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Option : (B)

x + y = 8 

⇒ y = 8 - x 

xy = x(8 - x) 

Let f(x) = 8x - x² 

Differentiating f(x) with respect to x, we get 

f’(x) = 8 - 2x 

Differentiating f’(x) with respect to x, we get 

f’’(x) = -2 < 0 

For maxima at x = c, 

f’(c) = 0 and f’’(c) < 0 

f’(x) = 0 ⇒ x = 4 

Also, 

f’’(4) = -2 < 0 

Hence, 

x = 4 is a point of maxima for f(x) and 

f(4) = 16 is the maximum value of f(x).

Option : (B)

xy = 1, x > 0, y > 0

⇒ y = \(\frac{1}{x}\)

x+y = x + \(\frac{1}{x}\)

Let f(x) =  x + \(\frac{1}{x}\),x > 0

Differentiating f(x) with respect to x, we get

 f'(x) =  1 - \(\frac{1}{x^2}\)

Also, 

Differentiating f’(x) with respect to x, we get

 f''(x) = \(\frac{2}{x^3}\)

For minima at x = c, 

f’(c) = 0 and f’’(c) < 0

⇒ 1 - \(\frac{1}{x^2}\) = 0

or x = 1 (Since x>0)

f’’(1) = 2 > 0 

Hence, 

x = 1 is a point of minima for f(x) and f(1) = 2 is the minimum value of f(x) for x > 0.

Option : (A)

f(x) = \(\frac{1}{4x^2+2x+1}\) achieves it’s maximum value when g(x) = 4x²+2x+1 achieves it’s minimum value.

Differentiating g(x) with respect to x, we get

g’(x) = 8x+2 

Differentiating g’(x) with respect to x, we get

g’’(x) = 8 

For minima at x = c, 

g’(c) = 0 and g’’(c) > 0 g’(x) = 0

⇒ x = \(-\frac{1}{4}\)

g''(\(-\frac{1}{4}\)) = 8>0

Hence,

x = \(-\frac{1}{4}\) is a point of minima for g(x) and g(\(-\frac{1}{4}\)) = \(\frac{3}{4}\)is the minimum value of g(x). 

Hence the maximum value of f(x) = \(\frac{1}{g(x)}\) = \(\frac{4}{3}\)

f(x) = 1 + 2sinx + 3cos²x, 

x ∈ [0,\(\frac{2\pi}{3}\)]

Differentiating f(x) with respect to x, we get

f’(x)=2cosx-6cosxsinx 

Also, 

Differentiating f’(x) with respect to x, we get

f’’(x) = - 2sinx - 6cos²x + 6sin²x 

=- 2sinx + 12sin²x - 6 (Since sin²x + cos²x = 1)

For extreme points, 

f’(x) = 0 

⇒ 2cosx(1 - 3sinx) = 0 

or x = \(\frac{\pi}{2}\)

or sin^-1(\(\frac{1}{3}\))

f"(\(\frac{\pi}{2}\)) = 4 > 0 and f"( sin^-1(\(\frac{1}{3}\)))

= \(-\frac{16}{3}\) < 0

Hence,

x = \(\frac{\pi}{2}\) is a point of minima and x = sin^-1(\(\frac{1}{3}\)) s a point of maxima.

Option : (B)

f(x) = x + \(\frac{1}{x}\), x > 0

Differentiating f(x) with respect to x, we get

f'(x) = 1 - \(\frac{1}{x^2}\)

Also, 

Differentiating f’(x) with respect to x, we get

f''(x) = \(\frac{2}{x^3}\)

For maxima at x = c, 

f’(c) = 0 and f’’(c)<0

⇒  1 - \(\frac{1}{x^2}\) = 0

or x = 1 (Since x>0)

f’’(1) = 2 > 0

Since, 

f’’(1)>0

Therefore, 

x = 1 is not a point of maxima and hence no maximum value of f(x) exists for x>0.

(d) (1/-3)\(\begin{pmatrix} 1 &0 \\[0.3em] 0&1 \\[0.3em] \end{pmatrix}\)

(d) 11

Given |A| = 4 and P is the adjoint matrix of A 

|P| = 4² = 16 

⇒ -2(3 – x) = 16 

⇒ -6 + 2x = 16 

⇒ 2x = 22 

⇒ x = 11

(b) adj(AB) = (adj A) (adj B)

i. Let the measure of the complementary angle be x°. 

∴ 20 + x = 90 

∴ 20 + x – 20 = 90 – 20 ….(Subtracting 20 from both sides) 

∴ x = 70

∴ The measure of the complement of an angle of measure 20° is 70°. 

ii. Let the measure of the complementary angle be x°. 

∴ 90 + x = 90 

∴ 90 + x – 90 = 90 – 90 ….(Subtracting 90 from both sides) 

∴ x = 0 

∴ The measure of the complement of an angle of measure 90° is 0°. 

iii. Let the measure of the complementary angle be a°. 

∴ x + a = 90 

∴ x + a – x = 90 – x ….(Subtracting x from both sides) 

∴ a = (90 – x) 

∴ The measure of the complement of an angle of measure x° is (90 – x)°.

1. 40 

2. 50 

3. 90 

Note: Here, the sum of the measures of ∠ABC and ∠PQR is 90 °. Therefore, they are complementary angles.

1. 40 

2. 50 

3. 90

Note: Here, the sum of the measures of ∠ABC and ∠PQR is 90 °. 

Therefore, they are complementary angles.

i. Let the measure of the complementary angle be x°.

∴ 40 + x = 90 

∴ 40 + x – 40 = 90 – 40 ….(Subtracting 40 from both sides) 

∴ x = 50 

∴ The measure of the complement of an angle of measure 40° is 50°.

ii. Let the measure of the complementary angle be x°. 

∴ 63 + x = 90 

∴ 63+x-63 = 90-63 ….(Subtracting 63 from both sides) 

∴ x = 27 

∴ The measure of the complement of an angle of measure 63° is 27°.

iii. Let the measure of the complementary angle be x°. 

∴ 45 + x = 90 

∴ 45+x-45 = 90-45 ….(Subtracting 45 from both sides) 

∴ x = 45 

∴ The measure of the complement of an angle of measure 45° is 45°.

iv. Let the measure of the complementary angle be x°. 

∴ 55 + x = 90 

∴ 55 + x - 55 = 90-55 ….(Subtracting 55 from both sides) 

∴ x = 35 

∴ The measure of the complement of an angle of measure 55° is 35°.

(y – 20)° and (y + 30)° are the measures of complementary angles. 

∴ (y – 20) + (y + 30) = 90 

∴ y + y + 30 – 20 = 90 

∴ 2y + 10 = 90 

∴ 2y = 90 – 10 

∴ 2y = 80

∴ y = 80/2

= 40

Measure of first angle = (y – 20)° = (40 – 20)° = 20° 

Measure of second angle = (y + 30)° = (40 + 30)° = 70° 

∴ The measure of the two angles is 20° and 70°.

i. Let the measure of the complementary angle be x°. 

∴ 40 + x = 90 

∴ 40 + x – 40 = 90 – 40 

….(Subtracting 40 from both sides) 

∴ x = 50 

∴ The measure of the complement of an angle of measure 40° is 50°.

ii. Let the measure of the complementary angle be x°.

∴ 63 + x = 90 

∴ 63+x-63 = 90-63 

….(Subtracting 63 from both sides) 

∴ x = 27 

∴ The measure of the complement of an angle of measure 63° is 27°.

iii. Let the measure of the complementary angle be x°. 

∴ 45 + x = 90 

∴ 45+x-45 = 90-45 

….(Subtracting 45 from both sides) 

∴ x = 45 

∴ The measure of the complement of an angle of measure 45° is 45°.

iv. Let the measure of the complementary angle be x°. 

∴ 55 + x = 90 

∴ 55 + x-55 = 90-55 

….(Subtracting 55 from both sides)

∴ x = 35 

∴ The measure of the complement of an angle of measure 55° is 35°.

v. Let the measure of the complementary angle be x°. 

∴ 20 + x = 90 

∴ 20 + x – 20 = 90 – 20 

….(Subtracting 20 from both sides) 

∴ x = 70 

∴ The measure of the complement of an angle of measure 20° is 70°.

vi. Let the measure of the complementary angle be x°. 

∴ 90 + x = 90 

∴ 90 + x – 90 = 90 – 90 

….(Subtracting 90 from both sides) 

∴ x = 0 

∴ The measure of the complement of an angle of measure 90° is 0°.

vii. Let the measure of the complementary angle be a°. 

∴ x + a = 90 

∴ x + a – x = 90 – x 

….(Subtracting x from both sides) 

∴ a = (90 – x) 

∴ The measure of the complement of an angle of measure x° is (90 – x)°.

1. ∠ABC or ∠CBA

2. Ray BA and ray BC

(i) In ∆’s ABM and PQN

AB = PQ (given)

AM = PN (given)

and BC = QR (given)

=> 1/2BC = 1/2QR

=> BM = QN

∴ ∆ABM ≅ ∆PQN

(by SSS congruency rule)

(ii) In ∆ABC and ∆PQR

∵ ∆ABM ≅ ∆PQN [proved in (i)]

=> ∠B = ∠Q (by c.p.c.t)

AB = PQ (given)

BC = QR (given)

∴ ∆ABC ≅ ∆PQR

(by SAS congruency rule)

1. Ray PL and ray PM 

2. Ray PN and ray PT

1. Straight angle 

2. 180°

(i) In ∆’s ABD and ACD, we have

AB = AC (given)

BD = DC (given)

and AD = AD (common)

∴ ∆ABD ≅ ∆ACD

(by SSS congruency rule)

(ii) In ∆’s ABP and ACP, we have

AB = AC (given)

∠BAP = ∠CAP

and AP = AP (common)

[∵ ∆ABD ≅ ∆ACD => ∠B AD = ∠CAD => ∠BAP = ∠CAP]

∴ ∆ABP ≅ ∆ACP

(by SAS congruency rule)

(iii) We have already proved in (i) that

∆ABD ≅ ∆CAD

=> ∠BAP = ∠CAP

=> AP bisects ∠A i.e. AP is the bisector of ∠A.

In ∆’s BDP and CDP, we have

BD = CD (given)

BP = CP [∵ ∆ABP = ∆ACP]

and DP = DP (common)

∴ ∆BDP ≅ ∆CDP

(by SSS congruency rule)

=> ∠BDP = ∠CDP
=> DP is the bisector of ∠D.

Hence, AP is the bisector of ∠A as well as ∠D.

(iv) In (iii), we have proved that 

∆BDP ≅ ∆CDP

=> BP = CP and ∠BPD = ∠CPD = 90°.

∴ ∠BPD and ∠CPD form a linear pair

=> DP is the perpendicular bisector of BC

Hence, AP is the perpendicular bisector of BC.

In ∆ABC

AB = AC (given)

∠B = ∠C = x (say)

x + x + 120° = 180°

=> 2x + 120 = 180°

=> 2x = 60° 

=> x = 30°

∠ADB = 180° – (60° + 30°)

=> ∠ADB = 90°

Let the measures of ∠A be x°. 

m∠A = m∠B = x° 

∠ACD is the exterior angle of ∆ABC 

∴ m∠ACD = m∠A + m∠B 

∴ 140 = x + x 

∴ 140 = 2x 

∴ 2x = 140

∴ x = 140/2

= 70

∴ The measures of the angles ∠A and ∠B is 70° each.

Since, ∠A and ∠B are supplementary angles. 

∴ m∠A + m∠B = 180 

∴ m∠A + x + 20 = 180 

∴ m∠A + x + 20 – 20 = 180 – 20 ….(Subtracting 20 from both sides) 

∴ m∠A + x = 160 

∴ m∠A + x – x = 160 – x ….(Subtracting x from both sides) 

∴ m∠A = (160 – x)° 

∴ The measure of ∠A is (160 – x)°.

1. Ray BA and ray BC 

2. Point B 

3. ∠ABC or ∠CBA

Let the measure of the complement of ∠A be x° and the measure of its supplementary angle be y°. 

m∠A + x = 90° 

∴ 70 + x = 90 

∴ 70 + x – 70 = 90 – 70 ….(Subtracting 70 from both sides) 

∴ x = 20 

Since, x and y are supplementary angles. 

∴ x + y = 180 

∴ 20 + y = 180 

∴ 20 + y – 20 = 180 – 20 ….(Subtracting 20 from both sides) 

∴ y = 160 

∴ The measure of supplement of the complement of ∠A is 160°.

No. 

Ray PM and Ray PT do not form a straight line and hence are not opposite rays.

Since, each angle of the rectangle is 90°. 

∴ Pairs of supplementary angles are: 

i. ∠P and ∠M 

ii. ∠P and ∠N 

iii. ∠P and ∠T 

iv. ∠M and ∠N 

v. ∠M and ∠T 

vi. ∠N and ∠T

Let the measure of one angle be x°. 

∴Measure of other angle = (x + 40)° 

x + (x + 40) = 90 …(Since, the two angles are complementary) 

∴ 2x + 40 – 40 = 90 – 40 ….(Subtracting 40 from both sides) 

∴ 2x = 50

∴ x = 50/2

∴ x = 25 

∴ x + 40 = 25 + 40 

= 65

∴ The measures of the two angles is 25° and 65°.

In ΔXYZ,

m∠X + m∠Y + m∠Z = 180° 

….(Sum of the measure of the angles of a triangle is 180°) 

∴ m∠X + 90 + m∠Z = 180 

∴ m∠X + 90 + m∠Z – 90 = 180 – 90 

….(Subtracting 90 from both sides) 

∴m∠X + m∠Z = 90° 

∴∠X and ∠Z make a pair of complementary angles.

In ∆ABC and ∆PQR

∠A = ∠Q, ∠B = ∠R then side AB should equal to QR

i.e., AB = QR

∴ ∆AOC ≅ ∆PQR (by ASA)

In ΔABD and ΔAEC

AB = AC (given) …(i)

∠ABC = ∠ACB …(ii) (angle opposite to equal sides are equal)

Also BD = EC (given) …(iii)

From (i), (ii) and (iii), we have

ΔABD = ΔAEC (by SAS congruency property)

⇒ AD = AE (by c.p.c.t)

Hence, ΔADE is an isosceles triangle.

Hence proved.

QX > PX > XR

(i) No, 

Two right angles would up to 180^o.,So the third angle becomes zero. This is not possible, so a triangle cannot have two right angles. [Since sum of angles in a triangle is 180^o ] 

(ii) No, 

A triangle can’t have 2 obtuse angles. Obtuse angle means more than 90^o So that the sum of the two sides will exceed 180^o which is not possible. As the sum of all three angles of a triangle is 180^o . 

(iii) Yes 

A triangle can have 2 acute angle. Acute angle means less the 90^o angle 

(iv) No, 

Having angles-more than 60^o make that sum more than 18^o Which is not possible [ The sum of all the internal angles of a triangle is 180^o ]

(v) No, 

Having all angles less than 60^o will make that sum less than 180^o which is not possible. [The sum of all the internal angles of a triangle is 180^o] 

(vi) Yes 

A triangle can have three angles are equal to 60^o . Then the sum of three angles equal to the 180^o Which is possible such triangles are called as equilateral triangle. [ The sum of all the internal angles of a triangle is 180^o ]

Yes, sum of two obtuse angles is greater than 180°.

No, two acute angles cannot make supplementary angles. Two right angles (90°) can make supplementary angles. One acute and one obtuse angles can make supplementary angles.

Let the measure of the required angle be x^o.

Then,

= x + x = 180^o

= 2x = 180^o

= x = 180/2

= x = 90^o

Hence, the required angle measures 90^o.

Given that,

PQ ‖ RS

∠AEF = 95°

∠BHS = 110°

∠ABC = x°

∠AEF = ∠AGH = 95°(Corresponding angles)

∠AGH + ∠HGB = 180°(Linear pair)

95° + ∠HGB = 180°

∠HGB = 85°

∠BHS + ∠BHG = 180°(Linear pair)

110° + ∠BHG = 180°

∠BHG = 70°

In BHG,

∠BHG + ∠HGB + ∠GBH = 180°

70° + 95° + ∠GBH = 180°

∠GBH = 25°

Thus,

∠ABC = ∠GBH = 25°

(A) 74°52'52"

74.87° = 74° + (0.87)°

= 74° + (0.87 x 60)'

= 74° + 52.2'

= 74° + 52' + (0.2 x 60)" = 74°52'12"

(i). Acute

No. If two angles are acute, means less than 90^o, the two angles cannot be supplementary. Because, their sum will be always less than 90^o.

(ii). Obtuse

No. If two angles are obtuse, means more than 90^o, the two angles cannot be supplementary. Because, their sum will be always more than 180^o.

(iii). Right

Yes. If two angles are right, means both measures 90^o, then two angles can form a supplementary pair.

∴90^o + 90^o = 180

(i) No, two acute angles cannot be supplementary. [∵ acute angles is ∠90°]

(ii) No, two obtuse angles cannot be supplementary. [∵ obtuse angles is ∠90°].

(iii) Yes, Two right angles always supplementary. [∵ right angles is = 90°].

(i) No, two obtuse angles cannot be supplementary

Because, the sum of two angles is greater than 90° so their sum will be greater than 180°

(ii) Yes, two right angles can be supplementary

Because, 90° + 90° = 180°

(iii) No, two acute angle cannot be supplementary

Because, the sum of two angles is less than 90° so their sum will also be less than 90°

(A) An acute angled triangle

Explanation:

According to the question,

The angles of a triangle are of the ratio 5 : 3 : 7

Let 5:3:7 be 5x, 3x and 7x

Using the angle sum property of a triangle,

5x + 3x +7x = 180

15x = 180

x = 12

Substituting the value of x, x = 12, in 5x, 3x and 7x we get,

5x = 5×12 = 60^o

3x = 3×12 = 36^o

7x = 7×12 = 84^o

Since all the angles are less than 90^o, the triangle is an acute angled triangle.

Therefore, option (A) is the correct answer.

(D) A right triangle

Explanation:

Let the angles of △ABC be ∠A, ∠B and ∠C

Given that ∠A= ∠B+∠C …(eq1)

But, in any △ABC,

Using angle sum property, we have,

∠A+∠B+∠C=180^o …(eq2)

From equations (eq1) and (eq2), we get

∠A+∠A=180^o

⇒2∠A=180^o

⇒∠A=180^o/2 = 90^o

⇒∠A = 90^o

Hence, we get that the triangle is a right triangle

Therefore, option (D) is the correct answer.

Correct option is  C) 280°

Correct option is (B) 130°

\(\angle A=80^\circ\)

\(\because\) \(\angle A+\angle B+\angle C=180^\circ\)

\(\Rightarrow\) \(\angle B+\angle C=180^\circ-\angle A\)

\(=180^\circ-80^\circ=100^\circ\)

Given that angle bisectors of \(\angle B\) and \(\angle C\) meet at O.

\(\therefore\) \(\angle OBC=\frac{\angle B}2\) & \(\angle OCB=\frac{\angle C}2\)

\(\therefore\) \(\angle OBC+\angle OCB=\frac{\angle B}2+\frac{\angle C}2\)

\(=\frac{\angle B+\angle C}2=\frac{100^\circ}2=50^\circ\)

\(\because\) \(\angle OBC+\angle OCB+\angle BOC=180^\circ\)

(Sum of angles in a triangle is \(180^\circ)\)

\(\Rightarrow\) \(\angle BOC=180^\circ-(\angle OBC+\angle OCB)\)

\(=180^\circ-50^\circ=130^\circ\)

Correct option is  B) 130°

(C) \((\frac{91}2)°\) 

30' = \((\frac{1}2)°\) 

∴ 45°30' = 45° + \((\frac{1}2)°\) = \((\frac{91}2)°\)

Let the angle be x° 

then its supplementary 180° – x° 

By problem, x° =180° – x°

x° + x° = 180° 

x° = 180°/2

x° = 90°

(B) 52 ½^o

Explanation:

According to the question,

Exterior angle of triangle= 105°

Let the two interior opposite angles of the triangle = x

We know that,

Exterior angle of a triangle = sum of interior opposite angles

Then, we have the equation,

105° = x + x

2x = 105°

x = 52.5°

x = 52½

Therefore, option (B) is the correct answer.

Correct option is  D) 7 : 6

(i) 105° 

Supplementary angle of 105° = 180° – 105° = 75° 

(ii) 95° 

Supplementary angle of 95° = 180° – 95° = 85 

(iii) 150° 

Supplementary angle of 150° = 180° – 150° = 30° 

(iv) 20° 

Supplementary angle of 20° = 20° = 180° – 20° = 60°