| (i) \[L+N+P\] |
| (ii) \[M+R\] |
| (iii) \[{{M}^{3}}+{{R}^{3}}\] |
A) i-1 ii-1 iii-\[-1\]
B) i-0 ii-\[-1\] iii-0
C) i-0 ii-0 iii-0
D) i\[-1\] ii-1 iii-1
Correct Answer: C
Solution :
Since, 1 and \[-1\] are zeroes of \[L{{x}^{4}}+M{{x}^{3}}+N{{x}^{2}}+Rx+P.\] \[\therefore \] \[L+M+N+R+P=0\] ?..(1) and \[L-M+N-R+P=0\] ?..(2) Adding (1) and (2), we get \[2L+2N+2P=0\,\,\Rightarrow \,\,L+N+P=0\] Subtracting (1) from (2), we get \[-2M-2R=0\] \[\Rightarrow \] \[M+R=0\] Now, \[{{(M+R)}^{3}}=0\] \[(\because \,\,\,M+R=0)\] \[\Rightarrow \] \[{{M}^{3}}+{{R}^{3}}+3MR(M+R)=0\] \[\Rightarrow \]\[{{M}^{3}}+{{R}^{3}}+3MR\times 0=0\] \[[\because \,\,M+R=0]\] \[\Rightarrow \] \[{{M}^{3}}+{{R}^{3}}=0\]
You need to login to perform this action.
You will be redirected in
3 sec
