‏إظهار الرسائل ذات التسميات mcq. إظهار كافة الرسائل
‏إظهار الرسائل ذات التسميات mcq. إظهار كافة الرسائل

السبت، 2 أكتوبر 2021

MCQ Questions : The Remainder Theorem and The Factor Theorem

  أكتوبر 02, 2021   , , , , ,   No comments   Edit

မှန်သော အဖြေကို ရွေးပေးရန် ဖြစ်ပါသည်။

  1. 1. The degree of a polynomial is the _____
    Explanation

    ကိန်းရှင် $x$ ပါဝင်သော polynomial တစ်ခုတွင် $x$ ၏ အကြီးဆုံးထပ်ညွှန်းကို ၎င်း polynomial ၏ degree (သို့) order ဟု ခေါ်သည်။
  2. When $9 x^{2}-6 x+2$ is divided by $x-3$, the remainder will be
    Explanation

    Let $f(x)=9 x^{2}-6 x+2$.

    When $f(x)$ is divided by $x-3$,

    The remainder = $f(3)=9 (3)^{2}-6 (3)+2=65$
  3. If a polynomial $f(x)$ is divided by a linear divisor $(x-a)$, then the remainder is
    Explanation

    Remainder Theorem ကို မေးထားခြင်း ဖြစ်ပါသည်။
  4. For what value of $k, x+4$ is a factor of the polynomial $x^{2}-x-(2 k-2) ?$
    Explanation

    $x+4$ is a factor of the polynomial $x^{2}-x-(2 k-2)$.

    By factor theorem,

    $(-4)^{2}-(-4)-(2 k-2)=0$

    $16+4-2(k-1)=0$

    $2(k-1)=20$

    $k=11$
  5. If the two roots of the equation $x^{3}-x^{2}-5 x+5=0$ are $\sqrt{5}$ and $-\sqrt{5}$, then the third root is
    Explanation

    The two roots of the equation $x^{3}-x^{2}-5 x+5=0$ are $\sqrt{5}$ and $-\sqrt{5}$.

    Let the third root be $k$.

    $\therefore \quad x^{3}-x^{2}-5 x+5=\left(x-\sqrt{5}\right)\left(x+\sqrt{5}\right)\left(x-k\right)$.

    $\therefore\quad \sqrt{5}\times \left(-\sqrt{5} \right)\times (-k)=5$

    $\quad\quad 5k=5$

    $\quad\quad k=1$
  6. What should be subtracted to the polynomial $x^{2}-16 x+30$, so that $x-15$ is the factor of the resulting polynomial?
    Explanation

    Let $f(x) = x^{2}-16 x+30$.

    When $f(x)$ is divided by $x-15$,

    The remainder = $f(15) = (15)^{2}-16 (15)+30 = 15(15 - 16 + 2) = 15$

    Hence, $15$ should be subtracted.
  7. The quadratic polynomial $a x^{2}+b x+c=0$ where $c=\dfrac{b^{2}}{4 a}$ has _____real roots.
    Explanation

    $a x^{2}+b x+c=0$ where $c=\dfrac{b^{2}}{4 a}$

    $\therefore\quad 4ac = b^2$

    $\therefore\quad b^2-4ac = 0$

    $\therefore\quad a x^{2}+b x+c=0$ has only one root.
  8. The remainder when $x^{4}-3 x^{2}+4 x+1$ is divided by $x^{2}-x-1$ is
    Explanation
  9. If the sum and the product of the roots of $x^{2}+b x+c=0$ are $3$ and $2$ , then
    Explanation

    The sum and the product of the roots of $x^{2}+b x+c=0$ are $3$ and $2$.

    Let the roots be $p$ and $q$.

    $\therefore\quad p+q = 3, \ pq=2$

    $\therefore\quad x^{2}+b x+c=(x-p)(x-q)$

    $\therefore\quad x^{2}+b x+c=x^{2}-(p+q) x+pq$

    $\therefore\quad x^{2}+b x+c=x^{2}-3x+2$

    $\therefore\quad b = -3, \ c = 2$
  10. Given that $f(x)=x^{4}+x^{2}+3 x-1$ and $g(x)=x^{2}-1$. If $f(x)=q(x) g(x)+r(x)$, which of the following is true?

    $\mathbf{I.}\ q(x)=x^2+2 \quad$ $\mathbf{II.}\ r(x)=3 x+1 \quad$ $\mathbf{III.}\ g(x)$ is a factor of $f(x)$
    Explanation



    $f(x)=x^{4}+x^{2}+3 x-1$ and $g(x)=x^{2}-1$

    $\quad\quad f(x)=q(x) g(x)+r(x)$ (given)

    $\therefore\quad f(x)=(x^2+2)(x^2-1)+3x+1$

    $\therefore\quad q(x)=x^2+2, \ r(x) = 3x+1$
  11. If $x-3$ is a factor of $x^{3}-(2+k) x^{2}+7 k$, then $k=$
    Explanation

    $x-3$ is a factor of $x^{3}-(2+k) x^{2}+7 k$.

    $\therefore\quad 3^{3}-(2+k) (3)^{2}+7 k=0$

    $\therefore\quad 27-9\left(2+k\right)+7k=0$

    $\therefore\quad k=\dfrac{9}{2} = 4.5$
  12. If $f(x)=6 x^{3}+13 x^{2}+p x+q$ is exactly divisible $2 x^{2}+7 x-4$, then
    Explanation

    $f(x)=6 x^{3}+13 x^{2}+p x+q$ is exactly divisible $2 x^{2}+7 x-4$

    $2 x^{2}+7 x-4=(2x-1)(x+4)$

    $\therefore\quad 6 \left(\dfrac{1}{2}\right)^{3}+13 \left(\dfrac{1}{2}\right)^{2}+p \left(\dfrac{1}{2}\right)+q=0 \ldots(1)$

    $\quad\quad 6 \left(-4\right)^{3}+13 \left(-4\right)^{2}+p \left(-4\right)+q=0\ldots(2)$

    $\therefore\quad p=-40,\ q=16$

    $\therefore\quad f(x)=6 x^{3}+13 x^{2}-40x+16$

    $\therefore\quad f \left(\dfrac{4}{3}\right)=6 \left(\dfrac{4}{3}\right)^{3}+13 \left(\dfrac{4}{3}\right)^{2}-40\left(\dfrac{4}{3}\right)+16$

    $\hspace{2.7cm} =\dfrac{128}{9}+\dfrac{208}{9}-\dfrac{160}{3}+16$

    $\hspace{2.7cm} =0$

    $\therefore\quad 3 x-4$ is a factor of $f(x)$.
  13. If $x^{2}+2 x-3$ is a factor of $f(x)=x^{4}+2 x^{3}-7 x^{2}+a x+b$, then
    Explanation

    $x^{2}+2 x-3$ is a factor of $f(x)=x^{4}+2 x^{3}-7 x^{2}+a x+b$

    Let $f(x) = g(x)(x^{2}+2 x-3)$

    $\therefore x^{4}+2 x^{3}-7 x^{2}+a x+b = g(x)(x+3)(x-1)$

    When $x=-3,\ (-3)^{4}+2 (-3)^{3}-7 (-3)^{2}+a (-3)+b = 0$

    When $x=1,\ (1)^{4}+2 (1)^{3}-7 (1)^{2}+a (1)+b = 0$

    $\therefore\quad 3a - b =-36 \ \ldots(1)$

    $\quad\quad a + b = 4 \ \ \quad\ldots(2)$

    $\therefore\quad a= -8, b=12$
  14. If $x^{2}+a x+b$ is divided by $x+c$, then remainder is
    Explanation

    $x^{2}+a x+b$ is divided by $x+c$, then

    the remainder $= (-c)^{2}+a (-c) +b = c^2-ac+b$
  15. If $x^{3}+9 x+5$ is divided by $x$, then the remainder is
    Explanation

    $x^{3}+9 x+5$ is divided by $x$, then

    the remainder $= (0)^{3}+9 (0)+5 = 5$
  16. If $(x-1)$ is a factor of $a x-a$, then the value of $a$ is
    Explanation

    Let $f(x) = ax - a$

    $f(1) = a - a = 0$

    $\therefore\quad x-1$ is a factor of $f(x)$ for all value of $a\in \mathbb{R}$.
  17. What is the remainder when $f(x)=x^{3}+3 x^{2}+k x+k$ is divided by $p x$ ?
    Explanation

    When $f(x)=x^{3}+3 x^{2}+k x+k$ is divided by $p x$,

    The remainder = $f(0) = k$
  18. If $x^{2}-3 x+2$ is a factor of the polynomial $f(x)$, then $f(2)=$
    Explanation

    $x^{2}-3 x+2$ is a factor of the polynomial $f(x)$.

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

    $\therefore\quad (x-1)$ and (x-2) are factors $f(x)$.

    $\therefore\quad f(2)=0$
  19. Given that $f(x)=q(x)\left(x^{2}-4\right)+3$. What is the remainder when $f(x)$ is divided by $x-2$ ?
    Explanation

    $f(x)=q(x)\left(x^{2}-4\right)+3$.

    When $f(x)$ is divided by $x-2$,

    The remainder $f(2)=q(2)\left(2^{2}-4\right)+3 =3$
  20. If $n$ is a positive integer, what is the remainder when $5 x^{2 n+1}+10 x^{2 n}-3 x^{2 n-1}+5$ is divided by $x+1$.
    Explanation

    Let $f(x)=5 x^{2 n+1}+10 x^{2 n}-3 x^{2 n-1}+5$ where $n$ is a positive integer.

    When $f(x)$ is divided $x + 1$,

    The remainder $= f(-1)$.

    $\hspace{2.5cm} =5 (-1)^{2 n+1}+10 (-1)^{2 n}-3 (-1)^{2 n-1}+5$

    $\hspace{2.5cm} =5 (-1)^{\text{odd integer}}+10 (-1)^{\text{even integer}}-3 (-1)^{\text{odd integer}}+5$ where $n\in \mathbb{J^+}$

    $\hspace{2.5cm} =5 (-1)+10 (1)-3 (-1) + 5$

    $\hspace{2.5cm} =13$

    21. Your Score:

الأربعاء، 4 أغسطس 2021

Chapter 4 : Functions - Multiple Choice Questions

  أغسطس 04, 2021   , , , , , ,   No comments   Edit

မှန်သော အဖြေကို ရွေးပေးရန် ဖြစ်ပါသည်။

  1. If $A=\{a, b, c\}$, then $n(A \times A)=$
    Explanation
    $\begin{aligned} A &=\{a, b, c\}\\\\ A \times A &=\{(a, a),(a, b),(a, c),\\\\ &\quad\ (b, a),(b, b),(b, c),\\\\ &\quad\ (c, a),(c, b),(c, c)\}\\\\ \therefore\quad n(A \times A)&=9\\\\ \text{Note that}\ & n(A \times A)=[n(A)]^{2} \end{aligned}$
  2. Given that $A=\{a\}$, then $A \times A$=
    Explanation
    $\begin{array}{l} A=\{a\} \\\\ A \times A=\{a\} \times\{a\} \\\\ \hspace{1.3cm}=\{(a, a)\} \end{array}$

    product sets ၏ အစု၀င်များကို orderd pair $(x,y)$ ပုံစံဖြင့်ရေးရသည်။
  3. Given that $f(x)=x^{2}+3 x+1 .$ If $f(a)=\dfrac{31}{4}$ where $a>0$, then what is the value of $a$ ?
    Explanation
    $\begin{array}{l} f(x)=x^{2}+3 x+1 \\\\ f(a)=\dfrac{31}{4} \\\\ a^{2}+2 a+1=\dfrac{31}{4} \\\\ 4 a^{2}+12 a-27=0\\\\ (2 a+9)(2 a-3)=0\\\\ a=-\dfrac{9}{2}\ \text{(or)}\ a=\dfrac{3}{2}\\\\ \end{array}$
    Since $a>0$, the correct solution is $a=\dfrac{3}{2} .$
  4. What is the domain of $f(x)=\dfrac{1}{x^{2}-4}$.
    Explanation
    $f(x)$ is not defined when
    $x^{2}-4=0$
    $x^{2}=4$
    $x=\pm 2$
    $\therefore$ dom $(f)=\mathbb{R} \smallsetminus\{-2,2\}$.
  5. Given that $A=\{x \mid x>0, x \in \mathbb{R}\}$ and function $f: A \rightarrow \mathbb{R}$ and $g: A \rightarrow \mathbb{R}$ ane defined as $f(x)=x-2$ and $g(x)=\dfrac{x^{2}-4}{x+2}$. Which of the following is(are) true?
    I. $f(2)=g(2)\quad$ II. $f=g\quad$ III. $f \ne g$
    Explanation
    $\operatorname{dom}(f)=\operatorname{dom}(g)=A=\{x \mid x>0, x \in \mathbb{R}\}$
    $\therefore \operatorname{dom}(f)$ and $\operatorname{dom}(g)$ are the set of positive real nembers.
    $f(x)=x-2$ and $g(x)=\dfrac{x^{2}-4}{x+2}=\dfrac{(x-2)(x+2)}{x+2}=x-2$ when $x \ne-2$
    Since $-2 \notin A$, we can say $f(x)=g(x)$ for all $x \in A$
  6. The graph of the function $y=a x^{2}+b x+c$ when $a=0$ is
    Explanation
    Generally $y=a x^{2}+b x+c$ is a quadratic function.
    But when $a=0, y=b x+c$ is a linear function and the graph is a straight line.
  7. What is the equation of horizontal asymptote of the curve $y=\dfrac{3}{x-1}+2 .$
    Explanation
    We have known that the graph $y=\dfrac{k}{x-p}+q$ has horizontal asymptote $y=q$ and vertical asymptote $x=p$.
    $\therefore$ The horizontal arympatote of $y=\dfrac{3}{x-1}+2$ is $y=2$.
  8. The furction $f(x)=\dfrac{3}{x-1}+2$ is not defined when
    Explanation
    A rational function is not defined when its denominator is equal to zero.
    $\therefore f(x)=\dfrac{3}{x-1}+2$ is not defined when
    $x-1=0 \text { or } x=1$,
  9. The vertical asymptote of the graph of function $y=\dfrac{-3 x+4}{x-2}$ is
    Explanation
    We have known that the graph $y=\dfrac{k}{x-p}+q$ has horizontal asymptote $y=q$ and vertical asymptote $x=p$.
    $y=-\dfrac{3 x+4}{x-2}=-\dfrac{2}{x-2}-3$
    $\therefore$ The vertical asymptote of $y=-\dfrac{3 x+4}{x-2}$ is $x=2$.
  10. Which of the following is one to one?
    Explanation
    See: Definition of one to one furction Chapter (4), Section $(4.3 .2)$
  11. If $f^{-1}(x)=\dfrac{x-3}{2}$, then $f(x)=\ldots$
    Explanation
    $f^{-1}(x)=\dfrac{x-3}{2}$
    Let $f(x)=y$, then
    $f^{-1}(y)=x$
    $\dfrac{y-3}{2}=x$
    $y=2 x+3$
    $\therefore f(x)=2 x+3$
  12. Given that $f(x)=\dfrac{4}{2-3 x}$, then the domain of $f^{-1}$ is
    Explanation
    $f(x)=\dfrac{4}{2-3 x}$
    If $ f^{-1}(x)=y$ then
    $f(y)=x$
    $\dfrac{4}{2-3 y}=x$
    $2-3 y=\dfrac{4}{x}$
    $3 y=-\dfrac{4}{x}+2$
    $y=\dfrac{2 x-4}{x}$
    $f^{-1}(x)=\dfrac{2 x-4}{3 x}$
    $\therefore f^{-1}$ exists when $x \ne 0 .$
  13. If $f(x)=\dfrac{3 x-1}{2 x+1}$, $f^{-1}(1)=\ldots$
    Explanation
    $f(x)=\dfrac{3 x-1}{2 x+1}$
    $f^{-1}(1)=a$
    $f(a)=1$
    $\dfrac{3 a-1}{2 a+1}=1$
    $3a-1=2 a+1$
    $a=2$
    $\therefore\ f^{-1}(1)=2$
  14. The function $f$ is given by $f(x)=10^{x}-2$, then $f^{-1}(2)=\cdots$
    Explanation
    $f(x)=10^{x}-2$
    Let $f^{-1}(2)=a$
    $f(a)=2$
    $10^{a}-2=2$
    $10^{a}=4$
    $a=\log _{10} 4$
    $\therefore\ f^{-1}(2)=\log _{10} 4$
  15. Given that $f(x)=x^{2}$, what is the domain of $f$ for which $f^{-1}$ exists?
    Explanation
    $f^{-1}$ exists if and only if $f$ is a one to one function.
    $f(x)$ is one to one only when $x \ge 0$.
    $\therefore \operatorname{dom}(f)=\{x \mid x \ge 0, x \in \mathbb{R}\}$.
  16. If $f(x)=x^{2}$ and $g(x)=2 x,(f \circ g)\left(-\dfrac{1}{2}\right)=\ldots$
    Explanation
    $f(x)=x^{2}$
    $g(x)=2 x$
    $(f \circ g)\left(-\dfrac{1}{2}\right)$
    $=f\left(g\left(-\dfrac{1}{2}\right)\right)$
    $=f\left(2\left(-\dfrac{1}{2}\right)\right)$
    $=f(-1)$
    $=(-1)^{2}$
    $=1$
  17. If $f(x)=x^{2}$ and $g(x)=\dfrac{2 x+1}{x-4}$, what is the domain of $g\circ f$ ?
    Explanation
    $\begin{aligned} f(x)&=x^{2} \\\\ g(x)&=\dfrac{2 x+1}{x-4} \\\\ (g \circ f)(x)&=g(f(x)) \\\\ &=g\left(x^{2}\right)\\\\ &=\dfrac{2 x^{2}+1}{x^{2}-4} \\\\ (g \circ f)(x) &\text { exists when } \\\\ x^{2}-4 &\neq 0 \\\\ x^{2} &\neq 4 \\\\ x &\neq \pm 2 \\\\ \therefore\ \operatorname{dom}(g \circ f)&=\mathbb{R} \smallsetminus\{\pm 2\} \end{aligned}$
  18. If $g(x)=\dfrac{x+2}{2 x-1}$ and $h(x)=2 x$, what is the range of $g \circ h ?$
    Explanation
    $\begin{aligned} g(x)&=\dfrac{x+2}{2 x-1} \\\\ h(x)&=2 x \\\\ (g \circ h)(x) &=g(h(x))\\\\ &=g(2 x) \\\\ &=\dfrac{2 x+2}{4 x-1} \\\\ &=\dfrac{5 / 2}{4 x-1}+\dfrac{1}{2} \\\\ \therefore \operatorname{ran}(g \cdot f)&=\left\{y \mid y \neq \dfrac{1}{2}, y \in \mathbb{R}\right\} \end{aligned}$
  19. The function $f: A \rightarrow B$ is onto function then the range of $f$ is
    Explanation
    A function $f$ is onto function when range of $f=$ codomain.
  20. If $f$ is a function an a set $A=\{1,2,3,4,5\}$ such that $f=\{(1,2),(2,3),(3,4),(4, x),(5,5)\}$ is a one to function, then $x=\ldots$
    Explanation
    To be one to ove furction $f$, $A$ must be related with $1$.

    21. Your Score:

السبت، 26 يونيو 2021

Quadratic Function : Multiple Choice Questions

  يونيو 26, 2021   , , , , , , ,   No comments   Edit
ဖုန်းဖြင့်ကြည့်သည့်အခါ စာများကို အပြည့်မမြင်ရလျှင် screen ကို ဘယ်ညာ ဆွဲကြည့်နိုင်ပါသည်။


Important Notes


Quadratic Function
(Standard Form)		 $f(x)=a x^{2}+b x+c, a \neq 0$
Graph Parabola
$a>0$ (opens upward)
$a<0$ (opens downward)
Axis of Symmetry $x=-\displaystyle\frac{b}{2 a}$
Vertex $\left(-\displaystyle\frac{b}{2 a}, f\left(-\displaystyle\frac{b}{2 a}\right)\right)=\left(-\displaystyle\frac{b}{2 a},-\displaystyle\frac{b^{2}-4 a c}{4 a}\right)$
y-intercept (0, c)
Discriminant $b^{2}-4 a c$
$b^{2}-4 a c>0 \Rightarrow$ two $x$ intercepts (cuts $x-$ axis at two points)
$b^{2}-4 a c=0 \Rightarrow$ one $x$ intercepts (touch $x$ -axis at one point $)$
$b^{2}-4 a c=0 \Rightarrow$ one $x$ intercepts (does not intersect $x$ -axis)
Quadratic Equation $a x^{2}+b x+c=0, a \neq 0$
Quadratic Formula $x=\displaystyle\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$
Quadratic Function
(Vertex Form)		 $f(x)=a(x-h)^{2}+k, a \neq 0$
Vertex $(h, k)$
Axis of Symmetry
(Vertex Form)		 $x=h$
Quadratic Function
(Intercept Form when
discriminant $>0$)		 $f(x)=a(x-p)(x-q), a \neq 0$
$X$ -intercept points $(p, 0)$ and $(q, 0)$
Axis of Symmetry $x=\displaystyle\frac{p+q}{2}$
Quadratic Inequality $a x^{2}+b x+c>0$
$a x^{2}+b x+c \geq 0$
$a x^{2}+b x+c<0$
$a x^{2}+b x+c \leq 0$
$a>0$ and $b^2-4ac<0$

 The graph does not cut the $x$ -axis.
$y<0 \Rightarrow$ solution set $=\varnothing$
$y=0 \Rightarrow$ solution set $=\varnothing$
$y>0 \Rightarrow$ solution set $=\mathbb{R}$
$a<0$ and $b^2-4ac<0$

 The graph does not cut the $x$ -axis.
$y<0 \Rightarrow$ solution set $=\mathbb{R}$
$y=0 \Rightarrow$ solution set $=\varnothing$
$y>0 \Rightarrow$ solution set $=\varnothing$
$a>0$ and $b^2-4ac=0$

 The graph touches the $x$ -axis.
$y<0 \Rightarrow$ solution set $=\varnothing$
$y=0 \Rightarrow$ solution set $=\left\{-\displaystyle\frac{b}{2 a}\right\}$
$y>0 \Rightarrow$ solution set $=\mathbb{R} \backslash\left\{-\displaystyle\frac{b}{2 a}\right\}$
$a<0$ and $b^2-4ac=0$

 The graph touches the $x$ -axis.
$y<0 \Rightarrow$ solution set $=\mathbb{R} \backslash\left\{-\displaystyle\frac{b}{2 a}\right\}$
$y=0 \Rightarrow$ solution set $=\left\{-\displaystyle\frac{b}{2 a}\right\}$
$y>0 \Rightarrow$ solution set $=\varnothing$
$a>0$ and $b^2-4ac>0$

 The graph cuts the $x$ -axis at two points.
$y<0 \Rightarrow$ solution set $=\{x \mid p<x<q\}$
$y=0 \Rightarrow$ solution set $=\{p, q\}$
$y>0 \Rightarrow$ solution set $=\{x \mid x<p$ or $x>q\}$
$a<0$ and $b^2-4ac>0$

 The graph cuts the $x$ -axis at two points.
$y<0 \Rightarrow$ solution set $=\{x \mid x<p$ or $x>q\}$
$y=0 \Rightarrow$ solution set $=\{p, q\}$
$y>0 \Rightarrow$ solution set $=\{x \mid p<x<q\}$

အထက်ဖော်ပြပါ quadratic function နှင့်ဆိုင်သော definitions နှင့် concepts များကိုသိရှိနားလည်ပြီးလျှင် အောက်ပါ MCQ များကို လေ့ကျင့် ဖြေဆိုနိုင်ပါပြီ။ ဖြေဆိုပြီးကြောင်း Submit လုပ်ပြီးလျှင် ရမှတ်နှင့် အဖြေမှန်ကိုပါ ပြပေးမည် ဖြစ်သည်။ ရှင်းလင်းချက်မပါဝင်ပါ။



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