dpp11.pdf

   

Q.1924log(TS) Solve for x: log2 (4  x) + log (4  x) . log  x 

1 

  1    2 log2  x    = 0. 2    2 

 7 3  24  [Ans. 0, , ] 4 2   [Sol.

   

log2 (4  x) + log (4  x) . log  x 

let

1 

  1    2 log2  x    = 0 2    2 

   

log(4 – x) = A & log  x 

1 

= B

2 

A2 + AB – 2B 2 = 0 A2 + 2AB – AB – 2B 2 = 0 A(A + 2B) – B(A + 2B) = 0 A = B or A = – 2B

   



log(4 – x) = log  x 



4–x=x+

1 2

   

1 x

x3 – 3x2 –

4

2

2x =

7



2

x=

7 4





x  (1 2) 2

4x2 – x3 + 1 –

1



2 



log(4 – x) = – 2 log  x 

4–x=

1 

(4 – x)(x2 +

1 4

 + x) = 1

 + 4x – x2 – 1 = 0

15 x

 = 0 4 x(4x2 – 12x – 15) = 0 x=0 or 4x2 – 12x – 15 = 0 x= Reject x =



3  24 2

12  144  240 8

;

hence x =

 7 3  24  x = 0, ,   ] 4 2  

 =

12  384

3  24 2

8

 =

12  4 24 8

 =

3  24 2

[6]

Q.17 [Sol.

Solve the system of equations 5(logxy + logyx) = 26 and xy = 64. [Ans. (32, 2) or (2, 32)] Let logxy = t hence 1st equation is 5(t +

1 t

[6]

) = 26

5t2 – 26t + 5 = 0 5t2 – 25t – t + 5 = 0 5t(t – 5) – (t – 5) = 0 (t – 5)(5t – 1) = 0 hence t = 5 or t = 1/5 if t = 5 logxy = 5 64



x5 = y =

if

x6 = 64 x = 2 and y = 32 t = 1/5 then logxy = 1/5 y = x1/5 x = y5



64 y

(use 2nd  equation)

x

 = y5



y6 = 64  y = 2 and x = 32 hence solutions are (32, 2) or (2, 32) Ans. ] r  4

4

r  4

4

  (2r  1)    cos 8   . r 1

  (2r  1)  Q.18 Prove that   sin  = 8   r 1  

[6]

Also find their exact numerical value. [Ans. 3/2] [Sol.

4 LHS = sin

 8

 sin 4

   

2 sin 4

 8

3

 sin 4

8

 sin 4

5 8

 sin 4

7 8

3 

  4  4     = 2 sin  cos   = 8   8 8   

   

21  2 sin 2

 8

cos 2

   8 

    1  3 3   = 21    = 2 ×  = 2 4  4 2   4  3 5 7 4   cos 4  cos 4  cos 4 RHS = cos    

= 21 

1

8

   

2 cos 4

sin 2

8

 8

 cos 4

8

3 

8

       4     cos 4   = 21  2 sin 2 cos 2    = 2 sin 8   8 8  8 8     

  1  3 = 21    =  ] 2   4

(ii) tan2

 2

tan Q.14

 =

 2

1  cos 

 =

1  cos  1

 =

3

4

1

  or –

1 5  = 4 9 1 5 1 3

Ans. ] [5]

If log1227 = a find the value of log616 in term of a. 3 log 2 3

[Sol.

2  log 2 3  = a 3 log2 3 = 2a + a log 23 (3 – a) log23 = 2a log23 =

Q.15

2a 3 a log 2 16

4

now

log616 = log 6  = 1  log 3 2 2

 but

1 + log23 = 1 +



log616 =

2a 3a

4(3  a ) 3a

 =

3 a 3a

 Ans. ]

sin x  cos x  1

Prove the identity,

sin x  cos x  1

=

1  sin x cos x

  x    , wherever it is defined. Starting with left  4 2 

 = tan 

[5]

hand side only. [Sol.

LHS =

=

Q.16

(sin 2 x )  (cos x  1) 2 (sin x  cos x  1) 2 cos x 1  sin x

 =

(1  sin x ) cos x

=

2 cos x  (cos 2 x  1) 2  2 sin x cos x  2 cos x  2 sin x cos( x 2)  sin( x 2)

 =

2 cos x (1  cos x ) 2(1  sin x )(1  cos x )

1  tan(x 2)

  x   =  =  = tan    ] cos( x 2)  sin( x 2) 1  tan( x 2)  4 2 

[5]

Find the exact value of cos 24° – cos 12° + cos 48° – cos 84°. [Ans. 1/2]

[Sol.

LHS = (cos 24° – cos 84°) – (cos 12° – cos 48°) 2 sin 54° sin 30° – 2 sin 30° sin 18° sin 54° – sin 18° cos 36° – sin 18° 5  1 4

 –

5  1 4

 =

1 2

  Ans. ]

Q.9

[Sol.

Q.10

Assuming that x and y are both + ve satisfying the equation log (x + y) = log x + log y find y interms of  x. Base of the logarithm is 10 everywhere. [3] x [Ans. y = ] x 1 log(x + y) = log x + log y log(x + y) = logxy x + y = xy  y(x – 1) = x x y=  Ans. ]  x 1 cos x  cos 3x

If x = 7.5° then find the value of

sin 3x  sin x

[3]

. [Ans. 2 –

[Sol.

Q.11

cos x  cos 3x sin 3x  sin x

2 sin 2 x sin x

 =

2 sin x cos 2 x

 = tan 2x = tan (2 × 7.5) = tan 15° = 2 –

Find the solutions of the equation, log

2 sin x

3]

3  Ans. ]

(1  cos x ) = 2 in the interval x  [0, 2]. [4] [Ans. /3]

[Sol.

2

2 sin x = 1 + cos x 2 cos2x + cos x – 1 = 0 1



cos x =

 but

x =   and



or – 1

2

5 3

 are rejected 

x=

 3

, ,



x=

Q.12

2 Given that log 2 (a  1)  = 16 find the value of log

[Sol.

Given

a

now

5 3

 3

a

 Ans. ]

32

1 (a  ) . a

log 2 (a 2  1) = 16

[Ans.

a

log

a

32

(

a2 1 a

[4]

31 32

]

)

a2 1 log 2 ( ) 1 a 1 a2 1 2 a  1)  log 2 a log ( a log ( )  =  = 2 2 a a a 16 16 a log 2 ( a 32 )





a

= Q.13

If cos  = (i)

[Sol.

1  1 1  31 31  16  =  =  Ans. ]     16  2  16  2  32 4 5

 find the values of 

cos 3

(ii) tan



[Ans. (i) –

2

(i) cos 3 = 4 cos3 – 3 cos 

= 4·

64 125

 –

12 5

 =

256  300 125

 = –

44 125

Ans.

44 125

; (ii)

1 3

 or –

1 3

]

[4]

Q.5

Suppose that for some angles x and y the equations 3a

sin2x + cos2y = 2

and

2 a2

2

cos x + sin y =

2 hold simultaneously. Determine the possible values of a.

[3]

[Ans. a = 1] [Sol.

2

2

sin x + cos y =

3a

....(1) and

2

2

2

cos x + sin y =

a2 2

....(2)

Adding (1) and (2) 2=

2

 +

a2



a2 + 3a – 4 = 0

2 a = – 4 (rejected), a = 1 Ans. ]

 Q.6

3a



(a + 4)(a – 1) = 0

Find the sum of all the solutions of the equation (log27x3)2 = log27x6.

[3]

[Ans. sum = 10] [Sol.

3

(log27x ) = log27x (3 log27x)2 = 6 log27x 3 log27x (3 log27x – 2) = 0



x=1

  Q.7

[Sol.

6

or

log27x =

2 3

2/3

x = (27)  = 9 sum = 1 + 9 = 10 Ans. ]

If –

10 y  10  y





 < x <  and y = log 10(tan x + sec x). Then the expression E = 2 2 2 the six trigonometric functions. find the trigonometric function. [Ans. tan x] y = log10(tan x + sec x)

 simplifies to one of  [3]

 1  sin x     cos x  

y = log10 

y

E=

=

Q.8

10  10 2

y

 1  sin x    cos x      1  sin 2 x  2 sin x  cos2 x cos x 1  sin x         =  = 2 cos x (1  sin x ) 2

2 sin 2 x  2 sin x 2 cos x (1  sin x )

 =

2 sin x (1  sin x ) 2 cos x (1  sin x )

 = tan x Ans. ]

If log 2 log 2 (log 2 x )  = 2 then find the number of digits in x. You may use log 102 = 0.3010. [Ans. 5 ]

[Sol.

log 2 log 2 (log 2 x )   = 2

log2(log2x) = 4  log2x = 16 x = 216  log10x = 16 log 2 = 16 × 0.3010 = 4.8160    Number of digits = 5 Ans. ]

[3]

M ATH EM ATIC S

BANSAL CLASSES Tar g et IIT JEE 200 7 CLASS : XIII (XYZ)

D a i ly Pra c t ic e Pro b l e m s  

DATE : 24-25//07/2006

TIME : 45 Min.

DPP. NO.-11

This is the test paper of Class-XI (J-Batch) held on 23-07-2007.  Take exactly 45 minutes. Q.1

If (sin x + cos x)2 + k sin x cos x = 1 holds  x  R then find the value of k. [Ans. k = – 2]

[Sol.

1 + sin 2x +

k  2

[3]

sin 2x = 1

 k  sin 2x 1   = 0 for this to be an identity  2 1

Q.2

k = – 2 Ans. ]

 3     + sin   x   + sin (32 + x) – 18 cos(19 – x) + cos(56  + x) – 9 sin(x + 17) 2     2  

is expressed in the form of a sin x + b cos x find the value of a + b. – sin x – cos x + sin x + 18 cos x + cos x + 9 sin x 18 cos x + 9 sin x = a sin x + b cos x a = 9, b = 18   a + b = 27 Ans.

Alternatively:



put x = 0 and x =

2

[Ans. 27]

[3]

 to get a and b directly]

3 statements are given below each of which is either True or False. State whether True or False with appropriate reasoning. Marks will be allotted only if appropriate reasoning is given. I (log3169)(log13243) = 10 II cos(cos ) = cos (cos 0°) III

[Hint:



 = 0

3 

   

Q.3

2

If the expression cos  x 

[Sol.

k 

cos x +

2 log13

I

log 3

1 cos x  ·

 =

3

[Ans. True; True; False]

2

5 log 3 log13

[3]

 = 10    True

cos(cos ) = cos(–1) = cos 1 and cos(cos 0°) = cos 1, hence True III If cos x > 0 then E  2 and if cos x < 0 then E  – 2   Hence False ] II

Q.4

Prove the identity cos4t =

[Sol.

cos4t =

=

1 4 3 8

(1 + cos 2t) 2 =

 +

1 2

cos 2t +

1 8

3 8 1 4

 +

1 2

cos 2t +

1 8

(1 + cos22t + 2 cos 2t) =

cos 4t]

[3]

cos 4t.

  1   1  cos 4 t  2 cos 2t  1  2 4    