taiwan-gsat 2020 Q4

taiwan-gsat · Other · ast__math-b 8 marks Discriminant and conditions for roots Proving no real roots exist for a given expression

Let $f ( x )$ be a quadratic polynomial function with real coefficients such that $f ( x ) = 0$ has no real roots. Select the correct options.
(1) $f ( 0 ) > 0$
(2) $f ( 1 ) f ( 2 ) > 0$
(3) If $f ( x ) - 1 = 0$ has real roots, then $f ( x ) - 2 = 0$ has real roots
(4) If $f ( x ) - 1 = 0$ has a double root, then $f ( x ) - \frac { 1 } { 2 } = 0$ has no real roots
(5) If $f ( x ) - 1 = 0$ has two distinct real roots, then $f ( x ) - \frac { 1 } { 2 } = 0$ has real roots

Let $f ( x )$ be a quadratic polynomial function with real coefficients such that $f ( x ) = 0$ has no real roots. Select the correct options.\\
(1) $f ( 0 ) > 0$\\
(2) $f ( 1 ) f ( 2 ) > 0$\\
(3) If $f ( x ) - 1 = 0$ has real roots, then $f ( x ) - 2 = 0$ has real roots\\
(4) If $f ( x ) - 1 = 0$ has a double root, then $f ( x ) - \frac { 1 } { 2 } = 0$ has no real roots\\
(5) If $f ( x ) - 1 = 0$ has two distinct real roots, then $f ( x ) - \frac { 1 } { 2 } = 0$ has real roots

Paper Questions

QA QB QC QD QI QII Q1 Q2 Q3 Q4 Q5 Q6