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JEE Mathematics Solved Question Paper 2014

Multiple Choice Questions

21.

According to Newton-Raphson method, the value of $\sqrt{12}$ . upto three places of decimal will be

3.463
3.462
3.467
None of these

22.

If $\frac{{(3 - i)}^{2}}{2 + i} = A + iB$ where A and B are real numbers, then A and Bare equal to

A = - 4, B = 2
A = 2, B = - 4
A = 2, B = 4
None of these

23.

The radical centre of the system of circles

x² + y² + 4x + 7 = 0,

2(x² + y²) + 3x + 5y + 9 = 0

and x² + y² + y = 0 is

(- 2, - 1)
(1, - 2)
(- 1, - 2)
None of these

24.

The sum of n terms of the series 1 + 5 + 12 + 22 + 35 + ... is

$\frac{n^{2} (n + 1)}{8}$
$\frac{n^{2} (n + 1)}{6}$
$\frac{n^{2} (n + 1)}{2}$
None of these

25.

If lines of regression are 3x+ 12y = 19 and 3y+ 9x = 46, then value of r_xy will be

0.289
- 0.289
0.209
None of these

26.

If 1, w and w² are the cube roots of unity, then the value of (1 - w + w²)(1 + w - w²) is equal to

27.

If geometric mean and harmonic mean of two numbers a and b are 16 and 64/5 respectively, then the value of a : b is

4 : 1
3 : 2
2 : 3
1 : 4

28.
If the sum offour numbers in GP is 60 and the arithmetic mean of the first and last numbers is 18, then the numbers are

3, 9, 27, 81

4, 8, 16, 32

2, 6, 18, 54

None of these

4, 8, 16, 32

$Let four terms mn a GP be {ar}^{3}, ar, \frac{a}{r} and \frac{a}{r^{3}} According to the given condition, {ar}^{3} + ar + \frac{a}{r} + \frac{a}{r^{3}} = 60 . . . (i) and \frac{{ar}^{3} + \frac{a}{r^{3}}}{2} = 18 \Rightarrow {ar}^{3} + \frac{a}{r^{3}} = 36 . . . (ii) Now, from Eq (i), we have (ar + \frac{a}{r}) + {ar}^{3} + \frac{a}{r^{3}} = 60 \Rightarrow a (r + \frac{1}{r}) + 36 = 60 [∵ from Eq . (ii)] \Rightarrow a (r + \frac{1}{r}) = 24 . . . (iii)$

$On dividing Eq . (iii) by Eq . (ii), we get \frac{a (r^{3} + \frac{1}{r^{3}})}{a (r + \frac{1}{r})} = \frac{36}{24} \Rightarrow \frac{(r + \frac{1}{r}) (r^{2} + \frac{1}{r^{2}} - 1)}{r + \frac{1}{r}} = \frac{3}{2} \Rightarrow 2 (r^{2} + \frac{1}{1^{2}} - 1) = 3 \Rightarrow \frac{2 (r^{4} + 1 - r^{2})}{r^{2}} = 3 \Rightarrow 2 r^{4} + 2 - 2 r^{2} = 3 r^{2} \Rightarrow 2 r^{4} - 5 r^{2} + 2 = 0 \Rightarrow 2 r^{4} - 4 r^{2} - r^{2} + 2 = 0 \Rightarrow 2 r^{2} (r^{2} - 2) - 1 (r^{2} - 2) = 0 \Rightarrow (r^{2} - 2) (2 r^{2} - 1) = 0 \Rightarrow r^{2} = 2, 2 r^{2} = 1 \Rightarrow r = \pm \sqrt{2}, r = \pm \frac{1}{\sqrt{2}}$

$On putting r = \sqrt{2} in Eq . (iii), we get a (\sqrt{2} + \frac{1}{\sqrt{2}}) = 24 \Rightarrow a (\frac{2 + 1}{\sqrt{2}}) = 24 \Rightarrow 3 a = 24 a \sqrt{2} \Rightarrow a = 8 \sqrt{2} ∴ Senes becomes 8 \sqrt{2} {(\sqrt{2})}^{3}, 8 \sqrt{2} (\sqrt{2}), \frac{8 \sqrt{2}}{\sqrt{2}} and \frac{8 \sqrt{2}}{{(\sqrt{2})}^{3}} i . e ., 32, 16, 8 and 4 . If we take r = \frac{1}{\sqrt{2}}, we get the seris 4, 8, 16 and 32 .$