MCQ 2511 Mark
If $x + 1$ is a factor of the polynomial $2 x^2+k x+1$, then the value of $'k'$ is:
Answer If $x+1$ is a factor of $p(x)=2 x^2+k x+1$, then $p(-1)=0$
$\Rightarrow 2 x^2+k x+1=0$
$\Rightarrow 2(-1)^2+k(-1)+1=0$
$\Rightarrow 2-k+1=0$
$\Rightarrow k=3$
View full question & answer→MCQ 2521 Mark
If $\text{x}+\frac{1}{\text{x}}=5,$ then $\text{x}^2+\frac{1}{\text{x}^2}=$
AnswerUsing, $(\text{a}+\text{b})^2=\text{a}^2+\text{b}^2+2\text{ab}$
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=\text{x}^2+\Big(\frac{1}{\text{x}^2}\Big)+2\text{x}\frac{1}{\text{x}}$
$\Rightarrow(5)^2=\text{x}^2+\Big(\frac{1}{\text{x}^2}\Big)+2$
$\Rightarrow\text{x}^2+\Big(\frac{1}{\text{x}^2}\Big)=25-2$
$\Rightarrow\text{x}^2+\Big(\frac{1}{\text{x}^2}\Big)=23$
View full question & answer→MCQ 2531 Mark
If $\text{x}^4+\frac{1}{\text{x}^4}=194,$ than $\text{x}^3+\frac{1}{\text{x}}^3=$
Answer$\Big(\text{x}^4+\frac{1}{\text{x}^4}\Big)=194$
$\Rightarrow(\text{x}^2)^2+\Big(\frac{1}{\text{x}^2}\Big)+2\times\text{x}^2\times\frac{1}{\text{x}^2}=194+2\times\text{x}^2\times\frac{1}{\text{x}^2}$
$\Rightarrow\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)^2=196$
$\Rightarrow\text{x}^2+\frac{1}{\text{x}^2}=\sqrt{196}=14$
Now,
$\Rightarrow(\text{x}^2)+\Big(\frac{1}{\text{x}^2}\Big)+2\times\text{x}\times\frac{1}{\text{x}}=14+2\times\text{x}\times\frac{1}{\text{x}}$
$\Rightarrow\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=16$
$\Rightarrow\text{x}+\frac{1}{\text{x}}=\sqrt{16}=4$
Now, $\Big(\text{x}+\frac{1}{\text{x}}\Big)^3=(4)^3$
$\Rightarrow(\text{x})^3+\Big(\frac{1}{\text{x}}\Big)^3+3\times\text{x}\times\frac{1}{\text{x}}\Big(\text{x}+\frac{1}{\text{x}}\Big)=64$
$\Rightarrow(\text{x}^3)+\Big(\frac{1}{\text{x}^3}\Big)+3(4)=64$
$\Rightarrow(\text{x}^3)+\Big(\frac{1}{\text{x}^3}\Big)=64-12 = 52$
View full question & answer→MCQ 2541 Mark
A polynomial of degree $3$ in $x$ has at most.
- ✓
$3$ terms
- B
$1$ terms
- C
$5$ terms
- D
$4$ terms
AnswerCorrect option: A. $3$ terms
$3$ terms of not more than the power of $3$
View full question & answer→MCQ 2551 Mark
If $\text{x}+\frac{1}{\text{x}}=3,$ than the value of $\text{x}^2+\frac{1}{\text{x}^2}$ is:
Answer$\text{x}+\frac{1}{\text{x}}=3,$
Squaring both sides, we get
$\text{x}^2+\frac{1}{\text{x}^2}+2\times\text{x}\times\frac{1}{\text{x}}=9$
$\Rightarrow\text{x}^2+\frac{1}{\text{x}^2}+2=9$
$\Rightarrow\text{x}^2+\frac{1}{\text{x}^2}=7$
View full question & answer→MCQ 2561 Mark
The zeros of the polynomial $p(x) = 3x^2 - 1$ are:
- A
$\frac{1}{3}\ \text{and}\ 3$
- B
$\frac{1}{\sqrt3}\ \text{and}\ \sqrt3$
- C
$\frac{-1}{\sqrt3}\ \text{and}\ \sqrt3$
- ✓
$\frac{1}{\sqrt3}\ \text{and}\ \frac{-1}{\sqrt3}$
AnswerCorrect option: D. $\frac{1}{\sqrt3}\ \text{and}\ \frac{-1}{\sqrt3}$
$p(x)=3 x^2-1$
$\text { Now, } p(x)=0$
$\Rightarrow 3 x^2-1=0$
$\Rightarrow 3 x^2=1$
$\Rightarrow\text{x}^2=\frac{1}{3}$
$\Rightarrow\text{x}=\frac{1}{\sqrt3}\ \text{and}\ -\frac{1}{\sqrt3}$
View full question & answer→MCQ 2571 Mark
When $p(x) = x^3+a x^2+2 x+a$ is divided by $x + a,$ the remainder is:
Answer$x+a=0$
$\Rightarrow x=-a$
By the remainder theorem, we know that when $p(x)$ is divided by $(x+a)$, the remainder is $p(-a)$.
Thus, we have: $P(-a)=(-a)^3+a \times(-a)^2+2 \times(-a)+a$
$=-a^3+a^3-2 a+a$
$=-a$
View full question & answer→MCQ 2581 Mark
The expression $(a-b)^3+(b-c)^3+(c-a)^3$ can be factorized as.
AnswerCorrect option: A. $3(a - b) (b - c) (c - a)$
Here, $a - b + b - c + c - a = 0$
Therefore, $(a-b)^3+(b-c)^3+(c-a)^3$
$= 3(a - b) (b - c) (c - a)$
View full question & answer→MCQ 2591 Mark
If $\text{x}+\frac{1}{\text{x}}=2,$ then $\text{x}^3+\frac{1}{\text{x}^3}=$
AnswerBy using identity,
$(a+b)^3=a^3+b^3+3 a b(a+b)$
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^3=\text{x}^3+\frac{1}{\text{x}^3}+3(\not\text{x})\frac{1}{\not\text{x}}\Big(\text{x}+\frac{1}{\text{x}}\Big)\\=\text{x}^3+\frac{1}{\text{x}^3}+3\Big(\text{x}+\frac{1}{\text{x}}\Big)$
Now $\text{x}+\frac{1}{\text{x}}=2$
$\Rightarrow(2)^3=\text{x}^3+\frac{1}{\text{x}^3}+3(2)$
$\Rightarrow\text{x}^3+\frac{1}{\text{x}^3}=(2)^3-3\times2=8-6=2$
$\Rightarrow\text{x}^2+\frac{1}{\text{x}^2}=23$
Hence, correct option is $(d).$
View full question & answer→MCQ 2601 Mark
Which of the following is a factor of $(x+y)^3-\left(x^3+y^3\right)?$
- A
$x y^2$
- B
$x^2+y^2+2 x y$
- ✓
$3xy$
- D
$x^2+y^2-x y$
Answer$\text { Given polynomial: }(x+y)^3-\left(x^3+y^3\right)$
$=\left(x^3+y^3+3 x^2 y+3 x y^2\right)-\left(x^3+y^3\right)\left[{\because }~(a+b)^3=\left(a^3+b^3+3 a^2 b+3 a b^2\right)\right]$
$=3 x^2 y+3 x y^2=3 x y(x+y)$
$\Rightarrow 3 x y \text { and }(x+y) \text { are factors of given polynomial. }$
View full question & answer→MCQ 2611 Mark
If $\Big(3\text{x}+\frac{1}{2}\Big)\Big(3\text{x}-\frac{1}{2}\Big)=9\text{x}^2-\text{p}$ then the value of $p$ is:
- A
$0$
- B
$-\frac{1}{4}$
- ✓
$\frac{1}{4}$
- D
$\frac{1}{2}$
AnswerCorrect option: C. $\frac{1}{4}$
$\Big(3\text{x}+\frac{1}{2}\Big)\Big(3\text{x}-\frac{1}{2}\Big)=9\text{x}^2-\text{p}$
$9\text{x}^2-\frac{1}{4}$ $\Big(\therefore\ \big(\text{a}^2-\text{b}^2\big)=(\text{a}+\text{b})(\text{a}-\text{b})\Big)$
$=9\text{x}^2-\text{p}$
$\Rightarrow\text{p}=\frac{1}{4}$
View full question & answer→MCQ 2621 Mark
The maximum number of terms in a polynomial of degree $10$ is:
AnswerThe maximum number of terms is one more than the power of the polynomial.Therefore, the maximum number of terms in a polynomial of degree $10$ is $(10 + 1) = 11$
View full question & answer→MCQ 2631 Mark
If $\text{x}^2+\frac{1}{\text{x}^2}=38,$ then the value of $\text{x}-\frac{1}{\text{x}}$ is:
Answer$\Big(\text{x}-\frac{1}{\text{x}}\Big)^2=\text{x}^2+\frac{1}{\text{x}^2}-2\times\text{x}\times\frac{1}{\text{x}}$
$\Rightarrow\Big(\text{x}-\frac{1}{\text{x}}\Big)^2=38-2$
$\Rightarrow\Big(\text{x}-\frac{1}{\text{x}}\Big)^2=36$
$\Rightarrow\text{x}-\frac{1}{\text{x}}=\pm6$
View full question & answer→MCQ 2641 Mark
Write the correct answer in the following: Degree of the polynomial $4 x^4+0 x^3+0 x^5+5 x+7$ is.
AnswerThe height power of the variable in a polynomial is called the degree of the polynomial. In this polynomial, the term with highest power of x is 4x$^4$. Highest power of x is 4, so the degree of the given polynomial is $4.$
View full question & answer→MCQ 2651 Mark
The value of $\Big(\text{x}-\frac{1}{\text{x}}\Big)\Big(\text{x}+\frac{1}{\text{x}}\Big)\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)$ is:
- ✓
$\text{x}^4-\frac{1}{\text{x}^4}$
- B
$\text{x}^2+\frac{1}{\text{x}^2}-2$
- C
$\text{x}^3+\frac{1}{\text{x}^3}+2$
- D
$\text{x}^4+\frac{1}{\text{x}^4}$
AnswerCorrect option: A. $\text{x}^4-\frac{1}{\text{x}^4}$
$\Big(\text{x}-\frac{1}{\text{x}}\Big)\Big(\text{x}+\frac{1}{\text{x}}\Big)\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)$
$=\Big(\text{x}^2-\frac{1}{\text{x}^2}\Big)\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)$ [Using identity $(a+b)(a-b)=a^2-b^2$]
$=\text{x}^4-\frac{1}{\text{x}^4}$ [Using identity $(a+b)(a-b)=a^2-b^2$]
View full question & answer→MCQ 2661 Mark
If P(x) = $x^3-1$, then the value of $p(1) + p(-1)$ is:
Answer$P(x)=x^3-1$,
Then the value of $P(1)+P(-1)$
$=(1)^3-1+(-1)^3-1$
$=1-1-1-1=1-3=-2$
View full question & answer→MCQ 2671 Mark
A polynomial containing one nonzero term is called a ______.
AnswerA polynomial containing one nonzero term is called a monomial.
$3 x, 5 x^2, y^3$
View full question & answer→MCQ 2681 Mark
The factors of $x^3-7 x-6$ are.
- A
$x(x - 6) (x - 1)$
- B
$(x - 1) (x - 3) (x + 2)$
- ✓
$(x + 1) (x + 2) (x - 3)$
- D
$\left(x^2-6\right)(x-1)$
AnswerCorrect option: C. $(x + 1) (x + 2) (x - 3)$
The given expression to be factorized is $x^3-7 x+6$
This can be written in the form
$x^3-7 x+6=x^3-(1+6) x+6$
$=x^3-x-6 x+6$
Take common $x$ from the first two terms and $-6$ from the last two terms.
Then we have,
$x^3+7 x+6=x\left(x^2-1\right)-6(x-1)$
$=x\left\{(x)^2-(1)^2\right\}-6\{x-1\}$
$=x(x+1)(x-1)-6(x-1)$
Finally, take common $(x-1)$ from the above expression,
$x^3-7 x+6=(x-1)\{(x+1)-6\}$
$=(x-1)\left(x^2+x-6\right)$
$=(x-1)\left(x^2+3 x-2 x-6\right)$
$=(x-1)\{x(x+3)-2(x+3)\}$
$=(x-1)(x+3)(x-2)$
View full question & answer→MCQ 2691 Mark
If $\text{p}(\text{x})=\text{x}^2-2\sqrt{2}\text{x}+1,$ than $\text{p}(2\sqrt{2})$ is equal to:
- A
$0$
- B
$4\sqrt{2}$
- C
$8\sqrt{2}+1$
- ✓
$1$
Answer$\text{p}(\text{x})=\text{x}^2-2\sqrt{2}\text{x}+1,$
$\Rightarrow\text{p}(2\sqrt{2})=(2\sqrt{2})^2-2\sqrt{2}(2\sqrt{2})+1$
$\Rightarrow\text{p}(2\sqrt{2})=8-8+1$
$\Rightarrow\text{p}(2\sqrt{2})=1$
View full question & answer→MCQ 2701 Mark
If $(x+y)^3-(x-y)^3-6 y\left(x^2-y^2\right)=k y^2$, then $k=$
AnswerWe have,
$=(x+y)^3-(x-y)^3-6 y\left(x^2-y^2\right)=k y^3$
$=(x+y-x+y)^3+3(x+y)(x-y)(x+y-x+y)-6 y\left(x^2-y^2\right)=k y^3$
$=2 y^3+6 y\left(x^2-y^2\right)-6 y\left(x^2-y^2\right)=k y^3$
$=8 y^3=k y^3$
$=k=8$
View full question & answer→MCQ 2711 Mark
One of the factors of $\left(25 x^2-1\right)+(1+5 x)^2$ is:
- A
$5x - 1$
- B
$5 - x$
- ✓
$10x$
- D
$5 + x$
Answer$\text { Now, }\left(25 x^2-1\right)+(1+5 x)^2$
$\left.=25 x^2-1+1+25 x^2+10 x \text { [using identity, }(a+b)^2=a^2+b^2+2 a b\right]$
$=50 x^2+10 x=10 x(5 x+1)$
Hence, one of the factor of given polynomial is $10 x$.
View full question & answer→MCQ 2721 Mark
When p(x) = $x^3+a x^2+2 x+a$ is divided by $(x + a)$, the remainder is:
Answer$p(x)=x^3+a x^2+2 x+a$
$x+a=0 \Rightarrow x=-a$
By the remainder theorem, we know that when $p(x)$ is divided by $(x+a)$, the remainder is $p(-a)$.
$\text { Now, } p(-a)=x^3+a x^2+2 x+a$
$=(-a)^3+a(-a)^2+2(-a)+a$
$=-a^3+a^3-2 a+a$
$=-a$
View full question & answer→MCQ 2731 Mark
$8$ is a polynomial of degree.
AnswerSince $8$ is a constant term.Therefore its degree is $0.$
View full question & answer→MCQ 2741 Mark
The value of $\frac{(2.3)^3-0.027}{(2.3)^20.69+0.09}.$
AnswerThe given expresstion is
$\frac{(2.3)^3-0.027}{(2.3)^20.69+0.09}$
This can be written in the form
$\frac{(2.3^3)-(0.3)^3}{(2.3)^2+2.3\times0.3+(0.3)^2}$
Assume $a = 2.3$ and $b = 0.3$. then the given expression can be rewritten as $\frac{\text{a}^3-\text{b}^3}{\text{a}^2+\text{ab}+\text{b}^2}$
Recall the formula for difference of two cubes
$a^3-b^3=(a-b)\left(a^2+a b+b^2\right)$
Using the above formula, the expression becomes $\frac{\text{(a}-\text{b})(\text{a}^2+\text{ab}+\text{b}^2)}{\text{a}^2+\text{ab}+\text{b}^2}$
Note that both a and b are positive, unequal. so, neither $a^3-b^3$ nor any factor of it can be zero.
Therefore we can cancel the term $\left(a^2+a b+b^2\right)$ from both numerator and denominator. then the expression becomes
$\frac{(\text{a}-\text{b})(\text{a}^2+\text{ab}+\text{b}^2)}{\text{a}^2+\text{ab}+\text{b}^2}=\text{a}-\text{b}$
$=2.3-0.3$
$=2$
View full question & answer→MCQ 2751 Mark
The expression $x^4+4$ can be factorized as:
- A
$\left(x^2-2 x-2\right)\left(x^2-2 x+2\right)$
- B
$\left(x^2+2\right)\left(x^2-2\right)$
- C
$\left(x^2+2 x+2\right)\left(x^2+2 x-2\right)$
- ✓
$\left(x^2+2 x+2\right)\left(x^2-2 x+2\right)$
AnswerCorrect option: D. $\left(x^2+2 x+2\right)\left(x^2-2 x+2\right)$
The given expression to be factorized is $x^4+4$This can be written in the form
$x^4+4=\left(x^2\right)^2+(2)^2+4 x^2-4 x^2$
$=\{(x^2)^2+2 \times x^2 \times 2+(2)^2\}-(2 x)^2$
$=\left(x^2+2\right)^2-(2 x)^2$
$=\left(x^2+2+2 x\right)\left(x^2+2-2 x\right)$
View full question & answer→MCQ 2761 Mark
Write the correct answer in the following: Which of the following is a factor of $(x+y)^3-\left(x^3+y^3\right)$?
- A
$x^2+y^2+2 x y$
- B
$x^2+y^2-x y$
- C
$x y^2$
- ✓
$3xy$
Answer$(x+y)^3-\left(x^3+y^3\right)=x^3+y^3+3 x y(x+y)-x^3-y^3$
${\left[(a+b)^3=a^3+b^3+3 a b(a+b)\right]}$
$=3 x y(x+y)$
So, $3 x y$ is a factor of $(x+y)^3-\left(x^3+y^3\right)$.
View full question & answer→MCQ 2771 Mark
Write the correct answer in the following: Zero of the zero polynomial is.
AnswerZero of the zero polynomial is any real number.
e.g., Let us consider zero polynomial be $0(x - k)$, where $k$ is a real number. For determining the zero, put $x - k = 0 ⇒ x = k$ Hence, zero of the zero polynomial be any real number.
View full question & answer→MCQ 2781 Mark
If $\text{x}^4+\frac{1}{\text{x}^4}=623,$ than $\text{x}+\frac{1}{\text{x}}=$
- A
$27$
- ✓
$3\sqrt{3}$
- C
$25$
- D
$-3\sqrt{3}$
AnswerCorrect option: B. $3\sqrt{3}$
$\Big(\text{x}^4+\frac{1}{\text{x}^4}\Big)=623$
$\Rightarrow(\text{x}^2)^2+\Big(\frac{1}{\text{x}^2}\Big)+2\times\text{x}^2\times\frac{1}{\text{x}^2}=623+2\times\text{x}^2\times\frac{1}{\text{x}^2}$
$\Rightarrow\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)^2=625$
$\Rightarrow\text{x}^2+\frac{1}{\text{x}^2}=\sqrt{625}=25$
Now,
$\Rightarrow(\text{x}^2)+\Big(\frac{1}{\text{x}^2}\Big)+2\times\text{x}\times\frac{1}{\text{x}}=25+2\times\text{x}\times\frac{1}{\text{x}}$
$\Rightarrow\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=27$
$\Rightarrow\text{x}+\frac{1}{\text{x}}=\sqrt{27}=3\sqrt{3}$
View full question & answer→MCQ 2791 Mark
Write the correct answer in the following:
If $49\text{x}^2 -\text{b}=\Big(7\text{x}+\frac{1}{2}\Big)\Big(7\text{x}-\frac{1}{2}\Big),$ the value of b is.
- A
$0$
- B
$\frac{1}{\sqrt2}$
- ✓
$\frac{1}{4}$
- D
$\frac{1}{2}$
AnswerCorrect option: C. $\frac{1}{4}$
$49\text{x}^2 -\text{b}=\Big(7\text{x}+\frac{1}{2}\Big)\Big(7\text{x}-\frac{1}{2}\Big)$
$\Rightarrow49\text{x}^2 -\text{b}=\Big(7\text{x}\Big)^2-\Big(\frac{1}{2}\Big)^2$
$49^2-\frac{1}{4} [\therefore(\text{a}+\text{b})(\text{a}-\text{b})=\text{a}^2-\text{b}^2]$
So, we get $\text{b}=\frac{1}{4}.$
View full question & answer→MCQ 2801 Mark
If $3x = a + b + c$, then the value of $(x-a)^3+(x-b)^3+(x-c)^3-3(x-a)(x-b)(x-c)$ is:
- A
$a + b + c$
- B
$(a - b)(b - c)(c - a)$
- ✓
$0$
- D
Answer$3 x=a+b+c$$\Rightarrow a+b+c-3 x=0$
$\Rightarrow 3 x-(a+b+c)=0$
$\Rightarrow(x-a)+(x-b)+(x-c)=0 ...(1)$
Using identity if $a+b+c=0$ then, $a^3+b^3+c^3-3 a b c=0$
If we take $\mathrm{x}-\mathrm{a}=\mathrm{A}, \mathrm{x}-\mathrm{b}=\mathrm{B}, \mathrm{x}-\mathrm{c}=\mathrm{C}$ in equation $(1)$, we get
$A+B+C=0$
$\Rightarrow A^3+B^3+C^3-3 A B C=0$
$\Rightarrow(x-a)^3+(x-b)^3+(x-c)^3-3(x-a)(x-b)(x-c)=0$
Hence, correct option is $(c).$
View full question & answer→MCQ 2811 Mark
The factors of $12 x^2-x-6$ are:
- ✓
$(3x + 2) (4x - 3)$
- B
$(12x - 1) (x + 6)$
- C
$(3x - 2) (4x + 3)$
- D
$(12x + 1) (x - 6)$
AnswerCorrect option: A. $(3x + 2) (4x - 3)$
$12 x^2-x-6$
$=12 x^2-9 x+8 x-6$
$=3 x(4 x-3)+2(4 x-3)$
$=(3 x+2)(4 x-3)$
View full question & answer→MCQ 2821 Mark
Which of the following is a polynomial?
AnswerCorrect option: C. $\text{y}$
A polynomial is an algebraic expression in which the variables involved have only non-negative integrals powers.
Option $(a)$, $(b)$ and $(d)$ have negative and non-integral powers,
So they are not polynomials.
View full question & answer→MCQ 2831 Mark
Which of the following is a binomial?
AnswerCorrect option: A. $x^2+4$
Clearly, $x^2+4$ is an expresstion having two non-zero terms.
So, it is a binomial.
View full question & answer→MCQ 2841 Mark
A polynomial containing three non-zero terms is called a ________.
AnswerA polynomial containing three non-zero terms is called a trinomial.
Example: $5 x^2+2 x+3, a x^2=b x+c, 3 x+2 y-3$
View full question & answer→MCQ 2851 Mark
The factors of $x^2-9$ is:
- A
$(x - 3) (x - 3)$
- ✓
$(x + 3) ( x + 3)$
- C
$(x + 3) (x - 3)$
- D
$(x - 3) (x + 9)$
AnswerCorrect option: B. $(x + 3) ( x + 3)$
$x^2-9$
$=x^2-3^2$
Using identity $\mathrm{a}^2-\mathrm{b}^2=(\mathrm{a}+\mathrm{b})(\mathrm{a}-\mathrm{b})$
$=(x+3)(x-3)$
View full question & answer→MCQ 2861 Mark
If x + a is a factor of $x^4-a^2 x^2+3 x-6 a$, then a is:
Answer$x+a$ is a factor of polynomial $f(x)=x^4-a^2 x^2+3 x-6 a$,
$\text { Then at } x=-a, p(x)=0$
$\Rightarrow(-a)^4-a^2(-a)^2+3(-a)-6 a=0$
$\Rightarrow a^4-a^4-3 a-6 a=0$
$\Rightarrow-9 a=0$
$\Rightarrow a=0$
View full question & answer→MCQ 2871 Mark
The factors of $x^3-1+y^3+3 x y$ are:
- ✓
$(x-1+y)\left(x^2+1+y^2+x+y-x y\right)$
- B
$(x+y+1)\left(x^2+y^2+1-x y-x-y\right)$
- C
$(x-1+y)\left(x^2-1-y^2+x+y+x y\right)$
- D
$3(x+y-1)\left(x^2+y^2-1\right)$
AnswerCorrect option: A. $(x-1+y)\left(x^2+1+y^2+x+y-x y\right)$
By using identity
$a^3+b^3+c^3-3 a b c=(a+b+c)\left(a^2+b^2+c^2-a b-b c-c a\right)$
We can write,
$x^3-1+y^3+3 x y$
$=\left(x^3\right)+(-1)^3+\left(y^3\right)-3(-1)(x)(y)$
$=[x+(-1)+y]\left[x^2+(-1)^2+y^2-x(-1)-y(-1)-x y\right]$
$=(x-1+y)\left(x^2+1+y^2+x+y-x y\right)$
Hence, correct option is $(a).$
View full question & answer→MCQ 2881 Mark
If $x - 1$ is the factor of p(x) = $x^3-23 x^2+k x-120$, then the value of $'k'$ is:
AnswerIf $x-1$ is the factor of $p(x)$, then
$p(1)=0$
$(1)^3-23(1)^2+k(1)-120=0$
$1-23+k-120=0$
$1-143+k=0$
$-142+k=0$
$k=142$
View full question & answer→MCQ 2891 Mark
The value of $\frac{(\text{a}^2-\text{b}^2)^3(\text{b}^2-\text{c}^2)+(\text{c}^2-\text{a}^2)^3}{(\text{a}-\text{b})^3+(\text{b}-\text{c})^3+(\text{c}-\text{a})^3}$ is:
- A
$3(a - b) (b - c) (c - a)$
- ✓
$3(a + b) (b + c) (c + a)$
- C
$3(a + b) (b + c) (c + a) (a - b) (b - c) (c - a)$
- D
AnswerCorrect option: B. $3(a + b) (b + c) (c + a)$
$\frac{(\text{a}^2-\text{b}^2)^3(\text{b}^2-\text{c}^2)+(\text{c}^2-\text{a}^2)^3}{(\text{a}-\text{b})^3+(\text{b}-\text{c})^3+(\text{c}-\text{a})^3}$
$=\frac{3(\text{a}^2-\text{b}^2)(\text{b}^2-\text{c}^2)(\text{c}^2-\text{a}^2)}{3(\text{a}-\text{b})(\text{b}-\text{c})(\text{c}-\text{a})}$ $[\text{Since}\text{ x}^3+\text{y}^3+\text{z}^3=3\text{xyz},\text{ if}\text{ x}+\text{y}+\text{z}=0]$
$=\frac{3(\text{a}-\text{b})(\text{a}+\text{b})(\text{b}-\text{c})(\text{b}+\text{c})(\text{c}-\text{a})(\text{c}+\text{a})}{3(\text{a}-\text{b})(\text{b}-\text{c}(\text{c})-\text{a})}$
$=3(\text{a}+\text{b})(\text{b}+\text{c})(\text{c}+\text{a})$
View full question & answer→MCQ 2901 Mark
Which of the following is a polynomial?
- A
$x^{-2}+x^{-1}+3$
- B
$x+x^{-1}+2$
- C
$x^{-1}$
- ✓
$0$
AnswerA polynomial is an algebraic expression in which the variables involved have only non-negative integrals powers.
Option $(a)$, $(b)$ and $(c)$ have negative and non-integral powers,
So they are not polynomials.
We know that, exery real number is a constant polynomial.
So, 0 being a real number is a polynomial.
View full question & answer→MCQ 2911 Mark
The factors of $8 a^3+b^3-6 a b+1$ are:
- A
$(2 a+b-1)\left(4 a^2+b^2+1-3 a b-2 a\right)$
- B
$(2 a-b+1)\left(4 a^2+b^2-4 a b+1-2 a+b\right)$
- ✓
$(2 a+b+1)\left(4 a^2+b^2+1-2 a b-b-2 a\right)$
- D
$(2 a-1+b)\left(4 a^2+1-4 a-b-2 a b\right)$
AnswerCorrect option: C. $(2 a+b+1)\left(4 a^2+b^2+1-2 a b-b-2 a\right)$
We know the identity
$a^3+b^3+c^3-3 a b c=(a+b+c)\left(a^2+b^2+c^2-a b-b c-c a\right)$
So by using identity, we can write given expression as
$(2 a)^3+(b)^3+(1)^3-3(2 a)(b)(1)$
$=(2 a+b+1)\left[(2 a)^2+b^2+1^2-2 a \times b-b \times 1-2 a \times 1\right]$
$=(2 a+b+1)\left(4 a^2+b^2+1-2 a b-b-2 a\right)$
Hence, correct option is $(c).$
View full question & answer→MCQ 2921 Mark
If $x + a$ is a factor of $x^4-a^2 x^2+3 x-6 a$, then $a =$
AnswerPut $x+a=0 \Rightarrow x=-a$
Then, value of $x^4-a^2 x^2+3 x-6 a$ at $x=-a$ is zero
Therefore, $(-a)^4-a^2(-a)^2+3(-a)-6 a=0$
$-9 a=0 \Rightarrow a=0$
View full question & answer→MCQ 2931 Mark
If $(x + 5)$ is a factor of = $x^3-20 x+5 k$ then $k = ?$
Answer$p(x)=x^3-20 x+5 k$
Now, $x+5=0 \Rightarrow x=(-5)$
By factor theorem,
$p(-5)=0$
$\Rightarrow(-5)^3-20(-5)+5 \mathrm{k}=0$
$\Rightarrow-125+100+5 \mathrm{k}=0$
$\Rightarrow-25+5 \mathrm{k}=0$
$\Rightarrow 5 \mathrm{k}=25$
$\Rightarrow \mathrm{k}=5$
View full question & answer→MCQ 2941 Mark
If $x^3-3 x^2 3 x-7=(x+1)\left(a x^2+b x+c\right)$, then $a+b+c=$
AnswerFirst multiply
$(x+1)\left(a x^2+b x+c\right)$
$=a x^3+b x^2+c x+a x^2+b x+c$
$=a x^3+b x^2+a x^2+c x+c$
$=a x^3+(b+a) x^2+(c+b) x+c$
Comparing it with
$x^3-3 x^2+3 x-7$
$a=1$
$b+a=-3 \Rightarrow b+1+-3 \Rightarrow b=-4$
$c+b=3 \Rightarrow c-4=3 \Rightarrow c=7$
$\mathrm{c}=-7$ should be 7
as if we put $x=-1$ in
$x^3-3 x^2+3 x-7$
$-1-3-3-7=14$ so $x+1$ can not be factor so $x+1$ will be factor if $x^3-3 x^2+3 x-7$ is actually
$x^3-3 x^2+3 x+7$
then $-1-3-3+7=0$
Hence, we can say that
$a=1$
$b=-1$
$c=7$
so, $a+b+c=4$
View full question & answer→MCQ 2951 Mark
If $x + y + z = 0$ then $x^3+y^3+z^3$ is:
AnswerCorrect option: A. $3xyz$
$x^3+y^3+z^3-3 x y z=(x+y+z)\left(x^2+y^2+z^2-x y-y z-z x\right)$
$\Rightarrow x^3+y^3+z^3-3 x y z=(0)\left(x^2+y^2+z^2 x y-y z-z x\right)$
$\Rightarrow x^3+y^3+z^3-3 x y z=0$
$\Rightarrow x^3+y^3+z^3=3 x y z \text { if } x+y+z=0, \text { then } x^3+y^3+z^3 \text { is } 3 x y z$
View full question & answer→MCQ 2961 Mark
If $a + b + c = 9$ and $ab + bc + ca = 23$, then $a^2+b^2+c^2=$
Answer$(a+b+c)^2=a^2+b^2+c^2+2 a b+2 b c+2 c a$Hence, $9^2=a^2+b^2+c^2+2 \times 23$
$\Rightarrow a^2+b^2+c^2=35$
View full question & answer→MCQ 2971 Mark
The zeros of the polynomial p(x) = $x^2+x-6$ are:
- A
$2, 3$
- B
$-2, 3$
- ✓
$2, -3$
- D
$-2, -3$
AnswerCorrect option: C. $2, -3$
Let $p(x)$ be a polynomial. If $\text{p}(\alpha)=0,$ then we say that $\alpha $ is a zero of a polynomial.$p(x) = x ^2 + x - 6$
Now, $p(x) = 0$
$⇒ x ^2 + x - 6$
$⇒ x ^2 + 3x - 2x - 6 = 0$
$⇒ x(x + 3) - 2(x + 3) = 0$
$⇒ (x - 2)(x + 3) = 0$
$⇒ (x - 2) = 0 or (x + 3) = 0$
$⇒ x = 2 or x = -3$
\therefore 2 and -3 are the zeroes of the polynomial $p(x).$
View full question & answer→MCQ 2981 Mark
$(207 \times 193) = ?$
- A
$39961$
- B
$38951$
- ✓
$39951$
- D
$39851$
AnswerCorrect option: C. $39951$
$207 \times 193 = (200 + 7) (200 - 7)$
$= (200)$^2$ - (7)$^2$$
$= 40000 - 49$
$= 39951$
View full question & answer→MCQ 2991 Mark
If $a^2+b^2+c^2-a b-b c-c a=0$, than.
- A
$c + a = b$
- B
$b + c = a$
- C
$a + b + c$
- ✓
$a = b = c$
AnswerCorrect option: D. $a = b = c$
$\text { Given: } a^2+b^2+c^2-a b-b c-c a=0$
$\Rightarrow 2\left(a^2+b^2+c^2-a b-b c-c a\right)=0$
$\Rightarrow\left(2 a^2+2 b^2+2 c^2-2 a b-2 b c-2 c a\right)=0$
$\Rightarrow\left(\left\{a^2+b^2-2 a b\right\}+\left\{b^2+c^2-2 b c\right\}+\left\{c^2+a^2-2 c a\right\}\right)=0$
$\Rightarrow(a-b)^2+(b-c)^2+(c-a)^2=0$
Now, since the sum of all squares is zero.
$\Rightarrow a - b = 0 $
$\Rightarrow a = b$
$\Rightarrow b - c = 0 $
$\Rightarrow b = c$
$\Rightarrow c - a = 0 $
$\Rightarrow c = a$
$\Rightarrow a = b = c$
View full question & answer→MCQ 3001 Mark
The product $(a+b)(a-b)\left(a^2-a b+b^2\right)\left(a^2+a b+b^2\right)$ is equal to:
- A
$a^6+b^6$
- ✓
$a^6-b^6$
- C
$a^3-b^3$
- D
$a^3+b^3$
AnswerCorrect option: B. $a^6-b^6$
$(a+b)(a-b)\left(a^2-a b+b^2\right)\left(a^2+a b+b^2\right)$
$\left(a^2-b^2\right)\left(a^2+b^2-a b\right)\left(a^2+b^2-a b\right)$
$=\left(a^2-b^2\right)$ $\Big\{$$\left(a^2+b^2\right)^2-(a b)^2$$\Big\}$
$=\left(a^2-b^2\right)\left\{a^4+b^4+2 a^2 b^2-a^2 b^2\right\}$
$=\left(a^2-b^2\right)\left\{a^4+b^4+a^2 b^2\right\}$
$=\left\{a^6+a^2 b^4+a^4 b^2-b^2 a^4-b^6-b^4 a^2\right\}$
$=a^6-b^6$
Hence, correct option is$ (b).$
View full question & answer→