cbse class 11 mathematics worksheet - relations functions (2) class 11...Β Β· cbse class 11...
TRANSCRIPT
www.stud
iestod
ay.co
m
Q.11) Find the domain of π(π₯) =1
β1βcos π₯
Sol.11) π(π₯) will be defined when 1 β cos π₯ > 0 for all π₯ β π β 1 β€ cos π₯ β€ 1 But cos π₯ cannot equal to 1 Since, 1 β cos π₯ > 0 and cos π₯ = 1 when π₯ = 2ππ (by general solution method β΄ Domain = π β {2ππ; π β π§} ans.
Q.12) Let π = {(2,4), (5,6), (8, β1), (10, β3)} π = {(2,5), (7,1), (8,4), (10,13), (11, β5)}. Find the domain of π + π, π β π & ππ.
Sol.12) Domain of π is π·π = {2,5,8,10}
Domain of π is π·π = {2,7,8,10,11}
Domain of π + π, π β π & ππ is always defined as
Domain of {π₯: π₯ β π·π β© π·π}
β΄ Domain of π + π, π β π & ππ = {2,8,10} ans.
Q.13) Draw the graph of π(π₯) = 1 + |π₯ β 2|
Sol.13) We have, π(π₯) = 1 + [π₯ β 2]
π(π₯) = {1 + (π₯ β 2): π₯ β 2 β₯ 0: π₯ β₯ 21 β (π₯ β 2): π₯ β 2 < 0: π₯ < 2
π(π₯) = {π₯ β 1: π₯ β₯ 2
βπ₯ + 1: π₯ < 2
Points (0,3), (1,2), (2,0), (3,2), (β1,4), (4,3)
Domain = R Range = [1,00]
Q.14) Let R be the relation on Set N (natural no.s) defined by π (roster form), domain, Range, Co-domain and allow diagram.
Sol.14) We have, Relation from N to N π = {(π, π): π + 3π = 12; π β π and π β π} (i) π = {(9,1), (6,2), (3,3)} (ii) Domain = {9,6,3} (iii) Range = {1,2,3} (iv) Co-domain = N (v) Arrow diagram : FIG 14
Q.15) Let π΄ = {1,2,3,4,6}, Let R is a relation on A defined by π = {(π, π): π ππ ππ₯πππ‘ππ¦ πππ£ππ ππππ ππ¦ π}. Find R, domain, Range, Co-domain and Arrow diagram.
Sol.15) We have, relation from A to A π = {(π, π): π ππ ππ₯πππ‘ππ¦ πππ£ππ ππππ ππ¦ π; π β π΄ , π β π΄} (i) π = {(1,1), (2,2), (3,3), (4,4), (6,6), (1,2), (1,4), (1,6), (2,4), (2,6), (3,6)} (ii) Domain = {1,2,3,4,6} (iii) Range = {1,2,3,4,6} (iv) Co-domain = A (v) Arrow diagram:
Downloaded from www.studiestoday.com
Copyright Β© www.studiestoday.com All rights reserved. No part of this publication may be reproduced, distributed, or transmitted in any
form or by any means, including photocopying, recording, or other electronic or mechanical methods, without the prior written permission.
www.stud
iestod
ay.co
m
Q.16) Given Arrow diagram. Find relation in set builder & Roster form also find Domain, Range,
co-domain.
Sol.16) π = {(25,5), (25, β5), (9,3), (9, β3), (4,2), (4, β2)} Clearly 25 = 52, 25 = (β52), 9 = 32 and so on β΄ π = {(π₯, π¦): π₯ = π¦2; π₯ β π πππ π¦ = π} Domain = {25,9,4} Range = {5, β5,3, β3,2, β2} Co-domain = π = {β2,5,4, β5,2,3, β3,2}
Q.17) find the domain of the function, π(π₯) = βπ₯ β 3 β 2βπ₯ β 4 β βπ₯ β 3 + 2βπ₯ β 4
Sol.17) π(π₯) is real for all values of π₯ such that
π₯ β 3 β 2βπ₯ β 4 β₯ 0; π₯ β 3 + 2βπ₯ β 4 β₯ 0 and π₯ β 4 β₯ 0
π₯ β 3 β₯ 2βπ₯ β 4; π₯ β 3 β₯ β2βπ₯ β 4 and π₯ β₯ 4 squaring squaring β π₯2 β 6π₯ + 9 β₯ 4(π₯ β 4); π₯2 β 6π₯ + 9 β₯ 4(π₯ β 4) and π₯ β₯ 4 β π₯2 β 10π₯ + 25 β₯ 0; π₯2 β 10π₯ + 25 β₯ 0 and π₯ β₯ 4 β (π₯ β 5)2 β₯ 0; (π₯ β 5)2 β₯ 0 and π₯ β₯ 4 β π₯ β 5 β₯ 0; π₯ β 5 β₯ 0 and π₯ β₯ 4 β π₯ β₯ 5; and π₯ β₯ 4 β΄ Domain π₯ β [5,00] ans.
Q.18) Find the domain of the function, π(π₯) = β
π₯+3
(2βπ₯)(π₯β5)
Sol.18 π(π₯) is real for all values of π₯ such that π₯+3
(2βπ₯)(π₯β5)β₯ 0 and (2 β π₯)(π₯ β 5) β 0
β (π₯+3)
(π₯β2)(π₯β5)β€ 0 and π₯ β 2 πππ π₯ β 5 β¦β¦.. {π πππ πβππππ}
β (π₯+3)(π₯β2)(π₯β5)
(π₯β2)2(π₯β5)2 β€ 0 β¦β¦ {ππ’ππ‘ππππ¦ & πππ£πππ ππ¦ (π₯ β 2)(π₯ β 5)}
β (π₯ + 3)(π₯ β 2)(π₯ β 5) β€ 0
Downloaded from www.studiestoday.com
Copyright Β© www.studiestoday.com All rights reserved. No part of this publication may be reproduced, distributed, or transmitted in any
form or by any means, including photocopying, recording, or other electronic or mechanical methods, without the prior written permission.
www.stud
iestod
ay.co
m
π₯ β [β00, β3] βͺ [2,5] but π₯ β 2 πππ π₯ β 5 β΄ Domain is π₯ β [β00, β3] βͺ [2,5] ans.
Q.19) π(π₯) = {
1 β π₯: π₯ < 01: π₯ = 0
π₯ + 1: π₯ > 0 draw graph of π(π₯).
Sol.19) For π₯ < 0; π(π₯) = 1 β π₯ Points (β1,2), (β2,3), (β3,4) For π₯ > 0; π(π₯) = π₯ + 1 Points (1,2), (2,3), (3,4) For π₯ = 0; π(π₯) = 1 Points (0,1)
Q.20) Find the domain of the function, π(π₯) =1
β[π₯]2β[π₯]β6.
Sol.20) We have, π(π₯) =1
β[π₯]2β[π₯]β6
π(π₯) will be defined when [π₯]2 β [π₯] β 6 > 0 Splitting middle term β [π₯]2 β 3[π₯] + 2[π₯] β 6 > 0 β ([π₯] β 3)([π₯] + 2) > 0 [π₯] < β2 ππ [π₯] > 3 β π₯ < β2 ππ π₯ β₯ 4 π₯ β (β00, β2) βͺ (4,00) ans.
Downloaded from www.studiestoday.com
Copyright Β© www.studiestoday.com All rights reserved. No part of this publication may be reproduced, distributed, or transmitted in any
form or by any means, including photocopying, recording, or other electronic or mechanical methods, without the prior written permission.