Etoos 1st Brain Busters Ques Ans

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Q-1. If we define = , find the natural number

that satisfies , then find .

Q-2. With respect to , where the first term is 25 and common difference is -2, Solve,

Q-3. Let's say that the two arithmetic progression for , , the sum of the first terms is and respectively. When , solve

BRAIN BUSTERS Mathematics for JEE Main & Advanced

20 Questions | Time: 60 mins


Q-4. In regards to the geometric progression where the term is , term is , and common ratio is real number, answer the following questions.(1) Express the general term to equation of (2) At which term, does this progression reaches over for the first time ? Note that, log .

Q-5. When progression is , the general term is

. Find . (Note that, are integers)

Q-6. If constants satisfies the equation limsin

, what is ?

Q-7. For constants satisfying lim

sin , find the value of .

Q-8. If the sum of an infinite SP is 20 and sum of their square is 100 then common ratio will be=(A)1/2 (B)1/4 (C)3/5 (D)1


Q-9. If in a , the altitudes from the vertices A, B, C on opposite sides are in H.P., then sinA, sinB, sinC are in-(A)G.P. (B)A.P. (C)Arithmetic-Geometric progression (D)H.P.

Q-10. The sum of the series (A) 300 (B) 125 (C) 425 (D)0

Q-11. The sum to infinity of the series

(A)2 (B)3 (C)4 (D)6

Q-12. lim

tan

tan sin

is

(A) (C) (B) (D)

Q-13. The value of limsin

cossin is

(A) (B)

(C) (D)

Q-14. Let and be the distinct roots of , then

lim

cos is equal to

(A) (B) (C)

(D)

Q-15. lim

cos

(A) does not exist (B) equals (C) equals (D) equals


Q-16. The function given by

can be made continuous at by defining as(A) (B) (C) (D)

Q-17. There are two functions defined on the positive

real numbers. Determine the range of real that makes the equation have at least one root between and .

Q-18. The equation tan has (A) no solution (B) at least one real solution in

(C) two real solutions in (D) Non of these

Q-19. If , and , the value of is (a) (b) (c) (d)

Q-20. If and , then the determinant

sin sincos log tancossin cos log cotlog cot log tan tan equals (a) 0 (b) log tan log cot (c) tan (d) none of these

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www.etoosindia.com for JEE Main & Advanced - 1 -

Q-1. If we define S = 1+212 +3( 12 )2++30( 12 )29, find the natural number a, b that satisfies S=a-( 12 )b, then find a+b.Solution :S=1+212 +3( 12 )2++30( 12 )29 multiplying 12 on both sides12 S=

12 +2( 12 ) 2+3( 12 )3++30( 12 )30

From -

12 S=1+

12 +( 12 ) 2+( 12 ) 3++( 12 ) 29-30( 12 )

30

12 S=

1-( 12 ) 301- 12

-30( 12 ) 30=2-2( 12 ) 30-30( 12 ) 30=2-32( 12 ) 30

S=4-64 ( 12 ) 30=4-( 12 ) 24 , a=4, b=24Answer : a+b=28

Q-2. With respect to {a n}, where the first term is 25 and common difference is -2, Solve, a 1+a 2+a 3++a 23Solution :Since the first term is 25 and the common difference is -2, when

n13, a n>0. when n14,a n


Q-3. Let's say that the two arithmetic progression for {a n}, {b n}, the sum of the first n terms is S n and T n respectively. When a 1+b 1=45, S 10+T 10= 500, solve a 10+b 10 Solution :Since, the two progressions {a n}, {b n} are arithmetic

S 10+T 10=10(a+a 10)

2 +10(b+b 10)

2 =5{(a 1+b 1)+(a 10+b 10)}=500

At this time, a 1+b 1=45, a 10+b 10=55Answer : 55

Q-4. In regards to the geometric progression {a n} where the 6 th term is 2 , 11 th term is 64 , and common ratio is real number, answer the following questions.(1) Express the general term a n to equation of n(2) At which term, does this progression reaches over 10000 for the first time ? Note that, log2=0.3010 .Solution :(1) If a is the first term and r is common ratio, from the given condition a 6=ar 5=2 a 11=ar 10=64

When r 5=32 and since r is real number r=2

substituting this value in to gives a= 116 , a n=arn-1= 116 2

n-1=2 n-5

(2) When at the n term the value gets bigger than 10000 for the first time, 2 n-5>10000 log2 n-5> log10000 (n-5) log2>4 (n-5)0.3010>4 n-5>13.2, n18.2 Therefore, from the 19 th term, it goes over 10000 for the first time

Answer : (1) a n=ar n-1= 116 2n-1=2 n-5 (2) 19 th

Q-5. When progression {an} is 0.23, 0.2323, 0.232323, 0.23232323, , the general term a n is an= 23a {b-(10

c) n} . Find a+b+c. (Note that, a, b, c are integers)

Solution :

an=23 1100 +23( 1100 )2+23( 1100 )3+ +23( 1100 )n =23100 (1-( 1100 )n)

1- 1100= 2399 (1-10

-2n)

a=99, b=1, c=-2 a+b+c=98Answer : 98


Q-6. If constants a, b satisfies the equation limx0

ae2x+ae-2x+bxsinx =1, what is a+b?

Solution :If x0, then (denominator)0 implies (numerator)0. 2a+b=0

limx0

ae2x+ae-2x+bxsinx =1

limsin

= limx0

a(e2x+e-2x-2)x2

sinxx

lim

sin lim

sin

= a(1+1)2

1 =4a=1

From , , a= 14 , b=-12

Answer : a+b=-14

Q-7. For constants p, q satisfying lim

sin , find the value of p+q.Solution :Since lim

x0sin2x=0, we get lim

x0px+q-1=0. q=1

From limx0

sin2xpx+q-1 = limx0

sin2xpx+1-1 , if we rationalize the denominator, we get

limx0

( px+1+1)sin2xpx = limx0{ sin2x2x 2x( px+1+1)px } =1 4p =2

p=2 p+q=2+1=3Answer : 3

Q-8. If the sum of an infinite SP is 20 and sum of their square is 100 then common ratio will be=(A)1/2 (B)1/4 (C)3/5 (D)1Solution :Let G.P. be a,ar,ar 2, S = a1-r =20(i)Now, a 2, a 2r 2, a 2r 4,

S = a2

1-r 2 =100 a

(1-r) a

(1+r) =100 a

1+r =5(ii)

Dividing (i) (ii)1+r1-r =4 1+r=4-4rr=

35

Answer : (C)


Q-9. If in a ABC, the altitudes from the vertices A, B, C on opposite sides are in H.P., then sinA, sinB, sinC are in-(A)G.P. (B)A.P. (C)Arithmetic-Geometric progression (D)H.P.Solution :Let p 1,p 2,p 3 be the altitudes

= 12 ap 1=12 bp 2=

12 cp 3

p 1= 2a ;p 2=2b ;p 3=

2c

Now, p 1,p 2,p 3are in HP

1p 1 ,1p 2 ,

1p 3 are in HP

a2 ,

b2 ,

c2 are in AP

a,b,c are in AP SinA,SinB,SinC are in AP Answer : (B)

Q-10. The sum of the series 1 3-2 3+3 3-+9 3=(A) 300 (B) 125 (C) 425 (D)0Solution :

(13+3 3++93)-(2 3+43+6 3+83)(13+2 3+33++9 3)-2(2 3+43+6 3+83)

[ 9(9+1)2 ] 2-2.23(1+23+3 3+43)(45) 2-16[ 4(4+1)2 ] 2

(45) 2-(40) 285(5)425

Answer : (C)

Q-11. The sum to infinity of the series 1+ 23 +632 +

1033 +

1434 +

(A)2 (B)3 (C)4 (D)6Solution :

Let,

S=1+ 23 +63 2 +

103 3 +

143 4 +

s-1= 23 +63 2 +

103 3 +

143 4 +(i)

S-13 =23 2 +

63 3 +

103 4 +

143 5 +(ii)

On subtracting Eq. (ii) from Eq. (i), we get

23 (S-1)=

23 +

43 2 +

43 3 +

43 4 +

S-1=1+ 23 +232 +

233 +

S=2+23

1- 13=2+1=3

Answer : (B)


Q-12. limx 2

[1- tan( x2 )](1-sinx)

[1+tan( x2 )](-2x)3 is

(A) (C) 18 (B) 0 (D) 132

Solution :

limx 2

[1-tan( x2 )](1-sinx)

[1+ tan( x2 )](-2x)3 Let x= 2 -h as x

2 , h0

limh0

1-tan( 4 -h2 )

1+tan( 4 -h2 )

(1-cosh)(2h) 3 = limh0 tan

h2

2sin 2 h28h 3

[tan( 4 -x)=1-tanx1+tanx ] = limh0

14

tan h2h2 2

(sin h2

h2

) 2 14

= 132 limh0(tan h2

h2

) limh0

(sin h2

h2

) 2 = 132

Answer : 132

Q-13. The value of limx0

(1-cos2x) sin 5xx 2sin3x is

(A) 103 (B) 310 (C)

65 (D)

56

Solution :

limx0

1-(1-2sin 2x)x 2

sin5xsin3x =

sin 2xx 24 2

sin5x

5xsin3x

3x 5x3x =2

53 =

103

Answer : 103

Q-14. Let and be the distinct roots of , then limx

1-cos(ax 2+bx+c)(x-) 2 is equal to

(A) a2

2 (-)2 (B) 0 (C) -a

2

2 (-)2 (D) 12 (-)

2

Solution :

limx

1-cos(ax 2+bx+c)(x-) 2 = limx

2sin 2( ax2+bx+c

2 )(x-) 2 = limx

2sin 2( a2 (x-)(x-))

( a2 )2(x-) 2(x-) 2

( a2 )2(x-) 2

= limx

a 22 (x-)

2 ( limx0

sinxx =1) =

a 22 (-)

2

Answer : a2

2 (-)2


Q-15. limx2

1-cos2(x-2)x-2

(A) does not exist (B) equals 2 (C) equals - 2 (D) equals 12Solution :limx2

1-cos2(x-2)x-2 lim

sin lim

sin

RHL at x=2 limh0

2|sin(2+h-2)|(2+h)-2 = limh0

2|sinh|h limh0

2 sin hh = 2

LHL at x=2

limh0

2|sin(2-h-2)|(2-h)-2 = limh0

2|sin(-h)|-h limh0

2 sin h-h =- 2

Limit does not exist.Answer : (A)

Q-16. The function given by f(x)= 1x -2

e 2x-1 can be made continuous at x=0 by defining f(0) as(A) 2 (B) -1 (C) 0 (D) 1 Solution :Now, lim

x0( 1x -

2e 2x-1 ) = limx0

2e 2x-1-2xx(e 2x-1) = limx0

2e 2x-2(e 2x-1)+2xe 2x

(using LHospital rule)f(x) is continuous at x=0, then lim

x0f(x)=f(0) 1=f(0)

Answer : (D)

Q-17. There are two functions f(x)= 2x+43 , g(x)=xa(a>0) defined on the positive real numbers.

Determine the range of real a that makes the equation g(f(x))-2x=0 have at least one root between 1 and 4. 1


Q-18. The equation 2tanx+5x-2=0 has (A) no solution (B) at least one real solution in [0, 4 ]

(C) two real solutions in [0, 4 ] (D) Non of theseSolution :2tanx+5x-2=0, tanx=- 52 x+1

If you draw graphs of y=tanx and y=- 52 x+1, you can find out that there is one solution in

[0, 4 ].

Answer : (B)

Q-19. If bc+ca+ab=18, and

1 a 2 a 31 b 2 b 31 c 2 c 3

=

1 1 1a b ca 2 b 2 c 2

, the value of is

(a) -1 (b) 0 (c) 9 (d) 18Solution :

we already know that

1 1 1a b ca 2 b 2 c 2

=(a-b)(b- c)(c-a)

Let 1=

1 a 2 a 31 b 2 b 31 c 2 c 3

Using R 3R 3-R 2 and R 2R 2-R 1, we get 1=

1 a 2 a 30 b 2-a 2 b 3-a 30 c 2-b 2 c 3-b 3

=(b-a)(c-b)| |b+a b 2+a 2+bac+b c 2+b 2+cb Applying C 2C 2-bC 1, we get 1=(b-a)(c-b)| |b+a a 2c+b c 2=(b-a)(c-b)| |b+a a 2c-a c 2-a 2 [using R 2R 2-R 1]= (b-a)(c-b)(c-a)| |b+a a 21 c+a =(a-b)(b-c)(c-a)(bc+ca+ab)Thus, =18Answer : =18

Q-20. If xR and nI, then the determinant

sin sincos log tan cossin cos log cotlog cot log tan tan equals (a) 0 (b) log tan x - log cot x (c) tan (/4-x) (d) none of theseSolution :

we can write as =

0 sinx-cosx log tanx-(sin x-cosx) 0 - log tanx

- log tanx log tanx 0

=(-1) 3

0 -(sinx-cosx) - log tanxsinx-cosx 0 log tanxlog tanx - log tanx 0

[taking -1 common from R1, R2 and R3 ]

=- [using reflection property ] 2=0 =0.Answer : (a)

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