Q 1 :

Let S={x∈(-π,π):x≠0,±π2}. The sum of all distinct solutions of the equation 3secx+cosec x+2(tanx-cotx)=0 in the set S is equal to     [2016]

  • -7π9

     

  • -2π9

     

  • 0

     

  • 5π9

     

(3)

3secx+cosecx+2(tanx-cotx)=0

⇒32sinx+12cosx=cos2x-sin2x

⇒cos(x-π3)=cos2x⇒x-π3=2nπ±2x

⇒x=2nπ3+π9  or  x=-2nπ-π3

For x∈S, n=0⇒x=π9,-π3

Now, n=1⇒x=7π9  and n=-1⇒x=-5π9

Hence, sum of all values of x=π9-π3+7π9-5π9=0



Q 2 :

For x∈(0,π), the equation sinx+2sin 2x-sin 3x=3 has                 [2014]

  • infinitely many solutions

     

  • three solutions

     

  • one solution

     

  • no solution

     

(4)

sinx+2sin 2x-sin 3x=3

⇒sinx+4sinxcosx-3sinx+4sin3x=3

⇒sinx(-2+4cosx+4sin2x)=3

⇒sinx(-2+4cosx+4-4cos2x)=3

2+4cosx-4cos2x=3sinx   [∵0≤sinx≤1]

⇒2-4(cos2x-2cosx·12+14)+1=3sinx

⇒3-4(cosx-12)2=3sinx    ∴ L.H.S≤3 and R.H.S≥3

Hence, the equation has no solution.



Q 3 :

The number of solutions of the pair of equations 

     2sin2θ-cos2θ=0

     2cos2θ-3sinθ=0

in the interval [0,2π] is                  [2007]

  • zero

     

  • one

     

  • two

     

  • four

     

(3)

2sin2θ-cos2θ=0 ⇒ 1-2cos2θ=0

⇒cos2θ=12 ⇒ 2θ=π3,5π3,7π3,11π3

⇒θ=π6,5π6,7π6,11π6              ...(i)

where θ∈[0,2π]

Also, 2cos2θ-3sinθ=0

⇒2sin2θ+3sinθ-2=0

⇒(2sinθ-1)(sinθ+2)=0 ⇒ sinθ=12   [∵sinθ≠-2]

⇒θ=π6,5π6                  ...(ii)

where θ∈[0,2π]

Combining (i) and (ii), we get θ=π6,5π6

Hence, there are two solutions.



Q 4 :

cos(α-β)=1 and  cos(α+β)=1e, where α,β∈[-π,π]. Pairs of α,β which satisfy both the equations is/are        [2005]

  • 0

     

  • 1

     

  • 2

     

  • 4

     

(4)

Given: cos(α-β)=1 and cos(α+β)=1e,  where α,β∈[-π,π]

Now, cos(α-β)=1⇒α-β=0⇒α=β

and cos(α+β)=1e⇒cos2α=1e

∵  0<1e<1

Now 2α∈[-2π,2π]

⇒There will be two values of 2α in [-2π,0] satisfying cos2α=1e and two values in [0,2π].

⇒There will be four values of α in [-π,π] and correspondingly four values of β. Hence there are four sets of (α,β).



Q 5 :

The number of integral values of k for which the equation 7cosx+5sinx=2k+1 has a solution is            [2002]

  • 4

     

  • 8

     

  • 10

     

  • 12

     

(2)

We know, -a2+b2≤acosθ+bsinθ≤a2+b2

⇒-74≤7cosx+5sinx≤74

⇒-74≤2k+1≤74⇒-8.6≤2k+1≤8.6

⇒-4.8≤k≤3.8

Hence, k can take only 8 integral values.



Q 6 :

The number of distinct solutions of the equation 54cos22x+cos4x+sin4x+cos6x+sin6x=2 in the interval [0,2π] is                 [2015]



(8)

54cos22x+cos4x+sin4x+cos6x+sin6x=2

⇒54cos22x+1-12sin22x+1-34sin22x=2

⇒54(cos22x-sin22x)=0⇒cos4x=0

⇒4x=(2n+1)π2  or  x=(2n+1)π8

For x∈[0,2π], n can take values 0 to 7

Hence, there are 8 solutions.



Q 7 :

The positive integer value of n>3 satisfying the equation 1sin(πn)=1sin(2πn)+1sin(3πn) is                       [2011]



(7)

1sinπn-1sin3πn=1sin2πn

⇒sin3πn-sinπnsinπnsin3πn=1sin2πn⇒2cos2πnsinπnsinπnsin3πn=1sin2πn

⇒2sin2πncos2πn=sin3πn⇒sin4πn-sin3πn=0

⇒2cos7π2nsinπ2n=0⇒cos7π2n=0 or sinπ2n=0

⇒7π2n=(2k+1)π2  or  π2n=2kπ, where k∈ℤ

⇒n=72k+1  or  n=14k

(n=14k not possible for any integral value of k)

As n>3; for k=0, we get n=7.



Q 8 :

Two parallel chords of a circle of radius 2 are at a distance 3+1 apart. If the chords subtend at the center, angles of πk and 2πk, where k>0, then the value of [k] is 

[Note: [k] denotes the largest integer less than or equal to k]                    [2010]



(3)

From the figure,

2cosπk+2cosπ2k=3+1

⇒2×2cos2π2k+2cosπ2k-2=3+1

⇒4cos2π2k+2cosπ2k-(3+3)=0

⇒cosπ2k=-2±4+16(3+3)8=-1±13+434

=-1±(23+1)4=32  or  -(3+12)

As π2k is an acute angle, cosπ2k=32=cosπ6⇒k=3



Q 9 :

The number of values of θ in the interval (-π2,π2) such that θ≠nπ5 for n=0,±1,±2 and tanθ=cot5θ as well as sin2θ=cos4θ is             [2010]



(3)

tanθ=cot5θ, θ≠nπ5

⇒cosθcos5θ-sin5θsinθ=0⇒cos6θ=0

⇒6θ=-5π2, -3π2, -π2, π2, 3π2, 5π2

⇒θ=-5π12, -π4, -π12, π12, π4, 5π12

Again sin2θ=cos4θ=1-2sin22θ

⇒2sin22θ+sin2θ-1=0⇒sin2θ=-1, 12

⇒2θ=-π2, π6, 5π6⇒θ=-π4, π12, 5π12

So, common solutions are θ=-π4, π12, 5π12

∴ Number of solutions =3



Q 10 :

The number of all possible values of θ where 0<θ<π, for which the system of equations

          (y+z)cos3θ=(xyz)sin3θ

           xsin3θ=2cos3θy+2sin3θz

           (xyz)sin3θ=(y+2z)cos3θ+ysin3θ

have a solution (x0,y0,z0) with y0z0≠0, is.                           [2010]



(3)

Given equations are

xyzsin3θ=(y+z)cos3θ    ...(i)

xyzsin3θ=2zcos3θ+2ysin3θ    ...(ii)

xyzsin3θ=(y+2z)cos3θ+ysin3θ    ...(iii)

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

(cos3θ-2sin3θ)y-(cos3θ)z=0    ...(iv)

On subtracting eq. (i) from (iii), we get

sin3θ y+(cos3θ)z=0    ...(v)

Eq. (iv) and (v) form a homogeneous system of linear equations.

But y≠0, z≠0

∴ cos3θ-2sin3θsin3θ=-cos3θcos3θ⇒cos3θ=sin3θ

⇒tan3θ=1⇒3θ=nπ+π4⇒θ=(4n+1)π12, n∈ℤ

For θ∈(0,π)⇒θ=π12, 5π12, 3π4

∴Three such solutions are possible.



Q 11 :

Let a,b,c be three non-zero real numbers such that the equation: 3acosx+2bsinx=c, x∈[-π2,π2], has two distinct real roots α and β with α+β=π3. Then, the value of ba is _______.                  [2018]



(0.5)

Given: 3acosx+2bsinx=c

which has two roots α and β, such that α+β=π3

∴ 3acosα+2bsinα=c    ...(i)

and 3acosβ+2bsinβ=c    ...(ii)

On subtracting equation (ii) from (i),

3a(cosα-cosβ)+2b(sinα-sinβ)=0

⇒-3a·2sinα+β2sinα-β2+2b·2cosα+β2sinα-β2=0

⇒-23asinπ6+4bcosπ6=0     (∵sinα-β2≠0)

⇒-23a×12+4b32=0⇒ba=12=0.5



Q 12 :

Let α and β be non-zero real numbers such that 2(cosβ-cosα)+cosαcosβ=1. Then which of the following is/are true?         [2017]

  • tan(α2)+3tan(β2)=0

     

  • 3tan(α2)+tan(β2)=0

     

  • tan(α2)-3tan(β2)=0

     

  • 3tan(α2)-tan(β2)=0

     

Select one or more options

(1, 3)

If we consider tanα2=x and tanβ2=y, then

2(cosβ-cosα)+cosαcosβ=1

⇒2[1-y21+y2-1-x21+x2]=1-(1-x2)(1-y2)(1+x2)(1+y2)

⇒2[(1+x2)(1-y2)-(1-x2)(1+y2)]=(1+x2)(1+y2)-(1-x2)(1-y2)

⇒4(x2-y2)=2(x2+y2)

⇒x2=3y2⇒x=±3y

⇒tanα2±3tanβ2=0



Q 13 :

The number of points in (-∞,∞), for which x2-xsinx-cosx=0, is                  [2013]

  • 6

     

  • 4

     

  • 2

     

  • 0

     

(3)

Let f(x)=x2-xsinx-cosx

∴ f'(x)=2x-xcosx=x(2-cosx)

∴ f is increasing on (0,∞) and decreasing on (-∞,0)

Also limx→∞f(x)=∞,  limx→-∞f(x)=∞, and f(0)=-1

∴ y=f(x) meets x-axis twice.

i.e., f(x)=0 has two points in (-∞,∞).



Q 14 :

For 0<θ<π2, the solution(s) of ∑m=16cosec(θ+(m-1)π4)cosec(θ+mπ4)=42 is (are)                [2009]

  • π4

     

  • π6

     

  • π12

     

  • 5π12

     

Select one or more options

(3, 4)

∑m=16cosec[θ+(m-1)π4]cosec[θ+mπ4]=42

⇒∑m=16sinπ4sin[θ+(m-1)π4]sin[θ+mπ4]=4

⇒∑m=16sin[(θ+mπ4)-(θ+(m-1)π4)]sin(θ+(m-1)π4)sin(θ+mπ4)=4

⇒∑m=16[sin(θ+mπ4)cos(θ+(m-1)π4)-cos(θ+mπ4)sin(θ+(m-1)π4)]sin(θ+(m-1)π4)sin(θ+mπ4)=4

⇒∑m=16[cot(θ+(m-1)π4)-cot(θ+mπ4)]=4

⇒[cotθ-cot(θ+π4)]+[cot(θ+π4)-cot(θ+2π4)]+⋯+[cot(θ+5π4)-cot(θ+6π4)]=4

⇒cotθ-cot(θ+3π2)=4⇒cotθ+tanθ=4

⇒cos2θ+sin2θ=4sinθcosθ

⇒sin2θ=12⇒2θ=π6 or 5π6⇒θ=π12 or 5π12



Q 15 :

Consider the following lists:                               [2022]

  List-I   List-II
(I) {x∈[-2π3,2π3]:cosx+sinx=1} (P) has two elements
(II) {x∈[-5π18,5π18]:3tan3x=1} (Q) has three elements
(III) {x∈[-6π5,6π5]:2cos(2x)=3} (R) has four elements
(IV) {x∈[-7π4,7π4]:sinx-cosx=1} (S) has five elements
    (T) has six elements

 

The correct option is:

  • (I) → (P); (II) → (S); (III) → (P); (IV) → (S)

     

  • (I) → (P); (II) → (P); (III) → (T); (IV) → (R)

     

  • (I) → (P); (II) → (P); (III) → (T); (IV) → (S)

     

  • (I) → (Q); (II) → (S); (III) → (P); (IV) → (R)

     

(2)

We have (I)

{x∈[-2π3,2π3]:cosx+sinx=1}

cosx+sinx=1

⇒12cosx+12sinx=12

⇒sin(π4+x)=12

⇒π4+x=nπ+(-1)nπ4

⇒x=nπ+(-1)nπ4-π4  ∴x has 2 elements →(P)

We have (II)

{x∈[-5π18,5π18]:3tan3x=1};    3tan 3x=1

⇒tan3x=13

x=(6n+1)π18, n∈ℤ⇒3x=nπ+π6

or  x=nπ3+π18  ∴ x has 2 elements →(P)

We have (III)

{x∈[-6π5,6π5]:2cos2x=3}

2cos(2x)=3

⇒cos2x=32⇒2x=2nπ±π6; n∈Z

or  x=nπ±π12; n∈Z

⇒x∈{±π12,-π±π12,π±π12}

∴ x has 6 elements →(T)

We have (IV)

{x∈[-7π4,7π4]:sinx-cosx=1}

sinx-cosx=1

⇒12sin(x)-12cos(x)=12

⇒sin(x-π4)=12

⇒x-π4=nπ+(-1)nπ4

⇒x=nπ+(-1)nπ4+π4

⇒x∈{π2,-3π2,-π,π}

∴ x has 4 elements →(R)



Q 16 :

Let f(x)=sin(πcosx) and g(x)=cos(2πsinx) be two functions defined for x>0. Define the following sets whose elements are written in the increasing order.

X={x:f(x)=0},Y={x:f'(x)=0}

Z={x:g(x)=0},W={x:g'(x)=0}

Column - I contains the sets X, Y, Z and W. Column - II contains some information regarding these sets.           [2019]

  Column I   Column II
(I) X (p) ⊇{π2,3π2,4π,7π}
(II) Y (q) an arithmetic progression
(III) Z (r) NOT an arithmetic progression
(IV) W (s) ⊇{π6,7π6,13π6}
    (t) ⊇{π3,2π3,π}
    (u) ⊇{π6,3π4}

 

Which of the following is the only CORRECT combination?

  • (IV), (p), (r), (s)

     

  • (III), (p), (q), (u)

     

  • (III), (r), (u)

     

  • (IV), (q), (t)

     

(1)

f(x)=0⇒sin(πcosx)=0⇒πcosx=nπ

⇒cosx=n⇒cosx=-1,0,1

⇒x=π2,π,3π2,2π,5π2,3π,7π2,4π,…

∴ X={π2,π,3π2,2π,5π2,3π,7π2,4π,…}

∴ (I)→P,Q

f'(x)=0⇒cos(πcosx)(-πsinx)=0

⇒cos(πcosx)=0, sinx=0

⇒πcosx=(2n-1)π2, x=nπ

⇒cosx=2n-12, x=π,2π,3π,…

⇒cosx=-12,12

⇒x=π3,2π3,4π3,5π3,7π3,8π3,10π3,11π3,13π3,…

∴ Y={π3,2π3,π,4π3,5π3,2π,…}

∴(II)→Q,T

g(x)=0⇒cos(2πsinx)=0

⇒2πsinx=(2n-1)π2⇒sinx=2n-14

⇒sinx=14,-14,34,-34

∴ Z={-sin-114,-sin-134,sin-114,sin-134}

∴ (III)→R

g'(x)=0⇒-sin(2πsinx)·2πcosx=0

⇒sin(2πsinx)=0, cosx=0

⇒2πsinx=nπ, x=(2n-1)π2

⇒sinx=n2, x=π2,3π2,5π2,7π2,…

⇒sinx=-1,-12,0,12,1

⇒x=π6,π2,5π6,π,3π2,11π6,2π,13π6,…

∴ W={π6,π2,5π6,π,3π2,11π6,2π,13π6,…}

∴ (IV)→P,R,S



Q 17 :

Let f(x)=sin(πcosx) and g(x)=cos(2πsinx) be two functions defined for x>0. Define the following sets whose elements are written in the increasing order.

X={x:f(x)=0},  Y={x:f'(x)=0}

Z={x:g(x)=0},  W={x:g'(x)=0}

Column - I contains the sets X, Y, Z and W. Column - II contains some information regarding these sets.             [2019]

  Column I   Column II
(I) X (p) ⊇{π2,3π2,4π,7π}
(II) Y (q) an arithmetic progression
(III) Z (r) NOT an arithmetic progression
(IV) W (s) ⊇{π6,7π6,13π6}
    (t) ⊇{π3,2π3,π}
    (u) ⊇{π6,3π4}

 

Which of the following is the only CORRECT combination?

  • (I), (q), (u)

     

  • (I), (p), (r)

     

  • (II), (r), (s)

     

  • (II), (q), (t)

     

(4)

f(x)=0⇒sin(πcosx)=0⇒πcosx=nπ

⇒cosx=n⇒cosx=-1,0,1

⇒x=π2,π,3π2,2π,5π2,3π,7π2,4π,…

∴ X={π2,π,3π2,2π,5π2,3π,7π2,4π,…}

∴ (I)→P,Q

f'(x)=0⇒cos(πcosx)(-πsinx)=0

⇒cos(πcosx)=0, sinx=0

⇒πcosx=(2n-1)π2, x=nπ

⇒cosx=2n-12, x=π,2π,3π,…

⇒cosx=-12,12

⇒x=π3,2π3,4π3,5π3,7π3,8π3,10π3,11π3,13π3,…

∴ Y={π3,2π3,π,4π3,5π3,2π,…}

∴ (II)→Q,T

g(x)=0⇒cos(2πsinx)=0

⇒2πsinx=(2n-1)π2⇒sinx=2n-14

⇒sinx=14,-14,34,-34

∴ Z={-sin-114,-sin-134,sin-114,sin-134}

∴ (III)→R

g'(x)=0⇒-sin(2πsinx)·2πcosx=0

⇒sin(2πsinx)=0, cosx=0

⇒2πsinx=nπ, x=(2n-1)π2

⇒sinx=n2, x=π2,3π2,5π2,7π2,…

⇒sinx=-1,-12,0,12,1

⇒x=π6,π2,5π6,π,3π2,11π6,2π,13π6,…

∴ W={π6,π2,5π6,π,3π2,11π6,2π,13π6,…}

∴ (IV)→P,R,S