Maths M.C.Q

By remainder theorem, when $p(x) = x^3 - ax^2 + x$ is divided by $(x - a),$ then the remainder $= p(a)$
Putting $x = a $ in $p(x),$ we get
$p(a) = a^3 - a × a^2 + a = ^3 - a^3 + a = a$
$\therefore$ Remainder $= a$

$-8$
$f(x) = x^3 + 6x^2 + 4x + k$
$f(-2) = 0$
$\therefore (-2)^3 + 6(-2)^2 + 4(-2) + k = 0$
$\therefore -8 + 6(4) + (-8) + k = 0$
$24 - 16 + k = 0$
$k + 8 = 0$
$k = -8$

$x + 1$ is a factor of $p(x) = 2x^2 + kx$
Then, $p(-1) = 0$
$i.e. 2(-1)^2 + k(-1) = 0$
$2 - k = 0$
$k = 2$

$(249)^2 - (248)^2 = (249 + 248)(249 - 248) [(a)^2 - (b)^2 = (a + b)(a - b)]$
$= (497)(1) = 497$

Let: $\text{p(x)} = 3\text{x}^2 - 1$
To find the zeroes of $p(x),$
We have:
$\text{p(x)}=0\Rightarrow3\text{x}^2-1=0$
$\Rightarrow3\text{x}^2=1$
$\Rightarrow\text{x}^2=\frac{1}{3}$
$\Rightarrow\text{x}=\pm\frac{1}{\sqrt{3}}$
$\Rightarrow\text{x}=\frac{1}{\sqrt{3}}\text{ and }\text{x}=\frac{-1}{\sqrt{3}}$

Given: $p(x) = (x - 1) (x + 1),$ then
$p(2) + p(1) - p(0)$
$= (2 - 1) (2 + 1) + (1 - 1) (1 + 1) - (0 - 1) (0 + 1)$
$= 1 × 3 + 0 × 2 -(-1) × 1$
$= 3 + 0 + 1$
$= 4$

Let $f(x) = px^2 + 5x + r$
Now,
If $x - 2$ and $\text{x}-\frac{1}{2}$ are factors of $f(x).$
Then at $x = 2$ and $\text{x}-\frac{1}{2}, f(x) = 0.$
So, $f(2) = 0, \text{f}\Big(\frac{1}{2}\Big)=0$
$⇒ p(2)^2 + 5(2) + r = 0$
And, $\text{p}\Big(\frac{1}{2}\Big)^2+5\Big(\frac{1}{2}\Big)+\text{r}=0$
$⇒ 4p + r + 10 = 0 ...(1)$
$And 4r + p + 10 = 0 ...(2)$
Subtracting equation $(2)$ from $(1),$ we have
$3p - 3r = 0$
$⇒ p = r$

$ \frac{\text{a}}{\text{b}}+\frac{\text{b}}{\text{a}}=-1$
$\Rightarrow\frac{\text{a}^2+\text{b}^2}{\text{ab}}=-1$
$⇒ a^2 + b^2 = -ab$
$⇒ a^2 + b^2 + ab = 0$
Thus, we have:
$(a^3 - b^3) = (a - b)(a^2 + b^2 + ab)$
$= (a - b) × 0$
$= 0$

Given: $\text{x}^3-\Big(\frac{1}{\text{x}^3}\Big)=14$
Let $x = a$ and $\frac{1}{\text{x}}=\text{b}$
Say, $\text{x}-\frac{1}{\text{x}}=\text{A}$
Then, $\text{a}^3-\text{b}^3=14$
$\Rightarrow(\text{a}-\text{b})(\text{a}^2+\text{ab}+\text{b}^2)=14$
$\Rightarrow(\text{a}-\text{b})(\{(\text{a}-\text{b})^2+2\text{ab}\}+2\text{ab})=14$
$\Rightarrow(\text{a}-\text{b})\{(\text{a}-\text{b})^2+3\text{ab}\}=14$
$\Rightarrow(\text{a}-\text{b})\{(\text{a}-\text{b})^2+3\}=14$
$\Rightarrow\text{A}(\text{A}^2+3)=14$
$\Rightarrow\text{A}(\text{A}^2+3)=14$
$\Rightarrow\text{A}^3+3\text{A}-14=0$
$\Rightarrow\text{A}^3-2\text{A}^2+2\text{A}^2-4\text{A}+7\text{A}-14=0$
$\Rightarrow\text{A}^2(\text{A}-2)+2\text{Y}(\text{Y}-2)+7(\text{Y}-2)=0$
$\Rightarrow(\text{A}-2)(\text{A}^2+2\text{A}+7)=0$
$\Rightarrow\text{A}-2=0,$
$\Rightarrow\text{A}=2$
$\Rightarrow\text{x}-\frac{1}{\text{x}}=2$

$3$ terms of not more than the power of $3$

Now, $a^3 + b^3 + c^3 = (a + b + c)(a^2 + b^2 + c^2 - ab - be - ca) + 3abc$
$[$Using identity, $a^3 + b^3 + c^3 – 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - be - ca)] = 0 + 3abc$
$\therefore a + b + c = 0,$ given$a^3 + b^3 + c^3 = 3abc$

On cubing we get.
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^3=\text{x}^3+\Big(\frac{1}{\text{x}^3}\Big)+3\times\text{x}\times\frac{1}{\text{x}}\Big(\text{x}+\frac{1}{\text{x}}\Big)$
$\Rightarrow27=\text{x}^3+\Big(\frac{1}{\text{x}^3}\Big)+3\times\text{3}$
$\Rightarrow\text{x}^3+\Big(\frac{1}{\text{x}^3}\Big)=27-9$
$\Rightarrow\text{x}^3+\Big(\frac{1}{\text{x}^3}\Big)=18$
Now, $\Big(\text{x}^3+\frac{1}{\text{x}^3}\Big)^2=\text{x}^6+\Big(\frac{1}{\text{x}^6}\Big)+2\times\text{x}^3\times\frac{1}{\text{x}^3}$
$\Rightarrow18^2=\text{x}^6+\Big(\frac{1}{\text{x}^6}\Big)+2$
$\text{x}^6+\Big(\frac{1}{\text{x}^6}\Big)=324-2=322$

$p(-1) = (-1)^3 + 3(-1)^2 + 3(-1) + 1$
$= -1 + 3 × 1 - 3 + 1$
$= -1 + 3 - 3 + 1$
$= 0$

Let: $p(x) = (x + 4)$
$\therefore p(-x) = (x + 4)$
$= -x + 4$
Thus, we have: $p(x) + p(-x) =\{(x + 4) + (-x + 4)\}$
$= 4 + 4$
$= 8$

$(x + 3)^3$
$= x^3 + (3)^3 + 3 × x × 3(x + 3)$
$= x^3 +27 + 9x^2 +27x$
$= x^3 + 9x^2 +27x + 27$
Therefore, the coefficient of $x,$ in the expansion of $(x + 3)^3$ is $27.$

$x - 3$ is a factor of $x^2 - ax - 15,$
then at $x = 3,$
$x^2 - ax - 15 = 0$
$i.e. (3)^2 - a(3) - 15 = 0$
$9 - 3a - 15 = 0$
$a = -2$

Given that,
$(3x - 1)^7 = a_7x^2 + a_5x^5 + \_\_\_ + a_1x + a_0$
Putting $x = 1$
We get$(3 × 1 - 1)^7 = a_6(1)^5 + a_5(1)^5 \_\_\_\_ + a_1(1) + a_0$
$⇒ a_6 + a_5 + \_\_\_\_ + a_1 + a_0 = 2^7 = 128$

If $p(x)$ is divided by $x + a,$ then the remainder is $p(-a).$
Here $p(x) = x^{51} + 51$ is divided by $x + 1,$ then
$x = -1$
Remainder $= p(-1) = (-1)^{51} + 51 = 50 = -1 + 51 = 50$

Finding a zero of $p(x)$ is the same as solving an equation $p(x) = 0.$
Now, $p(x) = 0 ⇒ 2x + 5 = 0,$
$2x = -5$
Which give us $\text{x}=-\frac{5}{2}.$
Therefore, $-\frac{5}{2}$ is the zero of the polynomial.

Volume of a cuboid of side $a, b$ and $c = abc$
Now, Volume $= 3x^2 - 27 ($given$)$
$abc = 3(x^2 - 9)$
$abc = 3(x - 3)(x + 3)$
So, possible dimensions are $3, x - 3$ and $x + 3$
Hence, correct option is $(b).$

$x^4 + 4$
$= x^4 + 4 + 4x^2 - 4x^2$
$= (x^4 + 4x^2 + 4) - 4x^2$
$= (x^2 + 2)^2 - (2x)^2$
$= (x^2 + 2 - 2x)(x^2 + 2 + 2x)$
$= (x^2 + 2x + 2)(x^2 - 2x + 2)$
Hence, correct option is $(a).$

We know, $(a - b)^2 = a^2 + b^2 + 2ab$
Here, $a = 75$ and $b = 25$
$⇒ (75 × 75) + (2 × 75 × 25) + (25 × 25)$
$= 75^2 + 2 × 75 × 25 + 25^2$
$= (75 + 25)^2$
$= (100)^2$
$= 10000$

 Assume $a = 0.013$ and $v = 0.007.$
Than 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 sum 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. So, neither $a^3 + b^3$ nor any factor of it can be zero.
Therefore we can cancel the term $(a^2 - ab + b^2)$ from both numerator and denominator. then the expression becomes
$\frac{(\text{a}+\text{b})(\text{a}^2-\text{ab}+\text{b}^2)}{\text{a}^2-\text{ab}+\text{b}^2}=\text{a}+\text{b}$
$=0.013+0.007$
$=0.02$

 A coefficient is a multiplicative factor in a term of a polynomial or any expression.
Therefore, the coefficient of $x^2$ in the polynomial $2 x^3+4 x^2+3 x+1$ is $4.$

 $x^2 + 5x + 4$ is a polynomial of deree $2.$
So, it is a quadratic polynomial.

 Given expression is $\left(x^2-1\right)\left(x^4+x^2+1\right)$
Let ${x}^2={A}$ and $1=\mathrm{B}$
Then, we have
$(A-B)\left(A^2+A B+B^2\right)$
$= A^3-B^3$
$= (X^2)^3-(1)^3$
$= X^6-1$
Hence, correct option is $(c).$

 Put $a = x + y$ and $b = x - y, $ then
$(x+y)^3-(x-y)^3=a^3-b^3$
$=(a-b)\left(a^2+b^2+a b\right)$
$=(x+y-x+y)\left[(x+y)^2+(x-y)^2+(x-y)(x+y)\right]$
$=2 y\left[2\left(x^2+y^2\right)+\left(x^2-y^2\right)\right]$
$=2 y\left[3 x^2+y^2\right]$

 Given: $x + y = 5 ⇒ x = 5 - y$
$x^3+y^3+15 x y-125$
$\text { Putting the value of } x \text {, we get }$
$(5-y)^3+y^3+15(5-y) y-125$
$=125-y^3-3 \times 5 \times y(5-y)+y^3+15(5-y) y-125$
$=125-y^3-75 y+15 y^2+y^3+75 y-15 y^2-125$
$=0$

 $x^2+4 y^2+4 y-4 x y-2 x-8$
$=x^2+(2 y)^2-2 \times x(2 y)+4 y-2 x-8$
$=(x-2 y)^2+4 y-2 x-8 \ldots(1)$
Now making eq$(1)$ a perfect square by adding $1$ and $-1$
$(x-2 y)^2+4 y-2 x-8=(x-2 y)^2+4 y-2 x-8+1-1$
$=(x-2 y)^2+(1)^2-2 \times(1) \times(x-2 y)-9$
$=(x-2 y-1)^2-(3)^2$
$=[(x-2 y-1)-3][x-2 y-1+3]$
$=(x-2 y-4)(x-2 y+2)$
Hence, correct option is $(a).$

 $(x-y)(x+y)=x^2-y^2\left[\text { by identity }(a+b)(a-b)=a^2-b^2\right]$
$\left(x^2-y^2\right)\left(x^2+y^2\right)=x^4-y^4$
$\left(x^4-y^4\right)\left(x^4+y^4\right)=x^8-y^8$
$\text { Now, }$
$(x-y)(x+y)\left(x^2+y^2\right)\left(x^4+y^4\right)$
$=\left(x^2-y^2\right)\left(x^2+y^2\right)\left(x^4+y^4\right)$
$=\left(x^4-y^4\right)\left(x^4+y^4\right)$
$=x^8-y^8$
Hence, correct option is $(b).$

$(x^2 - 4x - 21) = x^2 - 7x + 3x - 21$
$= x(x - 7) + 3(x - 7)$
$= (x - 7) (x + 3)$

 $(x+3)^3$
$=x^3+3^3+9 x(x+3)$
$=x^3+27+9 x^2+27 x$
So, the coefficient of $x$ in $(x+3)^3$ is $27.$