Maths M.C.Q

$x^2+k x+6=(x+2)(x+3)$
$\Rightarrow x^2+k x+6=x^2+(2+3) x+2 \times 3$
$\Rightarrow x^2+k x+6=x^2+5 x+6$
On comparing the terms,
We get $k=5$

We know that,
$(x+y+z)^2=x^2+y^2+z^2+2 x y+2 y z+2 z x$
$2 a^2+b^2+4+4 a b+8 a+4 b+4$
$=4 a^2+b^2+4+4 a b+8 a+4 b$
$=\left(2 a^2\right)+b^2+2^2+2(2 a) b+2(2 a)(2)+2(2 b)$
$=(2 a+b+2)^2$

$ x^{31} - 31$
Using remainder theorem.
$= (-1)^{31}- 31$
$= -1 - 31$
$= -32$

 When the polynomial $p(x)$ is divided by $q(x)$ i. e. $(\text{x}\pm\alpha)$ then $\text{p}(\mp\alpha)$ is the remainder.
If $\text{x}\pm\alpha$ is the factor of polynomial, then remainder is $'0'.$
So,
If $x^{51} + 51$ is divided $x + 1$.
Remainder $= (-1)^{51} + 51 = -1 + 51 = 50.$

 Now, $4x^2 + 8x + 3 = 4x^2  + 6x + 2x + 3 [$by splitting middle term$]$
$= 2x(2x + 3) + 1(2x + 3)$
$= (2x + 3) (2x + 1)$

 $\Big(\text{x}+\frac{1}{\text{x}}\Big)^3=\text{x}^3+\frac{1}{\text{x}^3}+3\Big(\text{x}+\frac{1}{\text{x}}\Big)$
$\Rightarrow\Big(\text{x}+\frac{1}{\text{x}}\Big)^3-3\Big(\text{x}+\frac{1}{\text{x}}\Big)=\text{x}^3+\frac{1}{\text{x}^3}$
$\Rightarrow\Big(\text{x}+\frac{1}{\text{x}}\Big)^3-3\Big(\text{x}+\frac{1}{\text{x}}\Big)=110$
Let $\text{x}+\frac{1}{\text{x}}=\text{t}$
$\Rightarrow\text{t}^3-3\text{t}-110=0$
t =5 is one of it's solution which is real, other two solutions are imaginary
$\Rightarrow\text{x}+\frac{1}{\text{x}}=5$
Hence, correct option is $(a).$

If $x+1$ is a factor of $x^n+1$
then, at $x=-1, x^n+1=0$
$(-1)^n+1=0$
$(-1)^n=-1$
$(-1)^{\mathrm{n}}$ will be equal to $-1$ if and only if $n$ is an odd integer.
If $n$ is even, then $(-1)^{\mathrm{n}}=1$
So, $n$ should be an odd integer.

$x^3-x^2 y-x y^2+y^3=x^3+y^3-x y(x+y)$
Now by identity $x^3+y^3=(x+y)\left(x^2+y^2-x y\right)$, we have
$x^3-x^2 y-x y^2+y^3=(x+y)\left(x^2+y^2-x y\right)-x y(x+y)$
$=(x+y)\left(x^2+y^2-x y-x y\right)$
$=(x+y)\left(x^2+y^2-2 x y\right)$
$=(x+y)(x-y)^2$
Hence, correct option is $(d).$

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

 $(x+3)^4$
$=(x+3)^2 \times(x+3)^2$
$=\left[x^2+6 x+9\right]\left[x^2+6 x+9\right]$
$=x^4+36 x^2+81+12 x^3+108 x+18 x^2$
$=x^4+12 x^3+54 x^2+108 x+81$
Therefore, the coefficient of $x^2$ is $54 .$

$\text{p}(\text{x})=\text{x}^2-2\sqrt2\text{x}+1$
$\text{p}(2\sqrt2)=(2\sqrt2)^2-2\sqrt2(2\sqrt2)+1$
$=8-8+1$
$=1$

Let $p(x) = 5x - 4x^2 + 3 ... (i)$
On putting $x= -1$ in eq. $(i),$ we get
$p(-1) = 5(-1) - 4(-1)^2 + 3 = -5 - 4 + 3 = -6$

$\text{a}^{\frac{1}{3}}+\text{b}^{\frac{1}{3}}+\text{c}^{\frac{1}{3}}=0$
$\Rightarrow\text{a}^{\frac{1}{3}}+\text{b}^{\frac{1}{3}}=-\text{c}^{\frac{1}{3}}$
$\Rightarrow\Big[(\text{a}^{\frac{1}{3}})(\text{b}^{\frac{1}{3}})\Big]^3=\Big(-\text{c}^{\frac{1}{3}}\Big)^3$
$\Rightarrow\text{a}+\text{b}+\Big[3\times\text{a}^{\frac{1}{3}}\times\text{b}^{\frac{1}{3}}\Big(\text{a}^{\frac{1}{3}}+\text{b}^{\frac{1}{3}}\Big)\Big]=-\text{c}$
$\Rightarrow\text{a}+\text{b}+3\times\text{a}^{\frac{1}{3}}\times\text{b}^{\frac{1}{3}}\Big(-\text{c}^{\frac{1}{3}}\Big)=-\text{c}$
$\Rightarrow\text{a}+\text{b}+\text{c}=3\times\text{a}^{\frac{1}{3}}\times\text{b}^{\frac{1}{3}}\times\text{c}^{\frac{1}{3}}$
$\Rightarrow(\text{a}+\text{b}+\text{c})^3=\Big(3\times\text{a}^{\frac{1}{3}}\times\text{b}^{\frac{1}{3}}\times\text{c}^{\frac{1}{3}}\Big)$
$\Rightarrow(\text{a}+\text{b}+\text{c})^3=27\text{abc}$

 $4 a^2+4 a-3$
To find the length and breadth, we will factorize the given polynomial.
$=4 a^2-6 a-2 a-3$
$=2 a(a+3)-1(2 a+3)$
$=(2 a+3)(2 a-1)$
Therefore, the Possible expressions for the length and breadth of the rectangle whose area is given by $4 a^2+4 a-3$ is ( $2 a+3$ ) and $(2 a-1)$.

 The zero of the polynomial $p(x)$ can be obtained by putting $p(x)$
$p(x) = 0$
$⇒ 2x + 5 = 0$
$⇒ 2x = -5$
$\Rightarrow\text{x}=\frac{-5}{2}$

 Now, $4 x^2+8 x+3=4 x^2+6 x+2 x+3 [$by splitting middle term$]$
$= 2x(2x + 3) + 1 (2x + 3)$
$= (2x + 3)(2x + 1)$

 $(a+b+c)^2=a^2+b^2+c^2+2 a b+2 b c+2 c a$
$\Rightarrow(9)^2=a^2+b^2+c^2+2(a b+b c+c a)$
$\Rightarrow(9)^2=a^2+b^2+c^2+2(23)$
$\Rightarrow a^2+b^2+c^2=81-46=35$
as we know that $a^3+b^3+c^3-3 a b c=(a+b+c)\left(a^2+b^2+c^2-a b-b c-c a\right)$
$\Rightarrow a^3+b^3+c^3-3 a b c=9 \times(35-23)$
$\Rightarrow a^3+b^3+c^3-3 a b c=108$

 A polynomial of degree $3$ is called a cubic polynomial.
Its general form is $ax^3 + bx^2 + cx + d.$

 The linear polynomial $(x - 1)$ is a factor of $xn + 1,$ only if,
$f(-1) = (-1)^n+ 1 = 0$ 
If n is odd integer, then $f(-1) = -1 + 1 = 0$

 $x^3+x^2+x+1=x^3(x+1)+1(x+3)=x^3+27~9 x^2+27 x=x^3+9 x^2+27 x+27$
therefore, the coefficient of $x$, in the expansion of $(x+3)^3$ is $27$ .

 $\Big(\text{x}-\frac{1}{\text{x}}\Big)^3=\text{x}^3-\frac{1}{\text{x}^3}-3\not\text{x}\frac{1}{\not\text{x}}\Big(\text{x}-\frac{1}{\text{x}}\Big)$
$\text{x}^3-\frac{1}{\text{x}^3}=\Big(\text{x}-\frac{1}{\text{x}}\Big)^3+3\Big(\text{x}-\frac{1}{\text{x}}\Big)$
$\Rightarrow\Big(\text{x}-\frac{1}{\text{x}}\Big)^3+3\Big(\text{x}-\frac{1}{\text{x}}\Big)-\text{x}^3-\frac{1}{\text{x}^3}=0$
$\Rightarrow\Big(\text{x}-\frac{1}{\text{x}}\Big)^3+3\Big(\text{x}-\frac{1}{\text{x}}\Big)-14=0$
Let $\Rightarrow\text{x}-\frac{1}{\text{x}}=\text{t}$
$\Rightarrow t^3+3 t-14=0$
$\Rightarrow t^3-2 t^2+2 t^2-4 t+7 t-14=0$
$\Rightarrow t(t-2)+2 t(t-2)+7(t-2)=0$
$\Rightarrow(t-2)(t+2 t+7)=0$
$\Rightarrow t^2+2 t+7=0 \text { has no real roots }$
So, $t = 2$ is a solution
$\Rightarrow\text{x}-\frac{1}{\text{x}}=2$
Hence, correct option is $(d).$

 $a+b+c=0$
$\Rightarrow a+b=-c$
$\Rightarrow(a+b)^3=(-c)^3$
$\Rightarrow a^3+b^3+3 a b(a+b)=-c^3$
$\Rightarrow a^3+b^3+3 a b(-c)=-c^3$
$\Rightarrow a^3+b^3+c^3=3 a b c$