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JEE Mathematics Solved Question Paper 2004

Multiple Choice Questions

61.

If G is the centroid of the $∆$ ABC, then GA + BG + GC is equal

62.

If the vectors $\hat{i} + 3 \hat{j} + 4 \hat{k}, λ \hat{i} - 4 \hat{j} + \hat{k}$ are orthogonal to each other, then $λ$ is equal to

63.

The vector c . (b + c) x (a + b + c) is equal to

c . b x a
0
c . a x b
a . c x b

64.

$If 3 \hat{i} + 3 \hat{j} + \sqrt{3} \hat{k}, \sqrt{3} \hat{i} + \sqrt{3} \hat{j} + λ \hat{k} are coplaner, then λ is equal to$

65.

If X is a poisson variate with P(X = 0) = 0.8, then the variance of X is

$\log_{e} (20)$
$\log_{10} (20)$
$\log_{e} (\frac{5}{4})$
0

66.
If the direction ratio of two lines are given by l + m + n = 0, mn - 2ln + lm = 0, then the angle between the lines is

$\frac{π}{4}$

$\frac{π}{3}$

$\frac{π}{2}$

0

$\frac{π}{2}$

$Given lines are l + m + n = 0 . . . (i) and mn - 2 \ln + lm = 0 . . . (ii) From equation (i) l = - (m + n) Putting inequation (ii), we get mn + 2 (m + n) n - (m + n) m = 0 \Rightarrow mn + 2 mn + 2 n^{2} - m^{2} - nm = 0 \Rightarrow 2 n^{2} - m^{2} + 2 nm = 0 \Rightarrow 2 {(\frac{n}{m})}^{2} + \frac{2 n}{m} - 1 = 0 This is a quadratic equation in (\frac{n}{m}) . ∴ \frac{n_{1} n_{2}}{m_{1} m_{2}} = \frac{- 1}{2} . . . . (iii) [where \frac{n_{1}}{m_{1}}, \frac{n_{2}}{m_{2}} are the roots of equation] From equation (i) m = - (n + l) Putting equation (ii), we get - (n + l) n - 2 \ln - l (n + l) \Rightarrow n^{2} + \ln + 2 \ln + \ln + l^{2} = 0$

$\Rightarrow l^{2} + 3 \ln + n^{2} = 0 \Rightarrow {(\frac{l}{n})}^{2} + \frac{3 l}{n} + 1 = 0 \Rightarrow \frac{l_{1} l_{2}}{n_{1} n_{2}} = 1 . . . (iv) [Where \frac{l_{1}}{n_{1}}, \frac{l_{2}}{n_{2}} are the roots of the equation] From equations (iii) and (iv), we get l_{1} l_{2} = - \frac{1}{2} m_{1} m_{2} = n_{1} n_{2} \Rightarrow \frac{l_{1} l_{2}}{1} = \frac{m_{1} m_{2}}{- 2} = \frac{n_{1} n_{2}}{1} = 1 (say) Now, l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2} = 1 - 2 + 1 = 0 ∴ \cos (θ) = 0 \Rightarrow θ = 90 °$