, then
A) both \[g(x)\]and \[h(x)\] are divisible by \[(x-1)\]
B) \[h(x)\] is divisible but \[g(x)\] is not divisible by \[x-1\]
C) \[g(x)\]is divisible but \[h(x)\] is not divisible by \[x-1\]
D) None of these
Correct Answer: A
Solution :
| \[f(x)=g({{x}^{3}})+xh({{x}^{3}})\] |
| Let \[{{f}_{1}}(x)=1+x+{{x}^{2}}\] |
| Clearly roots of \[{{f}_{1}}(x)=0\] are \[\omega ,{{\omega }^{2}},\]where \[\omega \]is non-real cube root of unity |
| \[\therefore f(\omega )=0,f({{\omega }^{2}})=0\] |
| \[\Rightarrow \]\[g({{\omega }^{3}})\omega h({{\omega }^{3}})=0\] and \[g({{\omega }^{6}})+{{\omega }^{2}}h({{\omega }^{6}})=0\] | |
| \[\Rightarrow \]\[g(1)+\omega h(1)=0\] | ? (i) |
| \[g(1)+{{\omega }^{2}}h(1)=0\] | ? (ii) |
| Adding (i) and (ii), \[2g(1)+h(1)(\omega +{{\omega }^{2}})=0\] | |
| \[\Rightarrow \]\[h(1)=2g(1)\] | |
| From (i), \[g(1)+2\omega g(1)=0\] | |
| \[\Rightarrow g(1)(1+2\omega )=0\] | |
| \[\Rightarrow g(1)=0\Rightarrow h(1)=0\] | |
| \[\Rightarrow \]\[g(x)\]and \[h(x)\] are divisible by \[(x-1)\] | |
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