Free Geometry Set I - 02 Practice Test - CAT
Question 1
The value of k, for which
(cos x + sin x)2 + k sin x cos x - 1 = 0 is an identity, is
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
Question 2
If tan(A−B)=1 , sec(A+B)=2√3, then the smallest positive value of B is,
SOLUTION
Solution : B
tan(A-B)=1
A−B=(2n+1)π4
n=0 since question is the smallest value possible
A−B=π4
A+B=2π−π6=11π6
Subtract both the equations
2B=11π6−π4
2B=(22−3)π12
B=19π24
Question 3
If sin4Aa+cos4Ab=1a+b, then the value of sin8Aa3+cos8Ab3 is equal to
SOLUTION
Solution : A
(a) It is given that sin4Aa+cos4Ab=1a+b
⇒(1−cos2A)24a+(1+cos2A)24b=1a+b
⇒b(a+b)(1−2cos2A+cos22A)+a(a+b)(1+2cos2A+cos22A)=4ab
⇒{b(a+b)+a(a+b)}cos22A+2(a+b)(a−b)cos2A
Question 4
SOLUTION
Solution : B
Question 5



SOLUTION
Solution : D
Question 6
For a positive integer n, let
fn(θ) = (tanθ2)(1 + sec θ)(1 + sec 2θ)(1 + sec 4θ) .........
(1 + sec 2nθ). Then
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
Question 7
Let n be a positive integer such that sin π2n + cos π2n = √n2. Then
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 ≤ 2√2 ⇒ n ≤ 8 .
Again √n2 = √2 sin(π4+π2n) > √2.1√2 = 1
(∵sin(π4+π2n)sinπ4)
∴ n > 4, Hence, 4 < n ≤ 8.
Question 8
If k = sinπ18.sin5π18.sin7π18, then the numerical value of k is
SOLUTION
Solution : B
(b) We have k = sinπ18.sin5π18.sin7π18
= cos(π2−π18) cos(π2−5π18) cos(π2−7π18)
= cosπ9 cos2π9 cos4π9 = sin23π923 sinπ9 = sin8π98sinπ9
= sin(π−π9)8sinπ9 = 18
Question 9
SOLUTION
Solution : A
Question 10
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
SOLUTION
Solution : C
(c) Given α < β < γ < δ and sin α = sin β = sin γ = sin δ = k. Also α, β, γ, δ 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(π2−α2) + 2 sin(π+α2) + 3 sin(3π2−α2)
= 4 sin α2 + 3 cos α2 - 2 sin α2 - cos α2 = 2 (sinα2+cosα2)
= 2 √(sin12α+cos12α)2 = 2 √1+sinα = 2√1+k.
Question 11
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
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.
Question 12
If tan θ = √32, the sum of the infinite series
1 + 2(1 - cosθ) + 3(1 - cosθ)2 + 4(1 - cosθ)3+........∞ is
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
Question 13
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 xn−1, sec xn)=
SOLUTION
Solution : A
(a) We have sin α sec x1 sec x2 + sin α sec x2 sec x3 +....... + sin α sec xn−1, sec xn)
=sin(x2−x1)cosx1 cosx2 + sin(x3−x2)cosx2 cosx3 + ........ + sin(xn−xn−1)cosxn−1 cosxn
= tan x2 - tan x1 + tan x3 - tan x2 + .......... + tan xn - tan xn−
= tan xn - tan x1 = sin(xn−xn−1)cosxn−1 cosx1 = sin(n−1)αcosxn cosx1 {(∵ xn = x1 + (n - 1)α)}
Question 14
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 α−β2
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 (α−β2) = 3√130130 = 3√130
Therefore cos (α−β2) = −3√130 , { ∵ π2 < α−β2 < 3π2 }
Question 15

SOLUTION
Solution : C
Option C is the correct answer.
Question 16
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
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
Question 17
3[sin4(3π2−α)+sin4(3π+α)] - [sin6(π2+α)+sin6(5π−α)] =
0
1
3
sin 4α + sin 6α
SOLUTION
Solution : B
(b) 3[sin4(3π2−α)+sin4(3π+α)] - [sin6(π2+α)+sin6(5π−α)]
=3(−cos α)4+(−sin α)4−2cos6α+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)
=3−6sin2 α cos2 α − 2 + 6sin2 α cos2 α=3−2=1
Trick: Put α=0 , the value of expressions remains 1 i.e., it is independent of α
Question 18
If α∈(0,π2) then √x2+x+tan2 α√x2+x is always greater than or equal to
SOLUTION
Solution : A
(a) √x2+x+tan2 α√x2+x≥2 tan α (A.M≤G.M).
Question 19
The maximum value of cosα1.cosα2........cosαn, under the restrictions 0≤α1,α2,.........αn ≤π2 and
cotα1.cotα2........cotαn = 1 is
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 ≤ 1
(cosα1.cosα2........cosαn)2 ≤ 12n
But each of cos αi, is positive.
∴ cosα1.cosα2........cosαn ≤ √12n = 12n/2.
Question 20
If A=sin8θ+cos14θ, then for all real value of θ
SOLUTION
Solution : B
Option B is the correct answer.
Question 21
If θ is an acute angle and sin θ=p−68−p, then p must satisfy
SOLUTION
Solution : B
(b) θ is an acute angle so 0∘≤θ<90∘
∴0≤p−68−p<1 => 0≤(p−6)<(8−p)=>6≤p<7.
Question 22
IfA,B,C be the angle of a triangle, the ∑cot A+cot Btan A+tan B
SOLUTION
Solution : A
Option A is the correct answer.
Question 23
If α+β=π2 and β+α=α, then tan α equals
SOLUTION
Solution : C
(c) α+β=π2=>tan β=cot α
tan(β+γ)=tanα=>tan α=tan β+tan γ1−tan β tan γ
=>tan α=cot α+tan γ1−cot α tan γ
=>tan α−tan γ=cot α+tan γ
=>tan α=tan β+2 tan γ
Question 24
If ∣∣∣asin2θ+bsinθ cosθ+c cos2θ−12(a+c)∣∣∣≤12k, then k2 is equal to
SOLUTION
Solution : A
(a) a sin2 θ+b sin θ cos θ+c cos2 θ−12(a+c)
= 12[−a cos 2 θ+b sin 2θ+c cos 2θ]
= 12[b sin 2 θ−(a−c)cos 2θ]
∵|b sin 2θ−(a−c)cos 2θ|≤√b2+(a−c)2
∴∣∣∣12b sin 2θ−(a−c)cos 2θ∣∣∣≤12√b2+(a−c)2
∴K=√b2+(a−c)2
Question 25
If for all real values of x,4x2+164x2−96 x sin α+5<132, then α lies in the interval.
SOLUTION
Solution : D
Option D is the correct answer.