A) Zero
B) one
C) Three
D) infinite
Correct Answer: D
Solution :
| Since \[{{a}_{1}}+{{a}_{2}}\cos 2x+{{a}_{3}}{{\sin }^{2}}x=0\] for all \[x\]putting \[x=0\] and \[x=\pi /2,\]\[\operatorname{we}\,get\,{{a}_{1}}+{{a}_{2}}=0..(1)\]\[\operatorname{and}\,{{a}_{1}}-{{a}_{2}}+{{a}_{3}}=0..(2)\] |
| \[\Rightarrow {{a}_{2}}=-{{a}_{2}}\,\operatorname{and}\,{{a}_{3}}=-2{{a}_{1}}\] |
| \[\therefore \]The given equation becomes \[{{a}_{1}}-{{a}_{1}}\cos 2x-2{{a}_{1}}{{\sin }^{2}}x=0,\forall x\]\[\Rightarrow {{a}_{1}}\left( 1-\cos 2x-2{{\sin }^{2}}x \right)=0,\forall x\]\[\Rightarrow {{a}_{1}}\left( 2-{{\sin }^{2}}x-2{{\sin }^{2}}x \right)=0,\forall x\] |
| The above is satisfied for all values of x\[{{a}_{1}}\] |
| Hence, infinite number of triplets\[\left( {{a}_{1}},-{{a}_{2}},-2{{a}_{1}} \right)\]are possible. |
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