1

JEE Advanced 2016 Paper 1 Offline

MCQ (Single Correct Answer)
Let $$ - {\pi \over 6} < \theta < - {\pi \over {12}}.$$ Suppose $${\alpha _1}$$ and $${B_1}$$ are the roots of the equation $${x^2} - 2x\sec \alpha + 1 = 0$$ and $${\alpha _2}$$ and $${\beta _2}$$ are the roots of the equation $${x^2} + 2x\,\tan \theta - 1 = 0.$$ $$If\,{\alpha _1} > {\beta _1}$$ and $${\alpha _2} > {\beta _2},$$ then $${\alpha _1} + {\beta _2}$$ equals
A
$$2\left( {\sec \theta - \tan \theta } \right)$$
B
$$2\,\sec \,\theta $$
C
$$ - 2\tan \theta $$
D
$$0$$
2

JEE Advanced 2014 Paper 2 Offline

MCQ (Single Correct Answer)
The quadratic equation $$p(x)$$ $$ = 0$$ with real coefficients has purely imaginary roots. Then the equation $$p(p(x))=0$$ has
A
one purely imaginary root
B
all real roots
C
two real and two purely imaginary roots
D
neither real nor purely imaginary roots
3

IIT-JEE 2012 Paper 2 Offline

MCQ (Single Correct Answer)

Let $$\alpha$$(a) and $$\beta$$(a) be the roots of the equation $$(\root 3 \of {1 + a} - 1){x^2} + (\sqrt {1 + a} - 1)x + (\root 6 \of {1 + a} - 1) = 0$$ where $$a > - 1$$. Then $$\mathop {\lim }\limits_{a \to {0^ + }} \alpha (a)$$ and $$\mathop {\lim }\limits_{a \to {0^ + }} \beta (a)$$ are

A
$$ - {5 \over 2}$$
B
$$ - {1 \over 2}$$
C
$$ - {7 \over 2}$$
D
$$ - {9 \over 2}$$

Explanation

Let a + 1 = t6. Thus, when a $$\to$$ 0, t $$\to$$ 1.

$$\therefore$$ $$({t^2} - 1){x^2} + ({t^3} - 1)x + (t - 1) = 0$$

$$ \Rightarrow (t - 1)\{ (t + 1){x^2} + ({t^2} + t + 1)x + 1\} = 0$$,

as t $$\to$$ 1

$$2{x^2} + 3x + 1 = 0$$

$$ \Rightarrow 2{x^2} + 2x + x + 1 = 0$$

$$ \Rightarrow (2x + 1)(x + 1) = 0$$

Thus, x = $$-$$1, $$-$$1/2

or, $$\mathop {\lim }\limits_{a \to {0^ + }} \alpha (a) = - {1 \over 2}$$

and $$\mathop {\lim }\limits_{a \to {0^ + }} \beta (a) = - 1$$

4

IIT-JEE 2011 Paper 2 Offline

MCQ (Single Correct Answer)
A value of $$b$$ for which the equations $$$\matrix{ {{x^2} + bx - 1 = 0} \cr {{x^2} + x + b = 0} \cr } $$$

have one root in common is

A
$$ - \sqrt 2 $$
B
$$ - i\sqrt 2 $$
C
$$i\sqrt 5 $$
D
$$\sqrt 2 $$

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