# 2.2.1: Exercises 2.2 - Mathematics

In Exercises (PageIndex{1}) - (PageIndex{12}), row and column vectors (vec{u}) and (vec{v}) are defined. Find the product (vec{u}vec{v}), where possible.

Exercise (PageIndex{1})

(-22)

Exercise (PageIndex{2})

(2)

Exercise (PageIndex{3})

(0)

Exercise (PageIndex{4})

(1)

Exercise (PageIndex{5})

(5)

Exercise (PageIndex{6})

(-21)

Exercise (PageIndex{7})

(15)

Exercise (PageIndex{8})

(-8)

Exercise (PageIndex{9})

(-2)

Exercise (PageIndex{10})

(23)

Exercise (PageIndex{11})

Not possible.

Exercise (PageIndex{12})

Not possible.

In Exercises (PageIndex{13}) - (PageIndex{27}), matrices (A) and (B) are defined.

1. Give the dimensions of (A) and (B). If the dimensions properly match, give the dimensions of (AB) and (BA).
2. Find the products (AB) and (BA), if possible.

Exercise (PageIndex{13})

(A=left[egin{array}{cc}{1}&{2}{-1}&{4}end{array} ight]) (B=left[egin{array}{cc}{2}&{5}{3}&{-1}end{array} ight])

(AB=left[egin{array}{cc}{8}&{3}{10}&{-9}end{array} ight])

(BA=left[egin{array}{cc}{-3}&{24}{4}&{2}end{array} ight])

Exercise (PageIndex{14})

(A=left[egin{array}{cc}{3}&{7}{2}&{5}end{array} ight]) (B=left[egin{array}{cc}{1}&{-1}{3}&{-3}end{array} ight])

(AB=left[egin{array}{cc}{24}&{-24}{17}&{-17}end{array} ight])

(BA=left[egin{array}{cc}{1}&{2}{3}&{6}end{array} ight])

Exercise (PageIndex{15})

(A=left[egin{array}{cc}{3}&{-1}{2}&{2}end{array} ight]) (B=left[egin{array}{ccc}{1}&{0}&{7}{4}&{2}&{9}end{array} ight])

(AB=left[egin{array}{ccc}{-1}&{-2}&{12}{10}&{4}&{32}end{array} ight])

(BA) is not possible.

Exercise (PageIndex{16})

(A=left[egin{array}{cc}{0}&{1}{1}&{-1}{-2}&{-4}end{array} ight]) (B=left[egin{array}{cc}{-2}&{0}{3}&{8}end{array} ight])

(AB=left[egin{array}{cc}{3}&{8}{-5}&{-8}{-8}&{-32}end{array} ight])

(BA) is not possible.

Exercise (PageIndex{17})

(A=left[egin{array}{ccc}{9}&{4}&{3}{9}&{-5}&{9}end{array} ight]) (B=left[egin{array}{cc}{-2}&{5}{-2}&{-1}end{array} ight])

(AB) is not possible.

(BA=left[egin{array}{ccc}{27}&{-33}&{39}{-27}&{-3}&{-15}end{array} ight])

Exercise (PageIndex{18})

(A=left[egin{array}{cc}{-2}&{-1}{9}&{-5}{3}&{-1}end{array} ight]) (B=left[egin{array}{ccc}{-5}&{6}&{-4}{0}&{6}&{-3}end{array} ight])

(AB=left[egin{array}{ccc}{10}&{-18}&{11}{-45}&{24}&{-21}{-15}&{12}&{-9}end{array} ight])

(BA=left[egin{array}{cc}{52}&{-21}{45}&{-27}end{array} ight])

Exercise (PageIndex{19})

(A=left[egin{array}{cc}{2}&{6}{6}&{2}{5}&{-1}end{array} ight]) (B=left[egin{array}{ccc}{-4}&{5}&{0}{-4}&{4}&{-4}end{array} ight])

(AB=left[egin{array}{ccc}{-32}&{34}&{-24}{-32}&{38}&{-8}{-16}&{21}&{4}end{array} ight])

(BA=left[egin{array}{cc}{22}&{-14}{-4}&{-12}end{array} ight])

Exercise (PageIndex{20})

(A=left[egin{array}{cc}{-5}&{2}{-5}&{-2}{-5}&{-4}end{array} ight]) (B=left[egin{array}{ccc}{0}&{-5}&{6}{-5}&{-3}&{-1}end{array} ight])

(AB=left[egin{array}{ccc}{-10}&{19}&{-32}{10}&{31}&{-28}{20}&{37}&{-26}end{array} ight])

(BA=left[egin{array}{cc}{-5}&{-14}{45}&{0}end{array} ight])

Exercise (PageIndex{21})

(A=left[egin{array}{cc}{8}&{-2}{4}&{5}{2}&{-5}end{array} ight]) (B=left[egin{array}{ccc}{-5}&{1}&{-5}{8}&{3}&{-2}end{array} ight])

(AB=left[egin{array}{ccc}{-56}&{2}&{-36}{20}&{19}&{-30}{-50}&{-13}&{0}end{array} ight])

(BA=left[egin{array}{cc}{-46}&{40}{72}&{9}end{array} ight])

Exercise (PageIndex{22})

(A=left[egin{array}{cc}{1}&{4}{7}&{6}end{array} ight]) (B=left[egin{array}{cccc}{1}&{-1}&{-5}&{5}{-2}&{1}&{3}&{-5}end{array} ight])

(AB=left[egin{array}{cccc}{-7}&{3}&{7}&{-15}{-5}&{-1}&{-17}&{5}end{array} ight])

(BA) is not possible.

Exercise (PageIndex{23})

(A=left[egin{array}{cc}{-1}&{5}{6}&{7}end{array} ight]) (B=left[egin{array}{cccc}{5}&{-3}&{-4}&{-4}{-2}&{-5}&{-5}&{-1}end{array} ight])

(AB=left[egin{array}{cccc}{-15}&{-22}&{-21}&{-1}{16}&{-53}&{-59}&{-31}end{array} ight])

(BA) is not possible.

Exercise (PageIndex{24})

(A=left[egin{array}{ccc}{-1}&{2}&{1}{-1}&{2}&{-1}{0}&{0}&{-2}end{array} ight]) (B=left[egin{array}{ccc}{0}&{0}&{-2}{1}&{2}&{-1}{1}&{0}&{0}end{array} ight])

(AB=left[egin{array}{ccc}{3}&{4}&{0}{1}&{4}&{0}{-2}&{0}&{0}end{array} ight])

(BA=left[egin{array}{ccc}{0}&{0}&{4}{-3}&{6}&{1}{-1}&{2}&{1}end{array} ight])

Exercise (PageIndex{25})

(A=left[egin{array}{ccc}{-1}&{1}&{1}{-1}&{-1}&{-2}{1}&{1}&{-2}end{array} ight]) (B=left[egin{array}{ccc}{-2}&{-2}&{-2}{0}&{-2}&{0}{-2}&{0}&{2}end{array} ight])

(AB=left[egin{array}{ccc}{0}&{0}&{4}{6}&{4}&{-2}{2}&{-4}&{-6}end{array} ight])

(BA=left[egin{array}{ccc}{2}&{-2}&{6}{2}&{2}&{4}{4}&{0}&{-6}end{array} ight])

Exercise (PageIndex{26})

(A=left[egin{array}{ccc}{-4}&{3}&{3}{-5}&{-1}&{-5}{-5}&{0}&{-1}end{array} ight]) (B=left[egin{array}{ccc}{0}&{5}&{0}{-5}&{-4}&{3}{5}&{-4}&{3}end{array} ight])

(AB=left[egin{array}{ccc}{0}&{-44}&{18}{-20}&{-1}&{-18}{-5}&{-21}&{-3}end{array} ight])

(BA=left[egin{array}{ccc}{-25}&{-5}&{-25}{25}&{-11}&{2}{-15}&{19}&{32}end{array} ight])

Exercise (PageIndex{27})

(A=left[egin{array}{ccc}{-4}&{-1}&{3}{2}&{-3}&{5}{1}&{5}&{3}end{array} ight]) (B=left[egin{array}{ccc}{-2}&{4}&{3}{-1}&{1}&{-1}{4}&{0}&{2}end{array} ight])

(AB=left[egin{array}{ccc}{21}&{-17}&{-5}{19}&{5}&{19}{5}&{9}&{4}end{array} ight])

(BA=left[egin{array}{ccc}{19}&{5}&{23}{5}&{-7}&{-1}{-14}&{6}&{18}end{array} ight])

In Exercises (PageIndex{28}) - (PageIndex{33}), a diagonal matrix (D) and a matrix (A) are given. Find the products (DA) and (AD), where possible.

Exercise (PageIndex{28})

(D=left[egin{array}{cc}{3}&{0}{0}&{-1}end{array} ight]) (A=left[egin{array}{cc}{2}&{4}{6}&{8}end{array} ight])

(DA=left[egin{array}{cc}{6}&{-4}{18}&{-8}end{array} ight])

Exercise (PageIndex{29})

(D=left[egin{array}{cc}{4}&{0}{0}&{-3}end{array} ight]) (A=left[egin{array}{cc}{1}&{2}{1}&{2}end{array} ight])

(DA=left[egin{array}{cc}{4}&{-6}{4}&{-6}end{array} ight])

Exercise (PageIndex{30})

(D=left[egin{array}{ccc}{-1}&{0}&{0}{0}&{2}&{0}{0}&{0}&{3}end{array} ight]) (A=left[egin{array}{ccc}{1}&{2}&{3}{4}&{5}&{6}{7}&{8}&{9}end{array} ight])

(DA=left[egin{array}{ccc}{-1}&{4}&{9}{-4}&{10}&{18}{-7}&{16}&{27}end{array} ight])

Exercise (PageIndex{31})

(D=left[egin{array}{ccc}{1}&{1}&{1}{2}&{2}&{2}{-3}&{-3}&{-3}end{array} ight]) (A=left[egin{array}{ccc}{2}&{0}&{0}{0}&{-3}&{0}{0}&{0}&{5}end{array} ight])

(DA=left[egin{array}{ccc}{2}&{2}&{2}{-6}&{-6}&{-6}{-15}&{-15}&{-15}end{array} ight])

Exercise (PageIndex{32})

(D=left[egin{array}{cc}{d_{1}}&{0}{0}&{d_{2}}end{array} ight]) (A=left[egin{array}{cc}{a}&{b}{c}&{d}end{array} ight])

(DA=left[egin{array}{cc}{d_{1}a}&{d_{1}b}{d_{2}c}&{d_{2}d}end{array} ight])

Exercise (PageIndex{33})

(D=left[egin{array}{ccc}{d_{1}}&{0}&{0}{0}&{d_{2}}&{0}{0}&{0}&{d_{3}}end{array} ight]) (A=left[egin{array}{ccc}{a}&{b}&{c}{d}&{e}&{f}{g}&{h}&{i}end{array} ight])

(DA=left[egin{array}{ccc}{d_{1}a}&{d_{1}b}&{d_{1}c}{d_{2}d}&{d_{2}e}&{d_{2}f}{d_{3}g}&{d_{3}h}&{d_{3}i}end{array} ight])

In Exercises (PageIndex{34}) - (PageIndex{39}), a matrix (A) and a vector (vec{x}) are given. Find the product (Avec{x}).

Exercise (PageIndex{34})

(A=left[egin{array}{cc}{2}&{3}{1}&{-1}end{array} ight]), (vec{x}=left[egin{array}{c}{4}{9}end{array} ight])

(Avec{x}=left[egin{array}{c}{35}{-5}end{array} ight])

Exercise (PageIndex{35})

(A=left[egin{array}{cc}{-1}&{4}{7}&{3}end{array} ight]), (vec{x}=left[egin{array}{c}{2}{-1}end{array} ight])

(Avec{x}=left[egin{array}{c}{-6}{11}end{array} ight])

Exercise (PageIndex{36})

(A=left[egin{array}{ccc}{2}&{0}&{3}{1}&{1}&{1}{3}&{-1}&{2}end{array} ight]), (vec{x}=left[egin{array}{c}{1}{4}{2}end{array} ight])

(Avec{x}=left[egin{array}{c}{8}{7}{3}end{array} ight])

Exercise (PageIndex{37})

(A=left[egin{array}{ccc}{-2}&{0}&{3}{1}&{1}&{-2}{4}&{2}&{-1}end{array} ight]), (vec{x}=left[egin{array}{c}{4}{3}{1}end{array} ight])

(Avec{x}=left[egin{array}{c}{-5}{5}{21}end{array} ight])

Exercise (PageIndex{38})

(A=left[egin{array}{cc}{2}&{-1}{4}&{3}end{array} ight]), (vec{x}=left[egin{array}{c}{x_{1}}{x_{2}}end{array} ight])

(Avec{x}=left[egin{array}{c}{2x_{1}-x_{2}}{4x_{1}+3x_{2}}end{array} ight])

Exercise (PageIndex{39})

(A=left[egin{array}{ccc}{1}&{2}&{3}{1}&{0}&{2}{2}&{3}&{1}end{array} ight]), (vec{x}=left[egin{array}{c}{x_{1}}{x_{2}}{x_{3}}end{array} ight])

(Avec{x}left[egin{array}{c}{x_{1}+2x_{2}+3x_{3}}{x_{1}+2x_{3}}{2x_{1}+3x_{2}+x_{3}}end{array} ight])

Exercise (PageIndex{40})

Let (A=left[egin{array}{cc}{0}&{1}{1}&{0}end{array} ight]). Find (A^{2}) and (A^{3}).

(A^{2}=left[egin{array}{cc}{1}&{0}{0}&{1}end{array} ight]); (A^{3}=left[egin{array}{cc}{0}&{1}{1}&{0}end{array} ight])

Exercise (PageIndex{41})

Let (A=left[egin{array}{cc}{2}&{0}{0}&{3}end{array} ight]). Find (A^{2}) and (A^{3}).

(A^{2}=left[egin{array}{cc}{4}&{0}{0}&{9}end{array} ight]); (A^{3}=left[egin{array}{cc}{8}&{0}{0}&{27}end{array} ight])

Exercise (PageIndex{42})

Let (A=left[egin{array}{ccc}{-1}&{0}&{0}{0}&{3}&{0}{0}&{0}&{5}end{array} ight]). Find (A^{2}) and (A^{3}).

(A^{2}=left[egin{array}{ccc}{1}&{0}&{0}{0}&{9}&{0}{0}&{0}&{25}end{array} ight]); (A^{3}=left[egin{array}{ccc}{-1}&{0}&{0}{0}&{27}&{0}{0}&{0}&{125}end{array} ight])

Exercise (PageIndex{43})

Let (A=left[egin{array}{ccc}{0}&{1}&{0}{0}&{0}&{1}{1}&{0}&{0}end{array} ight]). Find (A^{2}) and (A^{3}).

(A^{2}=left[egin{array}{ccc}{0}&{0}&{1}{1}&{0}&{0}{0}&{1}&{0}end{array} ight]); (A^{3}=left[egin{array}{ccc}{1}&{0}&{0}{0}&{1}&{0}{0}&{0}&{1}end{array} ight])

Exercise (PageIndex{44})

Let (A=left[egin{array}{ccc}{0}&{0}&{1}{0}&{0}&{0}{0}&{1}&{0}end{array} ight]). Find (A^{2}) and (A^{3}).

(A^{2}=left[egin{array}{ccc}{0}&{1}&{0}{0}&{0}&{0}{0}&{0}&{0}end{array} ight]); (A^{3}=left[egin{array}{ccc}{0}&{0}&{0}{0}&{0}&{0}{0}&{0}&{0}end{array} ight])

Exercise (PageIndex{45})

In the text, we state that ((A+B)^{2} eq A^{2}+2AB+B^{2}). We investigate that claim here.

1. Let (A=left[egin{array}{cc}{5}&{3}{-3}&{-2}end{array} ight]) and let (B=left[egin{array}{cc}{-5}&{-5}{-2}&{1}end{array} ight]). Compute (A+B).
3. Compute (A^{2}+2AB+B^{2}).
4. Are the results from (a) and (b) the same?
5. Carefully expand the expression ((A+B)^{2}=(A+B)(A+B)) and show why this is not equal to (A^{2}+2AB+B^{2}).
1. (left[egin{array}{cc}{0}&{-2}{-5}&{-1}end{array} ight])
2. (left[egin{array}{cc}{10}&{2}{5}&{11}end{array} ight])
3. (left[egin{array}{cc}{-11}&{-15}{37}&{32}end{array} ight])
4. No
5. ((A+B)(A+B)=AA+AB+BA+BB=A^{2}+AB+BA+B^{2})

(1) If a shopkeeper bought 5 metres of cloth at the rate of TK. 200 per metre and sold it at the rate of TK.225 per metre, how much did he gain as profit?

Solution:- Cost of 1 metre cloth = 200TK.

Therefore, cost of 5 metre cloth = (200࡫) = 1000 TK.

And, S.P of 1 metre cloth = 225TK.

S.P of 5 metre cloth = (5𴢹) Tk.

Therefore, Profit = (1125-1000) TK.

(2) If an orange seller bought 5 dozen oranges at the rate of TK.60 per four and sold them at the rate of TK.50 per four, how much did he lose?

Solution:- 1 dozen = 12 pc = 3 hali

Therefore, 4 pc= 1 hali orange cost= 60 TK.

15 hali orange cost = (60吋) TK.

And, S.P of hali orange 50 TK.

Therefore, S.P of 15 hali orange (15吮) TK.

(3) Rabi bought 50 kg rice at the of TK. 40 per kg and sols it at the rate of TK. 55 per kg. What is the amount of profit or loss?

Solution:- Cost price of 1 kg rice = 40TK.

Therefore, cost price of 50 kg rice = (50吤) TK.

And S.P of 1 kg rice = 44TK.

Therefore, S.P of 50 kg rice = (44吮) TK.

Therefore, Profit = (2200-2000) TK.

(4) The buying rate of Milk vita milk is TK. 52 per litre and the selling rate is TK. 55 per litre, what is the percentage of profit?

Solution:- Cost price of 1 litre of milk vita= 52 TK.

Therefore, profit = (55-52) = 3 TK.

(5) Some chocolates were bought at TK. 8 per piece, and sold them at TK. 8.50 per rice and them the profit was TK. 25. How many chocolates were bought?

Solution:- C. P of 1 chocolate = 8 TK.

Therefore. Profit = (8.50-8.00) = 0.50 TK.

Therefore, 0.5 RS profit in 1 chocolate.

Therefore, 1 RS profit in 1/0.5 chocolate.

Therefore, 25 RS profit in 25/0.5 chocolate.

(6) A Shopkeeper bought cloth at rate of is TK. 125 metre, and sold it at the rate of TK. 150 per metre. Then he gained TK.2000 as profit. How many metres of cloth did the shopkeeper buy?

Solution:- C P of cloth = 125 TK per metre

S P of cloth = 150 TK per metre

Therefore, 25 TK profit in 1 metre cloth.

Therefore, 1 TK profit in 1/20 cloth.

2000 TK profit in 2000/20 TK cloth.

(7) An item is bought at TK.190 and is sold at TK.175. What is the percentage of profit or loss?

Solution:- Item cost price = 190 TK.

Therefore, item selling price = 175 TK.

(8) The cost price of 25 metres of cloth is equal to the selling price of 20 metres of cloth. What is the percentage of profit or loss?

Solution:- Let, cost price 100RS.

Therefore, 25metre cloth C.P = 100TK.

Therefore, 1 metre cloth C.P = 100/25 TK.

Therefore, 20 metre cloth S.P = 100 TK.

1 metre cloth S.P = 100/20 TK.

Therefore, Profit = (5-4) = 1 TK.

Therefore, 4 TK profit – 1 TK.

Therefore, 1 TK profit – ¼ TK.
Therefore, 1000 TK. Profit 100ࡧ/4 TK.

(9) The buying price of 8 amloki is sold at TK.5, what is the percentage of profit or loss?

Solution:- 8 amloki cost price 5 TK.

Therefore, amloki cost price 5/8 TK.

Therefore, 1 amloki SP = 5/6 TK.

Therefore, Profit = (5/6-5/8) TK.

Therefore, 5/8 TK profit 5/24TK.

Therefore, 1 TK profit 5࡮/5吔 TK.

Therefore, 100 TK profit 5x8x100/5吔 TK

(11) If an object is sold at TK. 400, there is as much amount loss, there will be the profit amounting three times the loss, if it is sold at TK.480, what is the cost price of that object?

Solution:- From question, Let loss x TK.

Therefore, Cost price of object = (400+20) TK.

(12) If the selling price of a watch is TK. 625, the loss is 10%. What would be the price of the watch if a profit of 10% is to be gained by selling it?

Solution:- Let, Cost of watch 100 TK.

In 10% loss, SP of watch = (100-10)

And 10% profit, SP of watch = (100+10) Tk.

Therefore, 625TK. — 110𴩉/90 TK.

(13) Maisha bought 15 metres red ribbon at the rate of TK. 20 per metre. The rate of VAT is TK. 4 percent, and she gave TK. 500 to the shopkeeper. How much taka would he need for Indian 3000 Rupee?

Solution:- 1 metre red ribbon 20 TK.

Therefore, 15 metre red ribbon (15吐) TK.

Therefore, 100 TK. Vat – 4TK.

Therefore, 3000 TK. Vat = 4/100x 300

Total price of ribbon = (300+12)

Therefore, Shopkeeper return = (500-312) TK.

(14) Mr. Roy is a government officer. He will go to India to visit religious places. If Bangladeshi TK. 1 is equal to Indian 0.63 Rupee, how much Bangladeshi taka would he need for Indian 3000 Rupee?

Solution:- 0.63 Rupee= 1 TK.

Therefore, 3000 Rupee= 3000/0.6 TK.

(15) Nilim is a service holder. His monthly basic pay is TK. 22,250. The income tax of two lac fifty thousand taka of first slab of annual income is 0 taka. The rate of income tax of next annual income slab is TK. 10. How many taka does he has to pay as tax?

## Backwards Induction (Exercise 2.2.6) Analysis 1 by Terence Tao

I am new to the study of analysis and I decided to start with Terence's book in my endeavor. I want to show my "proof" of backwards induction since I have some difficulty in understanding this. I want to now if my proof is correct or have some error, because if have, a can't infer that. Any feedback is appreciated.

Let $n$ be a natural number, and let $P(m)$ be a property pertaining to the natural numbers such that whenever $P(m ext<++>)$ is true, then $P(m)$ is true. Suppose that $P(n)$ is also true. Prove that $P(m)$ is true for all natural numbers $m ≤ n$ this is known as the principle of backwards induction. (Hint: apply induction to the variable $n$ .)

First i want to show $P(m)$ is true $forall$ geq m$. H1:$forall mP(m ext<++>)implies P(m)$C:$P(m)$is true$forall$geq m$ .

geq m$means =m+a$ for some natural number $a$ , then $m=a=0$ for corollary 2.2.9. But $P(0)$ is true for H2, then the case $n=0$ is proved.

Suppose now that works for $n$ and prove $n ext<++>$ . then:

H1: $forall m$ $P(m ext<++>)implies P(m)$

H2: $P(n)implies P(m)$ $forall$ $ngeq m$

In H1, for $m=n$ we have $P(n ext<++>)implies P(n)$ and for H2 we now $P(n)implies P(m)$ (specifically for $n=m$ ), then $P(n ext<++>)implies P(m)$ for $n ext<++>>m$ . We need to prove that works for $n ext<++>=m$ but for that $P(n ext<++>)$ is true for H3. We conclude that $P(n ext<++>)implies P(m)$ $forall$ $n ext<++>geq m$ .

## NCERT Solutions for Class 6 Maths Exercise 2.2

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## 2.2.1: Exercises 2.2 - Mathematics

Therefore, let f(x) = 5x – 4x 2 + 3

(i) When x = 0

f(0) = 5(0)-4(0) 2 +3

= 3

(ii) When x = -1

f(x) = 5x−4x 2 +3

f(−1) = 5(−1)−4(−1) 2 +3

= −5–4+3

= −6

(iii) When x = 2

f(x) = 5x−4x 2 +3

f(2) = 5(2)−4(2) 2 +3

= 10–16+3

= −3

### Question 2: Find p(0), p(1) and p(2) for each of the following polynomials:

(i) p(y) = y 2 −y+1

(ii) p(t) = 2+t+2t 2 −t 3

(iii) p(x) = x 3

(iv) P(x) = (x−1)(x+1)

(i) p(y) = y 2 – y + 1

Given equation: p(y) = y 2 –y+1

Therefore, p(0) = (0) 2 −(0)+1 = 1

p(1) = (1) 2 –(1)+1 = 1

p(2) = (2) 2 –(2)+1 = 3

Hence, p(0) = 1, p(1) = 1, p(2) = 3, for the equation p(y) = y 2 –y+1

(ii) p(t) = 2 + t + 2t 2 − t 3

Given equation: p(t) = 2+t+2t 2 −t 3

Therefore, p(0) = 2+0+2(0) 2 –(0) 3 = 2

p(1) = 2+1+2(1) 2 –(1) 3 = 2+1+2–1 = 4

p(2) = 2+2+2(2) 2 –(2) 3 = 2+2+8–8 = 4

Hence, p(0) = 2,p(1) =4 , p(2) = 4, for the equation p(t)=2+t+2t 2 −t 3

(iii) p(x) = x 3

Given equation: p(x) = x 3

Therefore, p(0) = (0) 3 = 0

p(1) = (1) 3 = 1

p(2) = (2) 3 = 8

Hence, p(0) = 0,p(1) = 1, p(2) = 8, for the equation p(x)=x 3

(iv) p(x) = (x−1)(x+1)

Given equation: p(x) = (x–1)(x+1)

Therefore, p(0) = (0–1)(0+1) = (−1)(1) = –1

p(1) = (1–1)(1+1) = 0(2) = 0

p(2) = (2–1)(2+1) = 1(3) = 3

Hence, p(0)= 1, p(1) = 0, p(2) = 3, for the equation p(x) = (x−1)(x+1)

### Question 3: Verify whether the following are zeroes of the polynomial, indicated against them.

(i) p(x) = 3x+1, x=−1/3

(ii) p(x) = 5x–π, x = 4/5

(iii) p(x) = x 2 −1, x=1, −1

(iv) p(x) = (x+1)(x–2), x =−1, 2

(v) p(x) = x 2 , x = 0

(vi) p(x) = lx+m, x = −m/l

(vii) p(x) = 3x 2 −1, x = -1/√3 , 2/√3

(viii) p(x) = 2x+1, x = 1/2

(i) p(x)=3x+1, x=−1/3

Given: p(x)=3x+1 and x=−1/3

Therefore, substituting the value of x in equation p(x), we get.

For, x = -1/3

p(−1/3) = 3(-1/3)+1

= −1+1

= 0

Hence, p(x) of -1/3 = 0

(ii) p(x)=5x–π, x = 4/5

Given: p(x)=5x–π and x = 4/5

Therefore, substituting the value of x in equation p(x), we get.

For, x = 4/5

p(4/5) = 5(4/5)–π

= 4–π

Hence, p(x) of 4/5 ≠ 0

(iii) p(x)=x 2 −1, x=1, −1

Given: p(x)=x 2 −1 and x=1, −1

Therefore, substituting the value of x in equation p(x), we get.

For x = 1

p(1) = 1 2 −1

=1−1

= 0

For, x = -1

p(−1) = (-1) 2 −1

= 1−1

= 0

Hence, p(x) of 1 and -1 = 0

(iv) p(x) = (x+1)(x–2), x =−1, 2

Given: p(x) = (x+1)(x–2) and x =−1, 2

Therefore, substituting the value of x in equation p(x), we get.

For, x = −1

p(−1) = (−1+1)(−1–2)

= (0)(−3)

= 0

For, x = 2

p(2) = (2+1)(2–2)

= (3)(0)

= 0

Hence, p(x) of −1, 2 = 0

(v) p(x) = x 2 , x = 0

Given: p(x) = x 2 and x = 0

Therefore, substituting the value of x in equation p(x), we get.

For, x = 0

p(0) = 0 2 = 0

Hence, p(x) of 0 = 0

(vi) p(x) = lx+m, x = −m/l

Given: p(x) = lx+m and x = −m/l

Therefore, substituting the value of x in equation p(x), we get.

For, x = −m/l

p(-m/l)= l(-m/l)+m

= −m+m

= 0

Hence, p(x) of -m/l = 0

(vii) p(x) = 3x 2 −1, x = -1/√3 , 2/√3

Given: p(x) = 3x 2 −1 and x = -1/√3 , 2/√3

Therefore, substituting the value of x in equation p(x), we get.

For, x = -1/√3

p(-1/√3) = 3(-1/√3) 2 -1

= 3(1/3)-1

= 1-1

= 0

For, x = 2/√3

p(2/√3) = 3(2/√3) 2 -1

= 3(4/3)-1

= 4−1

=3 ≠ 0

Hence, p(x) of -1/√3 = 0

but, p(x) of 2/√3 ≠ 0

(viii) p(x) =2x+1, x = 1/2

Given: p(x) =2x+1 and x = 1/2

Therefore, substituting the value of x in equation p(x), we get.

For, x = 1/2

p(1/2) = 2(1/2)+1

= 1+1

= 2≠0

Hence, p(x) of 1/2 ≠ 0

### Question 4: Find the zero of the polynomials in each of the following cases:

(iii) p(x) = 2x+5

(vii) p(x) = cx+d, c ≠ 0, c, d are real numbers.

(i) p(x) = x+5

Given: p(x) = x+5

To find the zero, let p(x) = 0

p(x) = x+5

0 = x+5

x = −5

Therefore, the zero of the polynomial p(x) = x+5 is when x = -5

(ii) p(x) = x–5

Given: p(x) = x–5

p(x) = x−5

x−5 = 0

x = 5

Therefore, the zero of the polynomial p(x) = x–5 is when x = 5

(iii) p(x) = 2x+5

Given: p(x) = 2x+5

p(x) = 2x+5

2x+5 = 0

2x = −5

x = -5/2

Therefore, the zero of the polynomial p(x) = 2x+5 is when x = -5/2

(iv) p(x) = 3x–2

Given: p(x) = 3x–2

p(x) = 3x–2

3x−2 = 0

3x = 2

x = 2/3

Therefore, the zero of the polynomial p(x) = 3x–2 is when x = 2/3

(v) p(x) = 3x

Given: p(x) = 3x

p(x) = 3x

3x = 0

x = 0

Therefore, the zero of the polynomial p(x) = 3x is when x = 0

(vi) p(x) = ax, a0

Given: p(x) = ax, a≠ 0

p(x) = ax

ax = 0

x = 0

Therefore, the zero of the polynomial p(x) = ax is when x = 0

(vii) p(x) = cx+d, c ≠ 0, c, d are real numbers.

Given: p(x) = cx+d

p(x) = cx + d

cx+d =0

x = -d/c

Therefore, the zero of the polynomial p(x) = cx+d is when x = -d/c

## Polynomials : Exercise - 2.2 (Mathematics NCERT Class 10th)

Q.1 Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients :
(i)
(ii)
(iii)
(iv)
(v)
(vi)

Sol. (i) We have,

The value of is zero when the value of (x + 2) (x – 4) is zero, i.e.,
when x + 2 = 0 or x – 4 = 0 , i.e., when x = – 2 or x = 4.
So, The zeroes of are – 2 and 4.
Therefore , sum of the zeroes = (– 2) + 4 = 2

and product of zeroes = (– 2) (4) = – 8

(ii) We have,

The value of is zero when the value of
(2s – 1) (2s – 1) is zero, i.e., when 2s – 1 = 0 or 2s – 1 = 0,
i.e., when .
So, The zeroes of
Therefore, sum of the zeroes

and product of zeroes

(iii) We have, =

The value of is zero when the value of (3x + 1) (2x – 3) is zero, i.e., when 3x + 1 = 0 or 2x – 3 = 0, i.e, when
So, The zeroes of
Therefore, sum of the zeroes

and product of zeroes

(iv) We have, = 4u (u + 2)
The value of is zero when the value of 4u(u + 2) is zero, i.e., when u = 0 or u + 2 = 0, i.e., when u = 0 or u = – 2.
So, The zeroes of and 0 and – 2
Therefore, sum of the zeroes = 0 + (– 2) = – 2

and , product of zeroes = (0) (–2) = 0

(v) We have
The value of is zero when the value of is zero,
i.e., when = 0 i.e., when
So, The zeroes of
Therefore , sum of the zeroes =

and, product of the zeroes =

(vi) We have, =

The value of is zero when the value of (x + 1) (3x – 4) is zero, i.e., when x + 1 = 0 or 3x – 4 = 0, i.e., when x = – 1 or .
So, The zeroes of
Therefore , sum of the zeroes

and, product of the zeroes

Q.2 Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.
(i)
(ii)
(iii)
(iv) 1, 1
(v)
(vi) 4, 1
Sol. (i) Let the polynomial be , and its zeroes be . Then ,

and,
If a = 4, then b = – 1 and c = – 4.
Therefore, one quadratic polynomial which fits the given conditions is .

(ii) Let the polynomial be , and its zeroes be . Then,

and
If a = 3, then b
So, One quadratic polynomial which fits the given conditions is .

(iii) Let the polynomial be , and its zeroes be . Then,

and
If a = 1, then b = 0 and c =
So, one quadratic polynomial which fits the given conditions is .

(iv) Let the polynomial be and its zeroes be . Then,

a nd
If a = 1, then b = – 1 and c = 1.
So, one quadratic polynomial which fits the given conditions is .

(v) Let the polynomial be and its zeroes be . Then

and
If a = 4 then b = – 1 and c = 1.
So, one quadratic polynomial which fits the given conditions is .

(vi) Let the polynomial be and its zeroes be . Then,

If a = 1, then b = – 4 and c = 1
Therefore, one quadratic polynomial which fits the given conditions is .

## NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers Exercise 2.2

NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers Ex 2.2

Exercise 2.2

Ex 2.2 C lass 6 Maths Question 1.
Find the sum by suitable arrangement:
(a) 837 + 208 + 363
(b) 1962 + 453,+ 1538 + 647
Solution:
(a) 837 + 208 + 363 = (837 + 363) + 208
= 1200 + 208 [Using associative property]
= 1408

(b) 1962 + 453 + 1538 + 647
= (1962 + 1538) + (453 + 647)
= 3500 + 1100 = 4600

Ex 2.2 C lass 6 Maths Question 2.
Find the product by suitable arrangement:
(а) 2 x 1768 x 50
(b) 4 x 166 x 25
(c) 8 x 291 x 125
(d) 625 x 279 x 16
(e) 285 x 5 x 60
(f) 125 x 40 x 8 x 25
Solution:
(a) 2 x 1768 x 50 = (2 x 50) x 1768 = 176800
(b) 4 x 166 x 25 = 166 x (25 x 4) = 166 x 100 = 16600
(c) 8 x 291 x 125 = (8 x 125) x 291 = 1000 x 291 = 291000
(d) 625 x 279 x 16 = (625 x 16) x 279 = 10000 x 279 = 2790000
(e) 285 x 5 x 60 = 285 x (5 x 60) = 285 x 300 = (300 – 15)x 300 = 300 x 300 – 15 x 300 = 90000 – 4500 = 85500
(f) 125 x 40 x 8 x 25 = (125 x 8) x (40 x 25) = 1000 X 1000 = 1000000

Ex 2.2 C lass 6 Maths Question 3.
Find the value of the following:
(а) 297 x 17 + 297 x 3
(б) 54279 x 92 + 8 x 54279
(c) 81265 x 169 – 81265 x 69
(d) 3845 x 5 x 782 + 769 x 25 x 218
Solution:
(a) 297 x 17 x 297 x 3 = 297 x (17 + 3)
= 297 x 20 = 297 x 2 x 10
= 594 x 10 = 5940

(b) 54279 x 92 + 8 x 54279 = 54279 x (92 + 8)
= 54279 x 100 = 5427900

(c) 81265 x 169 – 81265 x 69
= 81265 x (169 – 69)
= 81265 x 100 = 8126500

(d) 3845 x 5 x 782 + 769 x 25 x 218 = 3845 x 5 x 782 + 769 x 5 x 5 x 218
= 3845 x 5 x 782 + (769 x 5) x 5 x 218
= 3845 x 5 x 782 + 3845 x 5 x 218
= 3845 x 5 x 782 + 3845 x 5 x 218
= 3845 x 5 x (782 + 218)
= 3845 x 5 x 1000
= 19225 x 1000
= 19225000

Ex 2.2 C lass 6 Maths Question 4.
Find the product using suitable properties.
(a) 738 x 103
(b) 854 x 102
(c) 258 x 1008
(d) 1005 x 168
Solution:
(a) 738 x 103 = 738 x (100 + 3)
= 738 x 100 + 738 x 3 [Using distributive property]
= 73800 + 2214 = 76014

(b) 854 x 102 = 854 x (100 + 2)
= 854 x 100 + 854 x 2 [Using distributive property]
= 85400 + 1708 = 87108

(c) 258 x 1008 = 258 x (1000 + 8)
= 258 x 1000 + 258 x 8 [Using distributive property]
= 258000 + 2064 = 260064

(d) 1005 x 168 = (1000 + 5) x 168
= 1000 x 168 + 5 x 168 [Using distributive property]
= 168000 + 840 = 168840

Ex 2.2 C lass 6 Maths Question 5.
A taxidriver filled his car petrol tank with 40 litres of petrol on Monday. The next day, he filled the tank with 50 litre of petrol. If the petrol cost ₹44 per litre, how much did he spend in all on petrol?
Solution:
Petrol filled on Monday = 40 litres
Cost of petrol = ₹44 per litre
Petrol filled on Tuesday = 50 litre
Cost of petrol = ₹44 pet litre
∴ Total money spent in all
= ₹(40 x 44 + 50 x 44)
= ₹(40 + 50) x 44 = ₹90 x 44 = ₹3960

Ex 2.2 C lass 6 Maths Question 6.
A vendor supplies 32 litres of milk to a hotel in the morning and 68 litres of milk in the evening. If the milk costs ₹15 per litre, how much money is due to the vendor per day?
Solution:
Milk supplied in the morning = 32 litres
Cost of milk = ₹15 per litre
Milk supplied in the evening = 68 litres
Cost of milk = ₹15 per litre

Ex 2.2 C lass 6 Maths Question 7.
Match the following:
(i) 425 x 136 = 425 x (6 + 30 + 100)
(ii) 2 x 49 x 50 = 2 x 50 x 49
(iii) 80 + 2005 + 20 = 80 + 20 + 2005
Hence (i) ↔ (c), (ii) ↔ (a) and (iii) ↔ (b)
∴ Money paid to the vendor
= ₹ (32 x 15 + 68 x 15)
= ₹(32 + 68) x 15
= ₹100 x 15
= ₹1500
(a) Commutativity under multiplication
(c) Distributivity of multiplication over addition

The basic idea is good. But after you've proven that $sum_^Nfrac1<2n+1>$ is smaller that the sum of the first $2N+1$ terms of your series, you're done. And all you need for that is that$frac1<2^2>+frac1<4^2>+cdots+frac1<(2N)^2><1.$This is equivalent to$frac1<1^2>+frac1<2^2>+cdots+frac1<4,$which in turn follows fromeginfrac1<1^2>+frac1<2^2>+cdots+frac1&<1+frac1<1 imes2>+frac1<2 imes3>+cdots+frac1<(N-1)N>&=1+1-frac12+frac12-frac13+cdots+frac1+frac1N&=2+frac1N&<4.end

Don't be swayed by the particular case. You have the following

Lemma Let $sum a_n$ be a series (resp. $sum b_n$ be a convergent series), then the intertwining $x_<2n>=a_n$ and $x_<2n+1>=b_n$ is convergent iff $sum a_n$ is so.

Proof Let us, for short, suppose the indexing starting from zero. Then, we get
$sum_^<2N+1>x_n=sum_^N a_n + sum_^N b_n$ this proves the equivalence $sum x_nmbox< converges >Longleftrightarrow sum a_nmbox < converges >$ end of proof

Here $1+dfrac<1><3>+dfrac<1><5>+dfrac<1><7>+dfrac<1><9>ldots$ diverges while $dfrac<1><2^2>+dfrac<1><4^2>+dfrac<1><6^2>+dfrac<1><8^2>ldots$ converges. Hence your series diverges.