Q 11 :

The value of  sin12°sin48°sin54° is equal to

  • 116

     

  • 132

     

  • 18

     

  • 14

     

(3)

Now,   sin12°sin48°sin54°=12(cos36°-cos60°)cos36°

=12[5+14-12][5+14]

=12[5-14][5+14]

=5-132=432=18



Q 12 :

The value of tan A + 2 tan 2A + 4 tan 4A + 8 cot 8A is

  • cot A

     

  • tan A

     

  • cos A

     

  • sin A

     

(1)

tanA+2tan2A+4tan4A+8(1-tan24A2tan4A)

=tanA+2tan2A+(4tan24A+4-4tan24Atan4A)

=tanA+2tan2A+4cot4A

=tanA+2tan2A+4(1-tan22A2tan2A)

=tanA+[2tan22A+2-2tan22Atan2A]

=tanA+2cot2A

=tanA+2(1-tan2A2tanA)

=tan2A+1-tan2AtanA=cotA



Q 13 :

The maximum value of cos2(π3-x)-cos2(π3+x) is

  • -32

     

  • 12

     

  • 32

     

  • 32

     

(3)

cos2(π3-x)-cos2(π3+x)

=[cos(π3-x)+cos(π3+x)][cos(π3-x)-cos(π3+x)]

=(2cosπ3cosx)(2sinπ3sinx)

=sin2π3sin2x=32sin2x

Hence, maximum value of the given expression is 32.



Q 14 :

If sum of all the solution of the equation  8cosx(cos(π6+x)·cos(π6-x)-12)=1 in [0,π] is kπ, then k is equal to

  • 23

     

  • 139

     

  • 89

     

  • 209

     

(2)

Key idea Apply the identities

cos(x+y)cos(x-y)=cos2x-sin2y  and  cos3x=4cos3x-3cosx

We have,

8cosx(cos(π6+x)cos(π6-x)-12)=1

⇒8cosx(cos2π6-sin2x-12)=1

⇒          8cosx(34-sin2x-12)=1

⇒8cosx(34-12-1+cos2x)=1

⇒         8cosx(-3+4cos2x4)=1

⇒               2(4cos3x-3cosx)=1

⇒           2cos3x=1⇒cos3x=12

⇒           3x=π3, 5π3, 7π3    [0≤3x≤3π]

⇒            x=π9, 5π9, 7π9

Now,    Sum=π9+5π9+7π9=13π9

⇒         kπ=13π9

∴             k=139



Q 15 :

If  sin2θ=4xy(x+y)2 is true, if and only if

  • x-y≠0

     

  • x=-y

     

  • x+y≠0

     

  • x≠0, y≠0

     

(3)

∵        sin2θ≤1

∴  4xy(x+y)2≤1

⇒ x2+y2+2xy-4xy≥0

⇒                       (x-y)2≥0

which is true for all real values of x and y, provided x+y≠0, otherwise 4xy(x+y)2 will be meaningless.



Q 16 :

If  A=sin2x+cos4x, then for all real x

  • 1316≤A≤1

     

  • 1≤A≤2

     

  • 34≤A≤1316

     

  • 34≤A≤1

     

(4)

A=sin2x+cos4x

⇒     A=1-cos2x+cos4x

=     cos4x-cos2x+14+34

=    (cos2x-12)2+34         ⋯(i)

where,  0≤(cos2x-12)2≤14     ⋯(ii)

∴     34≤A≤1



Q 17 :

If fk(x)=1k(sinkx+coskx), where x∈R, k≥1, then f4(x)-f6(x) is equal to

  • 16

     

  • 13

     

  • 14

     

  • 112

     

(4)

Given,  fk(x)=1k(sinkx+coskx), where x∈R and k≥1

Now,  f4(x)-f6(x)

      =14(sin4x+cos4x)-16(sin6x+cos6x)

       =14(1-2sin2x·cos2x)-16(1-3sin2x·cos2x)

       =14-16=112



Q 18 :

For 0<ϕ<π2, if x=∑n=0∞cos2nϕ,  y=∑n=0∞sin2nϕ  and  z=∑n=0∞cos2nϕ sin2nϕ, then

  • xyz=xz+y

     

  • xyz=xy-z

     

  • xyz=x+y+z

     

  • xyz=yz+x

     

(3)

x=11-cos2ϕ=1sin2ϕ,   y=1cos2ϕ

and                   z=11-sin2ϕ·cos2ϕ

Clearly,              1x+1y=1

⇒        xy=x+y  and  1z=1-1xy

⇒        xy=xyz-z

∴      xyz=xy+z=x+y+z



Q 19 :

The equation esinx-e-sinx-4=0 has

  • infinite number of real roots

     

  • no real roots

     

  • exactly one real root

     

  • exactly four real roots

     

(2)

Given equation is   

esinx-e-sinx=4⇒esinx-1esinx=4

Now, let y=esinx

Then, we get

y-1y=4⇒y2-4y-1=0

∴  y=4±16+42=2±5

⇒esinx=2±5

Since sine is a bounded function, i.e., -1≤sinx≤1.

Therefore,  we get e-1≤esinx≤e

⇒esinx∈[1e, e]

Also, it is obvious that 2+5>e

and 2-5<1e⇒2±5∉[1e, e]

So, esinx=2+5 is not possible for any x∈R,

and esinx=2-5 is also not possible for any x∈R

Hence, we can say that the given equation has no solution.



Q 20 :

The angles of a triangle are in A.P. The number of degrees in the least is to be the number of radians in the greatest as 60:π. Then the greatest angle is

  • 120°

     

  • 90°

     

  • 135°

     

  • 105°

     

(2)

Let the angles of the triangle be  (a-d)°,  a°,  (a+d)°

Their sum is a-d+a+a+d=180

⇒3a=180⇒a=60

Again, by the given condition,

a-d(a+d)π180=60π⇒3(a-d)a+d=1

⇒3a-3d=a+d

⇒2a=4d            ⇒   a=2d

∴  60=2d           ∴    d=30

Hence, the greatest angle =60+30=90°



Q 21 :

The value of  6(sin6θ+cos6θ)-9(sin4θ+cos4θ)+4  is



(1)

6(sin6θ+cos6θ)-9(sin4θ+cos4θ)+4

=6[(sin2θ+cos2θ)3-3sin2θcos2θ(sin2θ+cos2θ)]-9[(sin2θ+cos2θ)2-2sin2θcos2θ]+4

=6[1-3sin2θcos2θ]-9(1-2sin2θcos2θ)+4

=6-9+4=1



Q 22 :

The value of  tan15°+cot15°  is



(4)

tan15°+cot15°=tan15°+1tan15°

=1+tan215°tan15°

=2[12tan15°1+tan215°]

=2sin30°=212=4



Q 23 :

If  sinθ1+sinθ2+sinθ3=3, then cosθ1+cosθ2+cosθ3=



(0)

Since   sinθ∈[-1,1]

and  sinθ1+sinθ2+sinθ3=3

⇒ each sinθ=1

∴  each cosθ=0

Hence,  cosθ1+cosθ2+cosθ3=0



Q 24 :

The value of  cot54°tan36°+tan20°cot70°=



(2)

cot54°tan36°+tan20°cot70°

      =cot(90°-36°)tan36°+tan(90°-70°)cot70°

      =tan36°tan36°+cot70°cot70°=1+1=2



Q 25 :

If  xcosθ=ycos[θ-(2π3)]=zcos[θ+(2π3)],  then x+y+z is equal to



(0)

We have  xcosθ=ycos[θ-(2π3)]=zcos[θ+(2π3)]

Therefore, each ratio is equal to

x+y+zcosθ+cos[θ-(2π3)]+cos[θ+(2π3)]

         =x+y+zcosθ+2cosθcos(2π3)

          =x+y+z0

⇒x+y+z=0



Q 26 :

If 0°<θ<180°, then   2+2+2+⋯+2(1+cosθ) there being n number of 2's, is equal to

  • 2cosθ2n

     

  • 2cosθ2 n-1

     

  • 2cosθ2 n+1

     

  • none of these

     

(1)

2+2+2+⋯+2(1+cosθ), there being nnumbers of 2's

=2+2+2+⋯+2(2cos2θ2)

=2+2+2+⋯+2(1+cosθ2),

there being (n-1) number of 2's,

=⋯=2(1+cosθ2 n-1)=2cosθ2n



Q 27 :

If  cosecθ-cotθ=p, then the value of cosecθ is

  • 12(p+1p)

     

  • 12(1-1p)

     

  • (1+1p)

     

  • (1-1p)

     

(1)

Since cosecθ-cotθ=p

⇒       cosecθ+cotθ=1p

Adding (i) and (ii), we have

2cosecθ=p+1p

∴  cosecθ=12(p+1p)



Q 28 :

tan3x-tan2x-tanx  is equal to

  • tanx  tan2x  tan3x

     

  • -tanx  tan2x  tan3x

     

  • tanx  tan2x-tanx  tan3x-tan2x  tan3x

     

  • none of these

     

(1)

tan3x=tan(2x+x)=tan2x+tanx1-tan2x tanx

∴  tan3x-tan3x  tan2x  tanx=tan2x+tanx

∴  tan3x-tan2x-tanx=tan3x  tan2x  tanx



Q 29 :

The exact value of cosec10°+cosec50°−cosec70° is:

  • 4

     

  • 5

     

  • 6

     

  • 8

     

(3)

cosecθ+cosec(60°-θ)-cosec(60°+θ)    where θ=10°



Q 30 :

If 15sin4α+10cos4α=6, for α∈R, then the value of 27sec6α+8cosec6α is equal to:

  • 350

     

  • 250

     

  • 400

     

  • 500

     

(2)

15sin4α+10(1-sin2α)2=6.25sin4α-20sin2α+4=0

⇒25sin4α-10sin2α-10sin2α+4=0

⇒(5sin2α-2)=0

⇒sin2α=25



Q 31 :

Let f(θ)=11+(tanθ)π and s=∑θ=1°89°f(θ), then the value of 2s-25=______



(8)

f(θ)=(cosθ)x(cosθ)x+(sinθ)x,    f(θ)+f(π2-θ)=1



Q 32 :

The maximum value of the function f(x)=sinx1-cos2x+cosx1-sin2x+tanxsec2x-1+cotxcosec2x-1

wherever it is defined is k. Then |k| is

  • 4

     

  • – 2

     

  • 0

     

  • 2

     

(1)

f(x)=sinx1-cos2x+cosx1-sin2x+tanxsec2x-1+cotxcosec2x-1

=sinx|sinx|+cosx|cosx|+tanx|tanx|+cotx|cotx|={4,x∈1st quadrant-2,x∈2nd quadrant0,x∈3rd quadrant-2,x∈4th quadrant            

f(x)max=4