Q 31 :    

Let the range of the function f(x)=6+16 cos x·cos(π3x)·cos(π3+x)·sin 3x·cos 6x, xR be [α, β]. Then the distance of the point (α, β) from the line 3x + 4y + 12 = 0 is:          [2025]

  • 8

     

  • 11

     

  • 9

     

  • 10

     

(2)

f(x)=6+16(14cos 3x) sin 3x·cos 6x

                                           [ cos θ cos(π3θ)cos(π3+θ)=14cos 3θ]

                =6+4 cos 3x sin 3x cos 6x=6+sin 12x

  Range of sin x = [–1, 1]

  Range of f(x) is [5, 7]

 (α,β)(5,7)

   Distance of point (5, 7) from the line 3x + 4y + 12 = 0

                =|15+28+125|=11



Q 32 :    

Let the lines 3x – 4y – α= 0, 8x –11y – 33 = 0 and 2x – 3yλ = 0 be concurrent. If the image of the point (1, 2) in the line 2x – 3yλ = 0 is (5713,4013), then |αλ| is equal to          [2025]

  • 101

     

  • 113

     

  • 84

     

  • 91

     

(4)

As the three lines are concurrent,

      |34α8113323λ|=0

 3(11λ99)+4(8λ+66)α(24+22)=0

 33λ297+32λ+264+2α=0

 λ33+2α=0                                                    ...(i)

As image, (1, 2) w.r.t. 2x – 3yλ = 0 is (5713,4013)

       571312=401323=2(26+λ4+9)

 2213=2(4+λ)13  λ=7

Substitute λ = –7 in (i), we get

733+2α=0  a=13

  |αλ|=|13×(7)|=91



Q 33 :    

Two equal sides of an isosceles triangle are along – x + 2y = 4 and x + y = 4. If m is the slope of its third side, then the sum, of all possible distinct values of m, is:          [2025]

  • 210

     

  • 12

     

  • –6

     

  • 6

     

(4)

Slope of given lines are m1=12 and m2=1.

Since, AB = AC, then ABC = ACB

Now, angle between AB and BC = angle between AC and BC

m121+12(m)=1m1m

 2m12+m=m+1m1

 2m23m+1=m2+3m+2

 m26m1=0

Sum of roots = 6

   Required sum is 6.



Q 34 :    

Let the distance between two parallel lines be 5 units and a point P lie between the lines at a unit distance from one of them. An equilateral triangle PQR is formed such that Q lies on one of the parallel lines, while R lies on the other. Then (QR)2 is equal to _________.          [2025]



(28)

Let PRN=θ

 PR=cosec θ and PQ=4 sec(30°+θ)

For equilateral PQR.

Let PR = PQ

 cosec θ=4 sec(θ+30°)

 cos (θ+30°)=4 sin θ

 32cos θ12sin θ=4 sin θ

 tan θ=133

 cosecθ=28

Now, QR2=d2=cosec2θ=28.



Q 35 :    

If α=1+r=16(3)r1 C2r112, then the distance of the point (12,3) from the line αx3y+1=0 is _________.         [2025]



(5)

Given, α=1+r=16(3)r1 C2r112

                    =1+r=16C2r112(3i)2r13i

Let t=3i, then we have

α=1+13i[C112t+C312t3+...+C1112t11]

        =1+13i[(1+3i)12(13i)122]

       =1+13i[(2ω2)12(2ω)122]=1         [ ω3=0]

   Distance of (12,3) from x3y+1=0 is

          |123(3)+1(1)2+(3)2|=123+11+3=5.