Maths M.C.Q

$ 49\text{x}^2-\text{k}=\Big(7\text{x}+\frac{1}{3}\Big)\Big(7\text{x}-\frac{1}{3}\Big),$
$\Rightarrow\text{49}\text{x}^\text{2}-\text{k}=49\text{x}^2-\frac{1}{9} [$Using identity $(a + b) (a - b) = a^2 -b^2]$
On comapring $\text{k}=\frac{1}{9}$

 $p(x)=x^3-3 x^2+4 x+32$
$x+2=0 \Rightarrow x=-2$
By the renainder theorem, we know that when $p(x)$ is divided by
$(\mathrm{x}+2)$, the remainder is $\mathrm{p}(-2)$.
$\text { Now, } p(-2)=x^3-3 x^2+4 x+32$
$=(-2)^3-3(-2)^2+4(-2)+32$
$=-8-12-8+32$
$=4$

 $\frac{\text{x}}{\text{y}}+\frac{\text{y}}{\text{x}}=-1$
$\Rightarrow\frac{\text{x}^2+\text{y}^2}{\text{xy}}=-1$
$\Rightarrow x^2+y^2=-x y$
$\Rightarrow x^2+y^2+x y=0$
Thus, we have:
$\left(x^3-y^3\right)=(x-y)\left(x^2+y^2+x y\right)$
$=(x-y) \times 0$
$=0$

$10 x-4 x^2-3$
$p(x)-4 x^2+10 x-3$
$\Rightarrow p(0)+p(1)=\left[-4(0)^2+10(0)-3\right]+\left[-4(1)^2+10(1)-3\right]$
$\Rightarrow p(0)+p(1)=[0+0-3]+[-4+10-3]$
$\Rightarrow p(0)+p(1)=[-3]+[3]$
$\Rightarrow p(0)+p(1)=0$

$p(x) = x^3-x^2+x+1$,
$=\frac{\text{p}(-1)+\text{p}(1)}{2}$
$=\frac{(-1)^3-(-1)^2+(-1)+1+(1)^3-(1)^2+(1)+1}{2}$
$=\frac{-1-1-1+1+1-1+1+1}{2}$
$=\frac{0}{2}$
$=0$

Let $p(x)=3 x^3+8 x^2+8 x+3+5 k$ and $q(x)=x^2+x+1$
Now,
If $q(x)$ is a factor of $p(x)$, then arranging $p(x)$ in order to have $q(x)$ in common,
$p(x)=3 x^3+3 x^2+3 x+5 x^2+5 x+3+2-2+5 k[\text { Adding }+2,-2 \text { in } p(x)]$
$=3 x\left(x^2+x+1\right)+5\left(x^2+x+1\right)+5 k-2$
$p(x)=\left(x^2+x+1\right)(3 x+5)+5 k-2 \ldots(1)$
From equation $(1),$ we can see if we divide $p(x)$ by $q(x),$ then quotient will be $(3x + 5)$ and remainder will be $(5k - 2)$
But $q(x)$ is a factor of $p(x).$
So, remainder $= 0$
$⇒ 5k - 2 = 0$
$\Rightarrow\text{k}=\frac{2}{5}$

$\Big(\text{x}-\frac{1}{\text{x}}\Big)^2=\text{x}^2+\frac{1}{\text{x}^2}-2(\text{x})\frac{1}{\text{x}}$
$\Big(\text{x}-\frac{1}{\text{x}}\Big)^2=\text{x}^2+\frac{1}{\text{x}^2}-2$
$=102-2$ $\Big\{\text{x}^2+\frac{1}{\text{x}^2}=102\Big\}$
$=100$
$\Rightarrow\text{x}-\frac{1}{\text{x}}=\sqrt{100}$
$\Rightarrow\text{x}-\frac{1}{\text{x}}=10$
Hence, correct option is $(b).$

$\left(25 x^2-1\right)+(1+5 x)^2$
$=(5 x-1)(5 x+1)+(1+5 x)^2$
$=(5 x+1)[(5 x-1)+(1+5 x)]$
$=(5 x+1)(10 x)$
So, the factors of $\left(25 x^2-1\right)+(1+5 x)^2$ are $(5 x+1)$ and $10 x$

$104 × 96 = (100 + 4)(100 - 4)$
$= 100^2 - 4^2$
$= (10000 - 16)$
$= 9984$

Let $\text{a}^{\frac{1}{3}}=\text{A},\ \text{b}^{\frac{1}{3}}=\text{B}$ and $\text{c}^{\frac{1}{3}}=\text{C}$
Now, $A + B + C = 0 ($given$)$
$\text { If } A+B+C=0 \text {, then } A^3+B^3+C^3-3 A B C=0$
$\Rightarrow A^3+B^3+C^3-3 A B C=0$
$\Rightarrow A^3+B^3+C^3=3 A B C \ldots \text { (1) }$
$\begin{Bmatrix}\text{A}=\text{a}^{\frac{1}{3}},\ \text{B}=\text{b}^{\frac{1}{3}},\ \text{C}=\text{c}^{\frac{1}{3}}\\\text{A}^3=\text{a},\ \text{B}^3=\text{b},\ \text{C}^3=\text{c}\end{Bmatrix}$
Then, equation $(1)$ becomes
$\text{a}+\text{b}+\text{c}=3(\text{abc})^{\frac{1}{3}}$
Cubing both Sides of above equation, we get
$(a+b+c)^3=27 a b c$
Hence, correct option is $(b).$

$(3 x-1)^7=a_7 x^7+a_6 x^6+\ldots+a_1 x+a_0...(1)$
Putting $x=1$ in equation (1), we have
$[3(1)-1]^7=a_7+a_6+\ldots . .+a_1+a_0$
So,
$a_7+a_6+a_5+\ldots . .+a_1+a_0=2^7=128$

By using identity $(a+b)^2=a^2+b^2+2 a b$.
we have,
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=\text{x}^2+\Big(\frac{1}{\text{x}}\Big)^2+2\times\not\text{x}\times\frac{1}{\not\text{x}}$
$\Rightarrow\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=\text{x}^2+\frac{1}{\text{x}^2}+2$
$\Rightarrow(5)^2=\text{x}^2+\frac{1}{\text{x}^2}+2$ $\Big\{\text{x}+\frac{1}{\text{x}}=5\text{ given}\Big\}$
$\Rightarrow\text{x}^2+\frac{1}{\text{x}^2}=25-2$
$\Rightarrow\text{x}^2+\frac{1}{\text{x}^2}=23$
Hence, correct option is $(c).$

$\left(2-3 x^3\right)\left(x^2-5\right)$
$=2 x^2-10-3 x^4+15 x^2$
$=-3 x^4+17 x^2-10$
Therefore, the coefficient of $x^2$ is $17 .$

The given equation is
$x^3-3 x^2+3 x-7=(x+1)\left(a x^2+b x+c\right)$
This can be written as
$x^3-3 x^2+3 x-7=(x+1)\left(a x^2+b x+c\right)$
$=x^3-3 x^2+3 x-7=a x^3+b x^2+c x+a x^2+b x+c$
$=x^3-3 x^2+3 x-7=a x^3+(a+b) x^2+(b+c) x+c$
Comparing the cofficients on both sides of the equation.
We get,
$a = 1 ...(1)$
$a + b = 3 ...(2)$
$b + c = 3 ...(3)$
$c = -7 ...(4)$
Putting the value of a form $(1)$ in $(2)$
We get,
$1 + b = 3,$
$b = -3 - 1$
$b= -4$
So the value of $a, b$ and $c$ is $1, -4$ and $-7$ respectively.
Therefore,
$a + b + c = 1 - 4 - 7 = -10$
Hence, correct option is $(c).$

If $x - a$ is a factor then $x - a = 0$
Therefore, $x = a$
New substitute $x = a$ in polynomial
$\left(a^3\right)-3(a)^2 a+2 a^2(a)+b=0$
$a^3-3 a^3+2 a^3+b=0$
$-3 a^3+3 a^3+b=0$
$b=0$

$4 x^4+0 x^3+0 x^5+5 x+7$
$=4 x^4+5 x+7$
Here, the height power is $4.$
Therefore, the degree of given polynomial is $4.$

$x^{51}+51$
If $x^{51}+51$ is divided by $x + 1,$ then using remainder theorem
$(-1)^{51}+51$
$= -1 + 51$
$= 50$

We know that $(a+b+c)^2=a^2+b^2+c^2+2(a b+b c+c a)$
Here, $a+b+c=9, a b+b c+c a=23$
Thus, we have
$(9)^2=a^2+b^2+c^2+2(23)$
$81=a^2+b^2+c^2+46$
$a^2+b^2+c^2=81-46$
$a^2+b^2+c^2=35$
Hence, correct option is $(a).$

$(x + 5)$ is a factor or $p(x) = x^3 - 20x +5k$
$\therefore p(-5) = 0$
$⇒ (-5)^3 -20 × (-5) + 5k = 0$
$⇒ -125 + 100 + 5k = 0$
$⇒ 5k = 25$
$⇒ k = 5$

The value of $x^2+4 y^2+4 y-4 x y-2 x-8$ is.
$=x^2+4 y^2-4 x y+4 y-2 x-8$
$=(x-2 y)^2+2(2 y-x)-8$
$=(x-2 y)^2-2(x-2 y)-8$
Let $A=x-2 y$
Thus, $A^2-2 A-8$
$=(A-4)(A+2)$
Re-substitute the value of $A$.
$\text { Thus } x^2+4 y^2+4 y-4 x y-2 x-8$
$=(x-2 y-4)(x-2 y+2)$

Let: $\left(x^{100}+2 x^{99}+k\right)$
Now, $x+1=0 \Rightarrow x=-1$
$\therefore  p(-1)=0$
$\Rightarrow(1)^{100}+2 \times(-1)^{99}+k=0$
$\Rightarrow 1-2+k=0$
$\Rightarrow-1+k=0$
$\Rightarrow k=1$

A 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.
In option $(d),$
$\text{x}^2+\frac{2\text{x}^\frac{3}{2}}{\sqrt{\text{x}}}+6=\text{x}^2+2\text{x}^{\frac{3}{2}-\frac{1}{2}}+6$
$x^2 + 2x^1 + 6$ Which is a polynomial.

$0$ is a polynomial whose degree is not defined.

$y$ is a polynomial because it has a non- negative integral power $1.$

$305 × 308 = (300 + 5) (300 + 8)$
$= 300(300 + 8) + 5(300 + 8)$
$= 90000 + 2400 + 1500 + 40$
$= 93940$

$\text{x}+\frac{1}{\text{x}}=7,$
$\Rightarrow\Big(\text{x}+\frac{1}{\text{x}}\Big)^3=7^3$
$\Rightarrow\text{x}^3+\frac{1}{\text{x}^3}+3\times\text{x}\times\frac{1}{\text{x}}(\text{x}+\frac{1}{\text{x}})=343$
$\Rightarrow\text{x}^3+\frac{1}{\text{x}^3}+3\times7=343$
$\Rightarrow\text{x}^3+\frac{1}{\text{x}^3}=343-21$
$\Rightarrow\text{x}^3+\frac{1}{\text{x}^3}=322$

The coefficient of $x^3$ in $2 x+x^2-5 x^3+x^4$ is $-5 .$

$p(x)=x^{100}+2 x^{99}+k$
$x+1=0 \Rightarrow x=-1$
By the factor theorem, we know that when $p(x)$ is divided by $(x+1)$, the remainder is $p(-1)$.
Now, $p(-1)=(-1)^{100}+2(-1)^{99}+k$
$\Rightarrow 0=1-2+k \ldots($ Given that $\mathrm{p}(\mathrm{x})$ is divisible by $\mathrm{x}+1$.)
$\Rightarrow k=1$