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1

JEE Main 2013 (Offline)

MCQ (Single Correct Answer)
If the equations $${x^2} + 2x + 3 = 0$$ and $$a{x^2} + bx + c = 0,$$ $$a,\,b,\,c\, \in \,R,$$ have a common root, then $$a\,:b\,:c\,$$ is
A
$$1:2:3$$
B
$$3:2:1$$
C
$$1:3:2$$
D
$$3:1:2$$

Explanation

Given equations are

$$\,\,\,\,\,\,\,\,\,\,\,\,{x^2} + 2x + 3 = 0\,\,\,\,\,...\left( i \right)$$

$$\,\,\,\,\,\,\,\,\,\,\,\,a{x^2} + bx + c = 0\,\,\,...\left( {ii} \right)$$

Roots of equation $$(i)$$ are imaginary roots.

According to the question $$(ii)$$ will also have both roots same as $$(i).$$

Thus $${a \over 1} = {b \over 2} = {c \over 3} = \lambda \left( {say} \right)$$

$$ \Rightarrow a = \lambda ,b = 2\lambda ,c = 3\lambda $$

Hence, required ratio is $$1:2:3$$
2

JEE Main 2013 (Offline)

MCQ (Single Correct Answer)
The number of values of $$k$$, for which the system of equations : $$$\matrix{ {\left( {k + 1} \right)x + 8y = 4k} \cr {kx + \left( {k + 3} \right)y = 3k - 1} \cr } $$$
has no solution, is
A
infinite
B
1
C
2
D
3

Explanation

From the given system, we have

$${{k + 1} \over k} = {8 \over {k + 3}} \ne {{4k} \over {3k - 1}}$$

( as System has no solution)

$$ \Rightarrow {k^2} + 4k + 3 = 8k$$

$$ \Rightarrow k = 1,3$$

If $$k = 1$$ then $${8 \over {1 + 3}} \ne {{4.1} \over 2}$$ which is false

And if $$k = 3$$

Then $${8 \over 6} \ne {{4.3} \over {9 - 1}}$$ which is true, therefore $$k=3$$

Hence for only one value of $$k.$$ System has no solution.
3

JEE Main 2013 (Offline)

MCQ (Single Correct Answer)
The real number $$k$$ for which the equation, $$2{x^3} + 3x + k = 0$$ has two distinct real roots in $$\left[ {0,\,1} \right]$$
A
lies between 1 and 2
B
lies between 2 and 3
C
lies between $$ - 1$$ and 0
D
does not exist.

Explanation

$$f\left( x \right) = 2{x^3} + 3x + k$$

$$f'\left( x \right) = 6{x^2} + 3 > 0$$

$$\forall x \in R$$ $$\,\,\,\,\,\,$$ (as $$\,\,\,\,\,\,$$ $${x^2} > 0$$)

$$ \Rightarrow f\left( x \right)$$ is strictly increasing function

$$ \Rightarrow f\left( x \right) = 0$$ has only one real root, so two roots are not possible.
4

AIEEE 2012

MCQ (Single Correct Answer)
The equation $${e^{\sin x}} - {e^{ - \sin x}} - 4 = 0$$ has:
A
infinite number of real roots
B
no real roots
C
exactly one real root
D
exactly four real roots

Explanation

Given equation is $${e^{\sin x}} - {e^{ - \sin x}} - 4 = 0$$

Put $${e^{{\mathop{\rm sinx}\nolimits} \,}} = t$$ in the given equation,

we get $${t^2} - 4t - 1 = 0$$

$$ \Rightarrow t = {{4 \pm \sqrt {16 + 4} } \over 2}$$

$$\,\,\,\,\,\,\,\,\,\,\, = {{4 \pm \sqrt {20} } \over 2}$$

$$\,\,\,\,\,\,\,\,\,\,\, = {{4 \pm 2\sqrt 5 } \over 2}$$

$$\,\,\,\,\,\,\,\,\,\,\, = 2 \pm \sqrt 5 $$

$$ \Rightarrow {e^{\sin x}} = 2 \pm \sqrt 5 $$ $$\,\,\,\,\,$$ (as $$t = {e^{\sin x}}$$)

$$ \Rightarrow {e^{\sin x}} = 2 - \sqrt 5 $$ and

$$\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $${e^{\sin x}} = 2 + \sqrt 5 $$

$$ \Rightarrow {e^{\sin x}} = 2 - \sqrt 5 < 0$$

and $$\,\,\,\,\,\,\sin x = \ln \left( {2 + \sqrt 5 } \right) > 1$$ So, rejected

Hence given equation has no solution.

$$\therefore$$ The equation has no real roots.

Questions Asked from Quadratic Equation and Inequalities

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