Free Geometry Set I - 02 Practice Test - CAT

The value of k, for which
(cos x + sin x)2 + k sin x cos x - 1 = 0 is an identity, is

A. -1
B. -2
C. 0
D. 1

SOLUTION

Solution : B

(b) Given, (cos x + sin x)2 + k sin x cos x - 1 = 0, x
cos2x +sin2x + 2cos x sin x + k sin x cos x - 1 = 0,  x
(k + 2)cos x sin x = 0,  x
k = -2

If   tan(AB)=1 , sec(A+B)=23, then the smallest positive value of B is,

A. 2524π
B. 1924π
C. 1324π
D. 1124π

SOLUTION

Solution : B

tan(A-B)=1

AB=(2n+1)π4

n=0 since question is the smallest value possible

AB=π4

A+B=2ππ6=11π6

Subtract both the equations

2B=11π6π4

2B=(223)π12

B=19π24

If  sin4Aa+cos4Ab=1a+b, then the value of  sin8Aa3+cos8Ab3 is equal to

A. 1(a+b)3
B. a3b3(a+b)3
C. a2b2(a+b)2
D. None of these

SOLUTION

Solution : A

(a) It is given that  sin4Aa+cos4Ab=1a+b

(1cos2A)24a+(1+cos2A)24b=1a+b

b(a+b)(12cos2A+cos22A)+a(a+b)(1+2cos2A+cos22A)=4ab

{b(a+b)+a(a+b)}cos22A+2(a+b)(ab)cos2A

A. 15
B. 6
C. 1
D. 0

SOLUTION

Solution : B

A.
B.
C.
D. both b) and c)

SOLUTION

Solution : D

For a positive integer n, let
fn(θ) = (tanθ2)(1 + sec θ)(1 + sec 2θ)(1 + sec 4θ) .........
(1 + sec 2nθ). Then

A. f2(π16) = 1
B. f3(π32) = 1
C. f4(π64) = 1
D. All of above

SOLUTION

Solution : D

(a,b,c)  fn(θ) = sin(θ/2)cos(θ/2)[2cos2θ/2cosθ.2cos2θcos2θ.2cos22θcos4θ]
Combine first two factors, fn(θ) = sinθcosθ[2cos2θcos2θ.2cos22θcos4θ]
Again combine first two factors,
fn(θ) = tan 2θ[2cos22θcos4θ.......] = tan (2nθ)
f2(π16) = tan 4π16 = tan (π4)= 1
f3(π32) = tan 8π32 = tan (π4) = 1
f4(π64) = tan 16π64 = tan (π4) = 1
f5(π128) = tan 32π128 = tan (π4) = 1

Let n be a positive integer such that sin π2n + cos π2n = n2. Then

A. 6 n 8
B. 4 < n 8
C. 4 n < 8
D. 4 < n < 8

SOLUTION

Solution : B

(b) sin π2n + cos π2n = n2
2 (sinπ2n.cosπ4+cosπ2n.sinπ4)n2
2 sin(π4+π2n)n2
Since sin(π4+π2n)  1
n2 2 2 22 n 8 .
Again  n22 sin(π4+π2n) > 2.12 = 1
(sin(π4+π2n)sinπ4)
n > 4, Hence, 4 < n 8.

If k = sinπ18.sin5π18.sin7π18, then the numerical value of k is

A. 14
B. 18
C. 116
D. None of these

SOLUTION

Solution : B

(b) We have k = sinπ18.sin5π18.sin7π18
= cos(π2π18) cos(π25π18) cos(π27π18)
= cosπ9 cos2π9 cos4π9 = sin23π923 sinπ9 = sin8π98sinπ9
= sin(ππ9)8sinπ9 = 18

A. 0
B. 1
C. -1
D. 2

SOLUTION

Solution : A

If α, β, γ, δ are the smallest positive angles in ascending order of magnitude which have their sines equal to the positive
quantity k, then the value of 4 sin α2 + 3 sin β2+ 2 sin γ2 + sin δ2 is equal to

A. 21k
B. 121+k.
C. 21+k.
D. None of these

SOLUTION

Solution : C

(c) Given α < β < γ < δ and sin α = sin β = sin γ = sin δ = k. Also α, βγδ are smallest positive angles satisfying above two conditions.
We can take β = π - α, γ = 2π + α, δ = 3π - α
Given expression
= 4 sin α2 + 3 sin(π2α2) + 2 sin(π+α2) + 3 sin(3π2α2)
= 4 sin α2 + 3 cos α2 - 2 sin α2 - cos α2 = 2 (sinα2+cosα2)
= 2 (sin12α+cos12α)2 = 2 1+sinα = 21+k.

If (sec A + tan A)(sec B + tan B)(sec C + tan C) = (sec A - tan A)(sec B - tan B)(sec C - tan C), then each side is equal to

A. 1
B. -1
C. 0
D. None of these

SOLUTION

Solution : A

(a,b) If L=M, then L2 = LM or ML = M2
Both LM = ML = 1 as sec2A - tan2A = 1
L2 = M2 = 1.

If tan θ = 32, the sum of the infinite series
1 + 2(1 - cosθ) + 3(1 - cosθ)2 + 4(1 - cosθ)3+........ is

A. 23
B. 34
C. 522
D. 52

SOLUTION

Solution : D

(d)  1 + 2(1 - cosθ) + 3(1 - cosθ)2 + 4(1 - cosθ)3+........ to
= (1 - x)1, [Let (1 - cosθ) = x]
Series  = (1 - 1 + cosθ)2 = sec2θ
= (1 + tan2θ) = 1 + 32 = 52

If x1,x2,x3,......,xn are in A.P. whose common difference is a, then the value of sin α(sec x1 sec x2 + sec x2 sec x3 +....... + sec xn1, sec xn)=

A. sin(n1)αcosx1 cosxn
B. sinnαcosx1 cosxn
C. sin (n - 1)α cos x1 cos xn
D. sin nα cosx1 cosxn

SOLUTION

Solution : A

(a) We have sin α sec x1 sec x2 + sin α sec x2 sec x3 +....... + sin α sec xn1, sec xn)
=sin(x2x1)cosx1 cosx2sin(x3x2)cosx2 cosx3 + ........ + sin(xnxn1)cosxn1 cosxn
= tan x2 - tan x1 + tan x3 - tan x2 + .......... + tan xn - tan xn
= tan xn - tan x1sin(xnxn1)cosxn1 cosx1 = sin(n1)αcosxn cosx1  {( xn = x1 + (n - 1)α)}

Let α,β be such that π < (α - β) < 3π. If sin α + sin β = -2165 and cos α + cos β = -2765, then the value of \
cos αβ2

A. 665
B. 3130
C. 665
D. -3130

SOLUTION

Solution : D

(d) sin α + sin β = -2165 , cos α + cos β = -2765
Now (sin α + sin β)2 + (cos α + cos β)2 =  (2165)2(2765)2
2 + 2 sin α sin β + 2 cos α cos β = 441652 + 729652
2 + 2[cos (α - β)] = 1170(65)2 2.2 cos2(α+β2)=1170(65)2
cos (αβ2) = 3130130 = 3130
Therefore cos (αβ2) = 3130 , { π2 < αβ2 < 3π2 }

A. 1
B. 2
C. 0
D.

SOLUTION

Solution : C

Option C is the correct answer.

Let A0A1A2A3A4A5 be a regular hexagon inscribed in a circle of unit radius. Then the product of the lengths of the line segments A0A1, A0A2 and A0A4 is

A. 34
B. 33
C. 3
D. 332

SOLUTION

Solution : C

(c) Each triangle is an equilateral triangle

Hence A0A1 = 1
A0A20 = A0A21 + A1A20 - 2A0A1A1A2 cos 120
= 1 + 1 - 2.1.1(-12) = 3
A0A2 = 3 = A0A4
A0A1 × A0A2 × A0A4 = 1.3.3

3[sin4(3π2α)+sin4(3π+α)][sin6(π2+α)+sin6(5πα)]

A.

0

B.

1

C.

3

D.

sin 4α + sin 6α

SOLUTION

Solution : B

(b) 3[sin4(3π2α)+sin4(3π+α)] - [sin6(π2+α)+sin6(5πα)]
=3(cos α)4+(sin α)42cos6α+sin6α
=3(cos2 α + sin2 α)2  2sin2 α cos2 α  2(cos2 α + sin2 α)3  3sin2 α cos2 α (cos2 α + sin2 α)
=36sin2 α cos2 α  2 + 6sin2 α cos2 α=32=1
Trick: Put α=0 , the value of expressions remains 1 i.e., it is independent of α

If α(0,π2) then x2+x+tan2 αx2+x is always greater than or equal to

A. 2 tan α
B. 1
C. 2
D. sec2 α

SOLUTION

Solution : A

(a) x2+x+tan2 αx2+x2 tan α (A.MG.M).

The maximum value of cosα1.cosα2........cosαn, under the restrictions 0α1,α2,.........αn π2 and
cotα1.cotα2........cotαn = 1 is

A. 12n/2
B. 12n
C. 12n
D. 1

SOLUTION

Solution : A

(a) Here (cot α1).(cot α2)....(cot αn)= 1
cosα1.cosα2........cosαn =  sinα1.sinα2........sinαn
Now, (cosα1.cosα2........cosαn)2
= (cosα1.cosα2........cosαn) (cosα1.cosα2........cosαn)
= (cosα1.cosα2........cosαn) (sinα1.sinα2........sinαn)
= 12n sin 2α1.sin 2α2........sin 2αn
But each of sin 2αi
(cosα1.cosα2........cosαn)2  12n
But each of cos αi, is positive.
cosα1.cosα2........cosαn 12n = 12n/2.

If A=sin8θ+cos14θ, then for all real value of θ

A. A1
B. 0<A1
C. 1<2A3
D. None of these

SOLUTION

Solution : B

Option B is the correct answer.

If θ is an acute angle and sin θ=p68p, then p must satisfy

A. 6p<8
B. 6p<7
C. 3p4
D. 4p<7

SOLUTION

Solution : B

(b) θ is an acute angle so 0θ<90
0p68p<1   => 0(p6)<(8p)=>6p<7.

IfA,B,C be the angle of a triangle, the cot A+cot Btan A+tan B

A. 1
B. 2
C. -1
D. -2

SOLUTION

Solution : A

Option A is the correct answer.

If α+β=π2 and β+α=α, then tan α equals

A. 2(tan β+tan γ)
B. tan β+tan γ
C. tan β+2tan γ
D. 2tan β+tan γ

SOLUTION

Solution : C

(c) α+β=π2=>tan β=cot α
tan(β+γ)=tanα=>tan α=tan β+tan γ1tan β tan γ
=>tan α=cot α+tan γ1cot α tan γ
=>tan αtan γ=cot α+tan γ
=>tan α=tan β+2 tan γ

If asin2θ+bsinθ cosθ+c cos2θ12(a+c)12k, then k2 is equal to

A.  b2+(ac)2
B. a2+(bc)2
C. c2+(ab)2
D. None of these

SOLUTION

Solution : A

(a) a sin2 θ+b sin θ cos θ+c cos2 θ12(a+c)
= 12[a cos 2 θ+b sin 2θ+c cos 2θ]
= 12[b sin 2 θ(ac)cos 2θ]
|b sin 2θ(ac)cos 2θ|b2+(ac)2
12b sin 2θ(ac)cos 2θ12b2+(ac)2
K=b2+(ac)2

If for all real values of x,4x2+164x296 x sin α+5<132, then α lies in the interval.

A. (0,π3)
B. (π3,2π3)
C. (2π3,π)
D. (4π3,5π3)

SOLUTION

Solution : D

Option D is the correct answer.