If α, β, γ are are the zeros of the polynomial f(x) = x3 − px2 + qx − r,
Question: If $\alpha, \beta, y$ are are the zeros of the polynomial $f(x)=x^{3}-p x^{2}+q x-r$, then $\frac{1}{\alpha \beta}+\frac{1}{\beta \gamma}+\frac{1}{\gamma \alpha}=$ Solution: We have to find the value of $\frac{1}{\alpha \beta}+\frac{1}{\beta \gamma}+\frac{1}{\gamma \alpha}$ Given $\alpha, \beta, \gamma$ be the zeros of the polynomial $f(x)=x^{3}-p x^{2}+q x-r$ $\alpha+\beta+\gamma=\frac{-\text { Coefficient of } x^{2}}{\text { Coefficient of } x^{3}}$ $=\frac{-(p)}{1}$ $=p$ $\alpha \be...
Read More →If α, β, γ are the zeros of the polynomial f(x) = ax3 + bx2 + cx + d,
Question: If $\alpha, \beta, y$ are the zeros of the polynomial $f(x)=a x^{3}+b x^{2}+c x+d$, then $\alpha^{2}+\beta^{2}+y^{2}=$ (a) $\frac{b^{2}-a c}{a^{2}}$ (b) $\frac{b^{2}-2 a c}{a}$ (c) $\frac{b^{2}+2 a c}{b^{2}}$ (d) $\frac{b^{2}-2 a c}{a^{2}}$ Solution: We have to find the value of $\alpha^{2}+\beta^{2}+\gamma^{2}$ Given $\alpha, \beta, \gamma$ be the zeros of the polynomial $f(x)=a x^{3}+b x^{2}+c x+d$ $\alpha+\beta+\gamma=\frac{-\text { Coefficient of } x^{2}}{\text { Coefficient of } x...
Read More →If α, β, γ are the zeros of the polynomial f(x) = ax3 + bx2 + cx + d,
Question: If $\alpha, \beta, y$ are the zeros of the polynomial $f(x)=a x^{3}+b x^{2}+c x+d$, then $\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}=$ (a) $-\frac{b}{d}$ (b) $\frac{c}{d}$ (c) $-\frac{c}{d}$ (d) $-\frac{c}{a}$ Solution: We have to find the value of $\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}$ Given $\alpha, \beta, \gamma$ be the zeros of the polynomial $f(x)=a x^{3}+b x^{2}+c x+d$ We know that $\alpha \beta+\beta \gamma+\gamma \alpha=\frac{\text { Coefficient of } x}{\tex...
Read More →If (x2 + 1/x2) = 98, Find the value of
Question: If $\left(x^{2}+1 / x^{2}\right)=98$, Find the value of $x^{3}+1 / x^{3}$ Solution: Given, $\left(x^{2}+1 / x^{2}\right)=98$ We know that, $(x+y)^{2}=x^{2}+y^{2}+2 x y \ldots 1$ Substitute $\left(x^{2}+1 / x^{2}\right)=98$ in eq 1 $(x+1 / x)^{2}=x^{2}+1 / x^{2}+2 * x^{*} 1 / x$ $(x+1 / x)^{2}=x^{2}+1 / x^{2}+2$ $(x+1 / x)^{2}=98+2$ $(x+1 / x)^{2}=100$ $(x+1 / x)=\sqrt{100}$ $(x+1 / x)=\pm 10$ We need to find $x^{3}+1 / x^{3}$ So, $a^{3}+b^{3}=(a+b)\left(a^{2}+b^{2}-a b\right)$ $x^{3}+1...
Read More →If (x2 + 1/x2) = 51, Find the value of
Question: If $\left(x^{2}+1 / x^{2}\right)=51$, Find the value of $x^{3}-1 / x^{3}$ Solution: Given, $\left(x^{2}+1 / x^{2}\right)=51$ We know that, $(x-y)^{2}=x^{2}+y^{2}-2 x y \ldots 1$ Substitute $\left(x^{2}+1 / x^{2}\right)=51$ in eq 1 $(x-1 / x)^{2}=x^{2}+1 / x^{2}-2^{*} x^{*} 1 / x$ $(x-1 / x)^{2}=x^{2}+1 / x^{2}-2$ $(x-1 / x)^{2}=51-2$ $(x-1 / x)^{2}=49$ $(x-1 / x)=\sqrt{49}$ $(x-1 / x)=\pm 7$ We need to find $x^{3}-1 / x^{3}$ So, $a^{3}-b^{3}=(a-b)\left(a^{2}+b^{2}+a b\right)$ $x^{3}-1 ...
Read More →If (x − 1/x) = 5, Find the value of
Question: If $(x-1 / x)=5$, Find the value of $x^{3}-1 / x^{3}$ Solution: Given, If $(x-1 / x)=5$ We know that, $(a-b)^{3}=a^{3}-b^{3}-3 a b(a-b) \ldots 1$ Substitute $(x-1 / x)=5$ in eq 1 $(x-1 / x)^{3}=x^{3}-1 / x^{3}-3\left(x^{*} 1 / x\right)(x-1 / x)$ $5^{3}=x^{3}-1 / x^{3}-3(x-1 / x)$ $125=x^{3}-1 / x^{3}-\left(3^{*} 5\right)$ $125=x^{3}-1 / x^{3}-15$ $125+15=x^{3}-1 / x^{3}$ $x^{3}-1 / x^{3}=140$ Hence, the result is $x^{3}-1 / x^{3}=140$...
Read More →If one root of the polynomial f(x) = 5x2 + 13x + k is reciprocal of the other,
Question: If one root of the polynomial $f(x)=5 x^{2}+13 x+k$ is reciprocal of the other, then the value of $k$ is (a) 0 (b) 5 (c) $\frac{1}{6}$ (d) 6 Solution: If one zero of the polynomial $f(x)=5 x^{2}+13 x+k$ is reciprocal of the other. So $\beta=\frac{1}{\alpha} \Rightarrow \alpha \beta=1$ Now we have $\alpha \times \beta=\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}$ $=\frac{k}{5}$ Since $\alpha \beta=1$ Therefore we have $\alpha \beta=\frac{k}{5}$ $1=\frac{k}{5}$ $\Righta...
Read More →If (x − 1/x) = 7, Find the value of
Question: If $(x-1 / x)=7$, Find the value of $x^{3}-1 / x^{3}$ Solution: Given, If $(x-1 / x)=7$ We know that, $(a-b)^{3}=a^{3}-b^{3}-3 a b(a-b) \ldots 1$ Substitute $(x-1 / x)=7$ in eq 1 $(x-1 / x)^{3}=x^{3}-1 / x^{3}-3(x * 1 / x)(x-1 / x)$ $7^{3}=x^{3}-1 / x^{3}-3(x-1 / x)$ $343=x^{3}-1 / x^{3}-\left(3^{*} 7\right)$ $343=x^{3}-1 / x^{3}-21$ $343+21=x^{3}-1 / x^{3}$ $x^{3}-1 / x^{3}=364$ hence, the result is $x^{3}-1 / x^{3}=364$...
Read More →If Q.No. 14, c =
Question: If Q.No. 14,c= (a) $b$ (b) $2 b$ (c) $2 b^{2}$ (d) $-2 b$ Solution: We have to find the value ofc Given $f(x)=a x^{3}+b x-c$ is divisible by the polynomial $g(x)=x^{2}+b x+c$ We must have $b x-a c x+a b^{2} x+a b c-c=0$ for all $x$ $x\left(b-a c+a b^{2}\right)+c(a b-1)=0$ (1) $c(a b-1)=0$ Since $c \neq 0$, so $a b-1=0$ $a b=1$ Now in the equation (1) the condition is true for allx.So putx= 1 and also we haveab= 1 Therefore we have $b-a c+a b^{2}=0$ $b+a b^{2}-a c=0$ $b(1+a b)-a c=0$ Su...
Read More →If the polynomial f(x) = ax3 + bx − c is divisible
Question: If the polynomial $l(x)=a x^{3}+b x-c$ is divisible by the polynomial $g(x)=x^{2}+b x+c$, then $a b=$ (a) 1 (b) $\frac{1}{c}$ (c) $-1$ (d) $-\frac{1}{c}$ Solution: We have to find the value of $a b$ Given $f(x)=a x^{3}+b x-c$ is divisible by the polynomial $g(x)=x^{2}+b x+c$ We must have $b x-a c x+a b^{2} x+a b c-c=0$, for all $x$ So put $x=0$ in this equation $x\left(b-a c+a b^{2}\right)+c(a b-1)=0$ $c(a b-1)=0$ Since $c \neq 0$, so $a b-1=0$ $\Rightarrow a b=1$ Hence, the correct al...
Read More →If (x + 1/x) = 5, Find the value of x3 + 1/x3
Question: If $(x+1 / x)=5$, Find the value of $x^{3}+1 / x^{3}$ Solution: Given, $(x+1 / x)=5$ We know that, $(a+b)^{3}=a^{3}+b^{3}+3 a b(a+b)$ ... 1 Substitute (x + 1/x) = 5 in eq1 $(x+1 / x)^{3}=x^{3}+1 / x^{3}+3\left(x^{*} 1 / x\right)(x+1 / x)$ $5^{3}=x^{3}+1 / x^{3}+3(x * 1 / x)(x+1 / x)$ $125=x^{3}+1 / x^{3}+3(x+1 / x)$ $125=x^{3}+1 / x^{3}+3(5)$ $125=x^{3}+1 / x^{3}+15$ $125-15=x^{3}+1 / x^{3}$ $x^{3}+1 / x^{3}=110$ Hence, the result is $x^{3}+1 / x^{3}=110$...
Read More →If the product of two zeros of the polynomial f(x) = 2x3 + 6x2 − 4x + 9 is 3,
Question: If the product of two zeros of the polynomial $f(x)=2 x^{3}+6 x^{2}-4 x+9$ is 3 , then its third zero is (a) $\frac{3}{2}$ (b) $-\frac{3}{2}$ (C) $\frac{9}{2}$ (d) $-\frac{9}{2}$ Solution: Let $\alpha, \beta, \gamma$ be the zeros of polynomial $f(x)=2 x^{3}+6 x^{2}-4 x+9$ such that $\alpha \beta=3$ We have, $\alpha \beta \gamma=\frac{\text { Constant term }}{\text { Coefficient of } x^{3}}$ $=\frac{-9}{2}$ Putting $\alpha \beta=3$ in $\alpha \beta \gamma=\frac{-9}{2}$, we get $\alpha \...
Read More →If a - b = 4 and ab = 21, Find the value of a3 - b3
Question: If $a-b=4$ and $a b=21$, Find the value of $a^{3}-b^{3}$ Solution: Given, a - b = 4, ab = 21 we know that, $(a-b)^{3}=a^{3}-b^{3}-3 a b(a-b)$..........(1) substitute a - b = 4 , ab = 21 in eq 1 $\Rightarrow(4)^{3}=a^{3}-b^{3}-3(21)(4)$ $\Rightarrow 64=a^{3}-b^{3}-252$ $\Rightarrow 64+252=a^{3}-b^{3}$ $\Rightarrow 316=a^{3}-b^{3}$ Hence, the value of $a^{3}-b^{3}=316$...
Read More →If a + b = 10 and ab = 21, Find the value of a3 + b3
Question: If $a+b=10$ and $a b=21$, Find the value of $a^{3}+b^{3}$ Solution: Given, $a+b=10, a b=21$ we know that, $(a+b)^{3}=a^{3}+b^{3}+3 a b(a+b) \quad \ldots 1$ substitute $a+b=10, a b=21$ in eq 1 $\Rightarrow(10)^{3}=a^{3}+b^{3}+3(21)(10)$ $\Rightarrow 1000=a^{3}+b^{3}+630$ $\Rightarrow 1000-630=a^{3}+b^{3}$ $\Rightarrow 370=a^{3}+b^{3}$ Hence, the value of $a^{3}+b^{3}=370$...
Read More →If zeros of the polynomial f(x) = x3 − 3px2 + qx − r are in A.P.,
Question: If zeros of the polynomial $f(x)=x^{3}-3 p x^{2}+q x-r$ are in A.P., then (a) $2 p^{3}=p q-r$ (b) $2 p^{3}=p q+r$ (c) $p^{3}=p q-r$ (d) None of these Solution: Let $a-d, a, a+d$ be the zeros of the polynomial $f(x)=x^{3}-3 p x^{2}+q x-r$ then Sum of zeros $=\frac{-\text { Coefficient of } x^{2}}{\text { Coefficient of } x^{3}}$ $(a-d)+a+(a+d)=\frac{-(-3 p)}{1}$ $3 a=3 p$ $a=\frac{3}{3} p$ $a=p$ Since $a$ is a zero of the polynomial $f(x)$ Therefore, $f(a)=0$ $a^{3}-3 p a^{2}+q a-r=0$ S...
Read More →Simplify each of the following
Question: Simplify each of the following (a) $(x+3)^{3}+(x-3)^{3}$ (b) $(x / 2+y / 3)^{3}-(x / 2-y / 3)^{3}$ (c) $(x+2 / x)^{3}+(x-2 / x)^{3}$ (d) $(2 x-5 y)^{3}-(2 x+5 y)^{3}$ Solution: (a) $(x+3)^{3}+(x-3)^{3}$ The above equation is in the form of $a^{3}+b^{3}=(a+b)\left(a^{2}+b^{2}-a b\right)$ We know that, $a=(x+3), b=(x-3)$ By using $\left(a^{3}+b^{3}\right)$ formula $=(x+3+x-3)\left[(x+3)^{3}+(x-3)^{3}-(x+3)(x-3)\right]$ $=2 x\left[\left(x^{2}+3^{2}+2^{*} x^{*} 3\right)+\left(x^{2}+3^{2}-2...
Read More →If the product of zeros of the polynomial
Question: If the product of zeros of the polynomial $f(x) a x^{3}-6 x^{2}+11 x-6$ is 4 , then $a=$ (a) $\frac{3}{2}$ (b) $-\frac{3}{2}$ (c) $\frac{2}{3}$ (d) $-\frac{2}{3}$ Solution: Since $\alpha$ and $\beta$ are the zeros of quadratic polynomial $f(x)=a x^{3}-6 x^{2}+11 x-6$ $\alpha \beta=\frac{-\text { Constant term }}{\text { Coefficient of } x^{2}}$ So we have $4=-\left(\frac{-6}{a}\right)$ $4=\frac{6}{a}$ $4 a=6$ $a=\frac{6}{4}$ $a=\frac{3}{2}$ The value of $a$ is $\frac{3}{2}$ Hence, the ...
Read More →Find the cube of each of the following binomial expression
Question: Find the cube of each of the following binomial expression (a) $(1 / x+y / 3)$ (b) $\left(3 / x-2 / x^{2}\right)$ (c) $(2 x+3 / x)$ (d) $(4-1 / 3 x)$ Solution: (a) Given, $(1 / x+y / 3))^{3}$ The above equation is in the form of $(a+b)^{3}=a^{3}+b^{3}+3 a b(a+b)$ We know that, $a=1 / x, b=y / 3$ By using $(a+b)^{3}$ formula $(1 / x+y / 3))^{3}$ $=(1 / x)^{3}+(y / 3)^{3}+3(1 / x)(y / 3)(1 / x+y / 3)$ $=1 / x^{3}+y^{3} / 27+3 * 1 / x^{*} y / 3(1 / x+y / 3)$ $=1 / x^{3}+y^{3} / 27+y / x(1...
Read More →Figure 2.23 show the graph of the polynomial
Question: Figure $2.23$ show the graph of the polynomial $f(x)=a x^{2}+b x+c$ for which (a) $a0, b0$ and $c0$ (b) $a0, b0$ and $c0$ (c) $a0, b0$ and $c0$ (d) $a0, b0$ and $c0$ Solution: Clearly, $f(x)=a x^{2}+b x+c$ represent a parabola opening downwards. Therefore, $a0$ $y=a x^{2}+b x+c$ cuts $y$-axis at $\mathrm{P}$ which lies on $O Y$. Putting $x=0$ in $y=a x^{2}+b x+c$, we get $y=c$. So the coordinates $\mathrm{P}$ are $(0, c)$. Clearly, $P$ lies on $O Y$. Therefore $c0$ The vertex $\left(\f...
Read More →If the diagram in Fig. 2.22 shows the graph of the polynomial
Question: If the diagram in Fig. $2.22$ shows the graph of the polynomial $f(x)=a x^{2}+b x+c$, then (a) $a0, b0$ and $c0$ (b) $a0, b0$ and $c0$ (c) $a0, b0$ and $c0$ (d) $a0, b0$ and $c0$ Solution: Clearly, $f(x)=a x^{2}+b x+c$ represent a parabola opening upwards. Therefore, $a0 \quad y=a x^{2}+b x+c$ cuts $Y$ axis at $P$ which lies on $O Y$. Putting $x=0$ in $y=a x^{2}+b x+c$, we get $y=c$. So the coordinates of $P$ is $(0, c)$. Clearly, $\mathrm{P}$ lies on $O Y$. Therefore $c0$ Hence, the c...
Read More →If f(x) = ax2 + bx + c has no real zeros and a + b + c = 0,
Question: If $l(x)=a x^{2}+b x+c$ has no real zeros and $a+b+c=0$, then (a) $c=0$ (b) $c0$ (c) $c0$ (d) None of these Solution: If $f(x)=a x^{2}+b x+c$ has no real zeros and $a+b+c0$ then $c0$ Hence, the correct choice is $(c)$...
Read More →If α, β are the zeros of the polynomial f(x) = x2 − p(x + 1) − c
Question: If $\alpha, \beta$ are the zeros of the polynomial $f(x)=x^{2}-p(x+1)-c$ such that $(\alpha+1)(\beta+1)=0$, then $c=$ (a) 1 (b) 0 (c) $-1$ (d) 2 Solution: Since $\alpha$ and $\beta$ are the zeros of quadratic polynomial $f(x)=x^{2}-p(x+1)-c$ $f(x)=x^{2}-p x-p-c$ $\alpha+\beta=\frac{-\text { Coefficient of } x}{\text { Coefficient of } x^{2}}$ $=-\left(\frac{-p}{1}\right)$ $=p$ $\alpha \beta=\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}$ $=\frac{-p-c}{1}$ $=-p-c$ We hav...
Read More →Simplify each of the following expressions:
Question: Simplify each of the following expressions: (i) $(x+y+z)^{2}+(x+y / 2+2 / 3)^{2}-(x / 2+y / 3+z / 4)^{2}$ (ii) $(x+y-2 z)^{2}-x^{2}-y^{2}-3 z^{2}+4 x y$ (iii) $\left[x^{2}-x+1\right]^{2}-\left[x^{2}+x+1\right]^{2}$ Solution: (i) Expanding, we get $=\left[x^{2}+y^{2}+z^{2}+2 x y+2 y z+2 z x\right]+\left[x^{2}+\frac{y^{2}}{4}+\frac{z^{2}}{9}+2 x \frac{y}{2}+2 \frac{z x}{3}+\frac{y z}{3}\right]$ $-\left[\frac{x^{2}}{4}+\frac{y^{2}}{9}+\frac{z^{2}}{10}+\frac{x y}{3}+\frac{y z}{6}+\frac{x z...
Read More →If α, β are the zeros of polynomial f(x) = x2 − p (x + 1) − c, then (α + 1) (β + 1) =
Question: If $\alpha, \beta$ are the zeros of polynomial $f(x)=x^{2}-p(x+1)-c$, then $(\alpha+1)(\beta+1)=$ (a) $c-1$ (b) $1-c$ (c) $C$ (d) $1+c$ Solution: Since $\alpha$ and $\beta$ are the zeros of quadratic polynomial $f(x)=x^{2}-p(x+1)-c$ $=x^{2}-p x-p-c$ $\alpha+\beta=\frac{-\text { Coefficient of } x}{\text { Coefficient of } x^{2}}$ $=-\left(\frac{-p}{1}\right)$ $=p$ $\alpha \times \beta=\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}$ $=\frac{-p-c}{1}$ $=-p-c$ We have $(\a...
Read More →Find the value of the equation:
Question: Find the value of the equation: $4 x^{2}+y^{2}+25 z^{2}+4 x y-10 y z-20 z x$ when $x=4, y=3, z=2$ Solution: $4 x^{2}+y^{2}+25 z^{2}+4 x y-10 y z-20 z x$ $(2 x)^{2}+y^{2}+(-5 z)^{2}+2(2 x)(y)+2(y)(-5 z)+2(-5 z)(2 x)$ $(2 x+y-5 z)^{2}$ $(2(4)+3-5(2))^{2}$ $(8+3-10)^{2}$ $(1)^{2}$ 1 Hence value of the equation is equals to 1...
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