clinic class test 7 m e sol
DESCRIPTION
Clinic Class Test 7 M E SolTRANSCRIPT
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RESONANCE SOL050313 - 1
HINTS & SOLUTIONS
DATE : 05-03-2013COURSE NAME : VIKAAS (JA) & VIPUL (JB)
TEST - 7TARGET : JEE (IITs) 2014
COURSE CODE : CLINIC CLASSES Mathematics
(Fundamental of Mathematics)
1. (C)Sol. (x + 2y) + (x 3y) 6 = (x y 2) 5 + )2yx2(
Since x, y Qx + 2y = 2x + y 2 .........(i)x y 2 = 0 .........(ii)x 3y = 0 .........(iii)On solving (ii) and (iii) x = 3, y = 1 x + y = 4
2. (D)3. (B)Sol. 8x6x
2)2x( = 1Either |x 2| = 1 or x2 6x + 8 = 0x = 3, 1 x = 2, 4x = 2 is not possible x = 1, 3, 4
4. (C)Sol. P = log5 35 log5 245 log5 7 log5 6125
= log5 35 log5 245 log5 7 (log5 25 + log5 245) = log5 35 log5 245 log5 7 (2 + log5 245) = log5 245 (log5 35 log5 7) 2 log5 7
= log5 245 log5 49 = log5 49245
= log5 5 = 1
N = log2008 1 i.e N = 0
5. (C)
Sol. 47xlog
x
= 101 + logx
4
xlog7 logx = 1 + logx
7 logx + (logx)2 = 4 + 4 logx (logx)2 + 3 logx 4 = 0 logx = 4, 1 x = 104, 10 Number of solutions are = 2.
6. (D)
Sol. 7x . x)1x(3
3 = 73.33, no natural value of x.
7. (C)Sol. (x2 9) + ey + ix = 2y + 3i
(x2 9) + ey = 2y and x = 3ey = 2y and x = 3(e 2)y = 0 and x = 3y = 0 and x = 3x + y = 3
8. (C)Sol.
x [2, 1]
9. (B)Sol. 2/1)1x2x( + 2/1)1x2x(
= 22 )11x()11x(
= )11x( + | 11x |= )11x( )11x( = 2
10. (B)Sol. Put 2-x = t and 40.5 = 41/2 = 2
2t2 - 7t - 4 < 0 (t - 4) (2t + 1) < 0
421
t
But t = 2-x is +ive being exponential function hence
4221
x
2220 x
or
2
21
21
21
x
Since 121 -2 < x < ,2x
11. (D)
12. (C)Sol. P(1) = 5 2 a + b = 5 b a = 3 ......(i)
P(1) = 6 + a + b = 19 b + a = 13 .......(ii)solving (i) and (ii)b = 8, a = 5Hence remainder = P(2) = 10.
13. (A)Sol. We have |a b| = |a| |b| if a , b have the same sign and |a| |b|
i.e. x4 9 and x2 + 3 must be of same signi.e. (x4 9) (x2 + 3) > 0 (x2 3) (x2 + 3)2 > 0 x2 3 > 0 ...(1)and |x4 9| |x2 + 3|i.e. (x2 3) (x2 + 3) x2 + 3[using result (1)]i.e. (x2 + 3) (x2 4) 0 x2 4,gives x (,2) [2, ).
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RESONANCE SOL050313 - 2
14. (A)Sol. Domain x2 + 4x 5 0
x ( , 5] [1, )Case I :x ( , 5] [1, 3) ve < + ve alsways true x ( , 5] [1, 3) ... (1)Case IIx [3, ) .. (i)
x 3 < 5x4x2 x2 6x + 9 < x2 + 4x 5
x > 57
... (ii)
(i) (ii) x [3, ) ... (2)(1) (2) x ( , 5] [1, ) Ans. (A)
15. (C)
Sol. sin5
6 + i
5
6cos1
lies in 2nd quadrantand
56
sin56
cos1
= 10tan
102cot
53
cot
2nd quadrant 10
16. (AC)
Sol. 25 xlog +
x
5log x5 = 1
25 xlog + xlog1xlog1
5
5
= 1
t2 +
t1
t1 = 1 t2 + t3 t + 1 = 1 + t
t3 + t2 2t = 0t = 0 or t2 + t 2 = 0, (t + 2)(t 1) = 0 t = 1, t = 2
xi = 1, 5, 251
17. (ABC)
Sol. x = baab4
bab2
a2x
a2xa2x
= bab2bab2
a2xa2x
=
abab3
Similarly
baba3
b2xb2x
a2xa2x
+ b2xb2x
= 2)ba(
)ab3()ba3(
18. (ABC)Sol. Since the number is divisible by 5
b = 0, 5number is divisible by 3 iff a + b + 2 is divisible by 3(i) If b = 0, then a = 1, 4, 7(ii) If b = 5, then a = 2, 5, 8
19. (BC)Sol. m n = 4n2 3m2
(m n) (3m + 3n + 1) = 3 m2 3n2 + m n= 3m2 3n2 + 4n2 3m2= n2 .
Hence divisible by n2.(m n) (4m + 4n + 1) = 4m2 4n2 + 4n2 3m2 = m2.Hence divisible by m2.
20. (BC)Sol.(A) for two solution
||x 1| 2| = graph of f(x) = ||x 1| 2| (2, ) {0}(B) For three solution = 2(C) for four solution (0, 2)so integral values of = 1
21. (BCD)Sol. |z1 + z2|2 = |z1|2 + |z2|2
0zzzz 2121
2
1
2
1z
zz
z
0z
z
z
z
2
1
2
1
2
1z
zis purely imaginary
so amp
2
1z
z is may be 2
or 2
22. (A)Sol. |2 | [x] 1| | 2
||[x] 1| 2| 2 0 |[x] 1| 4 3 [x] 5 x [3, 6)