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Question

Consider a quadratic polynomial f(x) satisfying f(x)1xR , f(2) = 1 and f(3)=3. Then the value of f(5) is equal to-

A
20
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B
10
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C
19
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D
13
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Solution

The correct option is C 19
f(2)=1 f(3)=3
f(x)=ax2+bx+c quadratic polynomial
f(x)1 x R
f(x)=p(xq)2+1[ p(xq)20 for p>0]
f(2)=p(2q)2+1=1 p(2q)2=0 or =2 (p>0)
f(3)=p(3q)2+1=3 p+1=3 p=2
f(x)=2(x2)2+1
f(5)=2.32+1=19

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