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1978-16: Given that a\ast b=ab+a+b and a\circ b=ab+a+b. Find an expression not involving \ast \quad or\quad \circ for (a\ast b)\circ (a\ast c)if a,b,c are real numbers and the operations on the right are ordinary addition and multiplication of numbers
Solution
a\ast b=ab+a+b\\ a\ast c=ac+a+c\\ (a\ast b)\circ (a\ast c)=(a\ast b)+(a\ast c)+1\\ (a\ast b)\circ (a\ast c)=ab+a+b+ac+a+c+1\\ (a\ast b)\circ (a\ast c)=ab+ac+2a+b+c+1
1992-21: If the binary operation \ast is defined by m\ast n=mn+m+n for any real number m and n. Find the identity of the elements under this operation
Solution
operation of an element with an identity element gives that element
\therefore \quad m\ast I=m\\ I\ast n=n\\ using\quad m\ast I=m,\\ m\ast I=mI+m+I\\ equation\quad both\quad equation\quad of\quad m\ast I\\ m=mI+m+I\\ mI+I=0\\ I(m+1)=0\\ I=0
1993-24: A binary operation \ast is defined as a set of real numbers by x\ast y={ x }^{ y } for all real values of x and y. If x\ast 2=x, find the possible values of x
Solution
x\ast y={ x }^{ y }\\ \therefore \quad x\ast 2={ x }^{ 2 }\\ since\quad x\ast 2=x,\\ { x }^{ 2 }=x\\ { x }^{ 2 }-x=0\\ x(x-1)=0\\ x=0\quad or\quad 1\\
1995-22: If a\ast b=+\sqrt { ab } , evaluate 2\ast (12\ast 27)
Solution
a\ast b=+\sqrt { ab } \\ 12\ast 17=+\sqrt { 12\times 27 } \\ 12\ast 17=+\sqrt { 324 } \\ 12\ast 17=18\\ 2\ast (12\ast 27)=2\ast 18\\ 2\ast 18=+\sqrt { 2\times 18 } \\ 2\ast 18=+\sqrt { 36 } \\ 2\ast 18=6\\ \therefore \quad 2\ast (12\ast 27)=6
1997-21: Two binary operations \ast \quad and\quad \bigoplus are defined as m\ast n=mn-n-1 and m\bigoplus n=mn+n-2 for all real numbers m, n. Find the value of 3\bigoplus (4\ast 5)
Solution
m\ast n=mn-n-1\\ m\bigoplus n=mn+n-2\\ solving\quad the\quad bracket\quad first,\\ 4\ast 5=4\times 5-5-1\\ 4\ast 5=14\\ 3\bigoplus (4\ast 5)=3(4\ast 5)+(4\ast 5)-2\\ 3\bigoplus (4\ast 5)=3(14)+(14)-2\\ 3\bigoplus (4\ast 5)=54
1997-22: If x\ast y=x+y-xy, find x when (x\ast 2)+(x\ast 3)=68
Solution
x\ast y=x+y-xy\\ x\ast 2=x+2-2x\\ x\ast 3=x+3-3x\\ (x\ast 2)+(x\ast 3)=68\\ x+2-2x+x+3-3x=68\\ 5-3x=68\\ 3x=-63\\ x=-21
1998-21: The identity element with respect to the multiplication shown in the table is?
| \bigotimes | p | q | r | s |
| p | r | p | r | p |
| q | p | q | r | s |
| r | r | r | r | r |
| s | q | s | r | q |
Solution
Remember, when operated on; an identity element gives the element operated on
from the table p\bigotimes q=p,\quad q\bigotimes q=q,\quad r\bigotimes q=r,\quad s\bigotimes q=s\\ \therefore q\quad is\quad the\quad identity\quad element
1998-22: The binary operation \ast is defined by x\ast y=xy-y-x for all real values of x and y. If x\ast 3=2\ast x, find x
Solution
x\ast y=xy-y-x\\ x\ast 3=3x-3-x\\ 2\ast x=2x-x-2\\ x\ast 3=2\ast x\\ 3x-x-3=2x-x-2\\ 2x-3=x-2\\ x=1
1999-19: A binary operation \ast is defined by a\ast b=ab+a+b for any real number a and b, if the identity element is zero, find the inverse of 2 under this operation.
Solution
The operation of an element on its inverse gives identity element
2\ast { 2 }^{ -1 }=0\\ 2\ast 2^{ -1 }=2{ \times 2 }^{ -1 }+2+{ 2 }^{ -1 }\\ let{ \quad 2 }^{ -1 }\quad be\quad x\\ equating\quad both\quad equations,\\ 2x+2+x=0\\ 3x=-2\\ x=\frac { -2 }{ 3 } \\ \therefore \quad { 2 }^{ -1 }=\frac { -2 }{ 3 }
2000-17: A binary operation \ast is defined by a\ast b={ a }^{ b }, if a\ast 2=2-a, find the possible values of a
Solution
a\ast 2={ a }^{ b }\\ a\ast 2={ a }^{ 2 }\\ a\ast 2=2-a\\ \therefore \quad { a }^{ 2 }=2-a\\ { a }^{ 2 }+a-2=0\\ { a }^{ 2 }+2a-a-2=0\\ ({ a }^{ 2 }+2a)-(a+2)=0\\ a(a+2)-1(a+2)=0\\ (a+2)(a-1)=0\\ a=-2\quad or\quad 1\\
2000-22: Find the inverse of p under the binary operation \ast defined by p\ast q=p+q-pq, where p and q are real numbers and zero is the identity element.
Solution
p\ast q=p+q-pq\\ p\ast { p }^{ -1 }=0\\ p\ast { p }^{ -1 }=p+{ p }^{ -1 }-p\times { p }^{ -1 }\\ equating\quad both\quad equations\\ p+{ p }^{ -1 }-p{ \times p }^{ -1 }=0\\ { p }^{ -1 }(1-p)=-p\\ { p }^{ -1 }=\frac { -p }{ 1-p } \\ { p }^{ -1 }=\frac { p }{ p-1 } \\
2001-9:
| \bigotimes | k | l | m |
| k | l | m | k |
| l | m | k | l |
| m | k | l | m |
The identity element with respect to the multiplication shown in the table is?
Solution
From the table above, k\bigotimes m=k,\quad l\bigotimes m=l,\quad m\bigotimes m=m\\ \therefore \quad m\quad is\quad the\quad identity\quad element
2001-12: The binary operation \ast is defined on the set of real numbers by a\ast b=a+b+1. If the identity element is -1, find the inverse of the element 2 under \ast
Solution
a\ast b=a+b+1\\ let{ \quad 2 }^{ -1 }=x\\ 2\ast { 2 }^{ -1 }=-1\\ 2\ast { 2 }^{ -1 }=2+x+1\\ equating\quad both\quad equations\\ 2+x+1=-1\\ x=-4\\
2002-31: The binary operation \ast is defined on the set of integers p and q by p\ast q=pq+p+q. find 2\ast (3\ast 4)
Solution
p\ast q=pq+p+q\\ 3\ast 4=3\times 4+3+4\\ 3\ast 4=12+7\\ 3\ast 4=19\\ 2\ast (3\ast 4)=2\ast 19\\ 2\ast 19=2\times 19+2+19\\ 2\ast 19=38+21\\ \therefore \quad 2\ast (3\ast 4)=59
2004-21: If the operation \ast on the set of integers is defined by p\ast q=\sqrt { pq }, find the value of 4\ast (8\ast 32)
Solution
p\ast q=\sqrt { pq } \\ 8\ast 32=\sqrt { 8\times 32 } \\ 8\ast 32=\sqrt { 256 } \\ 8\ast 32=16\\ 4\ast (8\ast 32)=4\ast 16\\ 4\ast 16=\sqrt { 4\times 16 } \\ 4\ast 16=\sqrt { 64 } \\ 4\ast 16=\sqrt { 64 } \\ \therefore \quad 4\ast (8\ast 32)=8
2005-21: An operation \ast is defined on the set of real numbers by a\ast b=ab+2(a+b+1). Find the identity element
Solution
a\ast b=ab+2(a+b+1)\\ a\ast I=a\\ a\ast I=aI+2(a+I+1)\\ equating\quad both\quad equations\\ aI+2(a+I+1)=a\\ aI+2a+2I+2=a\\ I(2+a)+2a+2=a\\ I(2+a)=-2-a\\ I=\frac { -(2+a) }{ 2+a } \\ I=-1
2006-12: A binary operation \bigodot defined on the set of real numbers is such that x\bigodot y=\frac { xy }{ 6 } for all x,\quad y\quad \in \quad R. Find the inverse of 20 under this operation when the identity element is 6.
Solution
x\bigodot y=\frac { xy }{ 6 } \\ let\quad the\quad inverse\quad of\quad 20\quad be\quad k\\ 20\bigodot k=6\\ 20\bigodot k=\frac { 20k }{ 6 } \\ equating\quad both\quad equations,\\ \frac { 20k }{ 6 } =6\\ 20k=36\\ k=1\frac { 4 }{ 5 } \\
2006-15: A binary operation \ast on the set of rational numbers is defined as x\ast y=\frac { { x }^{ 2 }-{ y }^{ 2 } }{ 2xy }. Find -5\ast 3
Solution
x\ast y=\frac { { x }^{ 2 }-{ y }^{ 2 } }{ 2xy } \\ -5\ast 3=\frac { { (-5) }^{ 2 }-{ 3 }^{ 2 } }{ 2(-5)\times 3 } \\ -5\ast 3=\frac { 25-9 }{ -30 } \\ -5\ast 3=\frac { 14 }{ -30 } \\ -5\ast 3=\frac { -7 }{ 15 } \\
2007-30: A binary operation \triangle is defined by a\triangle b=a+b+1 for any real numbers a and b. Find the inverse of the real number 7 under the operation \triangle, if the identity element is -1
Solution
a\triangle b=a+b+1\\ let\quad the\quad inverse\quad of\quad 7\quad be\quad x\\ 7\triangle x=7+x+1\\ 7\triangle x=-1\\ equating\quad both\quad equation\\ 7+x+1=-1\\ x+8=-1\\ x=-9
2007-34: A binary operation \bigoplus on real numbers is defined by x\bigoplus y=xy+x+y for any two real numbers x and y. The value of \frac { -3 }{ 4 } \bigoplus 6 is?
Solution
x\bigoplus y=xy+x+y\\ \frac { -3 }{ 4 } \bigoplus 6=(\frac { -3 }{ 4 } )6+(\frac { -3 }{ 4 } )+6\\ \frac { -3 }{ 4 } \bigoplus 6=\frac { -9 }{ 2 } -\frac { 3 }{ 4 } +6\\ \frac { -3 }{ 4 } \bigoplus 6=\frac { 3 }{ 4 }
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Good day team, kindly help me with this problem:
Let * be a binary operation defined by a*b=axaxb-4. Evaluate (1*-3)*2
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