Quantitative Aptitude PYQs

Solution:

There are two AP with the different first terms and common difference.

For the first AP A:

Let the first term be a and common difference = d

Sum of n terms P = \(\frac{n}{2}[2a+(n-1)d]\)

and nth term an = a + (n - 1)d

For the second AP B:

Let the first term be M and common difference = D

Sum of n terms Q = \(\frac{n}{2}[2M+(n-1)D]\)

and nth term Mn = M + (n - 1)D

It is given that

\(\frac{Sum\ of\ n\ terms\ of\ first\ A.P}{Sum\ of\ n\ terms\ of\ second\ A.P }= \frac{5n+4}{9n+6}\)

⇒ \(\frac{\frac{n}{2}[2a+(n-1)d]}{\frac{n}{2}[2M+(n-1)D] }= \frac{5n+4}{9n+6}\)

⇒ \(\frac{[2a+(n-1)d]}{[2M+(n-1)D] }= \frac{5n+4}{9n+6} \)

⇒ \(\frac{a+\left ( \frac{n-1}{2} \right )d}{M+\left ( \frac{n-1}{2} \right )D}= \frac{5n+4}{9n+6} \).....(1)

Now, we need to find the ratio of the first term of A and B

\(\frac{1^{st}\ term\ of\ A}{1^{st}\ term\ of\ B}\)

= \(\frac{a}{M}\)

Hence, \(\frac{n-1}{2}=0\Rightarrow n=1\)

Putting n = 1 in equation (1), we get

\(\frac{a}{M}=\frac{5(1)+4}{9(1)+6}\)

⇒ \(\frac{a}{M}=\frac{9}{15}\)

⇒ \(\frac{a}{M}=\frac{3}{5}\)

⇒ \(\frac{1^{st}\ term\ of\ A}{1^{st}\ term\ of\ B} =\frac{3}{5}\)

Now, we need to find the ratio of the second term of A and B

\(\frac{2^{nd}\ term\ of\ A}{2^{nd}\ term\ of\ B}\)

= \(\frac{a+(2-1)d}{M+(2-1)D}\)

= \(\frac{a+d}{M+D}\)

Hence, \(\frac{n-1}{2}=1\Rightarrow n=3\)

Putting n = 3 in equation (1), we get

\(\frac{a+d}{M+D}=\frac{5(3)+4}{9(3)+6}\)

⇒ \(\frac{a+d}{M+D}=\frac{19}{33}\)

⇒ \(\frac{2^{nd}\ term\ of\ A}{2^{nd}\ term\ of\ B} =\frac{19}{33}\)

Let the first term of A = 3 and the second term of A = 19

then common difference d = 19 - 3 = 16

Let the first term of B = 5 and the second term of B = 33

then common difference D = 33 - 5 = 28

Hence, D > d

∴ The correct option is (1)

Concept:

General term of binomial expansion:

The general term in the expansion of (x + y)ⁿ is given by:

 T_r+1 = ⁿC_r x^r y^n-r

Calculation:

In the binomial expansion of (p + qx)9 :

the general term in the expansion of (p + qx)9 is given by:

 Tr+1 = 9Cr p9 (qx)9-r

The term with x³ when 9 - r = 3  __(i)

⇒ r = 9 - 3 = 6

⇒ The term with x³ = T7 = \(^9C_6 \ p^6 (qx)^{9-6}\) = \({9! \over 6!(9-6)!}\)p⁶q³x³

⇒ The term with x3 =  84 p6q3x3

⇒ Coefficient of x³ = 84p6q3 __(i)

The term with x6 when 9 - r = 6

⇒ r = 9 - 6 = 3

⇒ The term with x6 = T4 = \(^9C_3 \ p^3 (qx)^{9-3}\) 

⇒ The term with x6 = \({9! \over 3!(9-3)!}\)p3q6x6

⇒ Coefficient of x6 = 84p3q6 __(ii)

If the coefficients of x3 and x6 are equal

Equating equation (i) and (ii),

⇒  84p6q3 = 84p3q6

⇒ p³ = q³

⇒  p = q

∴ The correct option is (1).

Concept:

General term of binomial expansion:

The general term in the expansion of (x + y)n is given by:

 Tr+1 = nCr xr yn-r

Middle term in binomial expansion: The middle terms in the expansion of (x + y)n depend on the value of n.

• If n is even, there will be n + 1 term in the expansion. So, there is only one middle term and the middle term is (\(n \over 2\) + 1)^th term.
• If n is odd, there will be n +1 term in the expansion. So, there will be two middle terms which are  (\(n+1 \over 2\))th and (\(n+3 \over 2\))th term.

 

Calculation:

In the binomial expansion of (p + qx)9, there are 9 + 1 = 10 terms.

Since n is odd, there will be two middle terms.

⇒The middle terms are (\(9+1 \over 2\))th and (\(9+3 \over 2\))th term, that is, 5^th and 6^th terms.

In the expansion of (p + qx)9,  

5^th term = T₅ = T₄₊₁  

⇒ T4+1 = \(^9C_4 \ p^4 (qx)^{9-4}\) 

⇒ T4+1 = \({9! \over 4!(9-4)!} \ p^4 (qx)^{5}\) = 126p4q5x5

⇒ Coefficient of 5^th term = 126p4q5 __(i)

6^th term = T6 = T5+1  

⇒ T5+1 = \(^9C_5 \ p^5 (qx)^{9-5}\) 

⇒ T5+1 = \({9! \over 5!(9-5)!} \ p^5 (qx)^{4}\) = 126p5q4x4

⇒ Coefficient of 6^th term = 126p5q4 __(ii)

The ratio of the coefficients of middle terms in the expansion

= Coefficient of 5^th term: Coefficient of 6^th term 

Putting values from equations (i) and (ii),

The ratio of the coefficients of middle terms in the expansion = 126p5q4 : 126p4q5

⇒ The ratio of the coefficients of middle terms in the expansion = p : q

∴ The correct option is (2).

Concept:

General term of binomial expansion:

The general term in the expansion of (x + y)n is given by:

 Tr+1 = nCr xr yn-r

Calculation:

In the binomial expansion of (p + qx)9 :

the general term in the expansion of (p + qx)9 is given by:

 Tr+1 = 9Cr pr (qx)9-r

The term with x2 when 9 - r = 2  __(i)

⇒ r = 9 - 2 = 7

⇒ The term with x2 = T8 = \(^9C_7 \ p^7 (qx)^{9-7}\) 

⇒ The term with x2 = \({9! \over 7!(9-7)!}\)p7q2x2

⇒ The term with x2 =  36p7q2x2

⇒ Coefficient of x2 = 36p7q2__(i)

And the term with x4 when 9 - r = 4

⇒ r = 9 - 4 = 5

⇒ The term with x4 = T5+1 = \(^9C_5 \ p^5 (qx)^{9-5}\) 

⇒ The term with x4 = \({9! \over 5!(9-5)!}\)p5q4x4

⇒ The term with x4 = 126p5q4x4

⇒ Coefficient of x4 = 126p5q4 __(ii)

If the coefficients of x2 and x4 are equal

From (i) and (ii),

⇒  36p7q2 = 126p5q4

⇒ 2p2 = 7q2

⇒  p2 ∶ q2 = 7 ∶ 2

∴ The correct option is (2).

Concept:

The number of ways of choosing m objects out of n different objects = ⁿC_m

The number of ways of arranging n objects = n!

Explanation:

Given word 'QUESTION' which has 8 different letters

{4 vowels (U,E,I,O) and 4 consonants(Q,S,T,N)}

The number of ways of choosing two vowels from 4 vowels = 4C2 = 6

And the number of ways of choosing two vowels from 4 vowels = 4C2 = 6

⇒ The number of ways of choosing two vowels and two consonants from 4 vowels and 4 consonants = ⁴C₂ × 4C2 = 36

And the number of ways of arranging these four different letters = 4! = 24

⇒ Number of 4-letter words each of two vowels and two consonants with or without meaning = 36 × 24 = 864

∴ The correct option is (4).

Concept:

The number of ways of choosing m objects out of n different objects = nCm

The number of ways of arranging n objects = n!

Explanation:

Given word 'QUESTION' which has 8 different letters

{4 vowels (U,E,I,O) and 4 consonants(Q,S,T,N)}

If vowels occupy odd positions and consonants occupy even positions,

then 4 consonants are to be arranged in 4 positions and 4 vowels are to be arranged in 4 positions

As the number of ways of arranging 4 objects in 4 positions = 4! = 24

⇒ The number of ways arranging 4 vowels in 4 positions and 4 consonants in 4 positions = 24 × 24 = 576

Similarly, for the case when vowels occupy even positions and consonants occupy odd positions

So, the number of 8-letter words that can be formed such that consonants and vowels occupy alternate positions = 576 × 2 = 1152

∴ The correct answer is option (3).

Concept:

The number of ways of choosing m objects out of n different objects = nCm

The number of ways of arranging n objects = n!

Explanation:

Given word 'QUESTION' which has 8 different letters

{4 vowels (U,E,I,O) and 4 consonants(Q,S,T,N)}

If all consonants are together consider them as one string, which can be rearranged internally in 4! ways 

⇒ Number of ways of rearranging 4 consonants = 24

Now for forming a word, we have to arrange 4 vowels and one string of consonants 

The number of ways of arranging 5 objects = 5! = 120

So, the total number of words formed = 120 × 24 = 2880

∴ The correct answer is option (2).

Concept:

Multiplying the elements of any row with their respective cofactors gives the determinant of the matrix.

Calculation:

Given matrix A = \(\left(\begin{array}{lll}{a}_{11}&{a}_{12}&{a}_{13}\{a}_{21}&{a}_{22}&{a}_{23}\{a}_{31}&{a}_{32}&{a}_{33}\end{array}\right)\) 

Expanding along row 1,

|A| = Δ = a₁₁(a₂₂a₃₃ - a₂₃a₃₂) + a12(a23a31 - a21a33) + a

13(a21a32 - a22a31)

 Here,C11 = a22a33 - a23a32, C12 = a23a31 - a21a33 ,C13 = a21a32 - a22a31

where, C11, C12, C13 are the cofactors of a11, a12, a13 respectively.

Putting the values,

⇒ a11C11 + a12C12 + a13C13 = Δ

∴ The correct answer is option (3).

Concept:

Multiplying the elements of any row with the cofactors of another row is 0.

Calculation:

If C₁₁, C₁₂, C₁₃ be the cofactors of a₁₁, a₁₂, a₁₃ respectively,

⇒ C₁₁ = a₂₂a₃₃ - a₂₃a₃₂, C₁₂ = a₂₃a₃₁ - a₂₁a₃₃ ,C₁₃ = a₂₁a₃₂ - a₂₂a₃₁,

Putting the values in a₂₁C₁₁ + a₂₂C₁₂ + a₂₃C₁₃

⇒ a₂₁C₁₁ + a₂₂C₁₂ + a₂₃C₁₃ = a₂₁(a₂₂a₃₃ - a₂₃a₃₂) + a₂₂(a₂₃a₃₁ - a₂₁a₃₃) + a₂₃(a₂₁a₃₂ - a₂₂a₃₁)

⇒ a₂₁C₁₁ + a₂₂C₁₂ + a₂₃C₁₃ = a₂₁a₂₂a₃₃ - a₂₁a₂₃a₃₂ + a₂₂a₂₃a₃₁ - a₂₁a₂₂a₃₃ + a₂₃a₂₁a₃₂ - a₂₂a₂₃a₃₁

⇒ a₂₁C₁₁ + a₂₂C₁₂ + a₂₃C₁₃ = 0

∴ The correct answer is option (1).

Concept:

• By interchanging two rows or columns determinant changes to its negative value.
• The determinant of the transpose of a matrix is same as the determinant of A.

Calculation:

If  Δ = |A| = \(\left|\begin{array}{lll}\text{a}_{11} & \text{a}_{12} & \text{a}_{13} \ \text{a}_{21} & \text{a}_{22} & \text{a}_{23} \ \text{a}_{31} & \text{a}_{32} & \text{a}_{33}\end{array}\right|\),

Taking transpose of A,

⇒ |A^T| = \(\left|\begin{array}{lll}\text{a}_{11} & \text{a}_{21} & \text{a}_{31} \ \text{a}_{12} & \text{a}_{22} & \text{a}_{32} \ \text{a}_{13} & \text{a}_{23} & \text{a}_{33}\end{array}\right|\)  = Δ ,

Take C₁ ↔ C₃,

⇒  \(\left|\begin{array}{lll}\text{a}_{31} & \text{a}_{21} & \text{a}_{11} \ \text{a}_{32} & \text{a}_{22} & \text{a}_{12} \ \text{a}_{33} & \text{a}_{23} & \text{a}_{13}\end{array}\right|\) = - Δ 

Take C1 ↔ C2,

⇒  \(\left|\begin{array}{lll}\text{a}_{21} & \text{a}_{31} & \text{a}_{11} \ \text{a}_{22} & \text{a}_{32} & \text{a}_{12} \ \text{a}_{23} & \text{a}_{33} & \text{a}_{13}\end{array}\right|\) = -(- Δ) = Δ

Take R2 ↔ R3,

⇒  \(\left|\begin{array}{lll}\text{a}_{21} & \text{a}_{31} & \text{a}_{11} \text{a}_{23} & \text{a}_{33} & \text{a}_{13} \text{a}_{22} & \text{a}_{32} & \text{a}_{12} \end{array}\right|\) = - Δ

∴ The correct option is (4).

Concept:

• If f(x) is a function such that  f(x + y) = f(x) f(y),

           then the function will be of the form f(x) = a^x, a ∈ R.

• For a G.P., with first term a and common ratio r,

           Sum of first n terms = \(a({r^n - 1 \over r - 1})\), r > 1

Calculation:

Given, f(x) is a function satisfying f(x + y) = f(x) f(y) for all x, y ∈ N such that f(1) = 2 

Since,  f(x + y) = f(x) f(y) 

⇒ f(1 + 1) = f(1) f(1) =  2²

⇒ f(1 + 2) =  f(3)  = f(1) f(2) =  2 × 22 = 23

Similarly,

f(n) = 2n

So, \(\displaystyle\sum_{x=2}^n\)f(x) = ​​​​\(\displaystyle\sum_{x=2}^n\) 2^x

⇒ \(\displaystyle\sum_{x=2}^n\)f(x) = ​​​​(2² + 2³ + 2⁴ + ... + 2ⁿ) = 2044

which is a G.P. with first term 2² and common ratio 2.

and number of terms = n - 1

⇒ \(2^2({ 2^{n-1}-1 \over 2-1} )\) = 2044

⇒ 4(2^n-1 - 1) = 2044

⇒ 2^n-1 - 1 = 511

⇒ 2^n-1 = 512 = 2⁹

⇒ n -1 = 9

⇒ n = 10  

∴ The correct answer is option (3).

Concept:

• If f(x) is a function such that  f(x + y) = f(x) f(y),

           then the function will be of the form f(x) = ax, a ∈ R.​

Calculation:

Given, f(x) is a function satisfying f(x + y) = f(x) f(y) for all x, y ∈ N such that f(1) = 2 

Let f(x) = ax 

f(1) = a1 = 2 

⇒ f(x) = 2x

So, \(\displaystyle\sum_{x=1}^5\)f(2x - 1) = ​​​​\(\displaystyle\sum_{x=1}^5\)22x- 1

⇒ \(\displaystyle\sum_{x=1}^5\)f(x) = ​​​​(2 + 23 + 25 + 27 + 2⁹ )

⇒ \(\displaystyle\sum_{x=1}^5\)f(x) = 2 + ​​​​8 + 32 + 128 + 512

⇒ \(\displaystyle\sum_{x=2}^5\)f(x) = ​​​​682

∴ The correct answer is option (2).

Concept:

• If f(x) is a function such that  f(x + y) = f(x) f(y),

           then the function will be of the form f(x) = ax, a ∈ R.

• For a G.P., with first term a and common ratio r,

           Sum of first n terms = \(a({r^n - 1 \over r - 1})\), r > 1

Calculation:

Given, f(x) is a function satisfying f(x + y) = f(x) f(y) for all x, y ∈ N such that f(1) = 2 

Let f(x) = ax 

f(1) = a1 = 2 

⇒ f(x) = 2x

So, \(\displaystyle\sum_{x=1}^6\)2x f(x) = ​​​​\(\displaystyle\sum_{x=1}^6\)2x 2x

⇒ \(\displaystyle\sum_{x=1}^6\)2^x f(x) =  ​​​​\(\displaystyle\sum_{x=1}^6\)22x = ​​​​(22 + 24 + 2⁶ + 28 + 2¹⁰ + 212)

which is a G.P. with first term 22 and common ratio 4 and n = 6.

​⇒ \(\displaystyle\sum_{x=1}^6 \) 2x f(x)  = \(2^2({ 4^6-1 \over 4-1} )\) = \(4(4096 - 1) \over 3\)

⇒ \(\displaystyle\sum_{x=1}^6\) 2^x f(x) = 4 × 1365 = 5460

∴ The correct answer is option (4).

Concept:

n(A ∪ B ∪ C) = n(A) + n(B) + n(C) - n(A ∩ B) - n(B ∩ C) - n(C ∩ A) + n(A ∩ B ∩ C)

Explanation:

Given: x students (x < 6) got medals in all three sports and the medals went to a total of 15x students. It awarded 5x medals in basketball, (4x + 15) medals in football, and (x + 25) medals in volleyball.

Let A = students who got medals in basketball

B = students who got medals in football

C = students who got medals in volleyball

From given data,

⇒ n(A) = 5x, n(B) = 4x + 15, n(C) = x + 25, n(A ∪ B ∪ C) = 15x, n(A ∩ B ∩ C) = x

Putting in the formula,

n(A ∪ B ∪ C) = n(A) + n(B) + n(C) - n(A ∩ B) - n(B ∩ C) - n(C ∩ A) + n(A ∩ B ∩ C)

⇒ 15x = 5x + 4x + 15 + x + 25 - n(A ∩ B) - n(B ∩ C) - n(C ∩ A) + x

⇒ n(A ∩ B) + n(B ∩ C) + n(C ∩ A) = 11x - 15x + 40 

⇒ n(A ∩ B) + n(B ∩ C) + n(C ∩ A) = 40 - 4x  __(i)

From Venn diagram,

n(A ∩ B) + n(B ∩ C) + n(C ∩ A) = (p + s) + (r + s) + (q + r)

⇒ n(A ∩ B) + n(B ∩ C) + n(C ∩ A) = p + q + r + 3s

where (p + q + r) is the number of students who received medals in exactly two of the three sports.

and s = the number of students who received medals in all the three sports

⇒ The number of students who received medals in exactly two of the three sports

= n(A ∩ B) + n(B ∩ C) + n(C ∩ A) - 3x

Putting value from (i),

⇒ The number of students who received medals in exactly two of the three sports = 40 - 4x - 3x = 40 - 7x

∴ The correct option is (3).

Explanation: 

Given: x students (x < 6) got medals in all three sports and the medals went to a total of 15x students. It awarded 5x medals in basketball, (4x + 15) medals in football, and (x + 25) medals in volleyball.

Let A = students who got medals in basketball

B = students who got medals in football

C = students who got medals in volleyball

From given data,

⇒ n(A) = 5x, n(B) = 4x + 15, n(C) = x + 25, n(A ∪ B ∪ C) = 15x, n(A ∩ B ∩ C) = x

Putting in the formula,

n(A ∪ B ∪ C) = n(A) + n(B) + n(C) - n(A ∩ B) - n(B ∩ C) - n(C ∩ A) + n(A ∩ B ∩ C)

⇒ 15x = 5x + 4x + 15 + x + 25 - n(A ∩ B) - n(B ∩ C) - n(C ∩ A) + x

⇒ n(A ∩ B) + n(B ∩ C) + n(C ∩ A) = 11x - 15x + 40 

⇒ n(A ∩ B) + n(B ∩ C) + n(C ∩ A) = 40 - 4x  __(i)

From the Venn diagram,

n(A ∩ B) + n(B ∩ C) + n(C ∩ A) = (p + s) + (r + s) + (q + r)

⇒ n(A ∩ B) + n(B ∩ C) + n(C ∩ A) = p + q + r + 3s  ___(ii)

where (p + q + r) is the number of students who received medals in exactly two of the three sports.

and s = the number of students who received medals in all the three sports = x

⇒ p + q + r + s = the number of students who received medals in atleast two sports.

Putting the value of (i) in (ii),

⇒ 40 - 4x = p + q + r + 3s

⇒ 40 - 4x = p + q + r + s + 2x

⇒ p + q + r + s = 40 - 6x

⇒ The number of students who received medals in atleast two sports = 40 - 6x

∴ The correct option is (4).

Explanation: 

Given: x students (x < 6) got medals in all three sports and the medals went to a total of 15x students. It awarded 5x medals in basketball, (4x + 15) medals in football, and (x + 25) medals in volleyball.

Let A = students who got medals in basketball

B = students who got medals in football

C = students who got medals in volleyball

From given data,

⇒ n(A) = 5x, n(B) = 4x + 15, n(C) = x + 25, n(A ∪ B ∪ C) = 15x, n(A ∩ B ∩ C) = x

Putting in the formula,

n(A ∪ B ∪ C) = n(A) + n(B) + n(C) - n(A ∩ B) - n(B ∩ C) - n(C ∩ A) + n(A ∩ B ∩ C)

⇒ 15x = 5x + 4x + 15 + x + 25 - n(A ∩ B) - n(B ∩ C) - n(C ∩ A) + x

⇒ n(A ∩ B) + n(B ∩ C) + n(C ∩ A) = 11x - 15x + 40 

⇒ n(A ∩ B) + n(B ∩ C) + n(C ∩ A) = 40 - 4x  __(i)

From the Venn diagram,

n(A ∩ B) + n(B ∩ C) + n(C ∩ A) = (p + s) + (r + s) + (q + r)

⇒ n(A ∩ B) + n(B ∩ C) + n(C ∩ A) = p + q + r + 3s  ___(ii)

where (p + q + r) is the number of students who received medals in exactly two of the three sports.

and s = the number of students who received medals in all the three sports = x

⇒ p + q + r + s = the number of students who received medals in atleast two sports.

Putting the value of (i) in (ii),

⇒ 40 - 4x = p + q + r + 3s

⇒ 40 - 4x = p + q + r + s + 2x

⇒ p + q + r + s = 40 - 6x __(iii)

Now, from the vein diagram

the number of students who received medals in exactly one of three sports 

= n(A ∪ B ∪ C) - (p + q + r + s)

Putting the value from (iii)

the number of students who received medals in exactly one of three sports = 15x - (40 - 6x) = 21x - 40

∴ The correct option is (1).

Concept:

Every square matrix can be expressed as the sum of a symmetric and a skew-symmetric matrix uniquely, which is given as,

 A = \(1 \over 2\)(A + A^T) + \(1 \over 2\)(A - AT)

where \(1 \over 2\)(A + AT) is a symmetric matrix and \(1 \over 2\)(A - A^T)

​​Calculation:

Given

A = \(\left(\begin{array}{ccc}0 & \sin^2 \theta & \cos^2 \theta \ \cos^2 \theta & 0 & \sin^2 \theta \ \sin^2 \theta & \cos^2 \theta & 0\end{array}\right)\) and A = P + Q where P is a symmetric matrix and Q is a skew-symmetric matrix.

⇒ P = \(1 \over 2\)(A + A^T) 

⇒ P = \(1 \over 2\)[\(\left(\begin{array}{ccc}0 & \sin^2 \theta & \cos^2 \theta \ \cos^2 \theta & 0 & \sin^2 \theta \ \sin^2 \theta & \cos^2 \theta & 0\end{array}\right)\) + \(\left(\begin{array}{ccc}0 & \cos^2 \theta & \sin^2 \theta \ \sin^2 \theta & 0 & \cos^2 \theta \ \cos^2 \theta & \sin^2 \theta & 0\end{array}\right)\)]

⇒ P = \(1 \over 2\)(\left(\begin{array}{ccc}0 & \sin^2 \theta + \cos^2 \theta& \sin^2 \theta + \cos^2 \theta \ \sin^2 \theta + \cos^2 \theta & 0 & \sin^2 \theta + \cos^2 \theta \ \sin^2 \theta + \cos^2 \theta & \sin^2 \theta + \cos^2 \theta & 0\end{array}\right)\) 

⇒ P = \(1 \over 2\)(\left(\begin{array}{ccc}0 &1 &1 \ 1 & 0 &1 \ 1 & 1 & 0\end{array}\right)\)

⇒ P = \(\left(\begin{array}{ccc}0 & 1/2 & 1/2 \ 1/2 & 0 & 1/2 \ 1/2 & 1/2 & 0\end{array}\right)\)

∴ The correct answer is option (1).

Concept:

Every square matrix can be expressed as the sum of a symmetric and a skew-symmetric matrix uniquely, which is given as,

 A = \(1 \over 2\)(A + AT) + \(1 \over 2\)(A - AT)

where \(1 \over 2\)(A + AT) is a symmetric matrix and \(1 \over 2\)(A - AT)

​​Calculation:

Given

A = \(\left(\begin{array}{ccc}0 & \sin^2 θ & \cos^2 θ \ \cos^2 θ & 0 & \sin^2 θ \ \sin^2 θ & \cos^2 θ & 0\end{array}\right)\) and A = P + Q where P is a symmetric matrix and Q is a skew-symmetric matrix.

⇒ Q = \(1 \over 2\)(A - AT) 

⇒ P = \(1 \over 2\)[\(\left(\begin{array}{ccc}0 & \sin^2 θ & \cos^2 θ \ \cos^2 θ & 0 & \sin^2 θ \ \sin^2 θ & \cos^2 θ & 0\end{array}\right)\) - \(\left(\begin{array}{ccc}0 & \cos^2 θ & \sin^2 θ \ \sin^2 θ & 0 & \cos^2 θ \ \cos^2 θ & \sin^2 θ & 0\end{array}\right)\)]

⇒ P = \(1 \over 2\)(\left(\begin{array}{ccc}0 & \sin^2 θ - \cos^2 θ& \cos^2 θ - \sin^2 θ \ \cos^2 θ - \sin^2 θ & 0 & \sin^2 θ - \cos^2 θ \ \sin^2 θ - \cos^2 θ & \cos^2 θ - \sin^2 θ & 0\end{array}\right)\) 

⇒ P = \(1 \over 2\)(\left(\begin{array}{ccc}0 &-\cos 2θ &\cos 2θ \ \cos 2θ & 0 &-\cos 2θ \ -\cos 2θ & \cos 2θ & 0\end{array}\right)\)   {∵ cos²θ - sin²θ = cos 2θ }

⇒ P = cos 2θ\(\left(\begin{array}{ccc}0 & −1/2 & 1/2 \ 1/2 & 0 & −1/2 \ −1/2 & 1/2 & 0\end{array}\right)\) 

∴ The correct answer is option (4).

Formula used:

a³ + b³ = (a + b)(a² - ab + b²)

cos2θ + sin2θ = 1

sin 2θ = 2sin θ cos θ 

Calculation:

Given

A = \(\left(\begin{array}{ccc}0 & \sin^2 θ & \cos^2 θ \ \cos^2 θ & 0 & \sin^2 θ \ \sin^2 θ & \cos^2 θ & 0\end{array}\right)\) 

Expanding along R₁,

⇒ |A| = 0 - sin²θ (0 - sin4θ) + cos2θ (cos4θ  - 0)

⇒ |A| = cos6θ + sin6θ

⇒ |A| = (cos2θ)³ + (sin2θ)³

⇒ |A| = (cos2θ + sin2θ)(cos4θ - cos2θsin2θ + sin4θ)

⇒ |A| = (cos4θ + sin4θ - cos2θsin2θ)

⇒ |A| = (cos4θ + sin4θ + 2cos2θsin2θ - 3cos2θsin2θ)

⇒ |A| = ((cos2θ + sin2θ)² - 3cos2θsin2θ)

⇒ |A| = (1 - 3cos2θsin2θ)

⇒ |A| = \((1 - { 3\over 4} (\sin 2θ )^2\)

|A| is minimum when sin 2θ is maximum and maximum value of sin 2θ = 1

⇒ Minimum value of |A| = \((1 - { 3\over 4} (1) )\) = \(1 \over 4\)

∴ The correct option is (1).

Explanation:

Considering the given figure made with the help of information given in the question.

Here, Let lamp post be present at the middle of AB at point M and is standing perpendicularly out of the paper at the point O.

Consider the triangle OMB (right-angled at M)

tan 45° = \(\frac{OM}{BM}\)

⇒1 = \(\frac{OM}{8}\)

⇒OM = 8 m

Thus, Height of the lamp post = 8 m.