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1

MCQ (Single Correct Answer)

In a triangle $$PQR,\;\;\angle R = {\pi \over 2}.\,\,If\,\,\tan \,\left( {{P \over 2}} \right)$$ and $$ \tan \left( {{Q \over 2}} \right)$$ are the roots of $$a{x^2} + bx + c = 0,\,\,a \ne 0$$ then

A

$$a = b + c$$

B

$$c = a + b$$

C

$$b = c$$

D

$$b = a + c$$

$$\angle $$R = 90^{o} $$ \therefore $$ $$\angle $$P + $$\angle $$Q = 90^{o}

$$ \Rightarrow $$ $${P \over 2} + {Q \over 2} = {{90} \over 2} = 45$$^{o}

$$\tan \left( {{P \over 2}} \right),\tan \left( {{Q \over 2}} \right)$$ are the roots of $$a{x^2} + bx + c = 0$$

$$ \therefore $$ $$\tan \left( {{P \over 2}} \right) + \tan \left( {{Q \over 2}} \right) = - {b \over a},\,\,$$

and $$\tan \left( {{P \over 2}} \right).\tan \left( {{Q \over 2}} \right) = {c \over a}$$

$${{\tan \left( {{P \over 2}} \right) + \tan \left( {{Q \over 2}} \right)} \over {1 - \tan \left( {{P \over 2}} \right)\tan \left( {{Q \over 2}} \right)}}$$

$$ = \tan \left( {{P \over 2} + {Q \over 2}} \right) $$= tan 45^{o} = 1

$$ \Rightarrow {{ - {b \over a}} \over {1 - {c \over a}}} = 1$$

$$ \Rightarrow - {b \over a} = {a \over a} - {c \over a}$$

$$ \Rightarrow - b = a - c$$ or $$c = a + b.$$

$$ \Rightarrow $$ $${P \over 2} + {Q \over 2} = {{90} \over 2} = 45$$

$$\tan \left( {{P \over 2}} \right),\tan \left( {{Q \over 2}} \right)$$ are the roots of $$a{x^2} + bx + c = 0$$

$$ \therefore $$ $$\tan \left( {{P \over 2}} \right) + \tan \left( {{Q \over 2}} \right) = - {b \over a},\,\,$$

and $$\tan \left( {{P \over 2}} \right).\tan \left( {{Q \over 2}} \right) = {c \over a}$$

$${{\tan \left( {{P \over 2}} \right) + \tan \left( {{Q \over 2}} \right)} \over {1 - \tan \left( {{P \over 2}} \right)\tan \left( {{Q \over 2}} \right)}}$$

$$ = \tan \left( {{P \over 2} + {Q \over 2}} \right) $$= tan 45

$$ \Rightarrow {{ - {b \over a}} \over {1 - {c \over a}}} = 1$$

$$ \Rightarrow - {b \over a} = {a \over a} - {c \over a}$$

$$ \Rightarrow - b = a - c$$ or $$c = a + b.$$

2

MCQ (Single Correct Answer)

If one root of the equation $${x^2} + px + 12 = 0$$ is 4, while the equation $${x^2} + px + q = 0$$ has equal roots,

then the value of $$'q'$$ is

then the value of $$'q'$$ is

A

4

B

12

C

3

D

$${{49} \over 4}$$

$$4$$ is a root of $${x^2} + px + 12 = 0$$

$$ \Rightarrow 16 + 4p + 12 = 0$$

$$ \Rightarrow p = - 7$$

Now, the equation $${x^2} + px + q = 0$$

has equal roots.

$$\therefore$$ $${p^2} - 4q = 0$$ $$ \Rightarrow q = {{{p^2}} \over 4} = {{49} \over 4}$$

$$ \Rightarrow 16 + 4p + 12 = 0$$

$$ \Rightarrow p = - 7$$

Now, the equation $${x^2} + px + q = 0$$

has equal roots.

$$\therefore$$ $${p^2} - 4q = 0$$ $$ \Rightarrow q = {{{p^2}} \over 4} = {{49} \over 4}$$

3

MCQ (Single Correct Answer)

If $$\left( {1 - p} \right)$$ is a root of quadratic equation $${x^2} + px + \left( {1 - p} \right) = 0$$ then its root are

A

$$ - 1,2$$

B

$$ - 1,1$$

C

$$ 0,-1$$

D

$$0,1$$

Let the second root be $$\alpha .$$

Then $$\alpha + \left( {1 - p} \right) = - p \Rightarrow \alpha = - 1$$

Also $$\alpha .\left( {1 - p} \right) = 1 - p$$

$$ \Rightarrow \left( {\alpha - 1} \right)\left( {1 - p} \right) = 0$$

$$ \Rightarrow p = 1$$ [as $$\alpha = - 1$$]

$$\therefore$$ Roots are $$\alpha = - 1$$ and $$p-1=0$$

Then $$\alpha + \left( {1 - p} \right) = - p \Rightarrow \alpha = - 1$$

Also $$\alpha .\left( {1 - p} \right) = 1 - p$$

$$ \Rightarrow \left( {\alpha - 1} \right)\left( {1 - p} \right) = 0$$

$$ \Rightarrow p = 1$$ [as $$\alpha = - 1$$]

$$\therefore$$ Roots are $$\alpha = - 1$$ and $$p-1=0$$

4

MCQ (Single Correct Answer)

Let two numbers have arithmetic mean 9 and geometric mean 4. Then these numbers are the roots of the quadratic equation

A

$${x^2} - 18x - 16 = 0$$

B

$${x^2} - 18x + 16 = 0$$

C

$${x^2} + 18x - 16 = 0$$

D

$${x^2} + 18x + 16 = 0$$

Let two numbers be a and b then $${{a + b} \over 2} = 9$$

and $$\sqrt {ab} = 4$$

$$\therefore$$ Equation with roots $$a$$ and $$b$$ is

$${x^2} - \left( {a + b} \right)x + ab = 0$$

$$ \Rightarrow {x^2} - 18x + 16 = 0$$

and $$\sqrt {ab} = 4$$

$$\therefore$$ Equation with roots $$a$$ and $$b$$ is

$${x^2} - \left( {a + b} \right)x + ab = 0$$

$$ \Rightarrow {x^2} - 18x + 16 = 0$$

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Complex Numbers

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