1.

`a!=p , b!=q,c!=r` and `|(p,b,c),(a,q,c),(a,b,r)|=0` the value of `p/(p-a)+q/(q-b)+r/(r-c)=`A. 3B. 2C. 1D. 0

Answer» Correct Answer - B
We have,
`|(p,b,c),(p + q,q +b,2c),(a,b,r)|= 0`
`rArr |(p,b,c),(a -p,q -b,0),(a -p,0,r -c)| = 0 [("Applying " R_(2) rarr R_(2) - 2R_(1)),(R_(3) rarr R_(3) - R_(1))]`
`rArr p (q -b) (r -c) - (a - p) ( r-c) - (a -p) c (q -b) = 0`
`rArr p(q -b)(r -c) + b (p-a) (r -c) + c (p -a) (q -b) = 0`
`rArr (p)/(p -a) + (b)/(q -b) + (c)/(r -c) = 0 " " [("Dividing throughout by"),((p-q) (q -b) (r -c))]`
`rArr (p)/(p-q) + ((b)/(q -b) + 1) + ((c)/(r -c) + 1) = 2`
`rArr (p)/( p-q) + (q)/(q -b) + (r)/(r -c) = 2`


Discussion

No Comment Found