Joint Entrance Examination

Graduate Aptitude Test in Engineering

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1

MCQ (Single Correct Answer)

If the roots of the equation $${x^2} - bx + c = 0$$ be two consecutive integers, then $${b^2} - 4c$$ equals

A

$$-2$$

B

$$3$$

C

$$2$$

D

$$1$$

Let n and (n + 1) be the roots of x^{2} $$-$$ bx + c = 0.

Then, n + (n + 1) = b and n(n + 1) = c

$$\therefore$$ b^{2} $$-$$ 4c = (2n + 1)^{2} $$-$$ 4n(n + 1)

= 4n^{2} + 4n + 1 $$-$$ 4n^{2} $$-$$ 4n = 1

2

MCQ (Single Correct Answer)

The value of $$a$$ for which the sum of the squares of the roots of the equation

$${x^2} - \left( {a - 2} \right)x - a - 1 = 0$$ assume the least value is

$${x^2} - \left( {a - 2} \right)x - a - 1 = 0$$ assume the least value is

A

$$1$$

B

$$0$$

C

$$3$$

D

$$2$$

Given quadratic equation,

$${x^2} - \left( {a - 2} \right)x - a - 1 = 0$$

Let $$\alpha $$ and $$\beta $$ are the roots of the equation.

$$ \therefore $$ $$\alpha $$ + $$\beta $$ = $$a - 2$$

and $$\alpha $$$$\beta $$ = $$ - a - 1$$

Now $${\alpha ^2} + {\beta ^2} = {\left( {\alpha + \beta } \right)^2} - 2\alpha \beta $$

$$ \Rightarrow $$ $${\alpha ^2} + {\beta ^2} = {\left( {a - 2} \right)^2} + 2\left( {a + 1} \right)$$

$$ \Rightarrow $$ $${\alpha ^2} + {\beta ^2} = {a^2} - 2a + 6$$

$$ \Rightarrow $$ $${\alpha ^2} + {\beta ^2} = {\left( {a - 1} \right)^2} + 5$$

$$ \Rightarrow $$ The value of $${\alpha ^2} + {\beta ^2}$$ will be minimum, when $${a - 1}$$ = 0

$$ \Rightarrow $$ $${a = 1}$$

$${x^2} - \left( {a - 2} \right)x - a - 1 = 0$$

Let $$\alpha $$ and $$\beta $$ are the roots of the equation.

$$ \therefore $$ $$\alpha $$ + $$\beta $$ = $$a - 2$$

and $$\alpha $$$$\beta $$ = $$ - a - 1$$

Now $${\alpha ^2} + {\beta ^2} = {\left( {\alpha + \beta } \right)^2} - 2\alpha \beta $$

$$ \Rightarrow $$ $${\alpha ^2} + {\beta ^2} = {\left( {a - 2} \right)^2} + 2\left( {a + 1} \right)$$

$$ \Rightarrow $$ $${\alpha ^2} + {\beta ^2} = {a^2} - 2a + 6$$

$$ \Rightarrow $$ $${\alpha ^2} + {\beta ^2} = {\left( {a - 1} \right)^2} + 5$$

$$ \Rightarrow $$ The value of $${\alpha ^2} + {\beta ^2}$$ will be minimum, when $${a - 1}$$ = 0

$$ \Rightarrow $$ $${a = 1}$$

3

MCQ (Single Correct Answer)

If both the roots of the quadratic equation $${x^2} - 2kx + {k^2} + k - 5 = 0$$ are less than 5, then $$k$$ lies in the interval

A

$$\left( {5,6} \right]$$

B

$$\left( {6,\,\infty } \right)$$

C

$$\left( { - \infty ,\,4} \right)$$

D

$$\left[ {4,\,5} \right]$$

both roots are less than $$5,$$

then $$(i)$$ Discriminant $$ \ge 0$$

$$\left( {ii} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,p\left( 5 \right) > 0$$

$$\left( {iii} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{Sum\,\,of\,\,roots} \over 2} < 5$$

Hence $$\left( i \right)\,\,\,\,\,\,4{k^2} - 4\left( {{k^2} + k - 5} \right) \ge 0$$

$$4{k^2} - 4{k^2} - 4k + 20 \ge 0$$

$$4k \le 20 \Rightarrow k \le 5$$

$$\left( {ii} \right)\,\,\,\,\,f\left( 5 \right) > 0;25 - 10k + {k^2} + k - 5 > 0$$

or $${k^2} - 9k + 20 > 0$$

or $$k\left( {k - 4} \right) - 5\left( {k - 4} \right) > 0$$

or $$\left( {k - 5} \right)\left( {k - 4} \right) > 0$$

$$ \Rightarrow k \in \left( { - \infty ,4} \right) \cup \left( { - \infty ,5} \right)$$

$$\left( {iii} \right)\,\,\,\,\,\,{{Sum\,\,of\,\,roots} \over 2}$$

$$ = - {b \over {2a}} = {{2k} \over 2} < 5$$

The intersection of $$(i)$$, $$(ii)$$ & $$(iii)$$ gives

$$k \in \left( { - \infty ,4} \right).$$

4

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 $$c = a + b.$$

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