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JEE Mathematics Solved Question Paper 2007

Multiple Choice Questions

71.

If a = $\hat{i} + \hat{j} + \hat{k}$ , b = $2 \hat{i} - 4 \hat{k}$ and c = $\hat{i} + λ \hat{j} + 3 \hat{k}$ are coplanar, then the value of $λ$ , is

5/2
3/5
7/3
None of these

72.

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

$\cos^{- 1} (\sqrt{x}) + \sqrt{1 - x} (\sqrt{x} - 2) + c$
$\cos^{- 1} (\sqrt{x}) - \sqrt{1 - x} (\sqrt{x} - 2) + c$
$\cos^{- 1} (\sqrt{x}) + \sqrt{1 - x} (\sqrt{x} + 2) + c$
None of the above

73.

$\int \frac{\log (x + \sqrt{1 + x^{2}})}{\sqrt{1 + x^{2}}} dx$ is equal to

$\frac{1}{2} {[\log (x + \sqrt{1 + x^{2}})]}^{2} + c$
$\log {(x + \sqrt{1 + x^{2}})}^{2} + c$
$\log (x + \sqrt{1 + x^{2}}) + c$
None of the above

74.

$\int \frac{\sin^{8} (x) - \cos^{8} (x)}{1 - 2 \sin^{2} (x) \cos^{2} (x)} dx$ is equal to

$\sin (2 x) + c$
$- \frac{1}{2} \sin (2 x) + c$
$\frac{1}{2} \sin (2 x) + c$
$- \sin (2 x) + c$

75.

$\int \frac{dx}{(\sin (x) + \sin (2 x))}$ is equal to

$\frac{1}{6} \log (1 - \cos (x)) + \frac{1}{2} \log (\log (1 + \cos (x))) - \frac{2}{3} \log (1 + 2 \cos (x)) + c$
$6 \log (1 - \cos (x)) + 2 \log (\log (1 + \cos (x))) - \frac{2}{3} \log (1 + 2 \cos (x)) + c$
$6 \log (1 - \cos (x)) + \frac{1}{2} \log (\log (1 + \cos (x))) + \frac{2}{3} \log (1 + 2 \cos (x)) + c$
None of the above

76.

$\int [f (x) g'' (x) - f'' (x) g ((x))] dx$ is equal to

$\frac{f (x)}{g' (x)}$
f'(x)g(x) - f(x)g'(x)
f(x)g'(x) - f'(x)g(x)
f(x)g'(x) + f'(x)g(x)

77.

Correct value of $\int_{0}^{π} |\sin^{4} (x)| dx$ is

$\frac{8 π}{3}$
$\frac{2 π}{3}$
$\frac{4 π}{3}$
$\frac{3 π}{8}$

78.
$\int_{0}^{\frac{π}{2}} \log (\sin (x)) dx$ is equal to

$- \frac{π}{2} \log (2)$

$πlog (\frac{1}{2})$

$- \frac{π}{2} \log (\frac{1}{2})$

$\log (2)$

$- \frac{π}{2} \log (2)$

$Let I = \int_{0}^{\frac{π}{2}} \log (\sin (x)) dx . . . (i) = \int_{0}^{\frac{π}{2}} \log (\sin (\frac{π}{2} - x)) dx \Rightarrow I = \int_{0}^{\frac{π}{2}} \log (\cos (x)) dx . . . (ii) On adding Eqs . (i) and (ii), we get 2 I = \int_{0}^{\frac{π}{2}} \log (2 \sin (x) \cos (x) - \log (2)) dx = \int_{0}^{\frac{π}{2}} \log (\sin (2 x)) dx - \int_{0}^{\frac{π}{2}} \log (2) dx = \int_{0}^{π} \log (\sin (x)) dx - \int_{0}^{\frac{π}{2}} \log (2) dx = \int_{0}^{\frac{π}{2}} \log (\sin (x)) dx - \log (2) {[x]}_{0}^{\frac{π}{2}} = - \frac{π}{2} \log (2) - \log (2) [\frac{π}{2} - 0] ∴ I = - \frac{π}{2} \log (2)$