MATHS M.C.Q

$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=\text{x}^2+\frac{1}{\text{x}^2}+2$
$\text{x}+\frac{1}{\text{x}}=3$ (given)
$\Rightarrow\text{x}^2+\frac{1}{\text{x}^2}=(3)^2-2$
$\Rightarrow\text{x}^2+\frac{1}{\text{x}^2}=7\ ...(1)$
Cubing both side of equation (1). we have
$\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)^3=(7)^3$
$\Rightarrow(\text{x}^2)^3+\Big(\frac{1}{\text{x}^2}\Big)^3+3(\text{x}^2)\frac{1}{\text{x}^2}\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)=7^3$
$\Rightarrow\text{x}^6+\frac{1}{\text{x}^6}+3(7)=7^3$
$\Rightarrow\text{x}^6+\frac{1}{\text{x}^6}=343-21$
$\Rightarrow\text{x}^6+\frac{1}{\text{x}^6}=322$
Hence, correct option is $(d).$

$x^2-9$
$x^2-3^2$
$=(x+3)(x-3)\left[\text { Using identity } a^2-b^2=(a+b)(a-b)\right]$
Then the zeroes are $x + 3 = 0$ and $x - 3 = 0$
$\Rightarrow x = -3$ and $x = 3$

 It is given $(x+2)$ and $(x-1)$ are the factors of the polynomial $f(x)=x^3+10 x^2+m x+n$
i.e., $f(-2)=0$ and $f(1)=0$
New,
$f(-2)=(-2)^3+10(-2)^2+\mathrm{m}(-2)+\mathrm{n}=0$
$-8+40-2 \mathrm{~m}+\mathrm{n}=0$
$\Rightarrow-2 \mathrm{~m}+\mathrm{n}=-32$
$\Rightarrow 2 \mathrm{~m}-\mathrm{n}=32 \ldots . \text { (i) }$
$\mathrm{f}(1)=(1)^3+10(1)^2+\mathrm{m}(1)+\mathrm{n}=0$
$1+10+\mathrm{m}+\mathrm{n}=0$
$\mathrm{~m}+\mathrm{n}=-11 \ldots \text { (ii) }$
Solving equation $(i)$ and $(ii)$ we get,
$\mathrm{m}=7 \text { and } \mathrm{n}=-18$

 $(x-2)^2-(x+2)^2$
$= (x - 2 + x + 2) (x - 2 - x - 2) [$Using identity $a^2-b^2 = (a + b) (a - b)]$
$= (2x) (-4)$
$= -8x$
Then the zero is,
$-8 = 0$
$⇒ x = 0$

 Let $p(x) = 2 x^2+k x$
Since, $(x + 1)$ is a factor of $p(x)$, then
$p(-1) = 0$
$2(-1)^2 + k(-1) = 0$
$⇒ 2 - k = 0$
$⇒ k = 2$

 $P(x) = 5x - 4x^2 + 3$
$⇒ p(-1) = 5(-1) - 4(-1)^2 + 3$
$= -5 - 4 + 3$
$= -6$

 The general form of a polynomial is $a_n x^n$, where n is a natural number.
For zero polynomial $a =0 $.
Since the largest value of $n$ for which an is non-zero is negative infinity $($all the integers are bigger than negative infinity$).$
Therefore, the degree of zero polynomials is not defined.

 Given: $f(x) =px^3 + x^2 - 2x + q$
If $x + 1$ is a factor of $ f(x).$
Then $f(-1) = 0$
$p(-1)^3 + (-1)^2 - 2(1) + q = 0$
$-p + 1 + 2 + q = 0$
$-p + q = -3$
$p - q = 3 ......(i)$
Also, if $x - 1$ is a factor of $f(x),$ then
$p(1)^3 + (1)^2 - 2(1) + q = 0$
$p + 1 - 2 + q = 0$
$p + q = 1 ......(ii)$
$2p = 4$
$p = 2$
subtracting eq.$(ii)$ from eq. $(i),$
we get,
$-2q = 2$
$q = -1$
$q = -1$
Therefore, $p = 2, q = -1$

 $2x^2 + 7x - 4$
$= 2x^2 + 8x - x - 4$
$= 2x(x + 4) - 1(x + 4)$
$= (2x - 1)(x + 4)$
$2x - 1 = 0$ and $x + 4 = 0$
$\text{x}=\frac{1}{2}$ and $x = -4$
Therefore, one zero of the given polynomial is $\frac{1}{2}$

 $4x^3$ because monomial means only one term in an expression.

$ p(x) = x(x - 2) (x + 3)$
$⇒ x = 0$ and $x - 2 = 0$ and $x + 3 = 0$
$⇒ x = 0, x = 2$ and $x = -3$
Therefore, the zeroes are $0, 2, -3$

 In a polynomial, the power of the variable of each term should be a whole number.
Here, the power of variable $x$ is $3,$ which is a whole number.
Therefore, it is a polynomial.

 The given expression to be factorized is $a^2 - 1 - 2x - x^2$
Take common -1 from the last three terms and then we have
$a^2 - 1 - 2x - x^2 = a^2 - (1 + 2x + x^3)$
$= a^2 - {(1)^2 + 2.1 × x + (x)^2}$
$= a^2 - (1 + x)^2$
$= (a)^2- (1 + x)^2$
$= {a + (1 + x)} {a - (1 + x)}$
$= (a + 1 + x) (a - 1 - x)$
$= (a + x + 1) (a - x - 1)$

 A polynomial of degree $1$ is called a linear polynomial.
Options $(a),$ and $(c)$ have degree $2,$
so ther are quadratic polynomials.
option $(d)$ has a negative power, so it is not a polynomial.
The degree of $x + 1$ is $1,$ so it is a linear polynomial.

 As $(x^2 - 1)$ Is a factor of polynomial
$f(x^2) = ax^4 + bx^3 + cx^2 + dx + e$
Therefore,
$f(x) = 0$
And,
$f(1) = 0$
$a(1)^2 + b(1)^3 + c(1)^2 + d(1) + e = 0$
$⇒ a + b + c + d + e = 0$
And,
$f(-1) = 0$
$a(-1)^4 + b(-1)^3 + c(-1)^2 + d(-1) + e = 0$
$a - b - c - d + e = 0$
Hence, $a + c + e = b + d$

 By we know that $a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$
If $a + b + c = 0,$ then
$a^3 + b^3 + c^3 = 3abc$
In given expression,
Let $a - b = A, b - c = B, c - a = C$
Now, $a - b + b - c + c - a = 0$
i.e. $A + B + C = 0$
$⇒ A^3 + B^3 + C^3 = 3ABC$
$⇒ (a - b)^3 + (b - c)^3 + (c - a)^3 = 3(a - b)(b - c)(c - a)$
Hence, correct option is $(b).$

If both $x - a$ and $\text{x}-\frac{1}{2}$ are the factors of $\text{px}^2 + 5\text{x} +\text{r},$ than $f(2) = 0$
$\Rightarrow\text{p}(2)^2+5(2)+\text{r}=0$
$\Rightarrow4\text{p}+10+\text{r} = 0$
$\Rightarrow4\text{p}+\text{r}=-10\ ...(\text{i})$
Also, $\text{f}\Big(\frac{1}{2}\Big)=0$
$\Rightarrow\text{p}\Big(\frac{1}{2}\Big)^2+5\Big(\frac{1}{2}\Big)+\text{r}=0$
$\Rightarrow\frac{\text{p}}{4}+\frac{5}{2}+\text{r}=0$
$\Rightarrow\text{p}+10+4\text{r}=0$
$\Rightarrow\text{p}+4\text{r}=-10\ ...{\text{ii}}$
From eq. $(i)$ and eq. $(ii),$ we get
$4\text{p}+\text{r}=\text{p}+4\text{r}$
$\Rightarrow3\text{p}=3\text{r}$
$\Rightarrow\text{p}=\text{r}$

$ (x - a)^3 + (x - b)^3 + (x - c)^3 - 3(x - a) (x - b) (x - c)$
$= [x - a + x - b + x - c]$
$[(x - a^2) + (x - b^2) + (x - c^2) - (x - a) (x - b) - (x - b) (x - c) - (x - c) (x - a)]$
$= [3x - ({a + b + c})]$
$[(x - a)^2 + (x - b)^2 + (x - c)^2 - (x - a) (x - b) - (x - b) (x - c) - (x - c) (x -a)]$
$= 3x - 3x$
$[(x - a)^2 + (x - b)^2 + (x - c)^2 - (x - a) (x - b) - (x - b) (x - c) - (x - c) (x - a)]$
$= 0$
$[0] [(x - a)^2 + (x - b)^2 + (x - c)^2 - (x - a) (x - b) - (x - b) (x - c) - (x - c) (x - a)]$

$\sqrt{2}$ is a constant polynomial. The only term here is $\sqrt{2}$ which can be written as $\sqrt{2}\text{x}^\circ.$ So, the exponent of $x$ is zero. Therefore, the degree of the polynomial is $0.$

$p(x) = 5x - 4x^2 + 3$
Putting $x = -1$ in $p(x),$ we get
$p(-1) = 5 × (-1)-4 × (-1)^2 + 3 = -5 - 4 + 3 = -6$

Let $p(x) = 2x^2 + 7x - 4$
$= 2x^2 + 8x - x- 4 [$by splitting middle term$]$
$= 2x(x + 4) -1(x + 4)$
$=(2x - 1)(x + 4)$
For zeroes of $p(x),$ put $p(x) = 0 ⇒ (2x - 1)(x + 4) = 0$
$⇒ 2x - 1 = 0$ and $x + 4 = 0$
$\Rightarrow\text{x}=\frac{1}{2}$ and $\text{x}=-4$
Hence, one of the zeroes of the polynomial $p(x)$ is $\frac{1}{2}.$

 Zero 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.

 $\frac{(2.3)^3-0.027}{(2.3)^2+0.69+0.09}$
$=\frac{(2.3)^3-(0.3)^3}{(2.3)^2+(0.3)^3+(2.3)(0.3)}$
$=\frac{(2.3 - 0.3)\{(2.3)^2+(0.3)^2+(2.3)(0.3)\}}{((2.3)^2+(0.3)^2+(2.3)(0.3))}$
$=2.3 - 0.3$
$=2$
Hence, correct option is $(a).$

 $\frac{(0.87)^3+(0.13)^3}{(0.87)^2-(0.87\times0.13)+(0.13)^2}$
$=\frac{(0.87+0.13)[(0.87)^2-(0.87\times0.13)+(0.13)^2]}{(0.87)^2-(0.87\times0.13)+(0.13)^2}$
$=0.87+0.13$
$=1$

 Let $p(x)$ be a polynomial. If $\text{p}(\alpha)=0,$ then we say that $\alpha$ is a zero of a polynomial.
$p(x) = x^2 - 3x$
Now, $p(x) = 0$
$⇒ x^2 - 3x$
$⇒ x(x - 3) = 0$
$⇒ x = 0$ or $(x - 3) = 0$
$⇒ x = 0$ or $x = 3$
$\therefore 0$ and $3$ are the zeroes of the polynomial $p(x).$

 $p(x) = x^3 - ax^2 + x$
$x - a = 0 ⇒ x = a$
By the remainder theorem, we know that when $p(x)$ is divided by $(x - a)$, the remainder is $p(a).$
Now, $p(a) = a^3 - ax^2 + a$
$= a^3 - a^3 + a$
$= a$

 The given polynomial is $p(x) = 2x^2 + 5x - 4$
Putting $\text{x}=\frac{1}{2}$ in $p(x),$ we get
$\text{p}\Big(\frac{1}{2}\Big)=2\times\Big(\frac{1}{2}\Big)^2+5\times\frac{1}{2}-3$
$=\frac{1}{2}+\frac{5}{2}-3=3-3=0$
Putting $x = -3$ in $p(x),$ we get
$\text{p}(-3)=2\times(-3)^2+5\times(-3)-3$
$= 18-15-3$
$= 0$
Therefore, $x = -3$ is a zero of the polynomial $p(x)$
Thus, $\frac{1}{2}$ and $-3$ are the zeros of the given polynomial $p(x).$

 $\sqrt{2}$ is a constant term. Therefore, the degree of $\sqrt{2}$ is $0.$

 Let
$a - b = A$
$b - c = B$
$c - a = C$
Now $(A + B + C)^3 = A^3 + B^3 + C^3 + 3(A + B)(B + C)(C + A)$
$⇒ A^3 + B^3 + C^3 = (A + B + C)^3- 3(A + B)(B + C)(C + A)$
Now putting values of $A, B$ and $C.$ we get
$(\text{a} - \text{b})^3 + (\text{b} - \text{c})^3 + (\text{c} - \text{a})^3\\=(\not\text{a}-\not\text{b}+\not\text{b}-\not\text{c}+\not\text{c}-\not\text{a})^3\\-3(\text{a}-\not\text{b}+\not\text{b}-\text{c})(\text{b}-\not\text{c}+\not\text{c}-\text{a})(\text{c}-\not\text{a}+\not\text{a}-\text{b})$
$⇒ (a - b)^3+ (b - c)^3 + (c - a)^3 = 0 - 3 (a - c)(b - a)(c - b)$
$⇒ (a - b)^3+ (b - c)^3 + (c - a)^3 = 3(a - b)(b - c)(c - a)$
Hence, correct option is $(c).$

$(102)^3 = (100 + 2)^3$
$= (100)^3 + (2)^3 + 3 × 100 × 2(100 + 2)$
$= 1000000 + 8 + 60000 + 1200$
$= 1061208$

$a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$
If $a + b + c = 0,$ then
$a^3 + b^3 + c^3 - 3abc = 0$
$⇒ a^3 + b^3 + c^3 = 3abc ...(1)$
Now, consider $\frac{\text{a}^2}{\text{bc}}+\frac{\text{b}^2}{\text{ca}}+\frac{\text{c}^2}{\text{ab}}$
Multiplying dividing by $a. b.$ and $c$ in $\frac{\text{a}^2}{\text{bc}}.\frac{\text{b}^2}{\text{ca}} and \frac{\text{c}^2}{\text{ab}}$ respectively. we get
$\frac{\text{a}^3}{\text{abc}}+\frac{\text{b}^3}{\text{bca}}+\frac{\text{c}^3}{\text{cab}}$
$=\frac{\text{a}^3+\text{b}^3+\text{c}^3}{\text{abc}}$
$=\frac{3\text{abc}}{\text{abc}} ....[$From $(1)]$
$=3$
Hence, correct option is $(d).$

 $\text{p}(\text{x})=\text{ax}^2+2\text{x}+\text{b}$
$\Rightarrow\text{a}(-2)^2 + 2(-2) + b = 0$
$\Rightarrow4\text{a} - 4+\text{b}=0\ .... (\text{i})$
Also,
$\text{p}\Big(\frac{-1}{2}\Big)=0$
$\Rightarrow\text{a}\Big(\frac{-1}{2}\Big)^2+2\Big(\frac{-1}{2}\Big)+\text{b}=0$
$\Rightarrow\frac{\text{a}}{4}-1+\text{b}=0$
$\Rightarrow\text{a}-4+4\text{b}=0\ ...(\text{ii})$
Subtracting eq. $(ii)$ from eq. $(i),$ we get
$3\text{a}+0-3\text{b}=0$
$\Rightarrow3(\text{a}-\text{b})=0$
$\Rightarrow\text{a}-\text{b}=0$

 The given expression to be factorized is $x^3 - 1 + y^3 + 3xy$
This can be written in the form
$x^3 - 1 + y^3 + 3xy = (x)^2 + (-1)^3 + (y)^3 - 3(x) × (-1).(y)$
Recall the formula $a^3 + b^3 + c^3 - 3abc = (a + b + c) (a^2 + b^2 + c^2 - ab - bc - ca)$
Using the above formula, we have,
$x^3 - 1 + y^3 + 3xy$
$= (x + (-1) + y) {(x)^2 + (-1)^2 + (y)^2 - (x) × (-1) - (-1) × (y) - (y) × (x)}$
$= (x - 1 + y) (x^2 + 1 + y^2 + x + y - xy)$

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

 $p(x) = x^4 + 2x^3 - 3x^2 + x - 1$
$x - 2 = 0 ⇒ x = 2$
By the remainder theorem, we know that when $p(x)$ is divided by
$(x - 2)$, the remainder is $p(2).$
Now, $p(2) = x^4 + 2x^3 - 3x^2 + x - 1$
$= (2)^4 + 2(2)^3 - 3(2)^2 + 2 - 1$
$= 16 +16 - 12 + 2 - 1$
$= 21$

 Now, $(x + y)^3 = x^3+ 3^3= 3.x.3(x + 3)$
$[$Using identity, $(a + b)^3 = a^3 + b^3 + 3ab(a + b)]$
$= x^3 + 27 + 9x(x + 3)$
$= x^3 + 27 + 9x^2 + 27x$
Hence, the coefficient of $x$ in $(x + 3)^3$ is $27.$

 If both $x - 2$ and $\text{x}-\frac{1}{2}$ are the factors of $f(x) = px^2 + 5x + r,$ then $f(2) = 0$
$⇒ p(2)^2 + 5(2) + r = 0$
$⇒ 4p + 10 + r = 0$
$⇒ 4p + r = -10 .......(i)$
Also, $\text{f}\Big(\frac{1}{2}\Big)=0$
$\Rightarrow\text{p}\Big(\frac{1}{2}\Big)^2+5\Big(\frac{1}{2}\Big)+\text{r}=0$
$\Rightarrow\text{p}+10+4\text{r}=0$
$\Rightarrow\text{p}+4\text{r}=-10\ .....(\text{ii})$
From eq. $ (i)$ and eq. $(ii),$ we get
$4\text{p}+\text{r}=\text{p}+4\text{r}$
$\Rightarrow3\text{p}=3\text{r}$
$\Rightarrow\text{p}=\text{r}$

 $\text{x}^2+\frac{1}{\text{x}^2}=102,$
$\Rightarrow(\text{x}^2)+\Big(\frac{1}{\text{x}^2}\Big)-2\times\text{x}\times\frac{1}{\text{x}}=102-2\times\text{x}\times\frac{1}{\text{x}}$
$\Rightarrow\Big(\text{x}-\frac{1}{\text{x}}\Big)^2=100$
$\Rightarrow\text{x}-\frac{1}{\text{x}}=\sqrt{100}=10$