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Question

If a+x=b+y=c+z+1 where a,b,c,x,y,zare non-zero distinct real numbers, then xa+yx+ayb+yy+bzc+yz+c is equals to


A

y(a-b)

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B

0

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C

y(b-a)

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D

y(a-c)

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Solution

The correct option is A

y(a-b)


Step 1. Find The value of xa+yx+ayb+yy+bzc+yz+c:

Given that a,b,c,x,y,z are non zero-distinct real numbers

and a+x=b+y=c+z+1

Now,

xa+yx+ayb+yy+bzc+yz+c

Step 2.Operating (C3→C3-C1)

⇒xa+yayb+ybzc+yc

Step 3. Operating (C2→C2-C3)

⇒xyayybzyc

⇒yx1ay1bz1c

Step 4. Operating R2→R2–R1andR3→R3–R1

⇒yx1ay-x0b-az-x0c-a

=y[x×0-1(y-x)(c-a)-(b-a)(z-x)+a×0]=y[bz-bx-az+ax-(cy-ay-cx+ax)]=y[bz-bx-az-cy+ay+cx]=y[b(z-x)+a(y-z)+c(x-y)]=y[ba-c-1+a(c-b+1)+c(b-a)]=y[ab-bc-b+ac-ab+a+bc-ac]=y(a-b)

Hence, Option ‘A’ is Correct.


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