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JEE Mathematics Solved Question Paper 2008

Multiple Choice Questions

One possible condition for the three points (a, b), (b, a) and (a², - b²) to be collinear is

a - b = 2
a + b = 2
a = 1 + b
a = 1 - b

If the mth term and the nth term of an AP are respectively $\frac{1}{n}$ and $\frac{1}{m}$ , then the mnth term of the AP is

$\frac{1}{mn}$
$\frac{m}{n}$
1
$\frac{n}{m}$

$\int_{- \frac{π}{2}}^{\frac{π}{2}} \sin^{9} (x) \cos^{5} (x) d x$ equals

$\frac{1}{20}$
20
0
$\frac{1}{330}$

4.
The function f(x) = $\log (\frac{1 + x}{1 - x})$ satisfies the equation

f(x + 2) - 2f(x + 1) + f(x) = 0

f(x) + f(x + 1) = f{x(x + 1)}

f(x) + f(y) = $f (\frac{x + y}{1 + xy})$

f(x + y) = f(x)f(y)

f(x) + f(y) = $f (\frac{x + y}{1 + xy})$

Since, f(x) = $\log (\frac{1 + x}{1 - x})$

$∴ f (\frac{x + y}{1 + xy}) = \log (\frac{1 + \frac{x + y}{1 + xy}}{1 - \frac{x + y}{1 + xy}}) = \log (\frac{1 + xy + x + y}{1 + xy - x - y})$

$Now, f (x) + f (y) = \log (\frac{1 + x}{1 - x}) + \log (\frac{1 + y}{1 - y}) = \log (\frac{(1 + x) (1 + y)}{(1 - x) (1 - y)}) = \log (\frac{1 + x + y + xy}{1 + xy - x - y}) ∴ f (\frac{x + y}{1 + xy}) = f (x) + f (y)$

The value of (1 - w + w²)⁵ + (1 + w - w²)⁵, where w and w² are the complex cube roots of unity, is

0
32w
- 32
32

The value of $\lim_{n \to \infty} (\frac{1}{n + 1} + \frac{1}{n + 2} + . . . + \frac{1}{6 n})$ is

$\log (2)$
$\log (6)$
1
$\log (3)$

$\lim_{x \to \frac{π}{2}} \frac{a^{\cot (x)} - a^{\cos (x)}}{\cot (x) - \cos (x)}$

$\log_{e} (\frac{π}{2})$
$\log_{e} (2)$
$\log_{e} (a)$
a

The equation of the circle which passes through the points of intersection of the circles x² + y² - 6x = 0 and x² + y² - 6y = 0 and has its centre at $(\frac{3}{2}, \frac{3}{2})$ , is