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JEE Mathematics Solved Question Paper 2007

Multiple Choice Questions

The value of $\frac{2}{3!} + \frac{4}{5!} + \frac{6}{7!} + . . .$

e
2e
e²
$\frac{1}{e}$

$(1 + \cos (\frac{π}{8})) (1 + \cos (\frac{3 π}{8})) (1 + \cos (\frac{5 π}{8})) (1 + \cos (\frac{7 π}{8}))$ is equal to

$\frac{1}{2}$
$\frac{1}{8}$
$\cos (\frac{π}{8})$
$\frac{1}{4}$

If $\sqrt{3} \cos (θ) + \sin (θ) = \sqrt{2}$ , then the value of $θ$ is

$nπ + {(- 1)}^{n} \frac{π}{4}$
${(- 1)}^{n} \frac{π}{4} - \frac{π}{3}$
$nπ + \frac{π}{4} - \frac{π}{3}$
$nπ + {(- 1)}^{n} \frac{π}{4} - \frac{π}{3}$

If in a AABC, (s - a)(s - b) = s(s - c) then angle C is equal to

90°
45°
30°
60°

5.
The length of the shadows of a vertical pole of height h, thrown by the sun's rays at three different moments are h, 2h and 3h. The sum of the angles of elevation of the rays at these three moments is equal to

$\frac{π}{2}$

$\frac{π}{3}$

$\frac{π}{4}$

$\frac{π}{6}$

$\frac{π}{2}$

$In ∆ ABC, \tan (α) = \frac{AB}{BC} = \frac{h}{h} = 1 \Rightarrow \tan (α) = \tan (\frac{π}{4}) \Rightarrow α = \frac{π}{4} Now, in ∆ ABD \tan (β) = \frac{AB}{BD} = \frac{h}{2 h} \Rightarrow \tan (β) = \frac{1}{2} \Rightarrow β = \tan^{- 1} (\frac{1}{2}) and in ∆ ABE, \tan (γ) = \frac{AB}{BE} = \frac{h}{3 h} \Rightarrow \tan (γ) = \frac{1}{3} \Rightarrow γ = \tan^{- 1} (\frac{1}{3})$

$∴ Required sum of angles = α + β + γ = \frac{π}{4} + \tan^{- 1} (\frac{1}{2}) + \tan^{- 1} (\frac{1}{3}) = \frac{π}{4} + \tan^{- 1} (\frac{\frac{1}{2} + \frac{1}{3}}{1 - \frac{1}{2} \times \frac{1}{3}}) = \frac{π}{4} + \tan^{- 1} (\frac{\frac{5}{6}}{\frac{5}{6}}) = \frac{π}{4} + \tan^{- 1} (1) = \frac{π}{4} + \frac{π}{4} = \frac{π}{2}$