MCQ test, Set 10

Answer: (B) $-2xy$

$-2xy$: the powers of $x$ and $y$ are $1$ and $1$, non-negative integers, so it is a monomial.

Each of the others has a negative or fractional power.

Answer: (C) $6$

Degrees of the terms: $4, 1, 2, 6, 0$

The greatest is $6$.

Answer: (D) $-3$

Standard form: $-3x^{6} + 8x^{4} - 9x^{3} - 7$

$n = 6$ and $a_n$ is the coefficient of $x^{6}$: $a_n = -3$

Answer: (A) $6 - 9x^{2} + 7x + 4x^{3}$

$6 - 9x^{2} + 7x + 4x^{3}$ has highest power $3$, so it is a cubic polynomial.

Answer: (C) $4\sqrt{3}x - 3x^{-3}$

In $4\sqrt{3}x - 3x^{-3}$, the power $-3$ of $x$ is negative.

So it is not a polynomial. (A coefficient like $\sqrt{2}$ or $\displaystyle \frac{2}{3}$ is allowed.)

Answer: (B) $p = 5, q = 15, r = 45$

$p - 5 = 0 \Rightarrow p = 5$

$q - 3p = 0 \Rightarrow q - 15 = 0 \Rightarrow q = 15$

$r - 3q = 0 \Rightarrow r - 45 = 0 \Rightarrow r = 45$

Answer: (C) $4\sqrt{3}x^2 + 9x + 7\sqrt{3} = 0$

$4\sqrt{3}x^2 + 9x + 7\sqrt{3} = 0$: it has the form $ax^2 + bx + c = 0$ with $a \ne 0$.

The others have a negative or fractional power of $x$, or a degree other than $2$.

Answer: (D) $-8$

$p(-1) = -2(-1)^{2} + 5(-1) - 1$

$= -2 - 5 - 1$

$= -8$

Answer: (B) $\displaystyle \frac{9}{8}$

$\displaystyle -8x + 9 = 0 \Rightarrow -8x = -9 \Rightarrow x = \frac{9}{8}$

Answer: (A) $1, 7$

$x^{2} - 8x + 7 = x^{2} - x - 7x + 7$

$= x(x - 1) - 7(x - 1)$

$= (x - 1)(x - 7)$

$\Rightarrow x = 1, 7$

Answer: (B) $9$

$\Delta = b^2 - 4ac = (-5)^2 - 4(2)(2) = 9$

Answer: (D) $-6x + 11$

$-2x^{4} \div x^{2} = -2x^{2}$; $(-2x^{2}) \times (x^{2} - 2x - 2) = -2x^{4} + 4x^{3} + 4x^{2}$; subtract: $-5x^{3} + 11x^{2} + 2x + 9$

$-5x^{3} \div x^{2} = -5x$; $(-5x) \times (x^{2} - 2x - 2) = -5x^{3} + 10x^{2} + 10x$; subtract: $x^{2} - 8x + 9$

$x^{2} \div x^{2} = 1$; $1 \times (x^{2} - 2x - 2) = x^{2} - 2x - 2$; subtract: $-6x + 11$

Remainder $= -6x + 11$

Answer: (C) $8$

$\displaystyle 2x - 1 = 0 \Rightarrow x = \frac{1}{2}$; remainder $\displaystyle = f\left(\frac{1}{2}\right)$

$= 8$

Answer: (A) $-2$

$f(1) = (1)^{3} + 3(1)^{2} - 2(1) + a$

$= 1 + 3 - 2 + a$

$= a + 2$

$a + 2 = 0 \Rightarrow a = -2$

Answer: (D) $(\sqrt{2}x - 5)(2x - 3\sqrt{2})$

$2\sqrt{2}x^{2} - 16x + 15\sqrt{2} = 2\sqrt{2}x^{2} - 6x - 10x + 15\sqrt{2}$

$= \sqrt{2}x(2x - 3\sqrt{2}) - 5(2x - 3\sqrt{2}) = (\sqrt{2}x - 5)(2x - 3\sqrt{2})$

Answer: (D) $152$

Remainder $= f(-4)$ (remainder theorem)

$f(-4) = (-4)^{3} - 6(-4)^{2} - (-4) + 4$

$= -64 - 96 + 4 + 4$

$= -152$

The remainder is $-152$, so $152$ must be added.

Answer: (A) $a = 3, b = -4$

$f(-3) = 0 \Rightarrow 3a - b = 13$ ... (1)

$f(2) = 0 \Rightarrow 2a + b = 2$ ... (2)

$(1) + (2)$: $5a = 15$

$\Rightarrow a = 3$

Put $a = 3$ in (1): $9 - b = 13 \Rightarrow b = -4$

Answer: (D) $(x + 6 - \sqrt{42})(x + 6 + \sqrt{42})$

$\displaystyle x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-12 \pm \sqrt{144 - 4(1)(-6)}}{2(1)}$

$\displaystyle = \frac{-12 \pm \sqrt{168}}{2}$

$\displaystyle = \frac{-12 \pm 2\sqrt{42}}{2}$

$= -6 \pm \sqrt{42}$

$= -6 + \sqrt{42}, -6 - \sqrt{42}$

So $x^{2} + 12x - 6 = (x - \alpha)(x - \beta) = (x + 6 - \sqrt{42})(x + 6 + \sqrt{42})$

Answer: (B) $(x - 2)(x^{2} - 11x + 30)$

$x^{3} \div x = x^{2}$; $x^{2} \times (x - 2) = x^{3} - 2x^{2}$; subtract: $-11x^{2} + 52x - 60$

$-11x^{2} \div x = -11x$; $(-11x) \times (x - 2) = -11x^{2} + 22x$; subtract: $30x - 60$

$30x \div x = 30$; $30 \times (x - 2) = 30x - 60$; subtract: $0$

Quotient $= x^{2} - 11x + 30$

Answer: (A) $(x - 3)(x + 4)(x + 1)$

$f(-1) = -1 + 2 + 11 - 12 = 0$

Dividing and splitting the quotient gives $f(x) = (x - 3)(x + 4)(x + 1)$.

Answer: (B) $4x^{2} - 5x + 5$

Second row: $0 \quad -4 \quad 5 \quad -5$

Bottom row: $4 \quad -5 \quad 5 \;\big|\; -7$

Quotient $= 4x^{2} - 5x + 5$

Answer: (C) $x^{2} + 3x + 9$

$x^{3} - 27 = (x - 3)(x^{2} + 3x + 9)$

Signs are all $+$; the powers of $3$ go up as the powers of $x$ go down.

Answer: (A) $23$

$n = 45$: $x - a$ is a factor of $x^n - a^n$ (n odd).

So $39 - 16 = 23$ divides it.

Answer: (A) $\displaystyle \frac{3}{2}a^{2} + 6ax + 4x^{2}$

$\displaystyle \frac{1}{2}a^{3} \div \frac{1}{3}a = \frac{3}{2}a^{2}$; $\displaystyle \frac{3}{2}a^{2} \times (\frac{1}{3}a + x) = \frac{1}{2}a^{3} + \frac{3}{2}a^{2}x$; subtract: $\displaystyle 2a^{2}x + \frac{22}{3}ax^{2} + 4x^{3}$

$\displaystyle 2a^{2}x \div \frac{1}{3}a = 6ax$; $\displaystyle 6ax \times (\frac{1}{3}a + x) = 2a^{2}x + 6ax^{2}$; subtract: $\displaystyle \frac{4}{3}ax^{2} + 4x^{3}$

$\displaystyle \frac{4}{3}ax^{2} \div \frac{1}{3}a = 4x^{2}$; $\displaystyle 4x^{2} \times (\frac{1}{3}a + x) = \frac{4}{3}ax^{2} + 4x^{3}$; subtract: $0$

Answer: (C) $-x^{2} + 5x - 1$

Check each remainder with the remainder theorem: $f(-2) = -15$, $f(0) = -1$, $f(1) = 3$.

Only $-x^{2} + 5x - 1$ gives all three.