Algebra MCQs - Test Your Skills with Multiple Choice Questions

Q11 :

$x^4 - mx^3 + 2x^2 - 5x + 8 = 0$ , when divided by $x - 2$ , gives remainder as $3m$ . Then the value of $3m$ is equal to

A

$frac{-22}{8}$

B

-2

C

2

D

$frac{22}{8}$

View Answer

Correct Answer: C

2

Description:

$f(2) = 3m$

$Rightarrow 16 - 8m + 8 - 10 + 8 = 3m$

$Rightarrow 22 - 8m = 3m Rightarrow m =2$

Q12 :

If $x^m = y^n = z^p, xyz = 1 text{ and } m ne 0$ , then $frac{1}{m} + frac{1}{n} + frac{1}{p}$ is equal to

A

0

B

3

C

-2

D

-3

View Answer

Correct Answer: A

0

Description:

$text{Let } x^m = y^n = z^p = k$

$Rightarrow m log x = n log y = p log z = c$

$therefore log x = frac{c}{m}, log y = frac{c}{n}, log z = frac{c}{p}$

Now, given $frac{1}{m} + frac{1}{n} + frac{1}{p}$

$=frac{1}{c} (log x + log y + log z)$

$=frac{1}{c} log (xyz) = 0$

$[because log (xyz) = log 1 = 0]$

Q13 :

The value of $k$ , for which the system of equations $x + 2y + 7 = 0$ and $2x + ky + 14 = 0$ will have infinitely many solutions, is

A

2

B

4

C

6

D

8

View Answer

Correct Answer: B

4

Description:

For $k = 4$ , the equations are $x + 2y + 7 = 0$ and $2x + 4y + 14 = 0 = 2 (x + 2y + 7)$

$Bigg[because dfrac{a_1}{a_2} = dfrac{b_1}{b_2} = dfrac{c_1}{c_2}Bigg]$

Thus, the given equations have infinitely many solutions for $k = 4$

Q14 :

Roots of the equation $x^2 + x(2 - p^2) - 2p^2 = 0$ are

A

$-p^2 text{ and } -2$

B

$p^2 text{ and } -2$

C

$-p^2 text{ and } 2$

D

$p^2 text{ and } 2$

View Answer

Correct Answer: B

$p^2 text{ and } -2$

Description:

On the basis of quadratic equation, we solve the given equation for $x$ , we get

$x = frac{-Big(2 - p^2Big) pm sqrt{Big(2 - p^2Big)^2 + 8p^2}}{2}$

$= frac{-2 + p^2 pm (2 + p^2)}{2}$

$= p^2, -2$

Q15 :

If one root of the equation $ax^2 + bx + c = 0, a ne 0$ , is reciprocal of the other, then

A

$b = c$

B

$a = c$

C

$a = 0$

D

$b = 0$

View Answer

Correct Answer: B

$a = c$

Description:

Let the rots of the given equation be M and $M text{ and }frac{1}{text{M}}$
$therefore$ Product of the roots $= text{M } cdot frac{1}{text{M}} = frac{c}{a} = 1$

$therefore a = c$

Q16 :

If $alpha$ and $beta$ are the roots of the equation $x^2 - px + q = 0$ , then the equation whose roots are $alpha beta + alpha + beta$ and $alpha beta - alpha - beta$ is

A

$x^2 + qx - p = 0$

B

$x^2 + 2qx + p^2 + q^2 = 0$

C

$x^2 - 2qx + q^2 - p^2 = 0$

D

$x^2 + 2qx + p^2 - q^2 = 0$

View Answer

Correct Answer: C

$x^2 - 2qx + q^2 - p^2 = 0$

Description:

Use Given equation with $alpha$ and $beta$ are roots, then
$alpha + beta = frac{-(-p)}{1} = p$

$alphabeta = frac{q}{1} = q$

Now the quadratic equation of roots $alphabeta + alpha + beta text{ and } alpha beta - alpha - beta$ is

$[x - (alphabeta + alpha + beta)] [x - (alphabeta - alpha - beta)] = 0$

$Rightarrow (x - q - p) (x - q + p ) = 0$

$Rightarrow x^2 - 2qx + q^2 - p^2 = 0$

Q17 :

Roots of the equation $frac{(x - b)(x -c)}{(a - b)(a - c)} a^2 + frac{(x - a)(x - c)}{(b - c)(b - a)}b^2 = x^2$ are

A

1, 1

B

a, 0

C

b, 0

D

a, b

View Answer

Correct Answer: D

a, b

Description:

Apply the formula,

$x = frac{-b pm sqrt{text{D}}}{2a}$

Q18 :

Solution of $4^{2x} = frac{1}{32}$ is

A

$frac{5}{4}$

B

$frac{-5}{4}$

C

$frac{3}{4}$

D

$frac{-5}{2}$

View Answer

Correct Answer: B

$frac{-5}{4}$

Description:

Given $4^{2x} = frac{1}{32} = 2^{-5}$

$Rightarrow 2^{4x} = 2^{-5}$

$Rightarrow 4x = -5 Rightarrow x = frac{-5}{4}$

Q19 :

If $10^{2y} = 25$ , then to $10^{-y}$ equals

A

$-frac{1}{5}$

B

$frac{1}{625}$

C

$frac{1}{50}$

D

$frac{1}{5}$

View Answer

Correct Answer: D

$frac{1}{5}$

Description:

Given, $10^{2y} = 25$

$Rightarrow (10^y)^2 = 25$

$Rightarrow 10^y = 5$

$therefore frac{1}{10^y} = frac{1}{5} Big[because frac{1}{10^y} = 10^{-y}Big]$

$because 10^{-y} = frac{1}{5}$

Q20 :

If $x = 2^frac{1}{3} + 2^frac{1}{3}$ , then the value of $2x^3 - 6x$ will be:

A

5

B

6

C

8

D

10

View Answer

Correct Answer: A

5

Description:

Given

$x = 2^frac{1}{3} + 2^frac{1}{3}$

on cubing both side, we get

$x^3 = 2 + 2^{-1} + 3 times 2^{frac{1}{3}} times 2^{-frac{1}{3}} (2^frac{1}{3} + 2^{-frac{1}{3}})$

$Rightarrow x^3 = 2 + frac{1}{2} + 3 times 2^frac{1}{3} times frac{1}{2^{1/3}} Big(2^frac{1}{3} + frac{1}{2^{1/3}}Big)$

$Rightarrow 2x^3 - 6x = 5$

Algebra Questions Answers