1

IIT-JEE 2012 Paper 2 Offline

MCQ (Single Correct Answer)
Let $$PQR$$ be a triangle of area $$\Delta $$ with $$a=2$$, $$b = {7 \over 2}$$ and $$c = {5 \over 2}$$; where $$a, b,$$ and $$c$$ are the lengths of the sides of the triangle opposite to the angles at $$P.Q$$ and $$R$$ respectively. Then $${{2\sin P - \sin 2P} \over {2\sin P + \sin 2P}}$$ equals.
A
$${3 \over {4\Delta }}$$
B
$${45 \over {4\Delta }}$$
C
$${\left( {{3 \over {4\Delta }}} \right)^2}$$
D
$${\left( {{45 \over {4\Delta }}} \right)^2}$$
2

IIT-JEE 2010 Paper 1 Offline

MCQ (Single Correct Answer)
Let $$ABC$$ be a triangle such that $$\angle ACB = {\pi \over 6}$$ and let $$a, b$$ and $$c$$ denote the lengths of the sides opposite to $$A$$, $$B$$ and $$C$$ respectively. The value(s) of $$x$$ for which $$a = {x^2} + x + 1,\,\,\,b = {x^2} - 1\,\,\,$$ and $$c = 2x + 1$$ is (are)
A
$$ - \left( {2 + \sqrt 3 } \right)$$
B
$${1 + \sqrt 3 }$$
C
$${2 + \sqrt 3 }$$
D
$${4 \sqrt 3 }$$

Explanation

Using, $$\cos C = {{{a^2} + {b^2} - {c^2}} \over {2ab}}$$

$$ \Rightarrow {{\sqrt 3 } \over 2} = {{{{({x^2} + x + 1)}^2} + {{({x^2} - 1)}^2} - {{(2x + 1)}^2}} \over {2({x^2} + x + 1)({x^2} - 1)}}$$

$$ \Rightarrow (x + 2)(x + 1)(x - 1)x + {({x^2} - 1)^2} = \sqrt 3 ({x^2} + x + 1)({x^2} - 1)$$

$$ \Rightarrow {x^2} + 2x + ({x^2} - 1) = \sqrt 3 ({x^2} + x + 1)$$

$$ \Rightarrow (2 - \sqrt 3 ){x^2} + (2 - \sqrt 3 )x - (\sqrt 3 + 1) = 0$$

$$ \Rightarrow x = - (2 + \sqrt 3 )$$ and $$x = 1 + \sqrt 3 $$

But, $$x = - (2 + \sqrt 3 ) \Rightarrow c$$ is negative.

$$\therefore$$ $$x = 1 + \sqrt 3 $$ is the only solution.

Hence, (b) is the correct option.

3

IIT-JEE 2010 Paper 1 Offline

MCQ (Single Correct Answer)
If the angles $$A, B$$ and $$C$$ of a triangle are in an arithmetic progression and if $$a, b$$ and $$c$$ denote the lengths of the sides opposite to $$A, B$$ and $$C$$ respectively, then the value of the expression $${a \over c}\sin 2C + {c \over a}\sin 2A$$ is
A
$${1 \over 2}$$
B
$${{\sqrt 3 } \over 2}$$
C
$$1$$
D
$${\sqrt 3 }$$

Explanation

Since, A, B, C are in AP

$$\Rightarrow$$ 2B = A + C i.e., $$\angle$$B = 60$$^\circ$$

$$\therefore$$ $${a \over c}$$(2 sin C cos C) + $${c \over a}$$ (2 sin A cos A)

= 2k (a cos C + c cos A)

[using, $${a \over {\sin A}} = {b \over {\sin B}} = {c \over {\sin C}} = {1 \over k}$$]

= 2k (b)

= 2 sin B

[using, b = a cos C + c cos A]

= $$\sqrt3$$

4

IIT-JEE 2007

MCQ (Single Correct Answer)
Let $$ABCD$$ be a quadrilateral with area $$18$$, with side $$AB$$ parallel to the side $$CD$$ and $$2AB=CD$$. Let $$AD$$ be perpendicular to $$AB$$ and $$CD$$. If a circle is drawn inside the quadrilateral $$ABCD$$ touching all the sides, then its radius is
A
$$3$$
B
$$2$$
C
$${3 \over 2}$$
D
$$1$$

Joint Entrance Examination

JEE Main JEE Advanced WB JEE

Graduate Aptitude Test in Engineering

GATE CSE GATE ECE GATE EE GATE ME GATE CE GATE PI GATE IN

Medical

NEET

CBSE

Class 12