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**Question 1: 3sinx + 4cosx + r is always greater than or equal to 0. What is the smallest value ‘r’ can to take?****A. 5****B. -5****C. 4****D. 3**

**Answer : 5**

**Explanation- **

3sinx + 4cosx ≥ -r

3/5 = cosA ⇒ sinA = 4/5

5(sinx cosA + sinA cosx) ≥ -r

5(sin(x + A)) ≥ -r

5sin (x + A) ≥ -r

-1 ≤ sin (angle) ≤ 1

5sin (x + A) ≥ -5

r_{min} = 5

The question is "What is the smallest value ‘r’ can to take?"

Hence, the answer is 5.

Choice A is the correct answer.

**Question 2: Sin2014x + Cos2014x = 1, x in the range of [-5π, 5π], how many values can x take?****A. 0****B. 10****C. 21****D. 11****Answer: 21****Explanation-**

We know that Sin^{2}x + Cos^{2}x = 1 for all values of x.

If Sin x or Cos x is equal to –1 or 1, then Sin^{2014}x + Cos^{2014}x will be equal to 1.

Sin x is equal to –1 or 1 when x = –4.5π or –3.5π or –2.5π or –1.5π or –0.5π or 0.5π or 1.5π or 2.5π or 3.5π or 4.5π.

Cosx is equal to –1 or 1 when x = –5π or –4π or –3π or –2π or –π or 0 or π or 2π or 3π or 4π or 5π.

For all other values of x, Sin^{2014} x will be strictly lesser than Sin^{2}x.

For all other values of x, Cos^{2014} x will be strictly lesser than Cos^{2}x.

We know that Sin^{2}x + Cos^{2}x is equal to 1. Hence, Sin^{2014}x + Cos^{2014}x will never be equal to 1 for all other values of x. Thus there are 21 values.

The question is "How many values the 'x' can take?"

Hence, the answer is 21.

Choice C is the correct answer.

**Question 3: Consider a regular hexagon ABCDEF. There are towers placed at B and D. The angle of elevation from A to the tower at B is 30 degrees, and to the top of the tower at D is 45 degrees. What is the ratio of the heights of towers at B and D?****A. 1:√3****B. 1:2√3****C. 1:2****D. 3:4√3****Answer: 1:2√3****Explanation-**

Let the hexagon ABCDEF be of side ‘a’. Line AD = 2a. Let towers at B and D be B’B and D’D respectively.

From the given data we know that ∠B´AB = 30^{°} and ∠D´AB = 45^{°}. Keep in mind that the Towers B’B and D´D are not in the same plane as the hexagon.

In Triangle B’AB,

Tan∠B´AB = B´AB

⇒ B’B =

In Triangle D´AD, tan ∠D´AD

⇒ D’D = 2a

Ratio of heights =

Answer choice (B)

The question is "What is the ratio of heights?"

Hence, the answer is 1:2√3

Choice B is the correct answer.

**Question 4: Find the maximum and minimum value of 8 cos A + 15 sin A + 15****A. 11√2+15****B. 30; 8****C. 32; -2****D. 23; 8****Answer: 1:2√3**

**Explanation-**

Always look out for Pythagorean triplets, we know that (8,15,17) is one

∴ The expression becomes:

Let there be a angle B for which sin B = 8/17, cos B = 15/17

⇒ 17( sin B cos A + cos B sin A) + 15

17(sin(A+B)) + 15

We know that sin(A+B)_{max }= 1

sin(A+B)_{min} = -1

∴ Max value = 17 * 1 +15 = 32

Min value = 17 * -1 + 15 = -2

The question is **"to find the maximum and minimum value "**

Choice B is the correct answer.

**Question 5: : If cos A + cos ^{2} A = 1 and a sin^{12} A + b sin^{10} A + c sin^{8} A + d sin^{6} A - 1 = 0. Find the value of a+b/c+d**

**A. 4****B. 3****C. 6****D.1****Answer. 3****Explanation.**

Given,

Cos A = 1 - Cos^{2}A

⇒ Cos A = Sin^{2} A

⇒ Cos^{2}A = Sin^{4}A

⇒ 1 – Sin^{2} A = Sin^{4} A

⇒ 1 = Sin^{4} A + Sin^{2} A

⇒ 1^{3} = (Sin^{4}A + Sin^{2}A)^{3}

⇒ 1 = Sin^{12} A + Sin^{6}A + 3Sin^{8} A + 3Sin^{10} A

⇒ Sin^{12} A + Sin^{6}A + 3Sin^{8} A + 3Sin^{10} A – 1 = 0

On comparing,

a = 1, b = 3 , c = 3 , d = 1

Always look out for Pythagorean triplets, we know that (8,15,17) is one

The question is **"Find the value of a+b/c+d"**

Choice B is the correct answer.

**Question 6: In the below figure, the sheet of width W is folded along PQ such that R overlaps S Length of PQ can be written as :- **

**A. ****B. ****C. ****D. A****ny two of the above**** ****Answer: A ny two of the above**

**Explanation-**

If you are quick at observing , this question can be solved just by looking at the options as ,

∴ (d) Answer

Actually solving the question.

For R and S to overlap,

∆ PQR is congurent to ∆ PSQ

∴ ∠ QPR = ∝ ; ∠ SQP = ∠ PQR = 90 - ∝

⇒ ∠ RQT = 2∝ (180°-(180°-2∝))

We can draw the figure as :-

∴In ∆ QRT

QT = QR cos2∝

In ∆ PSQ,

SQ = PQ sin∝ = QR

∴ w = QT + SQ = PQ sin∝cos2∝ + PQ sin∝

= PQ sin∝ (1 + cos 2∝)

**(as shown earlier) **

The question is **"Length of PQ can be written as "**

Choice D is the correct answer.

**Question 7: Ram and Shyam are 10 km apart. They both see a hot air balloon passing in the sky making an angle of 60° and 30° respectively. What is the height at which the balloon could be flying?****A. ****B. 5√3****C. Both A and B****D. Can’t be determined****Answer: Both A and B****Explanation-**

tan 30° = h/x

⇒ h = (10 – x) √3

⇒ 30 – 3x = x ⇒ x = 30/4 = 15/2

⇒

Case 2 :

Also,

tan 30° =

⇒

⇒ x = 5

H = 5√3

The question is **"What is the height at which the balloon could be flying"**

Choice C is the correct answer.

**Question 8: A man standing on top of a tower sees a car coming towards the tower. If it takes 20 minutes for the angle of depression to change from 30° to 60°, what is the time remaining for the car to reach the tower?****A. 20√3 minutes****B. 10 minutes****C. 10√3 minutes****D. 5 minutes****Answer: 10 minutes**

**Explanation- **

**From the figure:- **

**tan 60° = ** h/b

⇒ a + b = 3b

⇒ 2b = a

⇒ b = a/2

Now since the car takes 20 minutes to travel a distance

Time taken to travel b = a/2 distance = 20/2 = 10 minutes

The question is **"What is the time remaining for the car to reach the tower?"**

Choice B is the correct answer.

**Question 9: A right angled triangle has a height ‘p’, base ‘b’ and hypotenuse ‘h’. Which of the following value can h2 not take, given that p and b are positive integers?****A. 74****B. 52****C. 13****D. 23**

**Answer: 23**

**Explanation-**

We know that,

h^{2} = p^{2} + b^{2} Given, p and b are positive integer, so h^{2} will be sum of two perfect squares.

We see

a) 7^{2} + 5^{2} = 74

b) 6^{2} + 4^{2} = 52

c) 3^{2} + 2^{2} = 13

d) Can’t be expressed as a sum of two perfect squares

The question is **"Which of the following value can h ^{2} not take, given that p and b are positive integers? "**

Choice D is the correct answer.

**Question 10: tan ∅ + sin ∅ = m, tan ∅ - sin ∅ = n, Find the value of m ^{2}- n^{2}**

**Answer: 4√mn**

**Explanation-**

Adding the two equations,

tan ∅ =

Subtracting the same,

sin ∅ =

Since, there are no available direct formula for relation between sin∅ ∅tan∅ but we know that

cosec^{2} ∅ – cos^{2} ∅ = 1

⇒

⇒((m^{2}-n^{2}))^{2} = 4(4mn)

⇒ m^{2} - n^{2} = 4√mn

The question is **"To find the value of m ^{2}= n^{2}"**

Choice B is the correct answer.

**Question 11: A student is standing with a banner at the top of a 100 m high college building. From a point on the ground, the angle of elevation of the top of the student is 60° and from the same point, the angle of elevation of the top of the tower is 45°. Find the height of the student.****A. 35 m****B. 73.2 m****C. 50 m****D. 75 m****Answer: 73.2 m****Explanation-**

Let BC be the height of the tower and DC be the height of the student.

In rt. ∆ ABC

AB = BC cot 45^{°}

AB = 100 m ……………(i)

In rt. ∆ ABD

AB = BD cot 60^{°}

AB = (BC + CD) cot 60^{°}**Equating (i) and (ii)**

(10 + CD)= 100√3

CD = 100√3 – 100

= 10(1.732 – 1) = 100 * 0.732 = 73.2 m

The question is **"To find the height of the student"**

Choice B is the correct answer.

**Question 12: If Cos x – Sin x = √2 Sin x, find the value of Cos x + Sin x:****A. √2 Cos x****B. √2 Cosec x****C. √2 Sec x****D. √2 Sin x Cos x**

**Answer: √2 Cos x**

**Explanation-**

Cos x – Sin x = √2 Sin x

⇒ Cos x = Sin x + √2 Sin x

⇒ Cos x = Sin x + √2 Sin x**⇒****⇒ ****⇒ ****⇒** Sin x = (√2 - 1) Cos x

The question is **"To find the value of Cos x + Sin x"**

Choice A is the correct answer.

**Question 13****: can be written as: ****A. 1/t****B. t****C. √t Sec x****D. ****Answer. t****Explanation.**

(Since, 1 = Cos^{2}x + Sin ^{2}x)

The question is **can be written as"**

Choice B is the correct answer.

**Question 14: A tall tree AB and a building CD are standing opposite to each other. A portion of the tree breaks off and falls on top of the building making an angle of 30°. After a while it falls again to the ground in front of the building, 4 m away from foot of the tree, making an angle of 45°. The height of the building is 6 m. Find the total height of the tree in meters before it broke.****A. 27√3 + 39****B. 12√3 + 10****C. 15√3 + 21****D. Insufficient Data****Answer: 15√3 + 21****Explanation-**

Let the broken portion of tree AA’ be x. Hence A’C = A’G = x

From the figure, total height of the tree = x + y + 6

Consider triangle A’BG, tan 45° =

Or BG = y + 6

Consider triangle A’C’C, tan 30° =

y√3 = y + 10

Therefore y = 5(√3 + 1)

Take sin 30° to find x

sin 30° =

1/2 = y/x

or x = 2y

Height of the tree = x + y = 6

= 2 * 5(√3 + 1) + 5(√3 + 1) + 6

= 10√3 + 10 + 5√3 + 5 + 6

= 15√3 + 21 meters

The question is **"To find the total height of the tree in meters before it broke"**

Choice C is the correct answer.

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