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Chapter 2 · સંબંધ અને વિધેય

11th_math_gm_index_part_1 · v1 · draft Source: Chapter2.zip · 4 sections, 132 questions
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#1 SUB

Question

òu (x + 1, y – 2) = (3, 1) íkku, x yLku y þkuÄku.

Answer

ynª, (x + 1, y 2) = (3, 1) nkuðkÚke,
x + 1 = 3, y – 2 = 1 ÚkkÞ.
\ x = 2, y = 3
#2 SUB

Question

òu P = {a, b, c}, Q = {r}, íkku P × Q yLku Q × P þkuÄku. þwt yk fkíkuorÍÞ økwýkfkh Mk{kLk Au ?

Answer

P × Q = {a, b, c} × {r}
= {(a, r), (b, r), (c, r)}...(1)
Q × P = {r} × {a, b, c}
= {(r, a), (r, b), (r, c)}...(2)
Ãkrhýk{ (1) yLku (2) ÃkhÚke òuE þfkÞ Au fu, ¢{Þwõík òuz (a, r) yLku (r, a) Mk{kLk Lk nkuðkÚke yk fkíkuorÍÞ økwýkfkh Mk{kLk LkÚke.
#3 SUB

Question

òu A = {1, 2, 3}, B = {3, 4}, C = {4, 5, 6}, íkku Lke[uLkk øký þkuÄku.
(i) A × (B C) (ii) (A × B) (A × C)
(iii) A × (B C) (iv) (A × B) (A × C)

Answer

(i) A × (B C) {kxu,
B C = {4}
\ A × (B C) = {1, 2, 3} × {4}
= {(1, 4), (2, 4), (3, 4)}
(ii) (A × B) (A × C)
A × B = {1, 2, 3} × {3, 4}
= {(1, 3), (1, 4), (2, 3), (2, 4), (3, 3),
(3, 4)}........(1)
A × C = {1, 2, 3} × {4, 5, 6}
= {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5),
(2, 6), (3, 4), (3, 5), (3, 6)}........(2)
\ (A × B) (A × C) = {(1, 4), (2, 4), (3, 4)}
(iii) A × (B C)
B C = {3, 4, 5, 6}
\ A × (B C) = {1, 2, 3} × {3, 4, 5, 6}
= {(1, 3), (1, 4), (1, 5), (1, 6), (2, 3),
(2, 4), (2, 5), (2, 6), (3, 3),
(3, 4), (3, 5), (3, 6)}
(iv) (A × B) (A × C)
Ãkrhýk{ (1) yLku (2)Lkku WÃkÞkuøk fhíkkt,
(A × B) (A × C)
= {(1, 3), (1, 4), (1, 5), (1, 6), (2, 3),
(2, 4), (2, 5), (2, 6), (3, 3),
(3, 4), (3, 5), (3, 6)}
#4 SUB

Question

òu P = {1, 2}, íkku P × P × P þkuÄku.

Answer

P × P × P = {1, 2} × {1, 2} × {1, 2}
= {(1, 1, 1), (1, 1, 2), (1, 2, 1), (1, 2, 2),
(2, 1, 1), (2, 1, 2), (2, 2, 1), (2, 2, 2)}
#5 SUB

Question

òu R ðkMíkrðf MktÏÞkykuLkku øký nkuÞ, íkku R × R yLku R × R × R þwt Ëþkoðþu ?

Answer

fkíkuorÍÞ økwýkfkh R × R = {(x, y) : x, y R} yu rîÃkrh{kýeÞ Þk{-Mk{ík÷Lkk «íÞuf ®çkËwLkwt rLkYÃký Ëþkoðu Au.
fkíkuorÍÞ økwýkfkh R × R × R = {(x, y, z) : x, y, z  R} yu rºkÃkrh{kýeÞ Þk{-Mk{ík÷Lkk «íÞuf ®çkËwLkwt rLkYÃký Ëþkoðu Au.
#6 SUB 🖼 1

Question

òu A × B = {(p, q), (p, r), (m, q), (m, r)}, íkku A yLku B þkuÄku.

Answer

øký A = A × BLke Ëhuf ¢{Þwõík òuzLkku «Úk{ ½xf
\ A = {p, m}
øký B = A × BLke Ëhuf ¢{Þwõík òuzLkku çkeòu ½xf
\ B = {q, r}
#7 SUB 🖼 12

Question

òu = , íkku x yLku y þkuÄku.

Answer

ynª, =
\ + 1 = , y =
\ = , =
\ x + 3 = 5 , 3y – 2 = 1
\ x = 2 , 3y = 3
\ y = 1
#8 SUB

Question

òu øký A{kt 3 ½xfku nkuÞ yLku B = {3, 4, 5}, íkku A × BLkk ½xfkuLke MktÏÞk þkuÄku.

Answer

ynª, n(A) = 3, n(B) = 3
\ n(A × B) = n(A) n(B) = 3 × 3 = 9 ÚkkÞ.
#9 SUB

Question

òu G = {7, 8}, H = {5, 4, 2}, íkku G × H yLku
H × G þkuÄku.

Answer

G × H = {7, 8} × {5, 4, 2}
= {(7, 5), (7, 4), (7, 2), (8, 5), (8, 4), (8, 2)}
H × G = {5, 4, 2} × {7, 8}
= {(5, 7), (5, 8), (4, 7), (4, 8), (2, 7), (2, 8)}
#10 SUB

Question

Lke[u ykÃku÷kt rðÄkLkku{ktÚke fÞwt rðÄkLk MkíÞ Au yLku fÞwt rðÄkLk yMkíÞ Au, íku sýkðku íkÚkk yMkíÞ rðÄkLk MkíÞ çkLku íku VheÚke ÷¾ku.
(i) òu P = {m, n}, Q = {n, m}, íkku P × Q = {(m, n), (n, m)}.

Answer

ykÃku÷ rðÄkLk yMkíÞ Au.
MkíÞ rðÄkLk : P × Q = {(m, n), (m, m), (n, n), (n, m)}
(ii) òu A yLku B yrhõík øký nkuÞ, íkku ßÞkt, x A íkÚkk y B nkuÞ íkuðe ík{k{ ¢{Þwõík òuz (x, y)Úke çkLkíkku yrhõík øký A × B Au.
ykÃku÷ rðÄkLk MkíÞ Au.
(iii) òu A = {1, 2}, B = {3, 4}, íkku A × (B f) = f
ykÃku÷ rðÄkLk MkíÞ Au.
ynª, B f = {3, 4} f = f Au.
\ A × (B f) = {1, 2} × f = f ÚkkÞ.
fu{ fu, çktLku{ktÚke fkuE Ãký yuf øký òu ¾k÷e øký nkuÞ, íkku íku{Lkku fkíkuorÍÞ økwýkfkh ¾k÷e øký s ÚkkÞ.
#11 SUB

Question

òu A = {–1, 1}, íkku A × A × A {u¤ðku.

Answer

A × A × A = {–1, 1} × {–1, 1} × {–1, 1}
= {(–1, –1, –1), (–1, –1, 1), (–1, 1, –1), (–1, 1, 1), (1, –1, –1), (1, –1, 1),
(1, 1, 1), (1, 1, –1)}
#12 SUB

Question

òu A × B = {(a, x), (a, y), (b, x), (b, y)}, íkku A yLku B þkuÄku.

Answer

ykÚke, (a, x) A × B
\ a A, x B
(a, y) A × B
\ a A, y B
(b, x) A × B
\ b A, x B
ykÚke, øký A = A × BLke «íÞuf ¢{Þwõík òuzLkku «Úk{ ½xf
= {a, b}
øký B = A × BLke «íÞuf ¢{Þwõík òuzLkku çkeòu ½xf
= {x, y}
#13 SUB

Question

Äkhku fu, A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6},
D = {5, 6, 7, 8},
íkku Lke[uLkkt Ãkrhýk{ku [fkMkku :
(i) A × (B C) = (A × B) (A × C)

Answer

zk.çkk. A × (B C) {kxu,
B C = f
\ A × (B C) = {1, 2} × f = f...(1)
s.çkk. (A × B) (A × C)

A × B = {1, 2} × {1, 2, 3, 4}

= {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1),

(2, 2), (2, 3), (2, 4)}

A × C = {1, 2} × {5, 6}

= {(1, 5), (1, 6), (2, 5), (2, 6)}

\ (A × B) (A × C) = f...(2)
\ Ãkrhýk{ (1) yLku (2) ÃkhÚke
zk.çkk. = s.çkk.
(ii) A × C yu B × DLkku WÃkøký Au.
ynª, A × C = {(1, 5), (1, 6), (2, 5), (2, 6)}

B × D = {(1, 5), (1, 6), (1, 7), (1, 8), (2, 5), (2, 6), (2, 7), (2, 8), (3, 5), (3, 6),
(3, 7), (3, 8), (4, 5), (4, 6), (4, 7), (4, 8)}

ynª, òuE þfkÞ Au fu, A × CLke Ëhuf ¢{Þwõík òuz B × D{kt ykðu÷ nkuðkÚke A × C yu B × DLkku WÃkøký Au.
#14 SUB

Question

òu A = {1, 2}, B = {3, 4}, íkku A × B þkuÄku. A × BLku fux÷k WÃkøkýku nþu ? íku ík{k{ WÃkøkýkuLke ÞkËe çkLkkðku.

Answer

A × B = {1, 2} × {3, 4} = {(1, 3), (1, 4), (2, 3), (2, 4)}
ynª, n(A × B) = 4 nkuðkÚke íkuLkk WÃkøkýkuLke MktÏÞk 24 = 16 Úkþu.
  • WÃkøkýku : f, A × B, {(1, 3)}, {(1, 4)}, {(2, 3)},
    {(2, 4)}, {(1, 3), (1, 4)}, {(1, 3), (2, 3)}, {(1, 3), (2, 4)}, {(1, 4), (2, 3)}, {(1, 4),
    (2, 4)}, {(2, 3), (2, 4)}, {(1, 3), (1, 4),
    (2, 3)}, {(1, 3), (1, 4), (2, 4)}, {(1, 4),
    (2, 3), (2, 4)}, {(2, 3), (2, 4), (1, 3)}
#15 SUB

Question

òu n(A) = 3, n(B) = 2 nkuÞ íkuðk çku øký A yLku B nkuÞ yLku r¼LLk ½xfku x, y, z {kxu (x, 1), (y, 2),
(
z, 1) yu A × BLkk ½xfku nkuÞ, íkku A yLku B þkuÄku.

Answer

ynª, (x, 1), (y, 2), (z, 1) A × B nkuðkÚke,
A = {x, y, z}, B = {1, 2} ÚkkÞ, fkhý fu n(A) = 3, n(B) = 2 ykÃku÷ Au.
#16 SUB

Question

òu fkíkuorÍÞ økwýkfkh A × ALkk ½xfkuLke MktÏÞk 9 nkuÞ yLku íku{ktLkk çku ½xfku (–1, 0) yLku (0, 1) nkuÞ, íkku A þkuÄku íkÚkk A × ALkk çkkfeLkk ½xfku ÷¾ku.

Answer

ynª, n(A × A) = 9 \ n(A) = 3 ÚkkÞ íkÚkk (–1, 0), (0, 1) A × A nkuðkÚke A = {–1, 0, 1} ÚkkÞ.
{kxu, A × ALkk çkkfeLkk ½xfku (–1, 1), (–1, –1),
(0, –1), (0, 0), (1, –1), (1, 0)
yLku (1, 1) Au.
S match 50% type: 42 Q ⤓ Export ZIP
#17 SUB 🖼 1

Question

òu A = {1, 2, 3, 4, 5, 6}, R = {(x, y) : y = x + 1} ÚkkÞ íku heíku MktçktÄ R Au, su AÚke A Ãkh ÔÞkÏÞkrÞík Au, íkku
(i) yk MktçktÄLku rfhý ykf]rík îkhk Ëþkoðku.
(ii) RLkku «Ëuþ, Mkn«Ëuþ íku{s rðMíkkh {u¤ðku.

Answer

ynª, x, y A = {1, 2, 3, 4, 5, 6} íkÚkk y = x + 1 Au.
x = 1, 2, 3, 4, 5 ÷uíkkt yLke ®f{íkku yLkw¢{u 2, 3, 4, 5, 6 {¤u. R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)}
RLkku «Ëuþ : {1, 2, 3, 4, 5},
rðMíkkh : {2, 3, 4, 5, 6}
Mkn«Ëuþ : {1, 2, 3, 4, 5, 6}
#18 SUB 🖼 1

Question

Lke[uLke ykf]rík{kt PÚke QLkku MktçktÄ Ëþkoðu÷ Au. yk MktçktÄLku (i) økwýÄ{oLke heíku, (ii) ÞkËeLke heíku ÷¾ku. íkuLkku «Ëuþ yLku rðMíkkh þkuÄku.

Answer

økwýÄ{oLke heík : R : {(x, y) : x = y2, x P, y Q}
ÞkËeLke heík : R : {(9, 3), (9, –3), (4, 2), (4, –2),
(25, 5), (25, –5)}
«Ëuþ : {4, 9, 25}, rðMíkkh : {–2, 2, –3, 3, –5, 5}
#19 SUB 🖼 1

Question

òu A = {1, 2}, B = {3, 4}, íkku AÚke BLkk MktçktÄkuLke MktÏÞk þkuÄku.

Answer

A × B = {(1, 3), (1, 4), (2, 3), (2, 4)}.
ynª, n(A) = 2, n(B) = 2 íkÚkk
n(A×B) = n(A) n(B) = 2 × 2 = 4 {kxu MktçktÄkuLke MktÏÞk = 24 = 16 ÚkkÞ.
#20 SUB

Question

A = {1, 2, 3, 4, ..., 14}, R = {(x, y) : 3xy = 0,
x, y A}. òu R yu AÚke ALkku MktçktÄ nkuÞ, íkku RLkku «Ëuþ, Mkn«Ëuþ yLku rðMíkkh {u¤ðku.

Answer

ynª, 3x y = 0, x, y {1, 2, 3, ..., 14}
\ 3x = y
ynª, x = 1, 2, 3, 4 ÷uíkkt, yLke ®f{íkku yLkw¢{u
3, 6, 9, 12 {¤u.
\ R = {(1, 3), (2, 6), (3, 9), (4, 12)} {¤u.
\ RLkku «Ëuþ : {1, 2, 3, 4}
(Ëhuf ¢{Þwõík òuzLkku «Úk{ ½xf)
\ RLkku rðMíkkh : {3, 6, 9, 12}
(Ëhuf ¢{Þwõík òuzLkku çkeòu ½xf)
\ RLkku Mkn«Ëuþ : øký A Ãkkuíku = {1, 2, 3, 4, ..., 14}
#21 SUB

Question

R = {(x, y) : y = x + 5, x yu 4Úke LkkLke «kf]ríkf MktÏÞk Au, x, y N} ÚkkÞ íku heíku yuf MktçktÄ N Ãkh ÔÞkÏÞkrÞík Au. RLku ÞkËeLke heíku ÷¾ku. RLkku «Ëuþ íku{s rðMíkkh {u¤ðku.

Answer

ynª, x = 1, 2, 3 ÷uíkkt, yLke ®f{íkku
y = x + 5 ÃkhÚke yLkw¢{u 6, 7, 8 {¤u.
\ R = {(1, 6), (2, 7), (3, 8)}
\ RLkku «Ëuþ : {1, 2, 3} íkÚkk RLkku rðMíkkh = {6, 7, 8}
#22 SUB

Question

A = {1, 2, 3, 5}, B = {4, 6, 9}, R = {(x, y) : x yLku yLkku íkVkðík yÞwø{ MktÏÞk Au. x A, y B} ÚkkÞ íku heíku MktçktÄ AÚke B Ãkh ÔÞkÏÞkrÞík Au. RLku ÞkËeLke heíku ÷¾ku.

Answer

ynª, x A = {1, 2, 3, 5}, y B = {4, 6, 9} íkÚkk R : x yLku y Lkku íkVkðík yÞwø{ MktÏÞk Au.
R : {(1, 4), (1, 6), (2, 9), (3, 4), (3, 6), (5, 4), (5, 6)}
#23 SUB 🖼 1

Question

Lke[uLke ykf]rík{kt PÚke QLkku MktçktÄ Ëþkoðu÷ Au. yk MktçktÄLku (i) økwýÄ{oLke heíku, (ii) ÞkËeLke heíku ÷¾ku. íkuLkku «Ëuþ yLku rðMíkkh þwt Úkþu ?

Answer

P = {5, 6, 7}, Q = {3, 4, 5}
(i) økwýÄ{oLke heík : R = {(x, y) : x P, y Q,
xy = 2}
(ii) ÞkËeLke heík : R = {(5, 3), (6, 4), (7, 5)}
RLkku «Ëuþ : {5, 6, 7},
RLkku rðMíkkh : {3, 4, 5}
#24 SUB

Question

òu A = {1, 2, 3, 4, 6}, R = {(a, b) : a, b A, b yu a ðzu rð¼kßÞ Au} íku heíku MktçktÄ R yu A Ãkh ÔÞkÏÞkrÞík Au. íkku RLku ÞkËeLke heíku ÷¾ku íkÚkk RLkku «Ëuþ yLku rðMíkkh {u¤ðku.

Answer

ynª, a, b A = {1, 2, 3, 4, 6} íkÚkk R : b yu a ðzu rð¼kßÞ Au.

R = { (1, 1), (1, 2), (1, 3), (1, 4), (1, 6), (2, 2),
(2, 4), (2, 6), (3, 3), (3, 6), (4, 4), (6, 6)}

«Ëuþ : {1, 2, 3, 4, 6}
rðMíkkh : {1, 2, 3, 4, 6}
#25 SUB

Question

R = {(x, x + 5) : x {0, 1, 2, 3, 4, 5}} ÚkkÞ íku heíku ÔÞkÏÞkrÞík MktçktÄLkku «Ëuþ íku{s rðMíkkh {u¤ðku.

Answer

ynª, x {0, 1, 2, 3, 4, 5} íkÚkk (x, x + 5) {kxuLke ¢{Þwõík òuz R : {(0, 5), (1, 6), (2, 7), (3, 8), (4, 9), (5, 10)} Au.
  • \ RLkku «Ëuþ : {0, 1, 2, 3, 4, 5} íkÚkk

RLkku rðMíkkh : {5, 6, 7, 8, 9, 10}

#26 SUB

Question

MktçktÄ R = {(x, x3) : x yu 10 fhíkkt LkkLke yrð¼kßÞ MktÏÞk Au}Lku ÞkËeLkk MðYÃk{kt ÷¾ku.

Answer

ynª, x = {2, 3, 5, 7} Au, {kxu (x, x3) {kxuLke ¢{Þwõík òuzku R : {(2, 8), (3, 27), (5, 125), (7, 343)} ÚkkÞ.
R = {(2, 23), (3, 33), (5, 53), (7, 73)}
#27 SUB

Question

òu A = {x, y, z}; B = {1, 2}, íkku A Úke B Lkk MktçktÄkuLke MktÏÞk þkuÄku.

Answer

ynª, n(A) = 3, n(B) = 2, íkku n(A × B) = n(A) n(B) ÃkhÚke n(A × B) = 6 ÚkkÞ. íkuÚke AÚke BLkk MktçktÄkuLke MktÏÞk = 26 = 64 ÚkkÞ.
#28 SUB 🖼 28

Question

R yu Z Ãkh R = {(a, b) : a, b Z, ab yu ÃkqýkOf Au} îkhk ÔÞkÏÞkrÞík Au. RLkku «Ëuþ yLku rðMíkkh þkuÄku.

Answer

ynª, Z = {..., –2, –1, 0, 1, 2, ...} Au íkÚkk
(a, b) Z {kxu a b yu ÃkqýkOf Au.
RLkku «Ëuþ yLku rðMíkkh çktLku Z Úkþu.
rðÄuÞ yu ÞtºkLke su{ Au, su yLkLÞ ykWxÃkwx Ëhuf RLkÃkwx {kxu ykÃku Au.
ÔÞkÏÞk : yrhõík øký A yLku B {kxu, MktçktÄ f îkhk øký ALkk «íÞuf ½xfLku Mktøkík øký B{kt yLkLÞ «rík®çkçk {¤u Au, íkku yk MktçktÄ fLku AÚke B ÃkhLkwt rðÄuÞ fnu Au. AÚke B ÃkhLkk rðÄuÞLku f : A B ÷¾kÞ Au.
òu a A, f(a) = b, b B íkku bLku f îkhk {¤íkwt aLkwt «rík®çkçk fnuðkÞ Au. aLku f îkhk bLkwt Ãkqðo«rík®çkçk fnuðkÞ Au.
yk{, rðÄuÞ yu rðrþ»x «fkhLkku MktçktÄ Au.
fkuE MktçktÄ yu rðÄuÞ íÞkhu s çkLku ßÞkhu, «ËuþLkku Ëhuf ½xf Mkn«Ëuþ{ktLkk {kºk yLku {kºk yuf s ½xf MkkÚku òuzkÞu÷ nkuÞ yLÞÚkk ykÃku÷ MktçktÄ yu rðÄuÞ çkLku Lknª.
WËknhý íkhefu,
ynª ykf]rík (i){kt Ëþkoðu÷ MktçktÄ R1 yu rðÄuÞ LkÚke, fkhý fu «Ëuþ{ktLkku ½xf 3 yu Mkn«ËuþLkk fkuE Ãký ½xf MkkÚku òuzkÞu÷ LkÚke.
ynª ykf]rík (ii){kt Ëþkoðu÷ MktçktÄ R2 yu rðÄuÞ Au, fkhý fu «Ëuþ{ktLkku Ëhuf ½xf yu Mkn«ËuþLkk yLkLÞ ½xf MkkÚku òuzkÞu÷ Au.
ynª ykf]rík (iii){kt Ëþkoðu÷ MktçktÄ R3 yu rðÄuÞ LkÚke, fkhý fu «Úk{ ½xf 1 yu (Mkn«ËuþLkkt çku) ½xfku MkkÚku òuzkÞu÷ Au.
Äkhku fu, f : A B nkuÞ, íkku øký ALku fLkku «Ëuþ fnu Au íkÚkk øký BLku fLkku Mkn«Ëuþ fnu Au íkÚkk ALkk çkÄk ½xfkuLkk «rík®çkçkLku fLkku rðMíkkh fnu Au.
ynª, MÃk»x Au fu rðMíkkh Mkn«Ëuþ.
òu y-yûkLku Mk{ktíkh Ëkuhu÷ hu¾k ykÃku÷ ð¢Lku yuf fhíkkt ðÄkhu ®çkËwyku{kt AuËu íkku ð¢ yu Võík MktçktÄ Ëþkoðu Au, Ãkhtíkw rðÄuÞ Lk Ëþkoðu.
Ãkhtíkw òu y-yûkLku Mk{ktíkh Ëkuhu÷ hu¾k ykÃku÷ ð¢Lku yuf s ®çkËw{kt AuËu íkku ð¢ yu rðÄuÞ Ëþkoðu Au.
(1) íkËuð rðÄuÞ (Identity Function) :
f : R R, x R {kxu f(x) = x Lku íkËuð rðÄuÞ fnu Au
ykLkku yÚko yu ÚkkÞ fu, su RLkÃkwx nkuÞ íku s ykWxÃkwx {¤u.
ynª, íkËuð rðÄuÞLkku «Ëuþ yLku rðMíkkh R Au.
íkËuð rðÄuÞLkku yk÷u¾ Wøk{®çkËw{ktÚke ÃkMkkh Úkíke hu¾k Ëþkoðu Au.
(2) y[¤ rðÄuÞ (Constant Function) :
f : R R, x R, f(x) = c, (ßÞkt, c yu fkuE y[¤ Au)Lku y[¤ rðÄuÞ fnu Au.
y[¤ rðÄuÞLkku «Ëuþ R yLku íkuLkku rðMíkkh {c} Au.
y[¤ rðÄuÞLkku yk÷u¾ X-yûkLku Mk{ktíkh hu¾k Au.
òu c > 0, íkku hu¾k X-yûkLke WÃkh Au.
òu c = 0, íkku hu¾k X-yûk Ãkh MktÃkkíke nþu.
òu c < 0, íkku hu¾k X-yûkLke Lke[u Au.
(3) çknwÃkËe rðÄuÞ (Polynomial Function) :
f : R R, x R {kxu
f(x) = a0 + a1x + a2x2 + ... + anxn, Lku n ½kíkLkwt çknwÃkËe rðÄuÞ fnu Au. ynª, n yu yLk]ý ÃkqýkOf Au.
a0, a1, a2, ..., an R íkÚkk an 0.
WËk. f (x) = x3x2 + 2, g(x) = x4 + x Au.
h(x) = + 2x LkÚke.
(4) Mkt{uÞ rðÄuÞ (Rational Function) :
g(x) 0 nkuÞ íkuðk «Ëuþ{kt ÔÞkÏÞkrÞík çknwÃkËe rðÄuÞ f(x) yLku g(x) {kxu Lku Mkt{uÞ rðÄuÞ fnu Au.
(5) {kLkktf rðÄuÞ (Modulus Function) :
f(x) = |x| , f : R RÚke ÔÞkÏÞkrÞík rðÄuÞLku {kLkktf rðÄuÞ fnu Au.
f(x) = |x| =
íkuðe s heíku,
|x – 1| = íkÚkk
|x + 2| = ÷¾e þfkÞ.
{kLkktf rðÄuÞLkku yk÷u¾ Y-yûk «íÞu Mktr{ík nkuÞ Au íkÚkk X-yûkLke WÃkh nkuÞ Au íkÚkk f(x) = |x|Lkku yk÷u¾ Wøk{®çkËw{ktÚke ÃkMkkh ÚkkÞ Au.
{kLkktf rðÄuÞLkku «Ëuþ R íkÚkk rðMíkkh [0, ) Au.
(6) r[nTLk rðÄuÞ (Signum Function) :
f : R R {kxu, f(x) =
ðzu ÔÞkÏÞkrÞík rðÄuÞLku r[nTLk rðÄuÞ fnu Au.
«Ëuþ R yLku rðMíkkh {–1, 0, 1} Au.
r[nTLk rðÄuÞ f(x) = ðzu Ãký ËþkoðkÞ Au.
(7) {n¥k{ ÃkqýkOf rðÄuÞ (Greatest Integer Function) :
rðÄuÞ f : R R, f(x) = [x] yÚkðk x,
x R Lku {n¥k{ ÃkqýkOf rðÄuÞ fnu Auu. ßÞkt, [x] yÚkðk
x yu xÚke LkkLkk yÚkðk xLku Mk{kLk nkuÞ íkuðk ík{k{ ÃkqýkOfku{kt MkkiÚke {kuxku ÃkqýkOf Ëþkoðu Au. íkuLku floor function íkhefu Ãký yku¤¾ðk{kt ykðu Au.
{n¥k{ ÃkqýkOf rðÄuÞLkku «Ëuþ R íkÚkk rðMíkkh Z Au.
fkuE Ãký x R Lkku {n¥k{ ÃkqýkOf Lke[u {wsçk þkuÄe þfkÞ Au :
[2.3] = 2, [– 3.2] = – 4, [5.99] = 5
[3] = 3, [0] = 0
fkuE Ãký ÃkqýkOf MktÏÞkLke {n¥k{ ÃkqýkOf íku MktÏÞk Ãkkuíku s ÚkkÞ.
(8) rMk®÷øk rðÄuÞ (Ceiling Function) :
rðÄuÞ f : R R, f(x) = x Lku rMk®÷øk rðÄuÞ fnu Au.
rMk®÷øk rðÄuÞLkku «Ëuþ R yLku rðMíkkh Z Au.
fkuE Ãký x R Lkku rMk®÷øk rðÄuÞ Lke[u {wsçk þkuÄe þfkÞ Au :
2.3 = 3, –3.2 = – 4, 5.99 = 5
3 = 3, 0 = 0
fkuE Ãký ÃkqýkOf MktÏÞkLkku rMk®÷øk rðÄuÞ íku MktÏÞk Ãkkuíku s ÚkkÞ.
(9) ÃkqýkOf ¼køk rðÄuÞ (Fractional Part function) :
rðÄuÞ f : R R, f(x) = {x}
= x – [x]
Lku ÃkqýkOf ¼køk rðÄuÞ fnu Au.
ÃkqýkOf ¼køk rðÄuÞLkku R yLku rðMíkkh [0, 1) Au.
fkuE Ãký x R Lkku ÃkqýkOf ¼køk rðÄuÞ Lke[u {wsçk þkuÄe þfkÞ :
{2.3} = 0.3, {– 3.2} = 0.8
{5.99} = 0.99, {3} = 0
fkuE Ãký ÃkqýkOf MktÏÞkLkku ÃkqýkOf ¼køk rðÄuÞLkwt {qÕÞ ‘þqLÞ’ {¤u.
Äkhku fu, f : D1 R íkÚkk g : D2 R çku ðkMíkrðf rðÄuÞku Au.

(i) (f + g) (x) = f(x) + g(x), x D1 D2

(ii) (fg) (x) = f(x) – g(x), x D1 D2

(iii) (fg) (x) = f(x) g(x), x D1 D2

(iv) (x) = ,

x (D1 D2) – {x : g(x) = 0}

(v) ðkMíkrðf rðÄuÞLkku y[¤ MkkÚku økwýkfkh : òu c yu fkuE y[¤ nkuÞ, íkku (c f) (x) = c f(x) ÚkkÞ.

#29 SUB

Question

N yu «kf]ríkf MktÏÞkykuLkku øký Au yLku íkuLke Ãkh ÔÞkÏÞkrÞík fkuE MktçktÄ R yuðku Au fu, R = {(x, y) :
y = 2x, x, y N}, íkku RLkku «Ëuþ, Mkn«Ëuþ yLku rðMíkkh þkuÄku. þwt yk MktçktÄ rðÄuÞ Au ?

Answer

ynª, x, y N íkÚkk y = 2x Au. {kxu ynª, «Ëuþ N, Mkn«Ëuþ N íkÚkk rðMíkkh yu Þwø{ «kf]ríkf MktÏÞk Úkþu.
ynª, «íÞuf x N {kxu yLkLÞ y N {¤u Au, {kxu ykÃku÷ MktçktÄ yu rðÄuÞ çkLkþu.
#30 SUB 🖼 13

Question

Lke[uLkkt WËknhýku{kt ykÃku÷ MktçktÄ [fkMkku yLku «íÞuf MktçktÄ rðÄuÞ Au fu Lknª íku fkhý ykÃke sýkðku.
(i) R = {(2, 1), (3, 1), (4, 2)}
(ii) R = {(2, 2), (2, 4), (3, 3), (4, 4)}
(iii) R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6), (6, 7)}

Answer

(i) R = {(2, 1), (3, 1), (4, 2)}Lku Lke[u {wsçk Ëþkoðe þfkÞ :

2

3

4

R

1

2

ynª, «Ëuþ{ktLkku Ëhuf ½xf Mkn«ËuþLkk yLkLÞ ½xf MkkÚku òuzkÞu÷ nkuðkÚke ykÃku÷ MktçktÄ yu rðÄuÞ çkLku.
(ii) R = {(2, 2), (2, 4), (3, 3), (4, 4)}

2

3

4

R

2

3

4

ynª, «Ëuþ{ktLkku ½xf 2 yu Mkn«ËuþLkk yuf fhíkkt ðÄw ½xf MkkÚku òuzkÞu÷ nkuðkÚke ykÃku÷ MktçktÄ yu rðÄuÞ çkLku Lknª.
(iii) R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6), (6, 7)}
ynª, «Ëuþ{ktLkku Ëhuf ½xf yu Mkn«ËuþLkk yLkLÞ ½xf MkkÚku òuzkÞu÷ nkuðkÚke ykÃku÷ MktçktÄ yu rðÄuÞ Au.
#31 SUB ▦ 1

Question

N yu «kf]ríkf MktÏÞkykuLkku øký Au f : N N, f(x) = 2x + 1 îkhk ÔÞkÏÞkrÞík ðkMíkrðf rðÄuÞ Au. yk ÔÞkÏÞkLke {ËËÚke Lke[uLkwt fku»xf Ãkqýo fhku :

x

1

2

3

4

5

6

7

y

f(1)=...

f(2)=...

f(3)=...

f(4)=...

f(5)=...

f(6)=...

f(7)=...

Answer

ynª, f(x) = 2x + 1 Au.
f(1) = 2(1) + 1 = 3, f(2) = 2(2) + 1 = 5, f(3) = 2(3) + 1 = 7
f(4) = 2(4) + 1 = 9, f(5) = 2(5) + 1 = 11, f(6) = 2(6) + 1 = 13,
f(7) = 2(7) + 1 = 15

x

1

2

3

4

5

6

7

y

f(1)=3

f(2)=5

f(3)=7

f(4)=9

f(5)=11

f(6)=13

f(7)=15

#32 SUB 🖼 1 ▦ 1

Question

f : R R, y = f(x) = x2, x R Úke ÔÞkÏÞkrÞík yuf rðÄuÞ Au. yk ÔÞkÏÞkLku ykÄkhu Lke[uLkwt fku»xf Ãkqýo fhku. yk rðÄuÞLkku «Ëuþ yLku rðMíkkh þwt Úkþu ? fLkku yk÷u¾ Ëkuhku.

x

– 4

– 3

– 2

– 1

0

1

2

3

4

y=f(x)=x2

Answer

ynª, y = f(x) = x2 Au.
\ f(– 4) = (– 4)2 = 16, f(– 3) = (– 3)2 = 9,
f(– 2) = (– 2)2 = 4, f(– 1) = (– 1)2 = 1,
f(0) = (0)2 = 0, f(1) = (1)2 = 1, f(2) = (2)2 = 4,
f(3) = (3)2 = 9, f(4) = (4)2 = 16 ÚkkÞ.

x

– 4

– 3

– 2

– 1

0

1

2

3

4

y=f(x)=x2

16

9

4

1

0

1

4

9

16

ynª, fLkku «Ëuþ R íkÚkk rðMíkkh [0, ) Úkþu.
y = f(x) = x2Lkku yk÷u¾ Lke[u {wsçk Au: su Ãkhð÷Þ Ëþkoðu Au.
#33 SUB 🖼 1

Question

f : R R, f(x) = x3, x RÚke ÔÞkÏÞkrÞík rðÄuÞLkku yk÷u¾ Ëkuhku.

Answer

f(x) = x3 {kxu,
f(0) = (0)3 = 0, f(1) = (1)3 = 1,
f(– 1) = (– 1)3 = – 1, f(2) = (2)3 = 8,
f(– 2) = (– 2)3 = – 8, f(–3) = –27
f(3) = 27 ðøkuhu ...
#34 SUB 🖼 4 ▦ 1

Question

f : R – {0} R, f(x) = , x R – {0}Úke ÔÞkÏÞkrÞík yuf rðÄuÞ ykÃku÷ Au. yk ÔÞkÏÞkLkk ykÄkhu Lke[uLkwt fku»xf Ãkqýo fhku. yk rðÄuÞLkku «Ëuþ yLku rðMíkkh þwt Úkþu ?

x

– 2

– 1.5

– 1

– 0.5

0.25

0.5

1

1.5

2

y = f(x) =

Answer

y = f(x) = ykÃku÷ Au.

x

– 2

– 1.5

– 1

– 0.5

0.25

0.5

1

1.5

2

y =

– 0.5

–0.67

– 1

– 2

4

2

1

0.67

0.5

fk[ku yk÷u¾ :
#35 SUB 🖼 5

Question

f(x) = x2, g(x) = 2x + 1 yu çku ðkMíkrðf rðÄuÞku nkuÞ, íkku (f + g)(x), (f g)(x), (fg)(x), (x) þkuÄku.

Answer

(f + g)(x) = f(x) + g(x) = x2 + 2x + 1
(fg)(x) = f(x) – g(x) = x2 – (2x + 1) = x2 – 2x – 1
(f g)(x) = f(x) g(x) = x2 (2x + 1) = 2x3 + x2
(x) = , g(x) 0, = , x
#36 SUB 🖼 14

Question

f(x) = , g(x) = x yu çku yLk]ý ðkMíkrðf MktÏÞkLkk øký Ãkh ÔÞkÏÞkrÞík rðÄuÞ nkuÞ, íkku (f + g)(x), (f g) (x), (f g)(x), (x) þkuÄku.

Answer

(f + g)(x) = f(x) + g(x) = + x
(fg)(x) = f(x) – g(x) = x
(f g)(x) = f(x) g(x) = x = x x = x+1 = x
(x) = , g(x) 0, = , x 0,
= x–1 = x, x 0
#37 SUB 🖼 3

Question

Lke[uLkk Ãkife fÞku MktçktÄ rðÄuÞ Au ? fkhý ykÃkku. òu íku rðÄuÞ nkuÞ, íkku íkuLkku «Ëuþ yLku rðMíkkh þkuÄku.
(i) {(2, 1), (5, 1), (8, 1), (11, 1), (14, 1), (17, 1)}

Answer

ynª, ykÃku÷ MktçktÄ yu rðÄuÞ Ëþkoðu Au, fkhý fu «Ëuþ{ktLkku Ëhuf ½xf yu Mkn«ËuþLkk yLkLÞ ½xf MkkÚku òuzkÞu÷ Au.
«Ëuþ : {2, 5, 8, 11, 14, 17}
rðMíkkh : {1}
(ii) {(2, 1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6), (14, 7)}
ynª, ykÃku÷ MktçktÄ yu rðÄuÞ Ëþkoðu Au, fkhý fu «Ëuþ{ktLkku Ëhuf ½xf yu Mkn«ËuþLkk yLkLÞ ½xf MkkÚku òuzkÞu÷ Au.
«Ëuþ : {2, 4, 6, 8, 10, 12, 14}
rðMíkkh : {1, 2, 3, 4, 5, 6, 7}
(iii) {(1, 3), (1, 5), (2, 5)}
ynª, ykÃku÷ MktçktÄ yu rðÄuÞ LkÚke, fkhý fu «Ëuþ{ktLkku ½xf 1 yu Mkn«ËuþLkk yuf fhíkkt ðÄw ½xf MkkÚku òuzkÞu÷ Au.
#38 SUB 🖼 4

Question

Lke[uLkkt ðkMíkrðf rðÄuÞkuLkk «Ëuþ yLku rðMíkkh þkuÄku.
(i) f(x) = – |x|

Answer

ynª, f(x) = |x|{kt xLke fkuE Ãký ðkMíkrðf ®f{ík {qfe þfkíke nkuðkÚke «Ëuþ R Úkþu.
íkÚkk |x|Lkku rðMíkkh [0, ) nkuðkÚke |x|Lkku rðMíkkh
(– , 0] ÚkkÞ.
(ii) f(x) =
«Ëuþ {u¤ððk {kxu,
\ 9 – x2 0
\ 9 > x2
\ x2 < 9
\ < 3
\ |x| < 3
\ – 3 < x < 3
\ «Ëuþ : x [– 3, 3]
rðMíkkh {u¤ððk {kxu,
x [– 3, 3] – 3 < x < 3
0 < x2 < 9
– 9 <x2 < 0
0 < 9 – x2 < 9
0 < < 3
0 < f(x) < 3
\ f(x) [0, 3]
\ f(x) = Lkku rðMíkkh [0, 3] Au.
#39 SUB

Question

f(x) = 2x – 5Úke ÔÞkÏÞkrÞík rðÄuÞ {kxu Lke[uLke ®f{íkku þkuÄku.
(i) f(0)

Answer

2(0) 5 = 0 5 = 5
(ii) f(7)
2(7) 5 = 14 5 = 9
(iii) f(– 3)
2(– 3) 5 = 6 5 = 11
#40 SUB 🖼 11

Question

rðÄuÞ t yu MkuÂÕMkÞMk{kt W»ýíkk{kLk yLku VuhLknex{kt W»ýíkk{kLk ðå[u YÃkktíkh fhíkwt Mkqºk, t(c) = + 32 îkhk ÔÞkÏÞkrÞík Au, íkku Lke[uLkkt {qÕÞku þkuÄku.

Answer

t(c) = + 32
(i) t(0)
(0) + 32 = 0 + 32 = 0
(ii) t(28)
(28) + 32 = + 32 = =
(iii) t(– 10)
(– 10) + 32 = –18 + 32 = 14
(iv) òu t(c) = 212 nkuÞ, íkku c þkuÄku.
\ 212 = c + 32
\ 212 – 32 = c
\ = c
\ c = 100
#41 SUB

Question

Lke[uLkk rðÄuÞkuLkk rðMíkkh þkuÄku.
(i) f(x) = 2 – 3x, x R, x > 0

Answer

heík 1 :
ynª, f(x) = 2 – 3x{kt x > 0Lke y÷øk y÷øk ®f{íkku {qfíkkt,

x

0.01

0.1

0.9

1

2

2.5

4

5

...ðøkuhu

f(x) =

1.97

1.7

– 0.7

– 1

– 4

– 5.5

– 10

– 13

...ðøkuhu

ynª, fku»xf ÃkhÚke òuE þfkÞ Au fu, x > 0 {kxu,
f(x) < 2 \ rðMíkkh : (– , 2) Úkþu.
heík 2 :
ynª, x > 0
\ 3x > 0
\ – 3x < 0
\ 2 – 3x < 0 + 2
\ 2 – 3x < 2
\ 2 – 3x (– , 2)
\ f(x) (– , 2)
(ii) f(x) = x2 + 2, x R
ynª, x2 0
\ x2 + 2 0 + 2
\ x2 + 2 2
\ rðMíkkh : [2, ) Úkþu.
(iii) f(x) = x, x R
ynª, f(x) = x yu íkËuð rðÄuÞ nkuðkÚke íkuLkku rðMíkkh R ÚkkÞ.
#42 SUB 🖼 1

Question

ðkMíkrðf MktÏÞk øký R Ãkh ÔÞkÏÞkrÞík ðkMíkrðf rðÄuÞ f : R R, f(x) = x + 10, íkku rðÄuÞ fLkku yk÷u¾ Ëkuhku.

Answer

f(x) = x + 10, x R ykÃku÷ Au. f(0) = 0 + 10 = 10,
f(1) = 1 + 10 = 11, f(–5) = –5 + 10 = 5, f(5) = 5 + 10 = 15 ðøkuhu ... ÷uíkkt yk÷u¾ Lke[u {wsçk {¤u :
#43 SUB

Question

òu R yu QÚke Q ÃkhLkku
R = {(a, b) : a, b Q yLku a b Z} ÚkkÞ íku heíku ÔÞkÏÞkrÞík MktçktÄ Au, íkku çkíkkðku fu,
(i) «íÞuf a Q {kxu (a, a) R
(ii) òu (a, b) R, íkku (b, a) R
(iii) òu (a, b) R yLku (b, c) R, íkku (a, c) R.

Answer

(i) Äkhku fu, a Q ÷uíkkt, {kxu a a = 0 Z. íkuÚke (a, a) R ÚkkÞ.
(ii) òu (a, b) R, íkku ab Z. íkuÚke ba Z. íkuÚke (b, a) R.
(iii) òu (a, b) R yLku (b, c) R, íkku ab Z.
bc Z. íkuÚke ac = (ab) + (bc) Z. íkuÚke (a, c) R ÚkkÞ.
#44 SUB 🖼 1

Question

f = {(1, 1), (2, 3), (0, – 1), (– 1, – 3)} ÚkkÞ íku heíku
Z Ãkh ÔÞkÏÞkrÞík Mkwhu¾ rðÄuÞ Au, íkku f(x) þkuÄku.

Answer

Äkhku fu, Mkwhu¾ rðÄuÞ f(x) = mx + c ÷uíkkt,
f(x) = mx + c íkÚkk f(1) = 1 f(2) = 3
\ f(1) = m(1) + c \ 3 = m(2) + c
\ 1 = m + c ...(1) \ 3 = 2m + c ...(2)
Ãkrhýk{ (1) yLku (2) ÃkhÚke,
m = 2Lku (1){kt {qfíkkt,
\ m + c = 1
\ c = 1 – 2
\ c = – 1
\ f(x) = mx + c ÃkhÚke,
f(x) = 2x – 1 ÚkkÞ.
#45 SUB 🖼 1

Question

f(x) = nkuÞ, íkku rðÄuÞLkku «Ëuþ þkuÄku.

Answer

ykÃku÷ rðÄuÞ f yu Mkt{uÞ rðÄuÞ Au. {kxu fkuE Ãký Mkt{uÞ rðÄuÞLkku «Ëuþ = R {x | AuË = 0},
ynª, x2 – 5x + 4 = 0 ÷uíkkt,
\ (x – 4) (x – 1) = 0
\ x = 4, x = 1
\ «Ëuþ : R – {1, 4}
#46 SUB 🖼 2

Question

f(x) = Úke ÔÞkÏÞkrÞík rðÄuÞLkku yk÷u¾
Ëkuhku.

Answer

x = 0, íkku f(0) = 1 Au.
x = – 1,
íkku f(– 1) = 1 – (– 1) = 2, x = 2,
íkku f(2) = 2 + 1 = 3, x = 3,
íkku f(3) = 3 + 1 = 4, x = – 2,
íkku f(– 2) = 1 – (– 2) = 3 ÚkkÞ.
#47 SUB 🖼 6

Question

MktçktÄ f yu f(x) = Úke ÔÞkÏÞkrÞík Au
yLku MktçktÄ g yu g(x) = Úke ÔÞkÏÞkrÞík
Au, íkku Mkkrçkík fhku fu, f yu rðÄuÞ Au yLku g yu rðÄuÞ LkÚke.

Answer

ynª, f(x) = ÃkhÚke,
ynª, ykf]rík ÃkhÚke òuE þfkÞ Au fu, «ËuþLkku Ëhuf ½xf Mkn«ËuþLkk yLkLÞ ({kºk yuf s) ½xf MkkÚku òuzkÞu÷ Au.
{kxu f yu rðÄuÞ Au.
g(x) =
ynª, ykf]rík ÃkhÚke òuE þfkÞ Au fu, «ËuþLkku ½xf 2 yu Mkn«ËuþLkk yuf fhíkkt ðÄw ½xf MkkÚku òuzkÞu÷k nkuðkÚke g yu rðÄuÞ LkÚke.
x2 = (2)2 = 4 = g(2)
3(2) = 6 = g(2)
#48 SUB 🖼 4

Question

f(x) = x2, íkku þkuÄku.

Answer

f(1.1) = (1.1)2 = 1.21, f(1) = (1)2 = 1
\ =
=
=
= = 2.1
#49 SUB 🖼 1

Question

rðÄuÞ f(x) = Lkku «Ëuþ þkuÄku.

Answer

f(x) =
«Ëuþ {u¤ððk {kxu,
AuË 0
\ x2 – 8x + 12 0
\ (x – 6) (x – 2) 0
\ x – 6 0 \ x – 2 0
\ x 6 \ x 2
\ x R – {2, 6}
\ «Ëuþ = R – {2, 6}
#50 SUB 🖼 3

Question

f(x) = Úke ÔÞkÏÞkrÞík rðÄuÞLkku «Ëuþ yLku rðMíkkh þkuÄku.

Answer

f(x) =
«Ëuþ {u¤ððk {kxu,
x – 1 > 0
\ x > 1
\ x [1, )
\ «Ëuþ : [1, )
rðMíkkh {u¤ððk {kxu,
x [1, ) x > 1
x – 1 > 0
> 0
f(x) > 0
f(x) [0, )
\ f Lkku rðMíkkh = [0, )
#51 SUB

Question

f(x) = |x – 1|Úke ÔÞkÏÞkrÞík rðÄuÞLkku «Ëuþ yLku rðMíkkh þkuÄku.

Answer

x R x 1 R
\ f(x) Lkku «Ëuþ R Au.
x 1 R |x 1| > 0
f(x) > 0
f(x) [0, )
\ f(x)Lkku rðMíkkh = [0, ) Au.
#52 SUB 🖼 5

Question

òu f = yu RÚke RLkwt rðÄuÞ
nkuÞ, íkku fLkku rðMíkkh þkuÄku.

Answer

f =
ynª, x R x2 > 0
1 + x2 > 1
> 0
nðu, x2 < x2 + 1
\ < 1
\ 0 < < 1
\ 0 < f(x) < 1
\ f(x) [0, 1)
\ f(x)Lkku rðMíkkh [0, 1) Au.
#53 SUB 🖼 5

Question

f, g : R R, f(x) = x + 1, g(x) = 2x – 3Úke ÔÞkÏÞkrÞík rðÄuÞ Au, íkku f + g, fg, þkuÄku.

Answer

(f + g) (x) = f(x) + g(x) = x + 1 + 2x 3 = 3x 2
(fg) (x) = f(x) – g(x) = x + 1 – (2x – 3)
= x + 1 – 2x + 3
= 4 – x
(x) = , g(x) 0, = , x
#54 SUB 🖼 1

Question

òu f : {(1, 1), (2, 3), (0, – 1), (– 1, – 3)} yu f(x) = ax + bÚke ÔÞkÏÞkrÞík MktçktÄ nkuÞ, íkku a yLku b þkuÄku.

Answer

(1, 1) : f(1) = 1 ÃkhÚke,
f(x) = ax + b
\ f(1) = a(1) + b
\ 1 = a + b ...(1)
(2, 3) : f(2) = 3 ÃkhÚke,
f(x) = ax + b
\ f(2) = a(2) + b
\ 3 = 2a + b ...(2)
(1) yLku (2) ÃkhÚke,
a = 2Lku Mk{e. (1){kt {qfíkkt,
\ a + b = 1
\ 2 + b = 1
#55 SUB

Question

R yu NÚke NLkku MktçktÄ Au. R : {(a, b) : a, b N yLku a = b2} ÚkkÞ íku heíku ÔÞkÏÞkrÞík Au, íkku þwt Lke[uLkkt rðÄkLkku MkíÞ Au ?
(i) a N {kxu (a, a) R
(ii) òu (a, b) R, íkku (b, a) R
(iii) òu (a, b) R, (b, c) R, íkku (a, c) R
«íÞuf rðÄkLk{kt ík{khk sðkçkLke MkíÞkÚkoíkk [fkMkku.

Answer

ynª, R = {(a, b) : a, b N, a = b2} ykÃku÷ Au.

(i) Äkhku fu, 3 N {kxu, 3 = (3)2 = 9 su þõÞ LkÚke, {kxu a N {kxu (a, a) R MkíÞ LkÚke.

(ii) a = 9, b = 3 ÷uíkkt, (9, 3) R, fkhý fu 32 = 9 Ãkhtíkw (3, 9) R, fkhý fu 92 3

\ ykÃku÷ rðÄkLk MkíÞ LkÚke.

(iii) a = 16, b = 4 ÷uíkkt, 42 = 16 ÚkkÞ.

\ (16, 4) R ÚkkÞ.

b = 4, c = 2 ÷uíkkt, 22 = 4 ÚkkÞ.

\ (4, 2) R Ãkhtíkw

(a, c) = (16, 2), R, fkhý fu 22 16

\ ykÃku÷ rðÄkLk MkíÞ LkÚke.

#56 SUB 🖼 1

Question

A = {1, 2, 3, 4}, B = {1, 5, 9, 11, 15, 16} yLku
f = {(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)}, íkku þwt Lke[uLkkt rðÄkLkku MkíÞ Au ?
(i) f yu AÚke B ÃkhLkku MktçktÄ Au.
(ii) f yu AÚke B ÃkhLkwt rðÄuÞ Au. ík{khk sðkçkLke MkíÞkÚkoíkk [fkMkku.

Answer

ynª, f yu A × BLkku WÃkøký nkuðkÚke f yu MktçktÄ Au, {kxu rðÄkLk (i) MkíÞ Au.
íkÚkk ykf]rík 5hÚke òuE þfkÞ Au fu, «ËuþLkku ½xf ‘2’ yu Mkn«ËuþLkk çku ½xfku MkkÚku òuzkÞu÷ Au, {kxu f yu rðÄuÞ LkÚke. {kxu rðÄkLk (ii) MkíÞ LkÚke.
#57 SUB

Question

f yu Z × ZLkku WÃkøký Au.
òu f = {ab, a + b) : a, b Z} Úke ÔÞkÏÞkrÞík Au, íkku þwt f yu ZÚke ZLkwt rðÄuÞ Au ? ík{khk sðkçkLke MkíÞkÚkoíkk [fkMkku.

Answer

f = {ab, a + b}, a, b Z
\ f (ab) = a + b
a = 1, b = 2 ÷uíkkt, f(1 × 2) = 1 + 2 \ f(2) = 3 {¤u.
a = –1, b = –2 ÷uíkkt, f(–1 × –2) = 1 2
\ f(2) = – 3 {¤u.
yk{, (2, 3) yLku (2, – 3) çku ¢{Þwõík òuz {¤u.
yk{, 2 yu çku ½xfku MkkÚku òuzkÞu÷ Au.
\ f yu Z Úke Z Lkwt rðÄuÞ LkÚke.
#58 SUB 🖼 1

Question

A = {9, 10, 11, 12, 13}, f : A N, f(n) = nLkku {n¥k{ yrð¼kßÞ yðÞð Au. fLkku rðMíkkh þkuÄku.

Answer

f = A N íkÚkk f(n) = nLkku {n¥k{ yrð¼kßÞ yðÞð
f(9) = 9Lkku {n¥k{ yrð¼kßÞ yðÞð = 3 íkuðe s heíku, f(10) = 5, f (9) = 3, f(11) = 11, f(12) = 3, f(13) = 13 {¤u.
\ fLkku rðMíkkh = {3, 5, 11, 13} ÚkkÞ.
2
«fhý Mkqr[
yøkkWLkk ð»ko{kt ÃkwAkÞu÷k JEE-Main «&LkkuLkwt «fhýðkh rð&÷u»ký
S src: NCERT Exemplar match 85% type: 14 Q ⤓ Export ZIP
#59 FB 1M 🖼 2

Question

f : N N, f (x) = x + 2Lkku rðMíkkh ________ Au.

Answer

{3, 4, 5, 6, ...}
f (1) = 1 + 2 = 3
f (2) = 2 + 2 = 4
f (3) = 3 + 2 = 5
f (4) = 4 + 2 = 6
\ f (x) Lkku rðMíkkh = {3, 4, 5, 6, ...}
#60 FB 1M

Question

f : N N, f (x) = 2xLkku rðMíkkh ________ Au.

Answer

{2, 4, 8, ...}
x N yLku N = {1, 2, 3, 4, ...}
f (x) = 2x
x = 1 ÷uíkkt,
\ f (1) = 21 = 2
x = 2 ÷uíkkt,
\ f (2) = 22 = 4
x = 3 ÷uíkkt,
\ f (3) = 23 = 8
\ rðMíkkh = {2, 4, 8, ...}
#61 FB 1M

Question

f : R R, f (x) = 5Lkku rðMíkkh ________ Au.

Answer

{5}
ynª, x R
\ f (– 2) = 5
f (0) = 5
f (1) = 5
\ f y[¤ rðÄuÞ Au.
\ f Lkku rðMíkkh {5} Au.
#62 FB 1M 🖼 4

Question

òu f : R+ R+, f (x) = x2 + 4 + 3 nkuÞ, íkku f(4) yLku f (16)Lkwt {qÕÞ ________ yLku ________ Au.

Answer

27, 275
f (x) = x2 + 4 + 3
f (4) = (4)2 + 4 + 3
= 16 + 4(2) + 3
f (16) = (16)2 + 4 + 3
= 256 + 4(4) + 3
#63 SUB 🖼 14

Question

òu f : R – {0} R, f (x) = + ax yLku
f = nkuÞ, íkku a þkuÄku.

Answer

3
ynª, f =
\ + =
\ 5 + =
\ = – 5
\ =
\ a = 3
#64 FB 1M

Question

íkËuð rðÄuÞLkku yk÷u¾ ________ nkuÞ Au.

Answer

hu¾k
#65 FB 1M 🖼 3

Question

òu f (x) = , x R íkku,
f (2011) = ________.

Answer

1
f ( x) =
=
=
f (x) = 1
\ f (2011) = 1
#66 FB 1M 🖼 22

Question

òu 2 f (x) – 3f = x2 (Q x 0) íkku f (2) = ________.

Answer

2f (x) 3f = x2 ...(1)
2f – 3f = (x Lke søÞkyu {qfíkkt)
\ 2f – 3f (x) =
\ 4f (x) – 6f = 2x2 ...(1)
\ 6f – 9f (x) = ...(2)
Mk{e. (1) & (2)Lkku Mkhðk¤ku fhíkkt,
\ – 5f (x) = 2x2 +
\ f (x) =
f (2) =
=
=
= =
#67 FB 1M 🖼 2

Question

rðÄuÞ f yu 2f (x) + f (1 – x) = x2, x R þhíkLkwt Ãkk÷Lk fhu, íkku f (x) = ________.

Answer

2f (x) + f (1 x) = x2 ...(1)
2f (1 – x) + f (1 – (1 – x)) = (1 – x)2
(x Lke søÞkyu (1 – x) {qfíkkt)
\ 2f (1 – x) + f (x) = (1 – x)2 ...(2)
Mk{e. (1) yLku (2) ÃkhÚke,
4f (x) + 2f (1 – x) – 2f (1 – x) – f (x) = 2x2 – (1 – x)2
\ 3f (x) = 2x2x2 + 2x – 1
4f (x) – 2f (1 – x) = 2x2
f (x) + 2f (1 – x) = (1 – x)2
3f (x) = 2x2 – (1 – x)2
= x2 + 2x – 1
\ f(x) =
#68 FB 1M

Question

Äkhku fu, f (x) = ax2 + bx + c Au.
òu f (x + 1) – f (x) = 8x + 3
íkku a = ________, b = ________.

Answer

a = 4, b = – 1
f (x) = ax2 + bx + c
f (x + 1) = a(x + 1)2 + b(x + 1) + c
\ f (x + 1) – f (x) = a(x + 1)2 + b(x + 1)
+ cax2bxc
\ 8x + 3 = a(x2 + 2x + 1) + bx + b + cax2bxc
= ax2 + 2ax + a + bx + b + cax2bxc
\ 8x + 3 = 2ax + a + b
\ 2a = 8 a + b = 3
\ a = 4 4 + b = 3
b = – 1
#69 FB 1M 🖼 3

Question

òu f (x) = íkku f (– x) = ________.

Answer

f (x) (Þwø{ rðÄuÞ)
f (x) =
f (– x) =
=
= f (x)
#70 FB 1M 🖼 1

Question

rðÄuÞ f (x) = Lkku «Ëuþ ________ Au.

Answer

R – {– 3, 3}
«Ëuþ {u¤ððk {kxu,
9 – x2 0
\ x2 9
\ x ± 3
\ x R – {– 3, 3}
#71 FB 1M 🖼 3

Question

f yu rðÄuÞ Au, íkku [ f (x) + f (– x)] yu ________ rðÄuÞ Au.

Answer

Þwø{ rðÄuÞ
g(x) = [ f (x) + f (– x)]
g(– x) = [ f (– x) + f (x)] = g(x)
\ g(x) yu Þwø{ rðÄuÞ Au.
#72 FB 1M 🖼 20

Question

òu f(x) = ex íkku, = ________.

Answer

f (– ab)
f (x) = ex
f (– a) = ea, f (b) = eb
= = ea + b
= e– (– ab)
= f (– ab)
S match 100% type: 60 Q ⤓ Export ZIP
#73 MCQ 1M

Question

òu (xy, x + y) = (6, 10), íkku x yLku y þkuÄku.

Options

  1. (A) x = – 8, y = 2
  2. (B) x = 8, y = 2
  3. (C) x = 8, y = – 2
  4. (D) x = – 8, y = – 2

Answer

(B) x = 8, y = 2

(xy, x + y) = (6, 10)
\ xy = 6
x + y = 10
2x = 16
\ x = 8
\ x = 8 Lku xy = 6{kt {qfíkkt,
\ 8 – y = 6
\ y = 2
#74 MCQ 1M

Question

òu A = {2, 4, 5}, B = {7, 8, 9}
íkku n(A × B) = _______.

Options

  1. (A) 6
  2. (B) 9
  3. (C) 3
  4. (D) 0

Answer

(B) 9

A = {2, 4, 5}, B = {7, 8, 9}
ynª, n(A) = 3, n(B) = 3
\ n(A × B) = n(A) × n(B)
= 3 × 3
= 9
#75 MCQ 1M

Question

òu A = {a, b}, B = {c, d}, C = {d, e}, íkku
{(a, c), (a, d), (a, e), (b, c), (b, d), (b, e)} yu ________ çkhkçkh Au.

Options

  1. (A) A (B C)
  2. (B) A (B C)
  3. (C) A × (B C)
  4. (D) A × (B C)

Answer

(C) A × (B C)

ynª, B C = {c, d} {d, e}
= {c, d, e}
A × (B C)
= {a, b} × {c, d, e}
= {(a, c), (a, d), (a, e), (b, c),
(b, d), (b, e)}
#76 MCQ 1M

Question

òu A = {1, 2, 4}, B = {2, 4, 5}, C = {2, 5} íkku,
(A – C) × (B – C) = ________.

Options

  1. (A) {(1, 2), (1, 5), (2, 5)}
  2. (B) {(1, 4)}
  3. (C) (1, 4)
  4. (D) ykÃku÷ Ãkife yuf Ãký Lknª

Answer

(B) {(1, 4)}

A – B = {1, 2, 4} – {2, 4, 5}
= {1}
B – C = {2, 4, 5} – {2, 5}
= {4}
(A – B) × (B – C) = {1} × {4}
= {(1, 4)}
#77 MCQ 1M

Question

òu (1, 3), (2, 5) yLku (3, 3) yu A × BLkk ½xfku nkuÞ yLku òu A × B{kt fw÷ 6 ½xfku Au íkku A × BLkk çkkfeLkk ½xfku {u¤ðku.

Options

  1. (A) (1, 5), (2, 3), (3, 5)
  2. (B) (5, 1), (3, 2), (5, 3)
  3. (C) (1, 5), (2, 3), (5, 3)
  4. (D) ykÃku÷ Ãkife yuf Ãký Lknª

Answer

(A) (1, 5), (2, 3), (3, 5)

ynª, (1, 3), (2, 5) yLku (3, 3)
yu A × BLkk ½xfku Au.
\ A = {1, 2, 3}, B = {3, 5}
\ A × B = {1, 2, 3} × {3, 5}
= {(1, 3), (1, 5), (2, 3),
(2, 5), (3, 3), (3, 5)}
\ A × BLkk çkkfe ½xfku
(1, 5), (2, 3), (3, 5) Au.
#78 MCQ 1M

Question

òu A = {1, 2, 3}, B = {3, 8} íkku,
(A B) × (A B) = ________ .

Options

  1. (A) {(3, 1), (3, 2), (3, 3), (3, 8)}
  2. (B) {(1, 3), (2, 3), (3, 3), (8, 3)}
  3. (C) {(1, 2), (2, 2), (3, 3), (8, 8)}
  4. (D) {(8, 3), (8, 2), (8, 1), (8, 8)}

Answer

(B) {(1, 3), (2, 3), (3, 3), (8, 3)}

A B = {1, 2, 3} {3, 8}
= {1, 2, 3, 8}
A B = {1, 2, 3} {3, 8}
= {3}
(A B) × (B C)
= {1, 2, 3, 8} × {3}
= {(1, 3), (2, 3), (3, 3), (8, 3)}
#79 MCQ 1M

Question

òu A = {2, 3, 5}, B = {2, 5, 6} íkku,
(A B) × (A B) = ________.

Options

  1. (A) {(3, 2), (3, 3), (3, 5)}
  2. (B) {(3, 2), (3, 5), (3, 6)}
  3. (C) {(3, 2), (3, 5)}
  4. (D) ykÃku÷ Ãkife yuf Ãký Lknª

Answer

(C) {(3, 2), (3, 5)}

A = {2, 3, 5}, B = {2, 5, 6}
A – B = {2, 3, 5} – {2, 5, 6}
= {3}
A B = {2, 3, 5} {2, 5, 6}
= {2, 5}
(A – B) × (A B)
= {3} × {2, 5}
= {(3, 2), (3, 5)}
#80 MCQ 1M

Question

òu n(A) = 4, n(B) = 3, n(A × B × C) = 24, íkku,
n(C) = _________.

Options

  1. (A) 228
  2. (B) 1
  3. (C) 12
  4. (D) 2

Answer

(D) 2

n(A × B × C) = 24
\ n(A) n(B) n(C) = 24
\ 4 × 3 × n(C) = 24
\ n(C) = 2
#81 MCQ 1M

Question

òu A = {1, 2, 3, 4}, B = {a, b} yLku f : A B íkku,
A × B = _________.

Options

  1. (A) {(a, 1), (3, b)}
  2. (B) {(a, 2), (4, b)}
  3. (C) {(1, a), (1, b), (2, a), (2, b), (3, a), (3, b), (4, a), (4, b)}
  4. (D) yuf Ãký Lknª

Answer

(C) {(1, a), (1, b), (2, a), (2, b), (3, a), (3, b), (4, a), (4, b)}

A × B = {1, 2, 3, 4} × {a, b}
= {(1, a), (1, b), (2, a),
(2, b), (3, a), (3, b),
(4, a), (4, b)}
#82 MCQ 1M

Question

òu çku øký A yLku B{kt 99 Mkk{kLÞ ½xfku Au. íkku
A × B yLku B × ALkk Mkk{kLÞ ½xfkuLke MktÏÞk {u¤ðku.

Options

  1. (A) 299
  2. (B) 992
  3. (C) 100
  4. (D) 18

Answer

(B) 992

n((A × B) (B × A))
= n((A B) × (B A))
= n(A B) n(B A)
= 99 × 99
= 992
#83 MCQ 1M 🖼 4

Question

òu A = {x | x2 – 5x + 6 = 0}, B = {2, 4},
C = {4, 5}
íkku A × (B C) = _________.

Options

  1. (A) {(2, 4), (3, 4)}
  2. (B) {(4, 2), (4, 3)}
  3. (C) {(2, 4), (3, 4), (4, 4)}
  4. (D) {(2, 2), (3, 3), (4, 4), (5, 5)}

Answer

(A) {(2, 4), (3, 4)}

A = {x | x2 – 5 + 6 = 0}
–3
– 6
– 2
\ x2 – 5 + 6 = 0
\ (x – 3) ( – 2) = 0
\ x = 3, = 2
\ A = {2, 3}
B = {2, 4}, C = {4, 5}
B C = {2, 4} {4, 5}
= {4}
A × (B C) = {2, 3} × {4}
= {(2, 4), (3, 4)}
#84 MCQ 1M

Question

òu A = {1, 2, 3, 4, 5}, B = {2, 3, 6, 7} íkku,
(A × B) (B × A)Lke MkÇÞ MktÏÞk _________ Au.

Options

  1. (A) 18
  2. (B) 6
  3. (C) 4
  4. (D) 0

Answer

(C) 4

ynª, øký A yLku øký BLkk
Mkk{kLÞ MkÇÞ 2 Au.
\ A × B yLku B × ALkk
Mkk{kLÞ MkÇÞ 22 {¤u.
\ n[(A × B) (B × A)] = 22
= 4
n(A × A) = n(A) n(A)
= n n = n2 {kxu A ÃkhLkk
çkÄk MktçktÄkuLke MktÏÞk = 2n2
#85 MCQ 1M

Question

Äkhku fu, n(A) = n, íkku A ÃkhLkk çkÄk MktçktÄkuLke MktÏÞk ________ Au.

Options

  1. (A) 2n
  2. (B) 2(n)!
  3. (C) 2n2
  4. (D) 2

Answer

(A) 2n

#86 MCQ 1M

Question

øký A = {1, 2, 3, 4, 5} ÃkhLkku MktçktÄ
R = {(x, y) : |x2y2| < 16} ðzu ÔÞkÏÞkrÞík Au,
íkku RLku ÞkËeLke heíku ________ ÷¾kÞ Au.

Options

  1. (A) {(1, 1), (2, 1), (3, 1), (4, 1), (2, 3)}
  2. (B) {(2, 2), (3, 2), (4, 2), (2, 4)}
  3. (C) {(3, 3), (3, 4), (5, 4), (4, 3), (3, 1)}
  4. (D) {(3, 3), (3, 4), (5, 4), (4, 3)}

Answer

(D) {(3, 3), (3, 4), (5, 4), (4, 3)}

A = {1, 2, 3, 4, 5} íkÚkk
|x2y2| < 16 ÚkkÞ íkuðe ¢{Þwõík òuz
{(3, 3), (3, 4), (5, 4), (4, 3)} Au.
= 3 × 3
= 9
A ÃkhLkk r¼LLk MktçktÄLke MktÏÞk = 29
#87 MCQ 1M

Question

òu A = {1, 2, 3}, íkku A ÃkhLkk r¼LLk MktçktÄLke MktÏÞk {u¤ðku.

Options

  1. (A) 29
  2. (B) 6
  3. (C) 8
  4. (D) yuf Ãký Lknª

Answer

(A) 29

#88 MCQ 1M

Question

«kf]ríkf MktÏÞk øký Ãkh MktçktÄ R yu {(a, b) | ab = 3} îkhk ÔÞkÏÞkrÞík nkuÞ íkku, R _________ .
rðÄuÞ, rðÄuÞLkk «fkh, «Ëuþ yLku rðMíkkh

Options

  1. (A) {(1, 4), (2, 5), (3, 6), .......}
  2. (B) {(4, 1), (5, 2), (6, 3), .......}
  3. (C) {(1, 3), (2, 6), (3, 9), .......}
  4. (D) ykÃku÷ Ãkife yuf Ãký Lknª

Answer

(B) {(4, 1), (5, 2), (6, 3), .......}

R = {(a, b) | a, b N, ab = 3}
= {(4, 1), (5, 2), (6, 3), (7, 4), ...}
#89 MCQ 1M 🖼 18

Question

f(x) = 2x2 – 1, g(x) = 1 – 3x Mk{kLk çkLku íku heíku x þkuÄku.
18. òu f (x) = 4x3 + 3x2 + 3x + 4, íkku x3 = ____.

Options

  1. (A) f (– x)
  2. (B)
  3. (C)
  4. (D) f (x)

Answer

(A) f (– x)

f(x) = 2x2 – 1,
g(x) = 1 – 3x
\ f(x) = g(x)
\ 2x2 – 1 = 1 – 3x
\ 2x2 + 3x – 2 = 0
\ 2x2 + 4xx – 2 = 0
\ 2x (x + 2) – (x + 2) = 0
+4
– 4
– 4
\ (x + 2) (2x – 1) = 0
\ x = – 2, x =
\ x =
#90 MCQ 1M

Question

òu f : R R {kxu rðÄuÞ f(x) = 2x + |x| heíku ÔÞkÏÞkrÞík nkuÞ íkku, f(2x) + f(– x) – f(x) = _________.

Options

  1. (A) 2x
  2. (B) 2|x|
  3. (C) – 2x
  4. (D) – 2|x|

Answer

(B) 2|x|

ynª, f(2x) = 2(2x) = 2(2x) + |2x|
= f (–x) = 2(– x) + |x|
= 4x + 2|x|
f(– x) = – 2x + |– x|
= – 2x + |x|
f(x) = 2x + |x|
\ f(2x) + f(– x) – f(x)
= 4x + 2|x| – 2x + |x| – 2x – |x|
= 2|x|
#91 MCQ 1M 🖼 5

Question

òu f(x) = cos[π2]x + cos[– π2]x íkku,

Options

  1. (A) f = 2
  2. (B) f(– π) = 2
  3. (C) f(π) = 1
  4. (D) f = – 1

Answer

(D) f = – 1

f(x) = cos[π2]x + cos[– π2]x
= cos9x + cos (– 10x)
f(x) = cos9x + cos 10x
( cos Þwø{ rðÄuÞ Au.)
\ f = cos + cos 5π
= 0 – 1
= – 1
#92 MCQ 1M 🖼 13

Question

f(x) = x3, íkku f(x) + f = ________.

Options

  1. (A) 2x3
  2. (B)
  3. (C) 1
  4. (D) 0

Answer

(D) 0

\ f(x) = x3
\ f =
=
= x3
\ f(x) + f
= x3 + x3
= 0
#93 MCQ 1M 🖼 14

Question

òu f(x) = 1 – nkuÞ, íkku f = ________.

Options

  1. (A)
  2. (B)
  3. (C)
  4. (D)

Answer

(A)

f(x) = 1 –
\ f = = 1 – x
\ f = f(1 – x)
= 1 –
=
=
=
#94 MCQ 1M

Question

çknwÃkËe f(x) = xn + 1 {kxu òu f(12) = 1729,
íkku f(15) = ?

Options

  1. (A) 1001
  2. (B) 2075
  3. (C) 3357
  4. (D) 3376

Answer

(D) 3376

f(x) = xn + 1 íkÚkk f(12) = 1729
\ 1729 = (12)n + 1
\ 1728 = (12)n = (12)3
\ n = 3
f(15) = (15)n + 1
= (15)3 + 1
= 3375 + 1
= 3376
#95 MCQ 1M 🖼 10

Question

òu f(x) yu rî½kík çknwÃkËe nkuÞ, f(0) = 4 íkÚkk
f(x + 3) – f(x) = 3x + 5, íkku íku rî½kík çknwÃkËe ________ Au.

Options

  1. (A) x2 + x + 24
  2. (B) 3x2 + x + 24
  3. (C) (x2 + 2x + 9)
  4. (D) (3x2 + x + 24)

Answer

(D) (3x2 + x + 24)

Äkhku fu, f(x) = ax2 + bx + c ÷uíkkt
íkÚkk f(0) = 4
\ f(0) = a(0)2 + b(0) + c
\ 4 = c íkÚkk
f(x + 3) – f(x) = 3x + 5 Au.
\ a(x + 3)2 + b(x + 3) + c
– (ax2 + bx + c) = 3x + 5
\ a(x2 + 6x + 9) + bx + 3b + c
ax2 bxc = 3x + 5
\ ax2 + 6ax + 9a + bx + 3b
ax2 bx = 3x + 5
\ 6ax + 9a + 3b = 3x + 5
çktLku çkksw xLkk Mknøkwýf yLku y[¤ ÃkËku Mkh¾kðíkkt,
\ 6a = 3 9a + 3b = 5
\ a = \ + 3b = 5
\ 3b = 5 –
\ 3b =
\ b =
\ çknwÃkËe : x2 + x + 4
= (3x2 + x + 24)
#96 MCQ 1M 🖼 1

Question

f : R R, f(x) =
íkku f(– 1) + f(2) + f(4) = ?

Options

  1. (A) – 13
  2. (B) 13
  3. (C) 7
  4. (D) – 7

Answer

(A) – 13

f(– 1) = 5(– 1) = – 5,
f(2) = (2)2 = 4,
f(4) = – 3(4) = – 12
fkhý fu, x 1 {kxu, f(x) = 5x,
1 < x 3 {kxu, f(x) = x2
íkÚkk x > 3 {kxu, f(x) = – 3x
ykÃku÷ Au.
\ f(– 1) + f(2) + f(4)
= – 5 + 4 – 12
= – 13
#97 MCQ 1M 🖼 6

Question

g(x) = sin [π2] x + sin [π2] x, ßÞkt, [ ] yu {n¥k{ ÃkqýkOf rðÄuÞ Ëþkoðu Au, íkku g = ________.

Options

  1. (A) 0
  2. (B) 1
  3. (C) 2
  4. (D) ykÃku÷ Ãkife yuf Ãký Lknª.

Answer

(B) 1

g(x) = sin [π2]x + sin [– π2]x
\ g(x) = sin [9.8596]x
+ sin [– 9.8596]x
= sin 9x + sin (– 10x)
\ g(x) = sin 9xsin 10x
( sin yÞwø{ rðÄuÞ
nkuðkÚke sin(– θ) = – sinθ)
\ g = sin sin
= sin sin 5π
= 1 – 0
= 1
\ u(– 10) = (–10) – 32
= – 18 – 32
= – 50
#98 MCQ 1M 🖼 1

Question

òu u(c) = – 32, íkku u(– 10) = ________.

Options

  1. (A) – 18
  2. (B) – 50
  3. (C) 50
  4. (D) 18

Answer

(C) 50

#99 MCQ 1M 🖼 7

Question

òu f(x) = , íkku = ______, a = – 1.

Options

  1. (A) 0
  2. (B) – 1
  3. (C) 1
  4. (D) Lk {¤u.

Answer

(A) 0

f(x) =
\ = ×
=
a = – 1 ÷uíkkt ...
= 0
#100 MCQ 1M 🖼 3

Question

f(x) = Lkku «Ëuþ yLku rðMíkkh yLkw¢{u ______ yLku _________ ÚkkÞ.

Options

  1. (A) R, [– 1, 1]
  2. (B) R – {3}, {1, – 1}
  3. (C) R+, R
  4. (D) yuf Ãký Lknª

Answer

(B) R – {3}, {1, – 1}

«Ëuþ = R – {3}
x > 3 {kxu, x – 3 > 0
|x – 3| = x – 3
f(x) = = 1
x < 3 {kxu, x – 3 < 0
|x – 3| = – (x – 3)
f(x) =
= – 1
\ rðMíkkh = {– 1, 1}
#101 MCQ 1M

Question

log |x2 – 9|Lkku «Ëuþ _________ Au.

Options

  1. (A) R
  2. (B) R – [– 3, 3]
  3. (C) R – {– 3, 3}
  4. (D) yuf Ãký Lknª

Answer

(C) R – {– 3, 3}

ynª, x = – 3, 3 {kxu, |x2 – 9| = 0
ykÚke, x = – 3 yLku 3 ykøk¤
log |x2 – 9|Lkwt yÂMíkíð LkÚke.
\ «Ëuþ = R – {– 3, 3}
#102 MCQ 1M 🖼 1

Question

f (x) = Lkku «Ëuþ _________ Au.

Options

  1. (A) R – {– 1, – 2}
  2. (B) (– 2, )
  3. (C) R – {– 1, – 2, – 3}
  4. (D) (– 3, ) – {– 1, – 2}

Answer

(D) (– 3, ) – {– 1, – 2}

ynª, x + 3 > 0
\ x > – 3
yLku x2 + 3x + 2 0
\ (x + 2)(x + 1) 0
\ x – 2, x – 1
\ x R – {– 2, – 1}
\ «Ëuþ = (– 3, ) – {– 2, – 1}
#103 MCQ 1M 🖼 1

Question

rðÄuÞ f(x) = Lkku «Ëuþ ________ Au.

Options

  1. (A) R – {1, – 2}
  2. (B) R – {– 1, – 2}
  3. (C) R – {1, 2}
  4. (D) {– 1, – 2}

Answer

(B) R – {– 1, – 2}

rðÄuÞ f íÞkhu s ÔÞkÏÞkrÞík çkLku
ßÞkhu AuË 0.
Äkhku fu, x2 + 3x + 2 = 0 ÷uíkkt,
\ (x + 2) (x + 1) = 0
\ x = – 2, x = – 1
\ «Ëuþ : R – {– 2, – 1}
#104 MCQ 1M 🖼 2

Question

rðÄuÞ h(x) = Lkku «Ëuþ ________ Au.

Options

  1. (A) (– 4, 4)
  2. (B) [– 4, 4)
  3. (C) [– 4, 4]
  4. (D) R – [– 4, 4]

Answer

(C) [– 4, 4]

h(x) = íÞkhu s
ÔÞkÏÞkrÞík ÚkkÞ ßÞkhu
16 – x2 0
\ x2 – 16 0
\ (x – 4) (x + 4) 0
\ x [– 4, 4]
#105 MCQ 1M 🖼 8

Question

ðkMíkrðf rðÄuÞ f(x) = Lkku rðMíkkh ________ Au.

Options

  1. (A) (1, )
  2. (B) (– , 1)
  3. (C) R
  4. (D) [1, )

Answer

(D) [1, )

heík : 1
ax2 + bx + c «fkhLke fkuE Ãký rî½kík çknwÃkËeLkku rðMíkkh Lke[u {wsçk þkuÄe þfkÞ Au :
ßÞkt, D = b2 – 4ac
ynª, x2 + 6x + 10Lku
ax2 + bx + c MkkÚku Mkh¾kðíkkt,
a = 1, b = 6, c = 10 íkÚkk a > 0
\ D = b2 – 4ac
= (36) – 4(1)(10) = – 4
\ rðMíkkh :
=
= [1, )
heík : 2
f(x) =
=
=
x R (x + 3)2 > 0
(x + 3)2 + 1 > 1
> 1
f(x) > 1
\ f(x)Lkku rðMíkkh [1, ) Au.
#106 MCQ 1M

Question

f(x) = [x] – xLkku rðMíkkh ________ Au.

Options

  1. (A) [0, 1]
  2. (B) (–1, 0]
  3. (C) R
  4. (D) (–1, 1)

Answer

(B) (–1, 0]

ynª, f (x) = [x] – x
= – (x – [x])
f (x) = [x] – x
ßÞkt, {x} yu yÃkqýkOf rðÄuÞ Au.
nðu, 0 < {x} < 1
–1 < – {x} < 0
–1 < f (x) < 0
f (x) (– 1, 0]
f (x)Lkku rðMíkkh (–1, 0]
#107 MCQ 1M

Question

f(x) = [x]Lkku rðMíkkh ________ Au.

Options

  1. (A) Z+
  2. (B) N
  3. (C) Z
  4. (D) Z

Answer

(D) Z

f(x) = [x],
{n¥k{ ÃkqýkOf rðÄuÞLkku rðMíkkh
Z (ÃkqýkOf MktÏÞk) nkuÞ Au.
#108 MCQ 1M 🖼 1

Question

f (x) = Lkku «Ëuþ {u¤ðku.

Options

  1. (A) (1, 2)
  2. (B) (– , – 2) (2, )
  3. (C) (– , – 2) (1, )
  4. (D) (– , ) – {– 1, ± 2}

Answer

(B) (– , – 2) (2, )

«Ëuþ {u¤ððk {kxu,
x – 1 0 x2 – 4 > 0
\ x 1 \ x2 > 4
\ | x | > 2
\ x (– , – 2)
(2, )
\ «Ëuþ : (– , – 2) (2, )
#109 MCQ 1M 🖼 4

Question

rðÄuÞ Lkku «Ëuþ _________ Au.

Options

  1. (A) (2, 3)
  2. (B) [2, 3]
  3. (C) [1, 2]
  4. (D) [1, 3]

Answer

(B) [2, 3]

log 0 yLku
> 0
\ 1 > 0
\ 5xx2 6
\ x2 – 5x + 6 0
\ (x – 3)(x – 2) 0
–3
+ 6
– 2
\ x [2, 3]
\ «Ëuþ : [2, 3]
#110 MCQ 1M 🖼 5

Question

rðÄuÞ Lkku «Ëuþ {u¤ðku.

Options

  1. (A) (– 3, 1)
  2. (B) [– 3, 1]
  3. (C) (– 3, 2]
  4. (D) [– 3, 1)

Answer

(C) (– 3, 2]

ynª, 2 – x 0 9 – x2 > 0
\ 2 x \ 9 > x2
\ x 2 \ x2 < 9
\ <
\ |x| < 3
\ – 3 < x < 3
\ x (– 3, 2]
\ «Ëuþ : (– 3, 2]
#111 MCQ 1M 🖼 3

Question

p(x) = Lkku «Ëuþ ________ Au.

Options

  1. (A) (0, 2)
  2. (B) [0, 3]
  3. (C) [0, 2]
  4. (D) R – [0, 2]

Answer

(C) [0, 2]

p(x) = íÞkhu s
ÔÞkÏÞkrÞík ÚkkÞ,
ßÞkhu 2xx2 0
\ x2 – 2x 0
\ x(x – 2) 0
\ x [0, 2] ÚkkÞ.
\ «Ëuþ : [0, 2]
#112 MCQ 1M 🖼 1

Question

rðÄuÞ Lkku «Ëuþ {u¤ðku.

Options

  1. (A) (– 1, 1)
  2. (B) (– 1, 1) – {0}
  3. (C) [– 1, 1]
  4. (D) [– 1, 1] – {0}

Answer

(D) [– 1, 1] – {0}

ynª, 1 + x 0 , 1 – x 0
\ x – 1 , \ 1 x
\ x 1
Ãkhtíkw, x 0
\ x [– 1, 1] – {0}
su ykÃku÷ rðÄuÞLkku «Ëuþ Au.
#113 MCQ 1M 🖼 4

Question

rðÄuÞ f (x) = + + Lkku «Ëuþ {u¤ðku.

Options

  1. (A) [– 4, )
  2. (B) [– 4, 4]
  3. (C) [0, 4]
  4. (D) [0, 1]

Answer

(D) [0, 1]

(i) xx2 0
\ x(1 – x) 0
\ x [0, 1]
(ii) 4 + x 0
\ x – 4
\ x [– 4, )
(iii) 4 – x 0
\ 4 x
\ x 4
\ x (– , 4]
ykÃku÷ ºkýuÞ «ËuþLkku AuËøký ÷uíkkt,
\ «Ëuþ = [– 4, ) [0, 1]
(– , 4]
= [0, 1]
#114 MCQ 1M 🖼 4

Question

rðÄuÞ f (x) = Lkku «Ëuþ çkLke þfu íkuðe ðkMíkrðf MktÏÞkykuLkku MkkiÚke {kuxku øký _________ Au.

Options

  1. (A) (0, 1) (0, )
  2. (B) (– 1, 0) (1, )
  3. (C) (– , – 1) (0, )
  4. (D) (– , 0) [1, )

Answer

(D) (– , 0) [1, )

ynª, 0
\ 0
\ x (– , 0) [1, )
ykÚke, x (– , 0) [1, ) ykÃku÷ yMk{íkkLkwt Mk{kÄkLk fhu Au.
#115 MCQ 1M 🖼 2

Question

rðÄuÞ f (x) = + Lkku «Ëuþ {u¤ðku.

Options

  1. (A) 1 < x <
  2. (B) < x <
  3. (C) < x < – 1
  4. (D) (– , ) – (– 1, 1)

Answer

(D) (– , ) – (– 1, 1)

ynª, x2 – 1 0 x2 + 1 0
\ x2 1 x R {kxu,
\ | x | 1 x2 + 1 0
\ x (– , – 1] [1, )
\ x (– , ) – (– 1, 1)
su ykÃku÷ rðÄuÞLkku «Ëuþ Au.
#116 MCQ 1M 🖼 2

Question

rðÄuÞ f (x) = Lkku «Ëuþ _________ Au.

Options

  1. (A) (– , – 1) (1, )
  2. (B) (– , – 1] (1, )
  3. (C) (– , – 1] [1, )
  4. (D) yuf Ãký Lknª

Answer

(A) (– , – 1) (1, )

ynª, x2 – 1 > 0
\ x2 > 1
\ > 1
\ | x | > 1
\ x > 1, x < – 1
\ x (– , – 1) (1, )
#117 MCQ 1M 🖼 4

Question

rðÄuÞ Lkku «Ëuþ _________ Au.

Options

  1. (A) (– , )
  2. (B) (– , 3 – ) (3 + , )
  3. (C) (– , 1] [5, )
  4. (D) [0, )

Answer

(C) (– , 1] [5, )

log (x2 – 6x + 6) 0
\ x2 – 6x + 6 1
\ x2 – 6x + 5 0
\ (x – 5)(x – 1) 0
\ «Ëuþ : (– , 1] [5, )
#118 MCQ 1M 🖼 9

Question

rðÄuÞ f(x) = Lkku «Ëuþ _________ Au.

Options

  1. (A)
  2. (B) [– 1, 0]
  3. (C) [0, 1]
  4. (D) [– 1, 1]

Answer

(A)

ynª, MÃkü Au fu,
– 1 1
Ãkhtíkw, 2 < ex < 3
\ 3 < ex + 1 < 4
\ < <
\ < f (x) <
\ f (x)Lkku «Ëuþ =
#119 MCQ 1M 🖼 1

Question

rðÄuÞ y = Lkku «Ëuþ _________ Au.

Options

  1. (A) (– , 0)
  2. (B) (– , 0]
  3. (C) (– , – 1)
  4. (D) (– , )

Answer

(A) (– , 0)

ynª, | x | – x > 0
\ | x | > x
x > xx > x
su þõÞ LkÚke. \ 0 > 2x
\ x < 0
\ x (– , 0)
\ «Ëuþ = (– , 0)
#120 MCQ 1M 🖼 10

Question

rðÄuÞ f (x) = Lkku «Ëuþ _________ Au.

Options

  1. (A)
  2. (B)
  3. (C) (– , 1]
  4. (D)

Answer

(D)

ynª, 5x – 3 – 2x2 0
\ 2x2 – 5x + 3 0
\ x2x + 0
\ (x – 1) 0
\ x
\ «Ëuþ :
#121 MCQ 1M 🖼 11

Question

rðÄuÞ f (x) = ,
x RLkku rðMíkkh _________ Au.

Options

  1. (A) [1, )
  2. (B)
  3. (C)
  4. (D)

Answer

(C)

ynª, y = f (x) =
\ y(x2 + x + 1) = x2 + x + 2
\ (y – 1)x2 + (y – 1)x
+ (y – 2) = 0
nðu, x R nkuðkÚke...
D > 0
\ (y – 1)2 – 4(y – 1)(y – 2) > 0
\ (y – 1)[y – 1 – 4(y – 2)] > 0
\ (y – 1)(y – 1 – 4y + 8) > 0
\ (y – 1)(– 3y + 7) > 0
\ – 3(y – 1) > 0
\ (y – 1) > 0
nðu, òu y = 1 nkuÞ, íkku
1 =
x2 + x + 1 = x2 + x + 2
1 = 2
su þõÞ LkÚke.
y 1
1 < y <
\ rðMíkkh =
#122 MCQ 1M 🖼 6

Question

òu rðÄuÞ f : R R, f (x) = Lkku rðMíkkh _________ Au.

Options

  1. (A) R
  2. (B) [0, 1)
  3. (C) R
  4. (D) R × R

Answer

(B) [0, 1)

f (x) =
\ y =
\ yx2 + y = x2
\ y = x2x2y
\ y = x2(1 – y)
\ x2 =
x R x2 0
0
\ y [0, 1)
\ rðMíkkh = [0, 1)
#123 MCQ 1M 🖼 3

Question

f (x) = Lkku rðMíkkh _________ Au.

Options

  1. (A) [5, 9]
  2. (B) (– , 5] [9, )
  3. (C) (5, 9)
  4. (D) yuf Ãký Lknª

Answer

(B) (– , 5] [9, )

ynª, y = f (x) =
\ y(x2 + 2x – 7) = x2 + 34x – 71
\ x2(y – 1) + x(2y – 34)
+ (–7y + 71) = 0
nðu, x R {kxu,
D > 0
\ (2y – 34)2 – 4(y – 1)(71 – 7y) > 0
\ 4y2 – 2(2y)(34) + (34)2
4(71y – 7y2 – 71 + 7y) > 0
\ 4y2 – 136y + 1156 – 312y
+ 28y2 + 284 > 0
\ 32y2 – 448y + 1440 > 0
\ y2 – 14y + 45 > 0
\ (y – 9)(y – 5) > 0
òu y = 1 nkuÞ, íkku
1 =
x2 + 2x – 7 = x2 + 34x – 71
71 – 7 = 32x
64 = 32x
x = 2
su þõÞ Au.
\ x (– , 5] [9, )
#124 MCQ 1M 🖼 4

Question

òu x yu ðkMíkrðf nkuÞ, íkku Mk{efhý Lke ®f{ík _________Lke ðå[u nkuÞ.

Options

  1. (A) 5 yLku 4
  2. (B) 5 yLku – 4
  3. (C) – 5 yLku 4
  4. (D) yuf Ãký Lknª

Answer

(C) – 5 yLku 4

y =
\ x2y + 2xy + 3y = x2 + 14x + 9
\ x2(y – 1) + (2y – 14)x
+ (3y – 9) = 0
nðu, D 0 ( x R)
\ (2y – 14)2 – 4(y – 1)(3y – 9) 0
\ 4(y – 7)2 – 4(y – 1)(3y – 9) 0
\ 4(y2 – 14y + 49 – 3y2 + 9y
+ 3y – 9) 0
\ –2y2 – 2y + 40 0
\ y2 + y – 20 0
\ (y + 5)(y – 4) 0
òu y = 1 nkuÞ, íkku
1 =
x2 + 2x + 3 = x2 + 14x – 9
– 6 = 12x
x =
su þõÞ Au.
\ y [–5, 4]
\ rðMíkkh – 5 yLku 4Lke ðå[u Au.
#125 MCQ 1M 🖼 2

Question

rðÄuÞ f (x) = + Lkku «Ëuþ _________ Au.

Options

  1. (A) (– 3, – 2.5) (– 2.5, – 2)
  2. (B) [– 2, 0) (0, 1)
  3. (C) (0, 1)
  4. (D) yuf Ãký Lknª

Answer

(B) [– 2, 0) (0, 1)

ynª, 1 – x > 0 x + 2 0
\ 1 > x \ x – 2
\ x < 1
\ [– 2, 0) (0, 1)
#126 MCQ 1M 🖼 4

Question

f (x) = log Lkku «Ëuþ _______ Au.

Options

  1. (A) [4, )
  2. (B) (– , 6]
  3. (C) [4, 6]
  4. (D) [6, 4]

Answer

(C) [4, 6]

ynª, + > 0
\ x – 4 0 6 – x 0
\ x 4 \ 6 x
\ x 6
\ «Ëuþ : [4, 6]
#127 MCQ 1M 🖼 1

Question

rðÄuÞ f (x) = log ((log10x)2 – 5 log10 x + 6)Lkku «Ëuþ _________ Au.

Options

  1. (A) (0, 10)2
  2. (B) (103, )
  3. (C) (102, 103)
  4. (D) (0, 102) (103, )

Answer

(D) (0, 102) (103, )

(log10 x)2 – 5 log10 x + 6 > 0
\ (log10 x – 3)(log10 x – 2) > 0
–3
+ 6
– 2
log10 x < 2 log10 x > 3
\ x < 102 x > 103
\ x (0, 102) (103, )
\ «Ëuþ = (0, 102) (103, )
#128 MCQ 1M 🖼 13

Question

òu f : [– 1, 1] R+ {0} Ãkh ÔÞkÏÞkrÞík rðÄuÞ nkuÞ íkÚkk (0, 0) yLku (x, f(x)) rþhku®çkËwðk¤k Mk{çkksw rºkfkuýLkwt ûkuºkV¤ [kuhMk yuf{ nkuÞ,
íkku f(x) = ________.

Options

  1. (A)
  2. (B)
  3. (C)
  4. (D)

Answer

(A)

Äkhku fu, ABC {kxu
A(0, 0) íkÚkk B(x, f(x)) ÷uíkkt,
AB =
AB =
nðu, Mk{çkksw ABCLkwt ûkuºkV¤
= (AB)2
\ = ×
\ 1 = x2 + (f(x))2
\ (f(x))2 = 1 – x2
\ f(x) =
#129 MCQ 1M 🖼 2

Question

f (x) = (tan x5) yu _________ rðÄuÞ Au.

Options

  1. (A) yÞwø{ rðÄuÞ
  2. (B) Þwø{ rðÄuÞ
  3. (C) yÞwø{ fu Þwø{ Ãkife yuf Ãký Lknª
  4. (D) {kLkktf rðÄuÞ

Answer

(A) yÞwø{ rðÄuÞ

f (x) =
= yÞwø{ × Þwø{
= yÞwø{ rðÄuÞ
#130 MCQ 1M 🖼 5

Question

f (x) = yu _________ rðÄuÞ Au.

Options

  1. (A) Þwø{ rðÄuÞ
  2. (B) yÞwø{ rðÄuÞ
  3. (C) yÞwø{ fu Þwø{ Ãkife yuf Ãký Lknª
  4. (D) {kLkktf rðÄuÞ

Answer

(B) yÞwø{ rðÄuÞ

f (x) =
f (– x) =
=
= –
f (– x) = – f (x)
\ f yu yÞwø{ rðÄuÞ Au.
#131 MCQ 1M 🖼 52

Question

f (x) = x yu _________ rðÄuÞ Au.

Options

  1. (A) yÞwø{ rðÄuÞ
  2. (B) Þwø{ rðÄuÞ
  3. (C) yÞwø{ fu Þwø{ Ãkife yuf Ãký Lknª
  4. (D) {kLkktf rðÄuÞ

Answer

(B) Þwø{ rðÄuÞ

f (x) = x
f (– x) = (– x)
= x
= f (x)
\ f (x) yu Þwø{ rðÄuÞ Au.
n(A) = 2
nðu (1, 1, 2) A3
(1, 1, 2) (A × A × A)
1 A, 1 A, 2 A
A = {1, 2}
yÞwø{ rðÄuÞ : òu rðÄuÞ f(–x) = –f(x) nkuÞ, íkku f yu yÞwø{ rðÄuÞ fnu Au.
Þwø{ rðÄuÞ : rðÄuÞ f(–x) = f(x) nkuÞ, íkku f yu Þwø{ rðÄuÞ fnu Au.
(i) f(x) = log (x + ) yu ________ rðÄuÞ Au.
(ii) f(x) = sin 2x yu ________ rðÄuÞ Au.
(iii) f(x) = yu ________ rðÄuÞ Au.
(iv) f(x) = x2 + 1 yu ________ rðÄuÞ Au.
Mk{sqíke :
(i) f(x) = log (x + )
f(–x) = log (–x + )
= log (–x + )
= log ×
= log
= log
= log
= log (x + )–1
= –log (x + )
= –f(x)
f(x) = log (x + ) yu yÞwø{ rðÄuÞ Au.
(ii) f(x) = sin 2x
f(–x) = sin 2(–x)
= –sin 2x
= –f(x)
f(x) = sin 2x yu yÞwø{ rðÄuÞ Au.
(iii) f(x) =
f(–x) =
=
= f(x)
f(x) = yu Þwø{ rðÄuÞ Au.
(iv) f(x) = x2 + 1
f(–x) = (–x)2 + 1
= x2 + 1
= f(x)
f(x) = x2 + 1 yu Þwø{ rðÄuÞ Au.
Lke[u ykÃku÷ rðÄuÞLkku «Ëuþ {u¤ðku.
(i) f(x) = +
(ii) f(x) =
(iii) f(x) =
Mk{sqíke :
(i) f(x) = +
«Ëuþ {u¤ððk {kxu,
x – 4 0 2 – x 0
x 4 x –2
x [4, ) x 2
x (–, 2]
«Ëuþ : x [2, 4]
(ii) f(x) =
«Ëuþ {u¤ððk {kxu,
2 + x 0
x –2
x [–2, ]
«Ëuþ : [–2, ]
(iii) f(x) =
«Ëuþ {u¤ððk {kxu,
x2 – 4 0
(x – 2) (x + 2) 0
«Ëuþ : x (–, –2] [2, ]
òu rðÄuÞLkku «Ëuþ R nkuÞ, íkku Lke[u ykÃku÷ rðÄuÞLkku rðMíkkh {u¤ðku.
(i) f(x) = x2 + 6x + 10
(ii) f(x) = 2 – 2xx2
(iii) f(x) = x2x – 6
Mk{sqíke :
(i) f(x) = x2 + 6x + 10
ytrík{ ÃkË =
=
= 9
f(x) = x2 + 6x + 9 + 1
= (x + 3)2 + 1
x R (x + 3)2 0
(x + 3)2 + 1 1
f(x) 1
f(x) [1, )
rðMíkkh : [1, ]
(ii) f(x) = 2 – 2xx2
ytrík{ ÃkË =
= 1
f(x) = 3 – 1 – 2xx2
= 3 – (x2 + 2x + 1)
= 3 – (x + 1)2
x R (x + 1)2 0
– (x + 1)2 0
3 – (x + 1)2 3
f(x) 3
f(x) (–, 3]
rðMíkkh : (–, 3]
(iii) f(x) = x2x – 6
ytrík{ ÃkË =
=
f(x) = x2x + – 6 –
=
nðu, x R 0
f(x)
f(x) ,
rðMíkkh : ,
Lke[uLke ykf]rík{kt PÚke QLkku MktçktÄ ËþkoÔÞku Au.
yk MktçktÄLku
(i) økwýÄ{oLke heíku ÷¾ku.
(ii) ÞkËeLke heíku ÷¾ku.
(iii) «Ëuþ yLku rðMíkkh þwt Úkþu ?
Mk{sqíke :
(i) R = {(x, y) : xy = 2, x P, y Q}
(ii) R = {(5, 3), (6, 4), (7, 5)}
(iii) «Ëuþ = {5, 6, 7}
rðMíkkh = {3, 4, 5}
òu A = {1, 2, 3, 4, 6}, R = {(a, b) : a, b A, b yu a ðzu rð¼kßÞ Au.} ÚkkÞ íku heíku MktçktÄ R yu A Ãkh ÔÞkÏÞkrÞík Au.
(i) RLku ÞkËeLke heíku ÷¾ku.
(ii) RLkku «Ëuþ {u¤ðku.
(iii) RLkku rðMíkkh {u¤ðku.
Mk{sqíke :
(i) R = {(1, 1), (2, 2), (3, 3), (4, 4), (6, 6), (2, 4),
(2, 6), (3, 6), (1, 3), (1, 4)}
(ii) RLkku «Ëuþ = {1, 2, 3, 4, 6}
(iii) RLkku rðMíkkh = {1, 2, 3, 4, 6}
#132 MCQ 1M

Question

òu n(A3) = 8 yLku (1, 1, 2) A3 íkku A = ______.

Options

  1. (A) {1, 2, 3}
  2. (B) {1, 2}
  3. (C) {1, 3}
  4. (D) {1, 8}

Answer

(B) {1, 2}

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