A) \[-\,1\]
B) \[1\]
C) \[2\]
D) \[4\]
Correct Answer: B
Solution :
[b] \[{{x}^{2}}=y+z\] \[\Rightarrow \] \[{{x}^{2}}+x=x+y+z\] \[\Rightarrow \] \[x\,(x+1)=x+y+z\] (i) Similarly, \[y\,(y+1)=x+y+z\] (ii) and \[z\,(z+1)=x+y+z\] (iii) \[\therefore \] \[\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\] \[=\frac{x}{x+y+z}+\frac{y}{x+y+z}+\frac{z}{x+y+z}\] \[=\frac{x+y+z}{x+y+z}=1\] |
You need to login to perform this action.
You will be redirected in
3 sec