CDS Elementary Mathematics 3 Sep 2023 Official Paper (Previous Year Paper)

Formula used:

a3 + b3 + c3 - 3abc = (a + b +c)(a2 + b2 + c2 - ab - bc - ac)

Calculation:

a2 - bc = α  ] × a

a3 - abc = aα        ------(1)

b2 - ac = β  ] × b

b3 - abc = bβ        ------(2)

c2 - ab = γ  ] × c

c3 - abc = cγ        ------(3)

Adding equation (1),(2) and (3)

a3 + b3 + c3 - 3abc = aα + bβ + cγ  

a2 - bc + b2 - ac + c2 - ab = α +  β +  γ

⇒ $\rm \frac{a \alpha+b \beta+c \gamma}{(a+b+c)(\alpha+\beta+\gamma)}$ = $\frac{a^3 + b^3 +c^3 -3abc}{(a+b+c)(a^2 +b^2 +c^2 - ab - bc -ca)}$

∴ $\frac{a^3 + b^3 +c^3 -3abc}{a^3 + b^3 +c^3 -3abc}$ = 1

Calculation:

(x - 1)³ = x³ - 3x² + 3x - 1

∴ Other factor will be x + 1

Confusion PointsFocus only on the constant terms of the expression (α - 4)x³ + βx² + (γ - 2)x. Try to find a factor to divide it by, such that subtracting it from -1 makes the expression equal to zero

Shortcut Trick

⇒ 11(x + y) = 55

⇒ x + y = 5 -----(1)

⇒ 9(x - y) = 45

⇒ x - y = 5 -----(2)

From equation (1) and (2)

⇒ x × y = 5 × 0 = 0

∴ The correct answer is 0.

Solution:-

<br>

Calculation:

Let 1st digit = x & 2nd digit = y

2-digit number = 10x + y

Reversing the order of the digits of the number = 10y + x

According to question

The sum of the number and the number obtained by reversing the order of the digits of the number = 55

⇒ 10x + y + 10y + x = 55

⇒ 11x + 11y = 55

⇒ x + y = 5 ---------(1)

The difference of the given number and the number obtained by reversing the order of the digits of the number = 45

⇒ 10x + y - 10y - x = 45

⇒ 9x - 9y = 45

⇒ x - y = 5 -----------(2)

Adding Equation (1) & (2)

⇒ x + y + x - y = 10

⇒ 2x = 10 ⇒ x = 5

Put the value of x in equation (1)

⇒ 5 + y = 5 ⇒ y = 0

The product of the digits = 5 × 0 = 0

∴ The correct answer is 0.

Shortcut Trick

Let Total Work be LCM of (10, 12, 15) = 60 units.

Efficiency of 2(A + B + C) = (60/10) + (60/12) + (60/15) = 6 + 5 + 4 = 15 units/day.

Combined Efficiency of (A + B + C) = 15 ÷ 2 = 7.5 units/day.

Time for Half work (30 units) = 30 ÷ 7.5 = 4 days.

∴ The correct answer is 4 days.

Alternate Method

Given:

Time taken by (A + B) = 10 days

Time taken by (B + C) = 12 days

Time taken by (C + A) = 15 days

Formula Used: Efficiency = Total Work ÷ Time

⇒ Total Efficiency of [ (A + B) + (B + C) + (C + A) ] = 6 + 5 + 4

⇒ 2(A + B + C) = 15 units/day

⇒ Efficiency of (A + B + C) = 15 ÷ 2 = 7.5 units/day

⇒ Half of the total work = 60 ÷ 2 = 30 units

⇒ Time taken by (A + B + C) = 30 ÷ 7.5 = 4 days

∴ The correct answer is 4 days.

Additional Information

Combined Efficiency Property

When individuals work in pairs, the sum of their efficiencies represents twice the combined efficiency of all individuals: 2(A+B+C) = Σ(Pair Efficiencies).

Work and Time Relationship

Total Work = Efficiency × Time. If work is halved, for a constant efficiency, the time required will also be halved.

LCM Method

Taking the Least Common Multiple (LCM) of time durations as total work simplifies fractional calculations into integer efficiencies.

Shortcut Trick 

R = 12/4 = 3% and T = 9/3 = 3 year

<br>

Interest = (10000 / 1000000) × 92727 =  927.27

∴ The correct answer is  Rs. 927.27.

Alternate Method 

Given:

Amount = Rs. 10,000

Rate,R = 12% 

Time ,n  = 9 months

Concept used:

For quarterly, Rate = R/4 & Time = n × 4

Calculation:

R = 12%/4 = 3% & n = 9 × 4 = 36 months ⇒  3 years

The interest of 1^st year = 10000 × 3% = 300

The interest of 2^nd year = 300 + 300 × 3% = 309

The interest of 3^rd year = 309 + 309 × 3% = 318.27

Total interest after 3 years = 300 + 309 + 318.27 = Rs. 927.27 

∴ The correct answer is  Rs. 927.27

Given:

Sinθ = (a + b)/2√ab

Concept:

-1 ≤ sinθ ≤ +1

Calculation:

-1 ≤ (a + b)/2√ab ≤ +1

We solve one of them

(a + b)/2√ab ≤ +1

⇒ a + b ≤ 2√ab

⇒ a + b - 2√ab ≤ 0

⇒ (√a - √b)²  ≤ 0

⇒ √a = √b = a = b

∴ The correct answer is a = b

Shortcut Trick We can solve this question by option to satisfy equation

put a = b

then Sinθ = (a + a)/ 2√(a × a)

Sinθ = 2a/2a = 1 (sinθ ≤ 1)

Given:

A, B, and C together can do a piece of work in 36 days

A and B together can do five times as much work as C alone

B and C together can do as much work as A alone

Calculation:

Ratios of Efficiency 

(A + B) : C = 5 : 1

⇒ A + B = 5a, C = a --(1)

⇒ A + B + C = 6a

(B + C) : A = 1 : 1

⇒ (B + C) = a, A = a

⇒ A + B + C = 2a 

Efficiency should be equal so 

 A + B + C = 2a × 3 = 6a  now 

 (B + C) = 3a, A = 3a --(2)

By solving (1) and (2)

A = 3a , C = a and B = 3a - a = 2a

Efficiency of (A + C) = n(B)

(3a + a) = n × 2a

n = 4a/2a = 2

∴ The correct answer is n = 2

Given:

2a/3 = 4b/5 =  3c/4

Calculation:

LCM of numerator value (2, 4, 3) is 12

$(\frac{2 a}{3}=\frac{4 b}{5}=\frac{3 c}{4})$ × $\frac{1}{12}$

a/18 = b/15 = c/16

a : b : c = 18 : 15 : 16

a = 18u, b = 15u, c = 16u

Put the value in $\frac{18}{a} \sqrt{a^2+c^2-b^2}$

$\frac{18}{18 u} \sqrt{(18u)^2+(16u)^2-(15u)^2}$

⇒ √355

∴ The correct answer is √355.

Given:

Sum of deviations of n numbers from 10 and 20 are a, b respectively. 

​​$\frac{b}{a}$ = - 4

Formula used:

Mean = A + $\frac{∑ d}{n}$

Where A is assumed to mean, 

∑d is sum of deviation, n = total terms

Calculation:

Mean = 10 + a/n       ----(1)

Mean = 20 + b/n       ----(2)

The mean is equal in equation (1) & (2). 

⇒ 10 + a/n = 20 + b/n

⇒ a/n - b/n = 10

⇒ a/n - (-4a)/n = 10       (∵ b/a = -4)

⇒ 5a/n = 10

⇒ a/n = 2

Put value of a/n in mean of equation (1)

Mean = 10 + 2 = 12

∴ The correct answer is 12.

Concept:

Median (M): The median is the middle number in a set to find the median, arrange all of the numbers (n) in the data set in ascending order, and then find the middle number. this is called median.

Formula used:

if n is odd then M = {(n + 1)/2}^th term

if n is even then M = {(n/2)th term + (n/2 + 1)th term} ÷ 2

Calculation:

Arrange the numbers in ascending order

1, 8, 12, 17, 21, 28, 35, 45, 54, 61

Total n = 10 then

We use M = {(n/2)th term + (n/2 + 1)th term} ÷ 2

(21 + 28)/2 = 24.5

2M + 5 = 2 × 24.5 + 5 = 54

∴ The correct answer is 54.

Calculation:

√ (x + 9) = x - 3

⇒ x + 9 = (x - 3)2

⇒ x + 9 = x2 - 6x + 9

⇒ x2 - 7x = 0

⇒ x(x - 7) = 0

For x = 0, the given equation doesn't satisfy.

For x = 7, the given equation satisfies.

∴ No. of real roots is one.

Concept used:

(a + b)² = a² + 2ab + b²

Calculation:

x = 97 + 56√3     ...(i)

Comparing the equation (i) with the given concept,

⇒ x = (a + b)2

⇒ 97 + 56√3 = a2 + 2ab + b2

So, 2ab = 56√3 = 2 × 7 × 4√3

Now, x = (7 + 4√3)²

⇒ √x = 7 + 4√3     ...(ii)

Comparing the equation (ii) with the given concept,

⇒ √x = (a + b)2

⇒ 7 + 4√3 = a2 + 2ab + b2

Here, 2ab = 2 × 2 × √3

Now, √x = (2 + √3)2

⇒ $\sqrt [4] {x}$  = 2 + √ 3

Similarly, $1\over \sqrt [4] {x}$ = 2 - √3

Now,

⇒ $\sqrt [4] {x}$ + $1 \over \sqrt[4]{x}$ = 2 + √ 3 + 2 - √3 = 4

∴ The answer is 4.

Given:

L : H = 3 : 2

Sum of two numbers = 45

Concept used:

Numbers are multiples of HCF.

Product of numbers = LCM × HCF

HCF (Highest Common Factor) is the largest number that can divide the given set of two or more numbers.

Calculation:

Let, the common factor of the ratio of L and H be m**.**

Now, LCM = 3m and HCF = 2m

According to the concept of HCF,

The numbers considered are 2ma and 2mb.

According to the question,

⇒ 2ma × 2mb = 3m × 2m

⇒ ab = 3/2     ---(i)

According to the question,

⇒ 2ma + 2mb = 45

⇒ 2m(a + b) = 45     ---(ii)

Now, using equations (i) and (ii), we can't find the value of all three variables a, b, and m.

∴ The answer cannot be determined due to insufficient data.

Given:

Distance between Home to Office = D

With S₁ = 3 km/hr, reaches 40 minutes early

With S₂ = 2 km/hr, reaches 40 minutes late

Formula Used:

Distance = Speed × Time

Calculation:

Let, original time to reach the office = T minutes

According to the question,

⇒ 3 × (T - 40) = 2 × (T + 40)

⇒ 3T - 120 = 2T + 80

⇒ 3T - 2T = 80 + 120

⇒ T = 200

Now, D = 3 × (T - 40)/60

⇒ D = 3 × (200 - 40)/60

⇒ D = 3 × 160/60

⇒ D = 3 × 8/3 = 8 km

∴ The distance is 8 km. 

Shortcut Trick 

According to the question,

⇒ 1 unit = 80 min

⇒ 2 unit = 80 × 2 = 160 min. = 8/3 hour

⇒ Distance = 3 × (8 / 3) = 8 km

∴ The distance is 8 km.

Calculation:

3x - 1 + 33 - x = 6

⇒ 3x - 1 + 33 - x = 3¹ + 3¹

Comparing the power of 3 in both side of the equation,

x - 1 = 1

⇒ x = 2

Similarly, 3 - x = 1

⇒ x = 2

So, x = 2

Now, 

⇒ 2x - 1 + 23 - x

⇒ 22 - 1 + 23 - 2 = 21 + 21 = 2 + 2 = 4

∴ The answer is 4. 

Shortcut Trick We know that 3 + 3 = 6

so put x = 2

3² - 1 + 33 - 2 = 6

than 

22 - 1 + 23 - 2 = 4

Given:

$x\left(a-b+\frac{a b}{a-b}\right)$ = $y\left(a+b-\frac{a b}{a+b}\right)$ and x + y = 2a³

Formula used:

a³ + b³ = (a + b) (a² + b² - ab)

(a + b)² = a² + b² + 2ab

Calculation:

We have

$\left(a-b+\frac{a b}{a-b}\right)$ = $\frac{(a\ -\ b )^2 \ +\ ab}{a\ -\ b}$ 

⇒ $\left(a-b+\frac{a b}{a-b}\right)$ = $\frac{a^2\ +\ b^2\ -\ ab }{a\ -\ b}$ 

Put the value of (a2 + b2 - ab) = (a3 + b3)/(a + b)

⇒ $\frac{a^2\ +\ b^2\ -\ ab }{a\ -\ b}$ = $\frac{a^3\ +\ b^3}{(a\ +\ b)(a\ - \ b)}$

Similarly,

$\left(a+b-\frac{a b}{a+b}\right)$ = $\frac{a^3\ -\ b^3}{(a\ +\ b)(a\ - \ b)}$

⇒ $\left(a+b-\frac{a b}{a+b}\right)$ = $\frac{a^3\ -\ b^3}{(a\ +\ b)(a\ - \ b)}$

Now,

$\frac{x}{y}\ =\ \frac{a^3\ -\ b^3}{(a\ +\ b)(a\ - \ b)}\ \times \frac{(a\ +\ b)(a\ - \ b)}{a^3\ +\ b^3}$

$⇒ \frac{x}{y} = \frac{a^3\ -\ b^3}{a^3\ +\ b^3}$

Here, x = a3 - b³ and y = a3 + b3 

⇒ x - y = (a3 - b³) - (a3 + b³) = -2b³

Hence the value of x - y is -2b³.

Given:

Expression 3√3x3 + 2√2y3 - 18xy + 6√6

Formula used:

a³ + b³ + c³ - 3abc = (a + b + c) (a² + b² + c² - ab - bc - ca)

Calculation:

The Expression 3√3x3 + 2√2y3 - 18xy + 6√6 can be written as

⇒ (√3x)³ + (√2y)³ + (√6)³ - 3 × √3x × √2y × √6

Here a = √3x, b = √2y and c = √6

Using a3 + b3 + c3 - 3abc = (a + b + c) (a2 + b2 + c2 - ab - bc - ca)

The factors can be written as,

⇒ (√3x + √2y + √6) (3x² + 2y² + 6 - √6xy - √12y - √18x)

Hence the required factor is 3x2 + 2y2 + 6 - √6xy - √12y - √18x.

Given:

The number = 125¹⁰⁰

log₁₀2 = 0.301

Formula Used:

log mⁿ = n log m

log (m/n) = log m - log n

Calculation:

Let x = 125¹⁰⁰

Taking log on both sides

⇒ log₁₀x = 100 log₁₀125

⇒ log10x = 100 log₁₀53

⇒ log₁₀x = 100 × 3 log₁₀5

⇒ log10x = 300log₁₀(10/2)

⇒ log₁₀x = 300 × (log₁₀10 - log₁₀2)

⇒ log10x = 300 × (1 - 0.301)

⇒ log₁₀x = 209.69

The number of digits = 209 + 1 = 210

Hence, the number of digits in the expansion of  125100 is 210.

Calculation:

Expression 1:

acx3 + bcx2 +  adx2 + acdx + bdx + bcd

(acx3 + bcx2) + (adx2 + bdx) + (acdx + bcd)

cx²(ax + b) + dx (ax + b) + cd(ax + b)

(ax + b)(cx² + dx + cd)

Expression 2:

dx3 + acx2 + bdx2 + bcx + acdx + bcd

(adx3 + bdx2) + (acx2 + bcx) + (acdx + bcd)

dx²(ax + b) + cx(ax + b) + cd(ax + b) 

(ax + b)(dx² + cx + cd)

We know that, HCF is the highest power of each common prime factor.

∴ HCF = ax + b

Given:

xn - pyn + qzn is divisible by x2 + abyz - bzx - axy

Calculation:

We have 

x2 + abyz - bzx - axy

This can be written as

⇒ x2 - bzx - axy + abyz

⇒ x(x - bz) - ay(x - bz)

⇒ x² + (bz - ay)x + abyz

⇒ (x - bz) (x - ay)

Therefore, bz & ay will be factor of (x2 + abyz - bzx - axy)

As, x2 + abyz - bzx - axy is factor of xn - pyn + qzn

We can write 

xn - pyn + qzⁿ = (x - bz) (x - ay) × k    -----(1)

Put x = bz

⇒ bⁿzⁿ - pyⁿ + qzⁿ = 0     -----(2)

Put x = ay

⇒ aⁿyⁿ - pyⁿ + qzⁿ = 0     -----(3)

Equation (3) - Equation (1) 

⇒ bⁿzⁿ - aⁿyⁿ = 0

⇒ bnzn = anyn

From equation (3)

⇒ bnzn - pyn + qzn = 0          (∵ bnzn = anyn)

⇒ 1 - (pyn/bnzn) + q/bn = 0

⇒ 1 - (pyⁿ/aⁿyⁿ) + q/bⁿ = 0    (∵ bnzn = anyn)

⇒ $\rm \frac{p}{a^n}-\frac{q}{b^n}$ = 1

Hence, the value of $\rm \frac{p}{a^n}-\frac{q}{b^n}$ = 1.

Given:

Statement 1: (a + b) ∝ (a - b)

Statement 2: a ∝ b

Calculation:

Statement: 1 (a + b) ∝ (a - b)

⇒ (a + b) = k (a - b), (where k is the proportionality constant)

On squaring both sides, we get 

(a + b)² = k² (a - b)²

⇒ $\frac{(a\ +\ b)^2}{(a\ -\ b)^2}$ = k²

⇒ $\frac{(a^2\ +\ b^2\ +\ 2ab)}{a^2\ +\ b^2\ -\ 2ab)}$ = k2

By using componendo & dividendo method, we get

⇒ $\frac{(a^2\ +\ b^2\ +\ 2ab \ +\ a^2\ +\ b^2\ -\ 2ab)}{(a^2\ +\ b^2\ +\ 2ab \ -\ a^2\ -\ b^2\ +\ 2ab)}$ = $\frac{k^2\ +\ 1}{k^2\ -\ 1}$

⇒ $\frac{2\ (a^2 \ + \ b^2)}{4ab}$ = $\frac{k^2\ +\ 1}{k^2\ -\ 1}$

⇒ (a² + b²) = $\frac{k^2\ +\ 1}{k^2\ -\ 1}$ (2ab) 

Thus, (a2 + b2) ∝ ab, Where $\frac{k^2\ +\ 1}{k^2\ -\ 1}$ & 2 is constant.

Hence, the statement 1 is correct.

Statement: 2 a ∝ b

⇒ a = kb, (where k is the proportionality constant)

⇒ a/b = k

By using componendo & dividendo method, we get

⇒ (a + b)/(a - b) = (k + 1)/(k - 1)

On squaring both sides,

⇒ (a + b)²/(a - b)² = (k + 1)²/(k - 1)²

⇒ (a² + b² + 2ab)/(a² + b² - 2ab) = (k + 1)2/(k - 1)2

Again using componendo & dividendo method, we get

⇒ 2 (a² + b²)/4ab = [(k + 1)2 + (k - 1)2]/[(k + 1)2 - (k - 1)2]

⇒ (a2 + b2) = ab [(k + 1)2 + (k - 1)2]/[(k + 1)2 - (k - 1)2]

Here, [(k + 1)2 + (k - 1)2]/[(k + 1)2 - (k - 1)2] is constant.

⇒ (a2 + b2) = ab

Therefore, (a2 - b2) ≠ ab 

Hence, the statement 2 is incorrect.

∴ Only 1st statement is correct.

Concept Used:

If $\frac{a+b}{a-b}$ = $\frac{c+d}{c-d}$ then $\frac{a}{b}$ = $\frac{c}{d}$

Calculation:

(3a + 6b + c + 2d) × (3a - 6b - c + 2d) = (3a - 6b + c - 2d) × (3a + 6b - c - 2d)

⇒ (3a + 6b + c + 2d)/(3a - 6b + c - 2d) = (3a + 6b - c - 2d)/(3a - 6b - c + 2d)

⇒ (3a + c) + (6b + 2d)/(3a + c) - (6b + 2d) = (3a - c) + (6b - 2d)/(3a - c) - (6b - 2d)

⇒ (3a + c)/(6b + 2d) = (3a - c)/(6b - 2d)

⇒ (3a + c)/(3a - c) = (6b + 2d)/(6b - 2d)

⇒ 3a/c = 6b/2d

⇒ a/c = b/d

⇒ ad = bc

∴ The correct option is 3 which is ad = bc.

Shortcut Trick 3 sin θ + 5 cos θ = 5

If in the given equation we put θ = 0 then LHS = RHS, 

Therefore,

⇒ 5 sin θ - 3 cos θ

⇒ 5 sin 0 - 3 cos 0    ---(Putting θ = 0)

⇒ 0 - 3 × 1 = - 3

∴ 5 sin θ - 3 cos θ is equal to - 3.

Alternate Method

Formula Used:

(a + b)² = a² + b² + 2ab 

sin2 θ = (1 - cos2 θ)

cos2 θ = (1 - sin2 θ)

Calculation:

3 sin θ + 5 cos θ = 5

Squaring on both sides, 

⇒ (3 sin θ + 5 cos θ)² = 5²

⇒ 9 sin² θ + 25 cos² θ + 30 sin θ . cos θ = 25

⇒ 9 (1 - cos2 θ) + 25 (1 - sin2 θ) + 30 sin θ . cos θ = 25

⇒ 9 - 9 cos2 θ + 25 - 25 sin2 θ + 30 sin θ . cos θ - 25 = 0

⇒ - 9 cos2 θ - 25 sin2 θ + 30 sin θ . cos θ + 9 = 0

⇒ 25 sin2 θ + 9 cos 2 θ - 2 × (5 sin θ × 3 cos θ) = 9

⇒ (5 sin θ - 3 cos θ)² = ± 3

∴ 5 sin θ - 3 cos θ is equal to - 3.

Given:

Condition 1: The combined age of a man & his wife is 6 times the combined age of their children.

Condition 2: Two years ago their combined age was 10 times the combined age of their children.

Condition 3: Six years later their combined age will be 3 times the combined age of their children.

Calculation:

Let the combined age of man & wife be P, and

The combined age of children is C.

According to the 1st Condition:

⇒ X = 6Y     -----(1)

According to the 2nd Condition:

2 years ago, 

⇒ P = X - 2 - 2 = X - 4 

⇒ C = (Y - 2N)   ---(N = No. of children)

⇒ X - 4 = 10 (Y - 2N)

⇒ X - 4 = 10Y - 20N

⇒ X - 10Y + 20N = 4     -----(2)

According to the 3rd Condition:

⇒ X + 12 = 3 (Y + 6N)

⇒ X + 12 = 3Y + 18N

⇒ X - 3Y - 18N = -12     -----(3)

Substituting equation 1 in 2 & 3, 

⇒ - 4Y + 20N = 4

⇒ 5N - Y = 1     -----(4)

⇒ 3Y -18N = - 12

⇒ Y - 6N = - 4     -----(5)

Adding equation 4 & 5,

⇒ - N = - 3

⇒ N = 3

∴ The number of children is 3.

Shortcut Trick

 Put x = 0, 

3 (sin x - cos x)4 + 6 (sin x + cos x)2 + 4 (sin x)6 + 4 (cos x)6

⇒ 3 (sin 0 - cos 0)⁴ + 6 (sin 0 + cos 0)2 + 4 (sin 0)6 + 4 (cos 0)6

⇒ 3 (-1)4 + 6 (1)2 + 4 (0)6 + 4 (1)6

⇒ 3 + 6 + 4 = 13

∴ 3 (sin x - cos x)4 + 6 (sin x + cos x)2 + 4 (sin x)6 + 4 (cos x)6 equal to 13.

Alternate MethodConcept Used:

(a + b)² = a² + b² + 2ab

(a - b)2 = a2 + b2 - 2ab

sin² θ + cos² θ = 1 

Calculation:

⇒ 3 [sin x - cos x]4 + 6 [sin x + cos x]2 + 4 [sin6 x + cos6 x]

⇒ 3 {(sin x - cos x)2}2 + 6 (sin2 x + cos2 x + 2 sin x cos x) + 4 [(sin2 x)3 + (cos2 x)3]

⇒ 3 {(sin2 x + cos2 x - 2 sin x cos x)2} + 6 (1 + 2 sin x cos x) + 4 [(sin2 x + cos2 x)(sin4 x + cos4 x - sin2 x cos2 x)]

⇒ 3 (1 - 2 sin x cos x)2 + 6 (1 + 2sin x cos x) + 4 [(sin2 x)2 + (cos2 x)2 + 2 sin2 x cos2 x - 3 sin2 x cos2 x]

⇒ 3 (1 + 4 sin2 x cos2 x - 4sin x cos x) + 6 (1 + 2sin x cos x) + 4 [(sin2 x + cos2 x)2 - 3 sin2 x cos2 x]

⇒ 3 + 12 sin2 x cos2 x - 12 sin x cos x + 6 + 12 sin x cos x + 4 - 12 sin2 x cos2 x

⇒ 13

∴ 3 (sin x - cos x)4 + 6 (sin x + cos x)2 + 4 (sin x)6 + 4 (cos x)6 equal to 13.

Shortcut Trick Here, 0 < θ < 45^∘

⇒ So possible value of θ = 30^∘

Putting θ = 30^∘ in the equation we get,

⇒ cot² 30^∘ - (√3 + 1) cot 30^∘ + √3

⇒ 3 - 3 - √3 + √3 = 0

⇒ So, sin 30^∘ + cos 30^∘ = $\frac{\sqrt3 + 1}{2}$

∴ The correct answer is $\frac{\sqrt3 + 1}{2}$.

Alternate Method

Given:

cot2 θ - ($√{3}$ + 1) cot θ + $√{3}$ = 0

$0<θ<\frac{\pi}{4}$

Calculation:

Since θ lies between 0⁰ and 45⁰, putting θ = 30⁰

⇒ cot2 θ - ($√{3}$ + 1) cot θ + $√{3}$ = cot² 30⁰ - ($√{3}$ + 1) cot 30⁰ + $√{3}$

⇒ cot2 θ - ($√{3}$ + 1) cot θ + $√{3}$ = (3) - ($√{3}$ + 1)$√{3}$ + $√{3}$

⇒ cot2 θ - ($√{3}$ + 1) cot θ + $√{3}$ = 3 - (3 + $√{3}$) + $√{3}$

⇒ cot2 θ - ($√{3}$ + 1) cot θ + $√{3}$ = 3 - 3 - $√{3}$ + $√{3}$ = 0

Thus, the given condition satisfies

⇒ sin θ + cos θ = sin 30⁰ + cos 30⁰

⇒ sin θ + cos θ = 1/2 + $√{3}$/2 = ($√{3}$ + 1)/2

∴ The correct answer is $\frac{√3\ +\ 1}{2}$.

Given:

tan 2θ tan 4θ - 1 = 0, $0<\theta<\frac{\pi}{2}$

Concept Used:

If tan A + tan B = 1

Then A + B = 90⁰

Calculation:

According to the question

⇒ tan 2θ tan 4θ - 1 = 0

⇒ tan 2θ tan 4θ = 1

⇒ 2θ + 4θ = 90⁰

⇒ θ = 15⁰ = $\frac{\pi }{12}$

∴ The value of θ is $\frac{\pi }{12}$ which satisfies the equation tan 2θ tan 4θ - 1 = 0.

Alternate Method We have to check for which value of θ, $0<θ<\frac{π}{2}$

The equation tan 2θ tan 4θ - 1 = 0 satisfies

Option (1): θ = $\frac{\pi }{12}$ = 15⁰

⇒ tan 2θ tan 4θ - 1 = tan 2(15⁰) tan 4(15⁰) - 1

⇒ tan 2θ tan 4θ - 1 = tan 30⁰ tan 60⁰ - 1

⇒ tan 2θ tan 4θ - 1 = ($\frac{1}{√3}$× √3) - 1 = 0

∴ The value of θ is $\frac{\pi }{12}$ which satisfies the equation tan 2θ tan 4θ - 1 = 0.

Given:

tan x = $\frac{\sin θ+\cos θ}{\sin θ-\cos θ}$, $\frac{\pi}{4}<θ<\frac{\pi}{2}$

Formula Used:

1. tan θ = $\frac{P}{B}$

2. sin θ = $\frac{P}{H}$ 

3. sin²θ + cos²θ = 1

Calculation:

By Pythagoras Theorem

⇒ H = $\sqrt{(sinθ+cosθ)^2+(sinθ-cosθ)^2}$

⇒ H = $\sqrt{sin^2θ+cos^2θ+2sinθ cosθ+sin^2θ+cos^2θ-2sinθ cosθ}$

⇒ H = $\sqrt{2(sin^2θ+cos^2θ)}$ = $\sqrt{2}$

According to the figure

⇒ $\sqrt{2}$ sin x = $\sqrt{2}$ × $\frac{sinθ+\cosθ}{\sqrt{2}}$ = sinθ + cosθ 

∴ The value of $\sqrt{2}$ sin x is equal to sinθ + cosθ.

Given:

(sin2 θ - 4 sin θ + 3) (4 - cos2 θ + 4 sin θ) = 0, 0 < θ < $\frac{\pi}{2}$

Calculation:

According to the question

(sin2 θ - 4 sin θ + 3) (4 - cos2 θ + 4 sin θ) = 0

⇒ (sin2 θ - 4 sin θ + 3) = 0 OR (4 - cos2 θ + 4 sin θ) = 0

Solving the first equation:

(sin2 θ - 4 sin θ + 3) = 0

⇒ sin2 θ - sin θ - 3 sin θ + 3 = 0

⇒ sin θ (sin θ - 1) - 3 (sin θ - 1) = 0

⇒ (sin θ - 1) (sin θ - 3) = 0

⇒ sin θ = 1 OR sin θ = 3

⇒ θ = 90⁰ = $\frac{\pi}{2}$ (∵ -1 < sin θ < 1, neglecting sin θ = 3)

Solving the another equation:

(4 - cos2 θ + 4 sin θ) = 0

⇒ 4 - (1 - sin2 θ) + 4 sin θ = 0

⇒ sin2 θ + 4 sin θ + 3 = 0

⇒ sin2 θ + sin θ + 3 sin θ + 3 = 0

⇒ sin θ (sin θ + 1) + 3 (sin θ + 1) = 0 

⇒ (sin θ + 1) (sin θ + 3) = 0

⇒ sin θ = -1 OR sin θ = -3

⇒ θ = 270⁰ = $\frac{3\pi}{2}$ (∵ -1 < sin θ < 1, neglecting sin θ = -3)

Again, according to the question 0 < θ < $\frac{\pi}{2}$

∴ The given equation does not satisfy for any value of θ, 0 < θ < $\frac{\pi}{2}$.

Given:

For every $θ \in\left(0, \frac{\pi}{2}\right)$

x sin3 θ + y cos3 θ = sin θ cos θ and x sin θ - y cos θ = 0

Formula Used:

1. (a + b)² = a² + b² + 2ab

2. (a - b)² = a² + b² - 2ab

Calculation:

Since every $θ \in\left(0, \frac{\pi}{2}\right)$, putting θ = 45⁰

According to the question

⇒ x sin3 θ + y cos3 θ = sin θ cos θ

⇒ x $(\frac{1}{√2})^3$ + y $(\frac{1}{√2})^3$ = $\frac{1}{√2}$ × $\frac{1}{√2}$

⇒ $\frac{x}{2√2}$ + $\frac{y}{2√2}$ = $\frac{1}{2}$

⇒ x + y = √2     -----(1)

Again, according to the question

⇒ x sin θ - y cos θ = 0

⇒ x $(\frac{1}{√2})$ - y $(\frac{1}{√2})$ = 0

⇒ x - y = 0        -----(2)

Squaring Equation(1) and (2) and adding

⇒ x² + y² + 2xy + x² + y² - 2xy = (√2)² + 0

⇒ 2(x² + y²) = 2

⇒ x² + y² = 1

∴ The value of x² + y² is 1.

Given:

The angle of elevation of the top of the flagstaff with the base of the tower is found to be α.

The angle of elevation of the top of the flagstaff is observed to be β.

And that of the top of the tower is observed to be γ.

Let H be the height of the top of the flagstaff from the base of the tower and h be the height of the tower.

Concept Used:

tan θ = P/B, Where

P = Perpendicular

B = Base

Calculation:

From the above figure, We know that

tan β = H/x and tan γ = h/x

⇒ tan β = H/x

⇒ x = H/tan β     -----(1)

⇒ tan γ = h/x

⇒ x = h/tan γ      -----(2)

From equation (1) & (2), We know that

⇒ H/tan β = h/tan γ  

⇒ H tan γ = h tan β

⇒ H tan γ - h tan β = 0

∴ The correct option is 1.

Given:

The angle of elevation of the top of the flagstaff with the base of the tower is found to be α.

The angle of elevation of the top of the flagstaff is observed to be β.

And that of the top of the tower is observed to be γ.

Let H be the height of the top of the flagstaff from the base of the tower and h be the height of the tower.

Concept Used:

cot θ = B/P, Where

P = Perpendicular

B = Base

Calculation:

From the above figure, We know that

cot α = (x + d)/H and cot β = x/H

⇒ cot α = (x + d)/H

⇒ x = H cot α - d     -----(1)

⇒ cot β = x/H

⇒ x = H cot β          -----(2)

From equation (1) & (2), We know that

H cot β = H cot α - d 

⇒ d = H cot α - H cot β

⇒ d = H (cot α - cot β)

∴ The correct option is 3.

Given: 

AB is a straight road leading to the foot P of a tower of height h.

Q is at distance x from P and R is at a distance y from Q.

The angle of elevation of the top of the tower at Q is twice of that at R.

Concept Used:

Use the formula: tan $2 θ=\frac{2 \tan θ}{1-\tan ^2 θ}$

tan θ = P/B, Where

P = Perpendicular

B = Base

Calculation:

From the above figure, We know that

tan θ = h/(x + y) and tan 2θ = h/x

⇒ tan 2θ = h/x

⇒ $\frac{2 \tan θ}{1-\tan ^2 θ}$ = h/x

⇒ 2 tan θ = (1 - tan2 θ) (h/x)

⇒ 2h/(x + y) = [1 - (h2/(x + y)2)] (h/x)

⇒ 2h/(x + y) = [(x + y)2 - h2]/(x + y)2 (h/x)

⇒ (2x) (x + y) = x2 + y2 + 2xy - h2 

⇒ 2x2 + 2xy = x2 + y2 + 2xy - h2 

⇒ x2 =  y2 - h2 

⇒ h2 = y2 - x2 

To find the height we need y greater than x.

Here, x < y

∴ The correct option is 2.

Given:

AB is a straight road leading to the foot P of a tower of height h.

Q is at distance x from P and R is at a distance y from Q.

The angle of elevation of the top of the tower at Q is twice of that at R.

Concept Used:

Use the formula: tan $2 θ=\frac{2 \tan θ}{1-\tan ^2 θ}$

tan θ = P/B, Where

P = Perpendicular

B = Base

Calculation:

From the above figure, We know that

tan θ = h/(x + y) and tan 2θ = h/x

⇒ tan 2θ = h/x

⇒ $\frac{2 \tan θ}{1-\tan ^2 θ}$ = h/x

⇒ 2 tan θ = (1 - tan² θ) (h/x)

⇒ 2h/(x + y) = [1 - (h²/(x + y)²)] (h/x)

⇒ 2h/(x + y) = [(x + y)2 - h2]/(x + y)2 (h/x)

⇒ (2x) (x + y) = x² + y² + 2xy - h² 

⇒ 2x² + 2xy = x2 + y2 + 2xy - h2 

⇒ x2 =  y2 - h2 

⇒ h2 = y2 - x2 

∴ The correct option is 4.

Shortcut Trick 

OD = OB = r

X² + 40² = 30² + (X + 10)²

Clearly X = 30

then OD² = 30²+40² 

OD = 50

Diameter = 2r = 100

Alternate Method 

Given:

AB = 60 cm and CD = 80 cm

The distance between AB and CD is 10 cm

Formula Used:

In right angles triangle

Hypotenuse2 = Base2 + Perpendicular2

Calculation:

Let OP be x cm

⇒ OQ = (x - 10) cm

OD = OB = Radius      -----(1)

In ΔOQD, by Pythagoras Theorem

OD2 = OQ2 + QD2       -----(2)

In ΔOPB, by Pythagoras Theorem

OB2 = OP2 + PB2        -----(3)

From (1), (2) and (3)

OQ2 + QD2 = OP2 + PB2 

⇒ (x - 10)2 + 402 = x2 + 302

⇒ x2 + 100 - 20x + 1600 = x2 + 900

⇒ 20x = 800

⇒ x = 40 cm

In ΔOPB, by Pythagoras Theorem

OB² = 40² + 30²

⇒ OB2 = 1600 + 900 = 2500

⇒ OB = 50 cm

Diameter = 2 × Radius = 2 × OB = 100 cm

∴ The diameter of the circle is 100 cm.

Given:

AB = 60 cm and CD = 80 cm

The distance between AB and CD is 10 cm

∠AOB = α and ∠COD = β

Formula Used:

In right angles triangle

1. Hypotenuse² = Base² + Perpendicular²

2. tan θ = $\frac{Perpendicular(Opposite\ Side)}{Base(Adjacent\ Side)}$

Calculation:

Let OP be x cm

⇒ OQ = (x - 10) cm

OD = OB = Radius      -----(1)

In ΔOQD, by Pythagoras Theorem

OD² = OQ² + QD²       -----(2)

In ΔOPB, by Pythagoras Theorem

OB² = OP² + PB²        -----(3)

From (1), (2) and (3)

OQ2 + QD2 = OP2 + PB2 

⇒ (x - 10)² + 40² = x² + 30²

⇒ x² + 100 - 20x + 1600 = x² + 900

⇒ 20x = 800

⇒ x = 40 cm

⇒ OQ = x - 10 = 30 cm

According to the question

∠AOB = α

⇒ In ΔPOB, ∠POB = α/2

⇒ tan (α/2) = PB/OP = 30/40

∠COD = β

⇒ In ΔQOD, ∠QOD = β/2

⇒ tan (β/2) = QD/OQ = 40/30

tan (β/2) - tan (α/2) = (40/30) - (30/40)

⇒ tan (β/2) - tan (α/2) = (1600 - 900)/1200 = 7/12

∴ The value of tan (β/2) - tan (α/2) is 7/12.

Calculation:

Let x = 1, y = 1, and z = 2

p = x4 - y2z2

⇒ p = 1⁴ - (1² × 2²) = 1 - 4 = (-3)

q = y4 - z2x2

⇒ q = 14 - (22 × 12) = 1 - 4 = (-3)

r = z4 - x2y2

⇒ r = 24 - (12 × 12) = 16 - 1 = 15

Now, px2 + qy2 + rz2 = (-3)(12) + (-3)(12) + 15(22)

⇒ px2 + qy2 + rz2 = (-3) + (-3) + 60 = 54

Option 01: (x2 + y2 + z2) (p + q + r)

⇒ (x2 + y2 + z2) (p + q + r) = (12 + 12 + 22) [(-3) + (-3) + 15]

⇒ (x2 + y2 + z2) (p + q + r) = 6 × 9 = 54

Thus, Option 01 satisfies the condition.

Thus, Option 01 is the correct answer.

Option 02: -(x2 + y2 + z2) (p + q + r)

⇒ -(x2 + y2 + z2) (p + q + r) = -(12 + 12 + 22) [(-3) + (-3) + 15]

⇒ -(x2 + y2 + z2) (p + q + r) = (-6) × 9 = (-54)

Thus, Option 02 does not satisfy the condition.

Option 03: (y2 + z2 - x2) (r - q - p)

⇒ (y2 + z2 - x2) (r - q - p) = (12 + 22 - 12) [15 - (-3) - (-3)]

⇒ (y2 + z2 - x2) (r - q - p) = 4 × 15 = 60

Thus, Option 03 does not satisfy the condition.

Option 04: (x2 + y2 - z2) (p - q - r)

⇒ (x2 + y2 - z2) (p - q - r) = (12 + 12 - 22) [(-3) - (-3) - 15]

⇒ (x2 + y2 - z2) (p - q - r) = (-2) × (-15) = 30

Thus, Option 04 does not satisfy the condition.

∴ The equation px2 + qy2 + rz2 is equal to (x2 + y2 + z2) (p + q + r).

Calculation:

Let x = 1, y = 1, and z = 2

p = x4 - y2z2

⇒ p = 14 - (12 × 22) = 1 - 4 = (-3)

q = y4 - z2x2

⇒ q = 14 - (22 × 12) = 1 - 4 = (-3)

r = z4 - x2y2

⇒ r = 24 - (12 × 12) = 16 - 1 = 15

Now,

x2 (px2 + qy2 + rz2) + qr - p2

⇒ 12 [(-3)12 + (-3)12 + 15(22)] + (-3)(15) - (-3)²

⇒ [(-3) + (-3) + 60] - 45 - 9

⇒ 54 - 54

⇒ 0

∴ The value of x2 (px2 + qy2 + rz2) + qr - p2 is 0.

Given:

A right circular cone covers two spheres.

Radius of small sphere = r

Radius of bigger sphere = R

Concept used:

The Angle-Angle (AA) Similarity Criterion for triangle similarity states that if two angles of one triangle are respectively equal to two angles of another, the triangles are similar. It proves that proves that the third angle of both triangles will also be equal by leveraging the angle sum property of a triangle.

The Side-Side-Side (SSS) Criterion for triangle similarity states that if the sides of one triangle are proportional to the sides of another (i.e., they maintain the same ratio), their corresponding angles will be equal, and the triangles will be similar.

Calculation:

Let, the height of the cone, AP = h; and AD = a

In ∆ ADE and ∆ AFG,

(i) ∠ ADE = ∠ AFG = 90º

(ii) ∠ DAE = ∠ FAG

So, according to the AA Similarity concept,

 ∆ ADE ≅  ∆ AFG

Now, according to the SSS similarity criterion,

⇒ DE/FG = AD/AF

⇒ r/R = a/(a + r + R)

⇒ R/r = (a + r +R)/a

⇒ R/r = a/a + (r + R)/a

⇒ R/r = 1 + (r + R)/a

⇒ (R - r)/r = (r + R)/a

⇒ a = (r + R) × r/(R - r)

⇒ a = r(r + R)/(R - r)

Now, h = AD + DO + OF + FP

⇒ (a + r + R + R) = (a + r + 2R)

⇒ $\frac {(r^2+ Rr + Rr - r^2 + 2R^2 - 2Rr)} {(R - r)}$ = $\frac {2R^2}{(R - r)}$

∴ The height of the cone is $\frac {2R^2}{(R - r)}$.

Given:

A right circular cone covers two spheres.

Radius of small sphere = r

Radius of bigger sphere = R

∠ CAB = 2θ 

Calculation:

Let, the radius of the base = BP

⇒ tanθ = BP/$\frac {2R^2} {(R - r)}$

⇒ 2R² × tanθ /(R - r) = BP

⇒ BP = 2R2 × tanθ /(R - r) 

∴ The answer is option 2.

Concept used:

Pythagoras Theorem states,

Hypotenuse² = Base² + Perpendicular²

Calculation:

According to the given data, this diagram is drawn.

Let, v and p are the centers of semi-circle S1 and circle S respectively.

In ΔpvC, ∠ pCv = 90° 

vp = Sum of the radius of S3 and S = (R + r)

pC = Radius of semi-circle S1 - Radius of circle S

⇒ DC - r = (2R - r)     [∵ DC = AC]

So, vp2 = pC2 + Cm2  

⇒ (R + r)2 = (2R - r)2 + R2

⇒ R2 + 2Rr + r2 = 4R2 - 4Rr + r2 + R2

⇒ ⇒ R2 + 2Rr + r2 = 5R2 - 4Rr + r2 

⇒ 5R2 - 4Rr + r2  - R2 - 2Rr - r2 = 0

⇒ 4R2 - 6Rr = 0

⇒ 4R2 = 6Rr 

⇒ 2R = 3r

∴ Option 4 is the correct answer.

Given:

Area of circle S = m

Area of semi-circle S1 = n

Concept used:

Area of a circle = πR²

Area of a semi-circle = (πR2)/2

Where, R = radius, and π = 22/7

Calculation:

Radius of circle S = r

Radius of semi-circle S1 = 2R

Let, v and p are the centers of semi-circle S1 and circle S respectively.

In ΔpvC, ∠ pCv = 90° 

vp = Sum of the radius of S3 and S = (R + r)

pC = Radius of semi-circle S1 - Radius of circle S

⇒ DC - r = (2R - r)     [∵ DC = AC]

So, vp2 = pC2 + Cm2  

⇒ (R + r)2 = (2R - r)2 + R2

⇒ R2 + 2Rr + r2 = 4R2 - 4Rr + r2 + R2

⇒ ⇒ R2 + 2Rr + r2 = 5R2 - 4Rr + r2 

⇒ 4R2 - 6Rr = 0

⇒ 2R = 3r

⇒ r = (2R)/3

Now, $\frac{m}{n} = \frac{\Pi r^{2}}{\frac{\Pi (2R)^{2}}{2}}$

⇒ $\frac{m}{n} = \frac{\Pi (\frac {2R}{3})^{2}}{\frac{\Pi (2R)^{2}}{2}}$ 

⇒ $\frac{m}{n} = \frac{\frac {4R^{2}}{9}}{\frac{4R^{2}}{2}}$

⇒ $m \times {\frac{4R^{2}}{2}} = n \times {\frac {4R^{2}}{9}}$

⇒ m/2 = n/9

⇒ 9m = 2n

∴ Option 1 is the correct answer.

Concept:

Componendo Dividendo Rule:

If (a + b ) : (a – b) = (c + d) : (c – d), than 

a : b = c : d

Calculation:

According to the question,

$\rm \frac{(x-a)(x-b)}{(x-m a)(x-m b)}=\frac{(x+a)(x+b)}{(x+m a)(x+m b)}$

⇒ $(x - a) (x - b) \over (x + a) (x + b)$ = $(x - ma) (x - mb) \over (x + ma) (x + mb)$

⇒ $x^2 - (a + b)x + ab \over x^2 + (a + b)x + ab$ = $x^2 - (ma + mb)x + m^2ab \over x^2 + (ma + mb)x + m^2ab$

By componendo dividendo method,

⇒ $x^2 + ab \over (a + b)x$ = $x^2 + m^2 ab \over (ma + mb)x$

⇒ $x^2 + ab \over x^2 + m^2 ab$ = $(a + b) \over (a + b) m$

⇒ $x^2 + ab \over x^2 + m^2 ab$ = $1 \over m$

∴ The equation $\rm \frac{x^2+a b}{x^2+m^2 a b}$  is equal to $1 \over m$.

Concept:

Componendo Dividendo Rule:

If (a + b ) : (a – b) = (c + d) : (c – d), than 

a : b = c : d

Calculation:

According to the question,

$\rm \frac{(x-a)(x-b)}{(x-m a)(x-m b)}=\frac{(x+a)(x+b)}{(x+m a)(x+m b)}$

⇒ $(x - a) (x - b) \over (x + a) (x + b)$ = $(x - ma) (x - mb) \over (x + ma) (x + mb)$

⇒ $x^2 - (a + b)x + ab \over x^2 + (a + b)x + ab$ = $x^2 - (ma + mb)x + m^2ab \over x^2 + (ma + mb)x + m^2ab$

By componendo dividendo method,

⇒ $x^2 + ab \over (a + b)x$ = $x^2 + m^2 ab \over (ma + mb)x$

⇒ $x^2 + ab \over x^2 + m^2 ab$ = $(a + b) \over (a + b) m$

⇒ $x^2 + ab \over x^2 + m^2 ab$ = $1 \over m$

⇒ mx² + mab = x² + m²ab

⇒ mx² - x² = m²ab - mab

⇒ x² (m - 1) = mab (m - 1)

⇒ x² = mab

⇒ x = $\pm \sqrt{\mathrm{mab}}$

∴ The value of x is equal to $\pm \sqrt{\mathrm{mab}}$.