Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.

JEE Mathematics Solved Question Paper 2008

Multiple Choice Questions

11.

$Given that a, b \in \{0, 1, 2, . . ., 9\} with a + b \neq 0 and that {(a + \frac{b}{10})}^{x} = {(\frac{a}{b} + \frac{b}{100})}^{y} = 1000 . Then, \frac{1}{x} - \frac{1}{y} is equal to$

1
$\frac{1}{2}$
$\frac{1}{3}$
$\frac{1}{4}$

12.

$If x = \frac{1}{2} (\sqrt{7} + \frac{1}{\sqrt{7}}), then \frac{\sqrt{x^{2} - 1}}{x - \sqrt{x^{2} - 1}} is equal to$

13.

$For any integer n \geq 1, the sum \sum_{k = 1}^{n} k (k + 2) is equal to$

$\frac{n (n + 1) (n + 2)}{6}$
$\frac{n (n + 1) (2 n + 1)}{6}$
$\frac{n (n + 1) (2 n + 7)}{6}$
$\frac{n (n + 1) (2 n + 9)}{6}$

14.

9 balls are to be placed in 9 boxes and 5 of the balls cannot fit into 3 small boxes. The number of ways of arranging one ball in each of the boxes is

18720
18270
17280
12780

15.

$If {}^{n}P_{r} = 30240 and {}^{n}C_{r} = 252, then the ordered pair (n, r) is equal to$

(12, 6)
(10, 5)
(9, 4)
(16, 7)

16.

$If {(1 + x + x^{2} + x^{3})}^{5} = \sum_{k = 0}^{15} a_{k} x^{k}, then {\sum a_{2}}_{k = 0}^{7} k = 0$

128
256
512
1024

17.
$If α = \frac{5}{2! 3} + \frac{5 . 7}{3! 3^{2}} + \frac{5 . 7 . 9}{4! 3^{3}} + . . ., then α^{2} + 4 α is equal to$

21

23

25

27

$Given that, α = \frac{5}{2! 3} + \frac{5 . 7}{3! 3^{2}} + \frac{5 . 7 . 9}{4! 3^{3}} + . . . . . . (i) We know that, {(1 + x)}^{n} = 1 + \frac{nx}{1!} + \frac{n (n - 1)}{2!} x^{2} + \frac{n (n - 1) (n - 2)}{3!} x^{3} + . . . . . . (ii) On compairing eqs . (i) and (ii), with respect to factorial n (n - 1) x^{2} = \frac{5}{3} . . . (iii) n (n - 1) (n - 2) x^{3} = \frac{5 . 7}{3^{2}} . . . (iv) and n (n - 1) (n - 2) (n - 3) x^{4} = \frac{5 . 7 . 9}{3^{3}} . . . (v) On dividing eq . (iv) by (iii) and eq . (v) by (iv), we get (n - 2) x = \frac{7}{3} . . . (vi) and (n - 3) x = 3 . . . (vii) Again, dividing eq . (vi) by (vii), we get \frac{n - 2}{n - 3} = \frac{7}{9}$

$\Rightarrow 9 n - 18 = 7 n - 21 \Rightarrow 2 n = - 3 \Rightarrow n = - \frac{3}{2} On putting the value of n in eq (vi, we get) (- \frac{3}{2} - 2) x = \frac{7}{3} \Rightarrow x = - \frac{2}{3} ∴ From eq . (ii) {(1 - \frac{2}{3})}^{- \frac{3}{2}} = 1 + 1 + \frac{5}{2! 3} + \frac{5 . 7}{3! 3^{2}} + . . . \Rightarrow 3^{\frac{3}{2}} - 2 = \frac{5}{2! 3} + \frac{5 . 7}{3! 3^{2}} + . . . \Rightarrow α = 3^{\frac{3}{2}} - 2 [from eq . (i)] Now, α^{2} + 4 α = {(3^{\frac{3}{2}} - 2)}^{2} + 4 (3^{\frac{3}{2}} - 2) = 27 + 4 - 4 3^{\frac{3}{2}} + 4 . 3^{\frac{3}{2}} - 8 = 23$