MCQ test, Set 2

Answer: (D) $\displaystyle -7x^{\frac{1}{2}}y$

The power $\displaystyle \frac{1}{2}$ of $x$ is not a whole number.

So $\displaystyle -7x^{\frac{1}{2}}y$ is not a monomial; the others have only non-negative integer powers.

Answer: (D) $6$

Degrees of the terms: $3, 6, 2, 0$

The greatest is $6$.

Answer: (C) $7$

Standard form: $7x^{4} - 5x^{3} - 9x^{2} + 6$

$n = 4$ and $a_n$ is the coefficient of $x^{4}$: $a_n = 7$

Answer: (B) $-4 + 5x^{3} - 3x$

$-4 + 5x^{3} - 3x$ has highest power $3$, so it is a cubic polynomial.

Answer: (C) $-4x^{3} - 5 + 2x^{-2}$

In $-4x^{3} - 5 + 2x^{-2}$, the power $-2$ of $x$ is negative.

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

Answer: (D) $a = -5, b = -10, c = 30$

$a + 5 = 0 \Rightarrow a = -5$

$b - 2a = 0 \Rightarrow b + 10 = 0 \Rightarrow b = -10$

$c + 3b = 0 \Rightarrow c - 30 = 0 \Rightarrow c = 30$

Answer: (A) $6x^{2} - 6x - 4 = 0$

$6x^{2} - 6x - 4 = 0$: 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: (A) $85$

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

$= 81 - 18 + 18 + 4$

$= 85$

Answer: (B) $5$

$p(4) = (4)^{2} - a(4) + 4$

$= 16 - 4a + 4$

$= -4a + 20$

$-4a + 20 = 0 \Rightarrow a = 5$

Answer: (D) $\displaystyle \frac{5}{4}, \frac{5}{2}$

$8x^{2} - 30x + 25 = 8x^{2} - 10x - 20x + 25$

$= 2x(4x - 5) - 5(4x - 5)$

$= (4x - 5)(2x - 5)$

$\displaystyle \Rightarrow x = \frac{5}{4}, \frac{5}{2}$

Answer: (C) $\displaystyle \frac{5 \pm \sqrt{31}}{2}$

$a = 2, b = -10, c = -3$

$\displaystyle x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{10 \pm \sqrt{100 - 4(2)(-3)}}{2(2)}$

$\displaystyle = \frac{5 + \sqrt{31}}{2}, \frac{5 - \sqrt{31}}{2}$

Answer: (B) $5$

$f(x) = g(x)q(x) + r(x)$: degree of $q(x)$ = degree of $f(x)$ $-$ degree of $g(x)$

$= 8 - 3 = 5$

Answer: (C) $-75$

$\displaystyle 2x + 3 = 0 \Rightarrow x = -\frac{3}{2}$; remainder $\displaystyle = f\left(-\frac{3}{2}\right)$

$= -75$

Answer: (A) $3$

$f(1) = (1)^{3} - 6(1)^{2} + a(1) + 2$

$= 1 - 6 + a + 2$

$= a - 3$

$a - 3 = 0 \Rightarrow a = 3$

Answer: (D) $2$

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

$\displaystyle f\left(-\frac{1}{2}\right) = 24\left(-\frac{1}{2}\right)^{4} - a\left(-\frac{1}{2}\right)^{3} - \left(-\frac{1}{2}\right)^{2} - \left(-\frac{1}{2}\right) - 2$

$\displaystyle = \frac{3}{2} + \frac{1}{8}a - \frac{1}{4} + \frac{1}{2} - 2$

$\displaystyle = \frac{1}{8}a - \frac{1}{4}$

$\displaystyle \frac{1}{8}a - \frac{1}{4} = 0 \Rightarrow a = 2$

Answer: (D) $(\sqrt{3}x - 5)(7x + 2\sqrt{3})$

$7\sqrt{3}x^{2} - 29x - 10\sqrt{3} = 7\sqrt{3}x^{2} + 6x - 35x - 10\sqrt{3}$

$= \sqrt{3}x(7x + 2\sqrt{3}) - 5(7x + 2\sqrt{3}) = (\sqrt{3}x - 5)(7x + 2\sqrt{3})$

Answer: (A) $x - 7$

$-4x^{3} \div x^{2} = -4x$; $(-4x) \times (x^{2} + 2) = -4x^{3} - 8x$; subtract: $x^{2} - x + 9$

$x^{2} \div x^{2} = 1$; $1 \times (x^{2} + 2) = x^{2} + 2$; subtract: $-x + 7$

Remainder $= -x + 7$, so $x - 7$ must be added.

Answer: (B) $a = 5, b = -2$

$f(2) = 0 \Rightarrow 2a + b = 8$ ... (1)

$f(-4) = 0 \Rightarrow 4a - b = 22$ ... (2)

$(1) + (2)$: $6a = 30$

$\Rightarrow a = 5$

Put $a = 5$ in (1): $10 + b = 8 \Rightarrow b = -2$

Answer: (B) $(x - 3 - \sqrt{6})(x - 3 + \sqrt{6})$

$\displaystyle x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{6 \pm \sqrt{36 - 4(1)(3)}}{2(1)}$

$\displaystyle = \frac{6 \pm \sqrt{24}}{2}$

$\displaystyle = \frac{6 \pm 2\sqrt{6}}{2}$

$= 3 \pm \sqrt{6}$

$= 3 + \sqrt{6}, 3 - \sqrt{6}$

So $x^{2} - 6x + 3 = (x - \alpha)(x - \beta) = (x - 3 - \sqrt{6})(x - 3 + \sqrt{6})$

Answer: (A) $(x - 4)(x^{2} - 6x + 5)$

$x^{3} \div x = x^{2}$; $x^{2} \times (x - 4) = x^{3} - 4x^{2}$; subtract: $-6x^{2} + 29x - 20$

$-6x^{2} \div x = -6x$; $(-6x) \times (x - 4) = -6x^{2} + 24x$; subtract: $5x - 20$

$5x \div x = 5$; $5 \times (x - 4) = 5x - 20$; subtract: $0$

Quotient $= x^{2} - 6x + 5$

Answer: (A) $(x - 5)(x + 2)(x + 3)$

$f(-2) = -8 + 38 - 30 = 0$

Dividing and splitting the quotient gives $f(x) = (x - 5)(x + 2)(x + 3)$.

Answer: (C) $12$

Second row: $0 \quad 1 \quad 1 \quad 7$

Bottom row: $1 \quad 1 \quad 7 \;\big|\; 12$

Remainder $= 12$

Answer: (C) $43$

$n = 155$: $x + a$ is a factor of $x^n + a^n$ (n odd).

So $31 + 12 = 43$ divides it.

Answer: (A) $\displaystyle \frac{1}{2}a^{2} - ax + 4x^{2}$

$\displaystyle a^{3} \div 2a = \frac{1}{2}a^{2}$; $\displaystyle \frac{1}{2}a^{2} \times (2a + x) = a^{3} + \frac{1}{2}a^{2}x$; subtract: $-2a^{2}x + 7ax^{2} + 4x^{3}$

$-2a^{2}x \div 2a = -ax$; $(-ax) \times (2a + x) = -2a^{2}x - ax^{2}$; subtract: $8ax^{2} + 4x^{3}$

$8ax^{2} \div 2a = 4x^{2}$; $4x^{2} \times (2a + x) = 8ax^{2} + 4x^{3}$; subtract: $0$

Answer: (B) $-2x^{2} - 6x + 8$

Check each remainder with the remainder theorem: $f(-1) = 12$, $f(1) = 0$, $f(2) = -12$.

Only $-2x^{2} - 6x + 8$ gives all three.