C}_{3}\) ways.
Therefore, in this case, required number of ways
= \({ }^{4} C_{3} \times{ }^{9} C_{4}\) + \({ }^{4} \mathrm{C}_{4} \times{ }^{9} \mathrm{C}_{3}\)
= 504 + 84 = 588.

(iii) Since atmost 3 girls are to be there in every committee, the committee can consist of
(a) 3 girls and 4 boys
(b) 2 girls and 5 boys
(c) 1 girl and 6 boys
(d) No girl and 7 boys
3 girls and 4 boys can be selected in \({ }^{4} C_{3} \times{ }^{9} C_{4}\) ways.
2 girls and 5 boys can be selected in \({ }^{4} C_{2} \times{ }^{9} C_{5}\) ways.
1 girl and 6 boys can be selected in \({ }^{4} C_{1} \times{ }^{9} C_{6}\) ways.
No girl and 7 boys can be selected in \({ }^{4} C_{0} \times{ }^{9} C_{7}\) ways.
Therefore, in this case, required number, of ways :
= \({ }^{4} \mathrm{C}_{3} \times{ }^{9} \mathrm{C}_{4}+{ }^{4} \mathrm{C}_{2} \times{ }^{9} \mathrm{C}_{5}+{ }^{4} \mathrm{C}_{1} \times{ }^{9} \mathrm{C}_{6}+{ }^{4} \mathrm{C}_{0} \times{ }^{9} \mathrm{C}_{7}\)

= \(\frac{4 !}{3 ! 1 !} \times \frac{9 !}{4 ! 5 !}+\frac{4 !}{2 ! 2 !} \times \frac{9 !}{5 ! 4 !}+\frac{4 !}{1 ! 3 !} \times \frac{9 !}{6 ! 3 !}+\frac{4 !}{0 ! 4 !} \times \frac{9 !}{7 ! 2 !}\)

= 504 + 756 + 336 + 36 = 1632.

Question 4.
If the different permutations of all the letter of the word EXAMINATION are listed as in a dictionary, how many words are there in this list before the first word starting with E?
Answer.
Words starting with A are formed with the letters 2I’s, 2N’s, A, E, X, M, T, O.
Number of words formed by these letters = \(\frac{10 !}{2 ! 2 !}\)

= \(\frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{2 \times 2}\) = 907200
Then the words starting with E, I, M, N, O, T, X will be formed.
∴ Number of words before the first word starting with E is formed = 907200.

Question 5.
How many 6-digit numbers can be formed from the digits 0, 1, 3, 5, 7 and 9 which are divisible by 10 and no digit is repeated?
Answer.
A number is divisible by 10 if its units digits is 0.
Therefore, 0 is fixed at the units place.
Therefore, there will be as many ways as there are ways of filling 5 vacant places in succession by the remaining 5 digits (i.e., 1, 3, 5, 7 and 9).
The 5 vacant places can be filled in 5! ways.
Hence, required number of 6-digit numbers = 5! = 120.

Question 6.
The English alphabet has 5 vowels and 21 consonants. How many words with two different vowels and 2 different consonants can be formed from the alphabet?
Answer.
2 different vowels and 2 different consonants are to be selected from the English alphabet.
Since there are 5 vowels in the English alphabet, number of ways of selecting 2 different vowels from the alphabet
= \({ }^{5} C_{2}=\frac{5 !}{2 ! 3 !}\) = 10
Since there are 21 consonants in the English alphabet, number of ways of selecting 2 different consonants from the alphabet
= \({ }^{21} \mathrm{C}_{2}=\frac{21 !}{2 ! 19 !}\)
= 210
Therefore, number of combinations of 2 different vowels and 2 different consonants = 10 × 210 = 2100.
Each of these 2100 combinations has 4 letters, which can be arranged among themselves in 4! ways.
Therefore, required number of words = 2100 × 4! = 50400.

Question 7.
In an examination, a question paper consists of 12 questions divided into two parts i.e., Part I and Part H, containing 5 and 7 questions, respectively. A student is required to attempt 8 questions in all, selecting at least 3 from each part. In how many ways can a student select the questions?
Answer.
Student may select 8 questions according to following scheme

= 10 × 7 + 5 × 35 + 5 × 35
= 70 + 175 +175 = 420 ways.

Question 8.
Determine the number of 5-card combinations out of a deck of 52 cards if each selection of 5 cards has exactly one king.
Answer.
From a deck of 52 cards, 5-card combinations have to be made in such a way that in each selection of 5 cards, there is exactly one king.
In a deck of 52 cards, there are 4 kings.
1 king can be selected out of 4 kings in \({ }^{4} \mathrm{C}_{1}\) ways.
4 cards out of the remaining 48 cards can be selected in \({ }^{48} \mathrm{C}_{4}\) ways.
Thus, the required number of 5-card combinations is \({ }^{4} \mathrm{C}_{1} \times{ }^{48} \mathrm{C}_{4}\).

Question 9.
It is required to seat 5 men and 4 women in a row so that the women occupy the even places. How many such arrangements are possible?
Answer.
5 men and 4 women are to be seated in a row such that the women occupy the even places.
The 5 men can be seated in 5! ways.
For each arrangement, the 4 women can be seated only at the cross marked places (so that women occupy the even places).
M × M × M × M × M
Therefore, the women can be seated in 4! ways.
Thus, possible number of arrangements = 4! × 5!
= 24 × 120 = 2880.

Question 10.
From a class of 25 students, 10 are to be chosen for an excursion party. There are 3 students who decide that either all of them will join or none of them will join. In how many ways can the excursion party be chosen?
Answer.
There are two cases :
(a) If the 3 students join the excursion party then the number of combinations will be P₁ = C (22, 7).
(b) If the 3 students do not join the excursion party. Then the number of combinations P₂ = C (22, 10).
If P is the combination of choosing the excursion party, then
P = P₁ + P₂
= C(22, 7) + C(22,10)
= \(\frac{22 !}{7 ! 15 !}+\frac{22 !}{10 ! 12 !}\)

= \(\frac{22 \times 21 \times 20 \times 19 \times 18 \times 17 \times 16 \times 15 !}{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \times 15 !}\) + \(\frac{22 \times 21 \times 20 \times 19 \times 18 \times 17 \times 16 \times 15 \times 14 \times 13 \times 12 !}{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \times 12 !}\)
= 170544 + 646646 = 817190.

Question 11.
In how many ways can the letters of the word ASSASSINATION be arranged so that all the S’s are together?
Answer.
In the given word ASSASSINATION, the letter A appears 3 times, S appears 4 times, I appears 2 times, N appears 2 times, and all the other letters appear only once.
Since all the words have to be arranged in such a way that all the Ss are together, SSSS is treated as a single object for the time being.
This single object together with the remaining 9 objects will account for 10 objects.
These 10 objects in which there are 3 As, 2 Is, and 2 Ns can be arranged in \(\frac{10 !}{3 ! 2 ! 2 !}\) ways.
Thus, required number of ways of arranging the letters of the given word = \(\frac{10 !}{3 ! 2 ! 2 !}\)
= \(\frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 !}{3 ! \times 2 \times 2}\)
= 151200.