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JEE Mathematics Solved Question Paper 2009

Multiple Choice Questions

If A(z₁), B(z₂), C(z₃) and P(z) represent complex numbers such that $|z_{1} - z| = |z_{2} - z| = |z_{3} - z|$ , then, A, B, C lies on

a straight line
a circle
a parabola
an ellipse

If the complex numbers z, z, and origin form vertices of an equilateral triangle, then the value of z₁² + z₂² will be

z₁z₂
z₁ + z₂
2z₁z₂
z₁ - z₂

Three numbers form an increasing GP. If the middle term is doubled, then the new numbers are in AP. The commonratio of the GP will be

$2 - \sqrt{3}$
$2 \pm \sqrt{3}$
$3 \sqrt{2}$
$2 + \sqrt{3}$

If the equations ax² + 2cx + b = 0 and ax² + 2bx + c = 0 $(b \neq c)$ have a common root, then the value of a + 4b + 4c will be

2
1
- 1
Non eof these

If one root of ax² + bx + c = 0 is twice the other root, then

b² = 9ac
2b² = 9ac
2b² = ac
b² = ac

The number of ways of distributing 8 distinct toys among 5 children will be

5⁸
8⁵
${}^{8}P_{5}$
40

The value of C₁ - 2 . C₂ + 3 . C₃ - 4 . C₄ + ... where C_r = ${}^{n}C_{r}$ will be

- 1
1
0
None of these

8.
In a $∆ ABC, if \tan (\frac{A}{2}) = \frac{5}{6} and \tan (\frac{C}{2}) = \frac{2}{5}$ , then the sides a, b, c are in

AP

GP

HP

None of these

$Given, \tan (\frac{A}{2}) = \frac{5}{6} and \tan (\frac{C}{2}) = \frac{2}{5} ∴ \tan (\frac{A}{2} + \frac{C}{2}) = \frac{\frac{5}{6} + \frac{2}{5}}{1 - \frac{5}{6} \times \frac{2}{5}} = \frac{37}{20} \Rightarrow \cot (\frac{B}{2}) = \frac{37}{20} \Rightarrow \tan (\frac{B}{2}) = \frac{37}{20} ∴ \sin (A) = \frac{2 \tan (\frac{A}{2})}{1 + \tan^{2} (\frac{A}{2})} = \frac{2 \times \frac{5}{6}}{1 + \frac{25}{36}} = \frac{60}{61} \sin (B) = \frac{2 \tan (\frac{B}{2})}{1 + \tan^{2} (\frac{B}{2})} = \frac{2 \times \frac{20}{37}}{1 + \frac{400}{1369}} = \frac{1480}{1769} \sin (C) = \frac{2 \tan (\frac{C}{2})}{1 + \tan^{2} (\frac{C}{2})} = \frac{2 \times \frac{2}{5}}{1 + \frac{4}{25}} = \frac{20}{29} Using sine rule, \frac{a}{\sin (A)} = \frac{b}{\sin (B)} = \frac{c}{\sin (C)} \Rightarrow \frac{a}{\frac{60}{61}} = \frac{b}{\frac{1480}{1769}} = \frac{c}{\frac{20}{29}} = k \Rightarrow a = \frac{60}{61} k, b = \frac{1480}{1769} k, c = \frac{20}{29} k Now, a + c = \frac{60}{61} k + \frac{20}{29} k = \frac{2960 k}{1769} \Rightarrow a + c = 2 b Hence, a, b and c are in AP .$