clinic class test 7 m e sol

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, ).

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)