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For a, b in R-{0), let f(x) = ax^2 + bx ...

For `a, b in R-{0)`, let `f(x) = ax^2 + bx + a `satisfies `fl(x +7/4) = f(7/4-x) AA x in R`Also the equation f(x) = 7x+a has only one real and distinct solution. The value of (a+b) is equal to

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Let f (x) =ax ^(2) +bx + c,a ne 0, such the f (-1-x)=f (-1+ x) AA x in R. Also given that f (x) =0 has no real roots and 4a + b gt 0. Let p =b-4a, q=2a +b, then pq is: