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(D) Given the function $f(x) = ax^{2} + bx + c$.We are given $f(1) = 3$,$f(-2) = \lambda$,$f(3) = 4$,and $f(0) + f(1) + f(-2) + f(3) = 14$.Substitutin

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$4$

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(D) Given the function $f(x) = ax^{2} + bx + c$.We are given $f(1) = 3$,$f(-2) = \lambda$,$f(3) = 4$,and $f(0) + f(1) + f(-2) + f(3) = 14$.Substituting the known values into the sum equation:$f(0) + 3 + \lambda + 4 = 14$$f(0) + 7 + \lambda = 14$$f(0) = 7 - \lambda$.Since $f(0) = a(0)^{2} + b(0) + c = c$,we have $c = 7 - \lambda$.Now,use the given values to form a system of equations:$f(1) = a + b + c = 3 \implies a + b = 3 - c = 3 - (7 - \lambda) = \lambda - 4$ (Equation $1$)$f(3) = 9a + 3b + c = 4 \implies 9a + 3b = 4 - c = 4 - (7 - \lambda) = \lambda - 3$ (Equation $2$)$f(-2) = 4a - 2b + c = \lambda \implies 4a - 2b = \lambda - c = \lambda - (7 - \lambda) = 2\lambda - 7$ (Equation $3$)From Equation $1$,$b = \lambda - 4 - a$. Substitute this into Equation $2$:$9a + 3(\lambda - 4 - a) = \lambda - 3$$9a + 3\lambda - 12 - 3a = \lambda - 3$$6a = -2\lambda + 9 \implies a = \frac{9 - 2\lambda}{6}$.Substitute $a$ into Equation $3$:$4(\frac{9 - 2\lambda}{6}) - 2b = 2\lambda - 7$$2(\frac{9 - 2\lambda}{3}) - 2b = 2\lambda - 7$$b = \frac{9 - 2\lambda}{3} - (\lambda - \frac{7}{2}) = \frac{18 - 4\lambda - 6\lambda + 21}{6} = \frac{39 - 10\lambda}{6}$.Using $a + b = \lambda - 4$:$\frac{9 - 2\lambda}{6} + \frac{39 - 10\lambda}{6} = \lambda - 4$$48 - 12\lambda = 6\lambda - 24$$18\lambda = 72 \implies \lambda = 4$.

Let $f(x) = ax^{2} + bx + c$ be such that $f(1) = 3$,$f(-2) = \lambda$,and $f(3) = 4$. If $f(0) + f(1) + f(-2) + f(3) = 14$,then $\lambda$ is equal to...

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