137880897 Maths Bansal Classes

8/13/2019 137880897 Maths Bansal Classes

1/12

MATHEMATICS

XI (P1-P7) PRACTICE PAPER [MATHS] Page # 2

PART-A

[STRAIGHT OBJECTIVE TYPE]

Q.1 to Q.13has four choices (A), (B), (C), (D) out of which ONLY ONEis correct.

Q.1 If r, s are the roots of Ax2+ Bx + C = 0 (A 0) and r2, s2are the roots of x2+ px + q = 0,then p is equal to

(A) 22

A

AC2B (B) 2

2

A

AC4B (C) 2

2

A

B2AC (D*) 2

2

A

BAC2

[Sol. We have r + s =A

B ; rs =

A

C

r2+ s2= p; r2s2= q ; Now, p = (r + s)2 2rs =A

C2

A

B2

2

p = 2

2

A

BAC2 Ans.]

Q.2 Which of the following is a graph of f (x) =2

1sin(2x)

(A) (B)

(C*) (D)

Q.3 The value of 'k' for which the equation x3+ kx2 + 3 = 0 and x2+ kx + 3 = 0 have a common root, is(A) 4 (B) 1 (C*) 4 (D) 1

[Sol. Let be a common root.then 3+ K2+ 3 = 0 ..............(1)and 2+ K+ 3 = 0 ...............(2)

Now, (1) (2) 3 3= 0 ; = 1 ; So, from (1), we get 1 + k + 3 = 0k = 4 ]

Q.4 If A (,2), then the value of

2

Asin1

2

Asin1

2

Asin1

2

Asin1

+

++is equal to

(A) tan4

A(B) cot

4

A(C) cot

4

A(D*) tan

4

A

[Sol. We have

2

Asin1

2

Asin1

2

Asin1

2

Asin1

+

++=

22

22

4

Asin

4

Acos

4

Asin

4

Acos

4

Asin

4

Acos

4

Asin

4

Acos

+

+

+

PRACTICE TEST MATHS


2/12

MATHEMATICS


=

4

Asin

4

Acos

4

Asin

4

Acos

4

Asin

4

Acos

4

Asin

4

Acos

+

++

=

+

+

+

4

Asin

4

Acos

4

Asin

4

Acos

4

Asin

4

Acos

4

Asin

4

Acos

=

4

Acos2

4

Asin2

= tan4

A ]

Q.5 If AD, BE, CF are medians of a triangle ABC and [(AD)2+ (BE)2+ (CF)2] : [(BC)2+ (CA)2+ (AB)2]

is equal to qp , where p and q are in lowest form then p + q equals

(A*) 7 (B) 10 (C) 6 (D) 15[Hint: Sum of the square of the median is 3/4 times sum of the square of the sides

p = 3; q = 4 p + q = 7 Ans. ]

Q.6 The value of x (0, 90) and satisfying cos x = sin 61 + sin 47 sin 25 sin 11, is(A*) 7 (B) 11 (C) 13 (D) 17

[Sol. RHS = 2 sin 54 cos 7 2 sin 18 cos 7 = 2 cos 7[sin 54 sin 18] = cos 7 x = 7 Ans. ]

Q.7 If in a triangle PQR, sin P, sin Q, sin R are in A.P., then(A) the altitudes are in A.P. (B*) the altitudes are in H.P.

(C) the medians are in G.P. (D) the medians are in A.P.[Sol. sinP, sinQ, sinR AP (given) sides are in AP (using sin law)

Let altitude be h1, h2, h3to sides p, q, r respectively.Now, = p h1 = q h2= r h3

p =1h

2 ; q =

2h

2 ; r =

3h

2

p, q, r AP 1h

2,

2h

2,

3h

2 AP

321 h

1,

h

1,

h

1 AP h1, h2, h3 are in HP]

Q.8 If , are the roots of the quadratic equation x2+ 2(1 cos 3) x 2 sin23= 0 (R),

then the maximum value of 2+ 2is equal to(A) 0 (B) 4 (C) 8 (D*) 16

[Sol. We have x + 2(1 cos 3 )x 2sin 3 = 02 2

; + = 2(1 cos 3) ; = 2sin23

Now, 2+ 2= (+ )2 2= 4 (1 cos 3)2+ 4 sin23= 4(1 2 cos 3+ cos23+ sin23)2+ 2= 4 (2 2 cos 3) ;Clearly maximum value of 2+ 2is 16. ]

Q.9 The number of integral value(s) of 'p' for which the equation99 cos 2 20 sin 2= 20p + 35, will have a solution is(A) 8 (B) 9 (C*) 10 (D) 11

[Sol. We have 99 cos 2 20 sin 2= 20p + 35 ....(1)As 101 99 cos 2 20 sin 2101 Equation (1) will have a solution,If 101 20p + 35 101 136 20p 66 68 p 33 Possible integral value(s) of 'p' are 6, 5, 4, 3, 2, 1, 0, 1, 2, 3Hence number of integral value(s) of 'p' = 10 ]

Q.10 A point 'P' is an arbitrary interior point of an equilateral triangle of side 4. If x, y, z are thedistances of 'P' from sides of the triangle then the value of (x + y + z)2is equal to(A) 3 (B*) 12 (C) 18 (D) 48

[Sol. We have ar. (ABC) = ar. (PBC) + ar. (PAC) + ar. (PAB)

2

)4(4

3=

2

1(4) (x) +

2

1(4) (y) +

2

1(4) (z)


3/12

MATHEMATICS


4 3 = 2 (x + y + z) x + y + z = 32

A

DB C

y

xP

zEF

Hence (x + y + z)2= 12 ]

Q.11 If , , are the roots of the cubic 2009x3+ 2x2+ 1 = 0, then the value of 2+ 2+ 2is equal to(A) 4 (B) 2 (C) 2 (D*) 4

[Sol. We have 2009 x3+ 2x2+ 1 = 0

...... (1)

Put x =t

1in equation (1), we get 3t

2009+ 2t

2+ 1 = 0

t3+ 2t + 2009 = 0

1/

1/

1/ ; So,

1

+ 1

+ 1

= 0

and 1

+ 1

+ 1

= 2

Now, we know that

2111

+

+

=

+

+

222111

+ 2

+

+

111

0 =

+

+

222111

+ (2 2) ; 2+ 2+ 2= 4 ]

[Alternative We have 2009 x3+ 2x2+ 1 = 0

Also, + + =2009

2

+ + = 0 ..... (1)

and =2009

1

Now, 21+ 21

+ 21= 222

222222

++ ....... (2)

But ( + + )2= 22+ 22+ 22+ 2( + + )Using (1), we get 22+ 22+ 22= 2 ( + + ) ..... (3)

Putting (3) in (2), we have 21

+ 2

1

+ 21

= 2)(

)(2

++=

2009

12009

22

= 4 ]


4/12

MATHEMATICS


Q.12 If Sn=1

1

1 2

1 23 3 3+

+

++...... +

1 2 3

1 2 33 3 3 3+ + + +

+ + + +

......

......

n

n, n = 1, 2, 3,...... Then Snis not greater than

(A) 1/2 (B) 1 (C*) 2 (D) 4

[Sol. Sn=.............

321

321

21

21

1

1333333

+++

+++

+

++ ; Tn= 3333 n...........321

n.............54321

+++

++++

=)1n(n

2

2

)1n(n

2 )1n(n

2 +=

+

+ Sn=

+

1n

1

n

12 ]

Q.13 Let f() = + 24 cos4sin + 24 sin4cos , then the value of

4

111f is equal to

(A)2

22 (B)

2

22 +(C)

2

22 (D*)

2

22 +

[Sol. We have

f() = )sin1(4sin 24 + )cos1(4cos 24 + = 22 )sin2( 22 )cos2(

= (2 sin2) (2 cos2) = cos2 sin2= cos 2

f

4

111 = cos 22

2

1; Now, cos2=

2

2cos1 +...........(1)

Put = 222

1in equation (1), we get cos222

2

1=

22

11+

=22

12+=

22

12+

2

2=

4

22 +

f

4

111 = cos 22

2

1=

2

22 + (If 0 < 1 (C) g( 1) > 0

Now, (1) (m + 1)2 4 (m 1) 0 m2 2m + 5 0, Rm .

(2) 2

1m +> 1 m > 3

and (3) 1 + (m + 1) + m 1 > 0 m > 2

1; Hence from (1), (2) & (3), we get m

,2

1]

[REASONING TYPE]

Q.17 & Q.18are Reasoning typequestions, contains Statement-1 (Assertion) and Statement-2 (Reason).Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONEis correct.

Q.17 Statement-1: In ABC, sin 2A + sin 2B + sin 2C is always positive.because

Statement-2: In ABC, sin 2A + sin 2B + sin2C = 8 sinA sinB sinC.(A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.(B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.

(C*) Statement-1 is true, statement-2 is false. (D) Statement-1 is false, statement-2 is true.[Sol. We know that in ABC, ; sin 2A + sin 2B + sin 2C = 4 sinA sinB sinCAs A,B,C(0, ) ; sin 2A + sin 2B + sin 2C is always positive Statement-1 is true, statement-2 is false]Q.18 Consider f(x) = x2 (a + b)x + 2, where a, b R.

Statement-1: If f(x) = 0 does not have two distinct real roots then the minimum value of a + b is 3.because

Statement-2: If ax2+ bx + c = 0 (a, b , c R and a 0) does not have two distinct real roots theneither f(x) 0 x R or f(x) 0 x R.

(A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.(B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.

(C) Statement-1 is true, statement-2 is false. (D*) Statement-1 is false, statement-2 is true.[Sol. We have f(x) = x2 (a + b)x + 2


6/12

MATHEMATICS


As f(x) = 0 does not have 2 distinct real roots and f(0) = 2 f(x) 0 x RIn particular f(1) 0 1 + (a + b) + 2 0 a + b 3Hence the least value of a + b is 3Statement-1 is false and Statement -2 is true. ]

[MULTIPLE OBJECTIVE TYPE]Q.19 to Q.25has four choices (A), (B), (C), (D) out of which ONE OR MORE THAN ONEis/are correct.

Q.19 If f() =

+

+= 4

neccos

4

)1n(eccos

6

1n

, where 0 < 0

)1x2()1x3(x

1x3

++

> 0 ........ (1)

+ + +

1 2

1 3

0 1 3

Now,)2x()1x5(

)1x2()5x(x2

++

< 0 ............ (2)

+ + +

2 1 5

0 1 2

5

+

Clearly, from (1) & (2), option(s) (A) & (C) are correct. ]


9/12

MATHEMATICS


PART-B

[MATCH THE COLUMN]

Q.1 is "Match the Column" type. Column-Icontains Fourentries and column-IIcontains Fourentries. Entry ofcolumn-I are to beuniquely matched with only one entry of column-II.

Q.1 Column-I Column-II

The solution of following inequalities :

(A) sin x cos3x > cos x sin3x, 0 x 2, is (P)

,4

3

4,

44

3,

(B) 4 sin2x 8 sin x + 3 0, 0 x 2, is (Q)

2,2

3{0}

(C) | tan x | 1 and x [ , ] is (R)

4

,0

(D) cos x sin x 1 and 0 x 2 is (S) 65,

6

[Ans. (A) R; (B) S; (C) P; (D) Q][Sol. (A) sin x cosx (cos2x sin2x) > 0

2

1sin 2x cos 2x > 0 sin 4x > 0; 0 < 4x < (R)

(B) Here, (2 sin x 1)(2 sin x 3) 0 ; but 2 sin x 3 is always negative.

2 sin x 1 0 sin x 1/2

from the figure, /6 x 5/6 (S)(C) 1 tan x 1. The value scheme for this is as shown below:

from the figure, 4

x 4

or x 4

3or

4

3x ]

x

,4

3

4,

44

3, (P)

(D) cos

+4

x >2

1. The value for this is as shown below:


10/12

MATHEMATICS


from the figure44

x4

+

and, in general 2n 4

x +

4

2n+

4

2n2

x 2n ; for n = 0

2

x 0; for n = 1,

2

3x 2 (Q) ]

Q.2 is "Match the Column" type.Column-Icontains Fourentries and column-IIcontainsFiveentries. Entry ofcolumn-I are to be matched with one or more than one entriesof column-II or vice versa.

Q.2 Column-I Column-II(A) If exactlyone root of the quadratic equation (P) 1

(a2 4a + 3)x2+ (a2 5a + 6)x + (a + 5) = 0lies at infinity then the possible integral value(s) of 'a' is equal to

(B) The smallest natural number 'n' so that (2 n) x2 8x 4 n < 0, x R (Q) 2is equal to

(C) If in ABC, A =2

then the maximum value of 4 sin B sin C is equal to (R) 3

(D) Number of values of 'k' so that the equation (S) 4(k 3) (k2 4) (k + 1) x2 (k3 5k2+ 6k) x + (k2 9) = 0has more than two unequal roots is equal to (T) 5

[Ans. (A) P ; (B) T ; (C) Q ; (D) P ][Sol.(A) If one root of the quadratic equation Ax2+ Bx + C = 0, lies at infinity then A = 0 and B 0.

Coefficient of x2= 0 a2 4a + 3 = 0 (a 1) (a 3) = 0 a = 1 ; a = 3But for a = 3, coefficient of x = 0; a = 3 (rejected think ?) a = 1

(B) We have (n 2)x2+ 8x + (n + 4) > 0. Rx n 2 > 0 and D < 0 64 4(n 2) (n + 4) < 0

n

2

+ 2n 24 > 0 (n + 6) (n 4) > 0 n > 4 as n N nsmallest= 5

(C) A =2

, so B + C =

2

Now, 4sin B sin C = 2(2sin B sin C) = 2[cos (B C) cos (B + C)] = 2 cos (B C) 2

Maximum value = 2, when B = C =4

(each).

(D) We must have (k 3) (k2 4) (k + 1) = 0, k3 5k2+ 6k = 0 and k2 9 = 0Hence clearly k = 3 only. ]

PART-C

[SUBJECTIVE]

Q.1 & Q.6are "Subjective" type questions. (The answer to each of the questions are upto 4 digit)

Q.1 If in a ABC , a = 6, b = 3 and cos(AB) = 4/5 then find its area. [Ans. 0009 sq. unit]

[Sol: tan

2

BA = 2/Ccot

ba

ba

+

(using Napier's Analogy)


11/12

MATHEMATICS


)BAcos(1

)BAcos(1

+

=

9

3cot C/2 (using tan=

+

2cos1

2cos1)

5/41

5/41

+

=3

1cot C/2 Now, = ab sinC

2Ccot

31

31 =

2Ccot = 1 3 6 sin 90

C = 90 = 9 ]

Q.2 Let A denotes the value of expression x4+ 4x3+ 2x2 4x + 7, when x = cot8

11

and B denotes the value of the expression

4tan

8cos12 +

+

4cot

8cos12 , when = 9

Find the value of (AB).[Ans 12]

[Sol. (A) We have

x = cot8

11= cot

+8

3= cot

8

3= 12

(x + 1)2= 2 x2+ 2x 1 = 0

Now, considerx4+ 4x3+ 2x2 4x + 7

= x243421)0(

2 )1x2x(

=

+ + 2x3+ 3x2 4x + 7

= 2x3+ 3x2 4x + 7 = 2x 43421)0(

2 )1x2x(=

+ x2 2x + 7 = x2 2x + 7 = 43421)0(

2 )1x2x(=

+ + 6

A = 6

(B) We have,

4tan

8cos12 +

+

4cot

8cos12 =

4tan

4sin22

2

+

4cot

4cos22

2

= 2 (cos24+ sin24) = 2

AB = 12 Ans.]

Q.3 Find the sum of all integers satisfying the inequalities

log5(x 3) + 2

1

log53 < 2

1

log5(2x2

6x + 7) and log3x + xlog 3 + xlog3

1 < 6.[Ans.0039 ]

[Sol. We have log5(x 3) < 2

1log5

+3

7x6x2 2

(x 3)2 3, so x (3, 10)

Also log3x + 2log3x log3x < 6 2log3x < 6 log3x < 3


12/12

MATHEMATICS


0 < x < 27 so x (0, 27) ; Possible integers are 4, 5, 6, 7, 8, 9. ;Hence sum of integers = 39.]

Q.4 Let A denotes the value of expression 4

15

2cos + cos

15

4 cos

15

7 cos

15

and B denotes the value of 8 cot ( + + ), wheretan , tan , tan are the real roots of the cubicx3 8(a b) x2+ (2a 3b) x 4(b + 1) = 0.

Find absolute value of (AB). [Ans. 0004 ][Sol.A) We have 4[(cos 24 + cos 48) (cos 84 + cos 12)] = 4[2 cos 36 cos 12 2 cos 48 cos 36]

= 8 cos 36 [cos 12 cos 48] = 8 cos 36 [2 sin 30 sin 18]

= 16 4

15+

2

1

4

15=

2

15 =

2

4= 2

(B) We have x 8(a b) x + (2a 3b) x 4(b + 1) = 03 2

tan tan tan

tan = 8(a b) ; tan tan = (2a 3b) andtan = 4(b + 1)

Now, tan ( + + ) =

tantan1tantan

=)b3a2(1

)1b(4)ba(8

+ =)b3a21(

)1bb2a2(4+

=)b3a21(

)1b3a2(4

+

= 4 cot ( + + ) = 4

1; 8 cot ( + + ) = 2Hence | AB | = 4 ]

Q.5 If k 1and k2are the two values of 'k' where k1< k2for which the expressionf(x, y) = x2+ 2xy + 4y2+ 2kx 6y + 3 can be resolved as a product of two linear factorsthen find the value of 5k2 4k1. [Ans. 0006 ]

[Sol. We have A = 1 ; B = 4 ; C = 3 ; F = 3 ; G = k ; H = 1

Now, ABC + 2FGH AF2 BG2 CH2= 0 k = 0, 2

3

k1= 2

3, k2= 0 5k2 4k1= 6 Ans. ]

Q.6 The set of values of c for which the equation 0c8x32x8cx4x 22 = has exactly twodistinct real solutions, is (a, b) then find the value of (b a). [Ans. 8]

[Sol. We havex2 4x c 0 D 0 16 + 4c0 4c 16 c 4

Let f(x) = x2 4x c ; x2 4x c c8x32x8 2 = 0

f(x) )x(f8 = 0 ; )x(f ( )8)x(f = 0 f(x) = 0 or f(x) = 8

Case-I f(x) = 0 has two distinct real solutions and f(x) = 8 has no real solution. For f(x) = 0, D > 0 and for f(x) = 8, D < 0 16 + 4c > 0 and 16 + 4(c + 8) < 0 c > 4 and c < 12 ; Q No common solution for 'c' exist. ; c

Case-II f(x) = 0 has no real roots but f(x) = 8 has two distinct real roots For f(x) = 0, D < 0 and for f(x) = 8, D > 0 16 + 4c < 0 and 16 + 4(c + 8) > 0 c < 4 and c > 12 c ( 12, 4)

Hence b a = 4 ( 12) = 4 + 12 = 8 ]

137880897 Maths Bansal Classes

Documents

Transcript of 137880897 Maths Bansal Classes