If a1 and a2are the zeros of the quadratic polynomial p(y)=y2−y−2 then find the polynomial whose zeros are2a1+1 and and 2a2+1.
A
k[y2−4y−5]
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B
k[y2+4y−5]
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C
k[y2+4y+5]
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D
k[y2−4y+5]
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Solution
The correct option is Ak[y2−4y−5] Since a1 and a2 are zeroes of p(y)=y2−y−2 a1+a2=1 and a1×a2=−2
Now polynomial with zeroes 2a1+1 and 2a2+1 is given by y2−[(2a1+1)+(2a2+1)]y+(2a1+1)(2a2+1)=y2−[2(a1+a2)+2]y+4a1a2+2a1+2a2+1=y2−[2×1+2]y+4×(−2)+2(1)+1=y2−4y−8+3=y2−4y−5
Thus, the required polynomial is k[y2−4y−5].