# JEE Main > Trigonometric Equation & Inequalities, Solution Of Triangles

Explore popular questions from Trigonometric Equation & Inequalities, Solution Of Triangles for JEE Main. This collection covers Trigonometric Equation & Inequalities, Solution Of Triangles previous year JEE Main questions hand picked by experienced teachers.

Physics
Chemistry
Mathematics
Q 1.

Correct4

Incorrect-1

The period of the function is

B

C

D

##### Explanation

Period of
Period of

Q 2.

Correct4

Incorrect-1

The period of the function is

A

B

C

##### Explanation

Period of and period of
L.C.M. of and

Q 3.

Correct4

Incorrect-1

Which of the following pieces of data does NOT uniquely determine an acute-angled triangle ABC (R being the radius of the circum-circle)

A

B

C

##### Explanation

and . So two sides and two angles are known. So is known. Therefore, two sides and included angle is known. So, is uniquely known in case .
If a, b, c are known the is uniquely known in case . So, sides a, b and angle A, B are known. So is known. Therefore two sides and included angle is known. So, is uniquely known in case .
So, only a side and an angle is known. So, is not uniquely known in case

Q 4.

Correct4

Incorrect-1

The sum of the radii of inscribed and circumscribed circles for an n sided regular polygon of side a, is

A

B

D

##### Explanation

and
.

Q 5.

Correct4

Incorrect-1

If the roots of the equation x2 - 2ax + a2 + a - 3 = 0 are real less than 3, then

a< 2

B

2 ≤ a ≤ 3

C

3 < a ≤ 4

D

a > 4

##### Explanation

s
f (x) = x2 - 2ax + a2 + a - 3 = 0, f (3) /> 0, α + β < 6, Δ ≥ 0.
⇒ a2 - 5a + 6 /> 0, a < 3, - 4a + 12 /> 0 ⇒ a < 2 or a /> 3, a < 3,
a < 3 ⇒ a < 2

Q 6.

Correct4

Incorrect-1

If a, b, c, d are positive real numbers such that a + b + c + d = 2, then M = (a + b) (c + d) satisfies the relation

A

0 ≤ M ≤ 1

B

1 ≤ M ≤ 2

C

2 ≤ M ≤ 3

3 ≤ M ≤ 4

##### Explanation

3 ≤ M ≤ 4
As A.M. ≥ G.M. for positive real numbers, we get
⇒ M ≤ I
(Putting values)
Also(a + b) (c + d) /> 0
[... a, b, c, d /> 0]
0 ≤ M ≤ 1

Q 7.

Correct4

Incorrect-1

If b > a, then the equation (x - a) (x - b) - 1 = 0 has

A

Both roots in (a, b)

B

Both roots in (-∞, a),

C

Both roots in (b, + ∞)

One root in (-∞, a) and the other in (b, +∞)

##### Explanation

The given equation is (x - a)(x - b) - 1 = 0, b /> a
Or x2 - (a + b) x + ab - 1 = 0

Since coeff. of x2 i.e. 1 /> 0 it represents upward parabola, intersecting x axis at two points. (corresponding to two real roots, D being +ve)
Also f = f = - 1 ⇒ curve is below x-axis at a and b ⇒ a
and b both lie between the roots.
Thus the graph of given eqn is as shown.
From graph it is clear that one root of the equation lies in
(- ∞, a) and other in (b, ∞).
.

Q 8.

Correct4

Incorrect-1

For the equation 3x2 + px + 3 = 0, p > 0, if one of the root is square of the other, then p is equal to

1/3

B

1

C

3

D

2/3

##### Explanation

Let α, α2 be the roots of 3x2 + px + 3 = 0
Now S = α + α2 = - p/3, p = α3 = 1 ⇒ α = 1, ω, ω2;
α + α2 = - p/3 ⇒ ω + ω2 = -p/3 ⇒ - 1= -p/3 ⇒ p = 3

Q 9.

Correct4

Incorrect-1

For all 'x', x2 + 2ax + 10 - 3a > 0, then the interval in which 'a' lies is

A

a< - 5

- 1 < a < 2

C

a> 5

D

2 < a < 5

##### Explanation

x2 + 2ax + 20 - 3x /> 0 x
⇒ D < 0 ⇒ 4a2 - 4(10 - 3a) < 0 ⇒ a2 + 3a - 10 < 0
⇒ (a + 5) (a - 2) < 0 ⇒ a ε (- 5, 2)

Q 10.

Correct4

Incorrect-1

If in ΔABC, ∠A = 900 and c, sin B, cos B are rational numbers, then

A

A is rational and b is irrational

B

A is irrational and b is rational

A and b both are rational

D

None of these

##### Explanation

Here C = 900 - B and sin C = cos B.
Also a =
⇒ a, b are rational

Q 11.

Correct4

Incorrect-1

If A is the area and 2s the sum of the sides of a triangle, then

A

A ≤

A ≤

C

A />

D

None of these

##### Explanation

We have 2s = a + b + c, A2 = s(s - a) (s - b) (s - c).
Now A.M ≥ G.M
≥ [s(s - a) (s - b) (s - c)]1/4
≥ [A2]1/4 ≥ A1/2 ⇒ A ≤.
Also ≥ [(s - a) (s - b) (s - c)]1/3
oror ⇒ A ≤

Q 12.

Correct4

Incorrect-1

ABC is a triangle whose medians AD and BE are perpendicular to each other. If AD = p and BE = q then area
of ΔABC is-

B

C

D

##### Explanation

AM = , MD = , BM =
BM2 = BM2 + DM2

Q 13.

Correct4

Incorrect-1

In a ΔABC if ∠A = 600, then ∠B - ∠C has value equal to

A

150

B

300

C

22.50

450

##### Explanation

(given)

Q tan
tancot 300
tan
= 150
⇒ B - C = 300

Q 14.

Correct4

Incorrect-1

If D is the mid-point of side BC of a triangle ABC and AD is perpendicular to AC, then

A

3b2 = a2 - c2

B

3a2 = b2 - 3c2

C

b2 = a2 - c2

a2+ b2= 5c2

##### Explanation

From the right-angled triangle CAD, we have
cos C =

⇒a2 + b2 - c2 = 4b2
⇒ a2 - c2 = 3b2

Q 15.

Correct4

Incorrect-1

In a triangle, if the sum of two sides is x and their product is y such that (x + z) (x - z) = y where z is the third side of the triangle then the triangle is

A

Equilateral

B

Tight angled

Obtuse angled

D

None of these

##### Explanation

Let ABC be the triangle with b + c = x and
bc = y then a = z, and from the given relations we have
(b + c + a) (b + c - a) = bc
⇒ b2 + c2 - a2 = - bc

⇒cos A = - = cos 1200
⇒ A = 1200 and the triangle is obtuse angled.
⇒ A is an obtuse angle.

Q 16.

Correct4

Incorrect-1

In ΔABC, a cos (B - C) + b cos (C - A) + c cos (A - B) (where a, b, c are sides of Δ) equals

B

C

D

None

##### Explanation

a = 2R sin A
b = 2R sin B
& c = 2R sin C

Q 17.

Correct4

Incorrect-1

In a triangle ABC, ∠A = 600, a = 4 and b = 3, then c is a root
of the equation

A

c2 + 3c + 7 = 0

B

c2 -3c + 7 = 0

C

c2 + 3c -7 = 0

c2 -3c -7 = 0

##### Explanation

cos 600 = ⇒ c2 -3c - 7 = 0

Q 18.

Correct4

Incorrect-1

In a triangle ABC, if + = then C is equal to -

A

300

B

600

C

750

900

##### Explanation

The given relation can be written as
=
⇒ (a + b + 2c) (a + b + c) = 3(a + c) (b + c)
⇒ (a + b)2 + 3c(a + b) + 2c2 = 3(ab + ac + bc+ c2)
⇒ a2 + b2 - c2 = ab
cos C =
= =
⇒ C = 600.

Q 19.

Correct4

Incorrect-1

If 4 cos A cos B + sin 2A + sin 2B + sin 2C = 4 then ΔABC is

A

Right angle

B

Isosceles

Right angled isosceles

D

None of these

##### Explanation

Q 4 cos A cos B + 4 sin A sin B sin C = 4
sin C = ≤1 ... (1)
⇒ cos (A - B) ≥ 1
⇒ cos (A - B) = 1
True if A = B ... (2)
Put (2) in (1)
sin C = 1
C = 900

Q 20.

Correct4

Incorrect-1

Thesumof all solutions of the equation is

A

C

D

None of these

##### Explanation

Here, or
or
where ;
The required sum = 30π.

Q 21.

Correct4

Incorrect-1

In any triangle ABC if , then the triangle isC

A

Right angled

B

Equilateral

Isosceles

D

None of these

##### Explanation

= sinA

Triangle is isosceles.

Q 22.

Correct4

Incorrect-1

If in a ΔABC, the sides b, a, c are in A.P. then-

A

acos2+ c cos2=

c cos2 + b cos2=

C

bcos2 + a cos2=

D

2 sin= cos

##### Explanation

b + c = 2a
Option B :
⇒ c cos2 B/2 + b cos2 C/2 = 3a/2
⇒ 2c cos2 B/2 + 2b cos2 C/2 = 3a
⇒ c (1 + cos B) + b(1 + cos C) = 3a
⇒ b + c + c cos B + b cos C = 3a
⇒ b + c + a = 3a
⇒b + c = 2a which is in A.P.

Q 23.

Correct4

Incorrect-1

If p1, p2, p3 are respectively the perpendiculars from the vertices of a Δ to the opposite sides, then -

A

++= r

B

p1p2p3=

++=

D

++=

##### Explanation

p1= , p2=, p3=
++=
== ...(i)
p1p2p3 === ......(ii)
Also ++=
(a cos A + b cos B + c cos C)
=(4 sinA sin B sin C) ==.......(iii)
and ++
=×+×+×
= 2Δ= 2Δ
= 2(a2 + b2 + c2). =

Q 24.

Correct4

Incorrect-1

In a right angled triangle acute angle α, β satisfy tan α + tan β + tan2α + tan2β = 4, if hypotenuse is of length d then area of Δ is-

A

d2

B

C

d2

##### Explanation

α + β = 900
tan α + cot α + tan2 β + cot2 β = 4
Let tan α = t
t ++ t2 + = 4
t = 1 satisfy
α = β = 450
So area is

Q 25.

Correct4

Incorrect-1

If cos A + cos B+ cos C = , then (ratio of inradius (r) and R is circumradius) is a root of the equation -

A

- 2 + = 0

B

- 2 - 1 = 0

+ 2 - 1 = 0

D

+ 3 - 1 = 0

##### Explanation

cos A + cos B + cos C =
⇒ 1 + 4 sinA/2 sin B/2 sinC/2 =
⇒1 + = = - 1