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JEE Mathematics Solved Question Paper 2003

Multiple Choice Questions

21.

$\int_{0}^{π} \frac{xdx}{a^{2} \cos^{2} (x) + b^{2} \sin^{2} (x)} dx$ is equal to

$\frac{π}{2 ab}$
$\frac{π}{ab}$
$\frac{π^{2}}{2 ab}$
$\frac{π^{2}}{ab}$

22.

The differential equation for which sin^-1(x) + sin^-1(y) = c is given by

$\sqrt{1 - x^{2}} dy + \sqrt{1 - y^{2}} dx = 0$
$\sqrt{1 - x^{2}} dx + \sqrt{1 - y^{2}} dy = 0$
$\sqrt{1 - x^{2}} dx - \sqrt{1 - y^{2}} dy = 0$
$\sqrt{1 - x^{2}} dy - \sqrt{1 - y^{2}} dx = 0$

23.

$\int e^{x} (\frac{1 + \sin (x)}{1 + \cos (x)}) dx$ is equal to

$e^{x} \sec^{2} (\frac{x}{2}) + c$
$e^{x} \tan (\frac{x}{2})$ + c
$e^{x} \sec (\frac{x}{2}) + c$
$e^{x} \tan (x) + c$

24.

$\int \sqrt{1 + \sin (\frac{x}{4})} dx$ is equal to

$8 (\sin (\frac{x}{8}) + \cos (\frac{x}{8})) + C$
$8 (\sin (\frac{x}{8}) - \cos (\frac{x}{8})) + C$
$8 (\cos (\frac{x}{8}) - \sin (\frac{x}{8})) + C$
$\frac{1}{8} (\sin (\frac{x}{8}) - \cos (\frac{x}{8})) + C$

25.

$\int_{0}^{\infty} \frac{xdx}{(1 + x) (1 + x^{2})}$ is equal to

$\frac{π}{2}$
0
1
$\frac{π}{4}$

26.

If I_n = $\int {(\log (x))}^{n} dx$ , then I_n + nI_{n - 1} is equal to

${(xlog (x))}^{n}$
$x {(\log (x))}^{n}$
$n {(\log (x))}^{n}$
${(\log (x))}^{n - 1}$

27.
The area included between the parabolas x² = 4y and y² = 4x is (in square units)

$\frac{4}{3}$

$\frac{1}{3}$

$\frac{16}{3}$

$\frac{8}{3}$

$\frac{16}{3}$

$y^{2} = 4 x . . . (i) x^{2} = 4 y . . . (ii) From Eqs . (i) and (ii), we get {(\frac{x^{2}}{4})}^{2} = 4 x \Rightarrow x^{4} - 64 x = 0 \Rightarrow x (x^{3} - 4^{3}) = 0 ∴ x = 0, x = 4 ∴ Required area = |\int_{0}^{4} \sqrt{4 x} dx - \int_{0}^{4} \frac{x^{2}}{4} dx| = |{\{2 (\frac{2}{3}) x^{\frac{3}{2}}\}}_{0}^{4} - \frac{1}{4} {\{\frac{x^{3}}{3}\}}_{0}^{4}| = \frac{32}{3} - \frac{16}{3} = \frac{16}{3} sq unit$