Exams > Cat > Quantitaitve Aptitude
GEOMETRY SET I MCQs
Total Questions : 110
| Page 4 of 11 pages
Answer: Option A. -> 2 tan α
:
A
(a) √x2+x+tan2α√x2+x≥2tanα (A.M≤G.M).
:
A
(a) √x2+x+tan2α√x2+x≥2tanα (A.M≤G.M).
Answer: Option C. -> tan β+2tan γ
:
C
(c) α+β=π2=>tanβ=cotα
tan(β+γ)=tanα=>tanα=tanβ+tanγ1−tanβtanγ
=>tanα=cotα+tanγ1−cotαtanγ
=>tanα−tanγ=cotα+tanγ
=>tanα=tanβ+2tanγ
:
C
(c) α+β=π2=>tanβ=cotα
tan(β+γ)=tanα=>tanα=tanβ+tanγ1−tanβtanγ
=>tanα=cotα+tanγ1−cotαtanγ
=>tanα−tanγ=cotα+tanγ
=>tanα=tanβ+2tanγ
Answer: Option A. -> sin(n−1)αcosx1 cosxn
:
A
(a) We havesin α sec x1 sec x2 + sin α sec x2 sec x3 +....... + sin α sec xn−1, sec xn)
=sin(x2−x1)cosx1cosx2 +sin(x3−x2)cosx2cosx3 + ........ +sin(xn−xn−1)cosxn−1cosxn
= tan x2 -tan x1 +tan x3 -tan x2 + .......... +tan xn -tan xn−
=tan xn -tan x1 =sin(xn−xn−1)cosxn−1cosx1 = sin(n−1)αcosxncosx1 {(∵ xn = x1 + (n - 1)α)}
:
A
(a) We havesin α sec x1 sec x2 + sin α sec x2 sec x3 +....... + sin α sec xn−1, sec xn)
=sin(x2−x1)cosx1cosx2 +sin(x3−x2)cosx2cosx3 + ........ +sin(xn−xn−1)cosxn−1cosxn
= tan x2 -tan x1 +tan x3 -tan x2 + .......... +tan xn -tan xn−
=tan xn -tan x1 =sin(xn−xn−1)cosxn−1cosx1 = sin(n−1)αcosxncosx1 {(∵ xn = x1 + (n - 1)α)}
Answer: Option B. -> 4 < n ≤ 8
:
B
(b)sin π2n + cosπ2n = √n2
⇒ √2 (sinπ2n.cosπ4+cosπ2n.sinπ4) =√n2
⇒√2 sin(π4+π2n) =√n2
Sincesin(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.
:
B
(b)sin π2n + cosπ2n = √n2
⇒ √2 (sinπ2n.cosπ4+cosπ2n.sinπ4) =√n2
⇒√2 sin(π4+π2n) =√n2
Sincesin(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.
Answer: Option D. -> 52
:
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
:
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
Answer: Option B. -> 1924π
:
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
:
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
Answer: Option B. -> 1
:
B
(b) 3[sin4(3π2−α)+sin4(3π+α)] - [sin6(π2+α)+sin6(5π−α)]
=3(−cosα)4+(−sinα)4−2cos6α+sin6α
=3(cos2α+sin2α)2−2sin2αcos2α−2(cos2α+sin2α)3−3sin2αcos2α(cos2α+sin2α)
=3−6sin2αcos2α−2+6sin2αcos2α=3−2=1
Trick: Put α=0 , the value of expressions remains 1 i.e., it is independent of α
:
B
(b) 3[sin4(3π2−α)+sin4(3π+α)] - [sin6(π2+α)+sin6(5π−α)]
=3(−cosα)4+(−sinα)4−2cos6α+sin6α
=3(cos2α+sin2α)2−2sin2αcos2α−2(cos2α+sin2α)3−3sin2αcos2α(cos2α+sin2α)
=3−6sin2αcos2α−2+6sin2αcos2α=3−2=1
Trick: Put α=0 , the value of expressions remains 1 i.e., it is independent of α
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