Mixed elements MCQ Quiz - Objective Question with Answer for Mixed elements - Download Free PDF

Last updated on Apr 27, 2025

Latest Mixed elements MCQ Objective Questions

Mixed elements Question 1:

Let f(x)=|x3sinxcosx610pp2p3|, where p is a constant, then d3dx3(f(x)) at x = 0 is

  1. p
  2. p + p2
  3. p + p3
  4. independent of p

Answer (Detailed Solution Below)

Option 4 : independent of p

Mixed elements Question 1 Detailed Solution

Formula used:

ddxsinx = cosx

ddxsinx = sinx

ddxn = nxn  1

Calculation:

f(x)=|x3sinxcosx610pp2p3|

f(x) = x3(- 1 × p3 -  0 × p2) - sinx(6 × p3 - 0 × p) + cosx(6p2 + p)

⇒ f(x) =  - p3x3 - sinx.6p3 + cosx(6p2 + p)

⇒ f'(x) = -3p3x2 - cosx.6p3 - sinx(6p2 + p)

⇒ f''(x) = -6p3x + sinx.6p3 - cosx.(6p2 + p)

⇒ f'''(x) = -6p3 + cosx.6p3 + sinx.(6p2 + p)

Put x = 0 in above equation, we will get

 f'''(0) = -6p3 + cos0.6p3 + sin0.(6p2 + p)

⇒ f'''(0) = -6p+ 6p3

⇒ f'''(0) = 0

We can see that d3dx3(f(x)) at x = 0 is independent of p.

Mixed elements Question 2:

If A=[α22α] and det (A3) = 125, then α is equal to

  1. ± 1
  2. ± 2
  3. ± 3
  4. ± 5

Answer (Detailed Solution Below)

Option 3 : ± 3

Mixed elements Question 2 Detailed Solution

Concept:

1. If A=[a11a12a21a22] then determinant of A is given by: |A| = a11 × a22 – a21 × a12

2. |An| = |A|n

Calculation:

Given that,

A=[α22α] and det{A3} = 125

⇒ |A| = α × α – 2 × 2

⇒ |A| = α2 - 4     ...1)

Given |A3| = 125

⇒ |A|3 = 125     ...[∵ |An| = |A|n]

⇒ |A| = (125)1/3 = 5     ...2)

From equations 1 and 2

⇒ α2 - 4 = 5

⇒ α2 = 9

⇒ α = ± 3

Top Mixed elements MCQ Objective Questions

If A=[α22α] and det (A3) = 125, then α is equal to

  1. ± 1
  2. ± 2
  3. ± 3
  4. ± 5

Answer (Detailed Solution Below)

Option 3 : ± 3

Mixed elements Question 3 Detailed Solution

Download Solution PDF

Concept:

1. If A=[a11a12a21a22] then determinant of A is given by: |A| = a11 × a22 – a21 × a12

2. |An| = |A|n

Calculation:

Given that,

A=[α22α] and det{A3} = 125

⇒ |A| = α × α – 2 × 2

⇒ |A| = α2 - 4     ...1)

Given |A3| = 125

⇒ |A|3 = 125     ...[∵ |An| = |A|n]

⇒ |A| = (125)1/3 = 5     ...2)

From equations 1 and 2

⇒ α2 - 4 = 5

⇒ α2 = 9

⇒ α = ± 3

Let f(x)=|x3sinxcosx610pp2p3|, where p is a constant, then d3dx3(f(x)) at x = 0 is

  1. p
  2. p + p2
  3. p + p3
  4. independent of p

Answer (Detailed Solution Below)

Option 4 : independent of p

Mixed elements Question 4 Detailed Solution

Download Solution PDF

Formula used:

ddxsinx = cosx

ddxsinx = sinx

ddxn = nxn  1

Calculation:

f(x)=|x3sinxcosx610pp2p3|

f(x) = x3(- 1 × p3 -  0 × p2) - sinx(6 × p3 - 0 × p) + cosx(6p2 + p)

⇒ f(x) =  - p3x3 - sinx.6p3 + cosx(6p2 + p)

⇒ f'(x) = -3p3x2 - cosx.6p3 - sinx(6p2 + p)

⇒ f''(x) = -6p3x + sinx.6p3 - cosx.(6p2 + p)

⇒ f'''(x) = -6p3 + cosx.6p3 + sinx.(6p2 + p)

Put x = 0 in above equation, we will get

 f'''(0) = -6p3 + cos0.6p3 + sin0.(6p2 + p)

⇒ f'''(0) = -6p+ 6p3

⇒ f'''(0) = 0

We can see that d3dx3(f(x)) at x = 0 is independent of p.

If A=[α22α] and det (A3) = 125, then α is equal to

  1. ± 1
  2. ± 2
  3. ± 3
  4. ± 5
  5. None of these

Answer (Detailed Solution Below)

Option 3 : ± 3

Mixed elements Question 5 Detailed Solution

Download Solution PDF

Concept:

1. If A=[a11a12a21a22] then determinant of A is given by: |A| = a11 × a22 – a21 × a12

2. |An| = |A|n

Calculation:

Given that,

A=[α22α] and det{A3} = 125

⇒ |A| = α × α – 2 × 2

⇒ |A| = α2 - 4     ...1)

Given |A3| = 125

⇒ |A|3 = 125     ...[∵ |An| = |A|n]

⇒ |A| = (125)1/3 = 5     ...2)

From equation 1 and 2

⇒ α2 - 4 = 5

⇒ α2 = 9

⇒ α = ± 3

Mixed elements Question 6:

If A=[α22α] and det (A3) = 125, then α is equal to

  1. ± 1
  2. ± 2
  3. ± 3
  4. ± 5

Answer (Detailed Solution Below)

Option 3 : ± 3

Mixed elements Question 6 Detailed Solution

Concept:

1. If A=[a11a12a21a22] then determinant of A is given by: |A| = a11 × a22 – a21 × a12

2. |An| = |A|n

Calculation:

Given that,

A=[α22α] and det{A3} = 125

⇒ |A| = α × α – 2 × 2

⇒ |A| = α2 - 4     ...1)

Given |A3| = 125

⇒ |A|3 = 125     ...[∵ |An| = |A|n]

⇒ |A| = (125)1/3 = 5     ...2)

From equations 1 and 2

⇒ α2 - 4 = 5

⇒ α2 = 9

⇒ α = ± 3

Mixed elements Question 7:

Let f(x)=|x3sinxcosx610pp2p3|, where p is a constant, then d3dx3(f(x)) at x = 0 is

  1. p
  2. p + p2
  3. p + p3
  4. independent of p

Answer (Detailed Solution Below)

Option 4 : independent of p

Mixed elements Question 7 Detailed Solution

Formula used:

ddxsinx = cosx

ddxsinx = sinx

ddxn = nxn  1

Calculation:

f(x)=|x3sinxcosx610pp2p3|

f(x) = x3(- 1 × p3 -  0 × p2) - sinx(6 × p3 - 0 × p) + cosx(6p2 + p)

⇒ f(x) =  - p3x3 - sinx.6p3 + cosx(6p2 + p)

⇒ f'(x) = -3p3x2 - cosx.6p3 - sinx(6p2 + p)

⇒ f''(x) = -6p3x + sinx.6p3 - cosx.(6p2 + p)

⇒ f'''(x) = -6p3 + cosx.6p3 + sinx.(6p2 + p)

Put x = 0 in above equation, we will get

 f'''(0) = -6p3 + cos0.6p3 + sin0.(6p2 + p)

⇒ f'''(0) = -6p+ 6p3

⇒ f'''(0) = 0

We can see that d3dx3(f(x)) at x = 0 is independent of p.

Mixed elements Question 8:

If A=[α22α] and det (A3) = 125, then α is equal to

  1. ± 1
  2. ± 2
  3. ± 3
  4. ± 5
  5. None of these

Answer (Detailed Solution Below)

Option 3 : ± 3

Mixed elements Question 8 Detailed Solution

Concept:

1. If A=[a11a12a21a22] then determinant of A is given by: |A| = a11 × a22 – a21 × a12

2. |An| = |A|n

Calculation:

Given that,

A=[α22α] and det{A3} = 125

⇒ |A| = α × α – 2 × 2

⇒ |A| = α2 - 4     ...1)

Given |A3| = 125

⇒ |A|3 = 125     ...[∵ |An| = |A|n]

⇒ |A| = (125)1/3 = 5     ...2)

From equation 1 and 2

⇒ α2 - 4 = 5

⇒ α2 = 9

⇒ α = ± 3

Mixed elements Question 9:

Let f(x)=|x3sinxcosx610pp2p3|, where p is a constant, then d3dx3(f(x)) at x = 0 is

  1. independent of p
  2. p + p2
  3. p + p3
  4. More than one of the above 
  5. None of the above

Answer (Detailed Solution Below)

Option 1 : independent of p

Mixed elements Question 9 Detailed Solution

Formula used:

ddxsinx = cosx

ddxsinx = sinx

ddxn = nxn  1

Calculation:

f(x)=|x3sinxcosx610pp2p3|

f(x) = x3(- 1 × p3 -  0 × p2) - sinx(6 × p3 - 0 × p) + cosx(6p2 + p)

⇒ f(x) =  - p3x3 - sinx.6p3 + cosx(6p2 + p)

⇒ f'(x) = -3p3x2 - cosx.6p3 - sinx(6p2 + p)

⇒ f''(x) = -6p3x + sinx.6p3 - cosx.(6p2 + p)

⇒ f'''(x) = -6p3 + cosx.6p3 + sinx.(6p2 + p)

Put x = 0 in above equation, we will get

 f'''(0) = -6p3 + cos0.6p3 + sin0.(6p2 + p)

⇒ f'''(0) = -6p+ 6p3

⇒ f'''(0) = 0

We can see that d3dx3(f(x)) at x = 0 is independent of p.

Mixed elements Question 10:

Let f(x)=|x3sinxcosx610pp2p3|, where p is a constant, then d3dx3(f(x)) at x = 0 is

  1. independent of p
  2. p + p2
  3. p + p3
  4. More than one of the above 

Answer (Detailed Solution Below)

Option 1 : independent of p

Mixed elements Question 10 Detailed Solution

Formula used:

ddxsinx = cosx

ddxsinx = sinx

ddxn = nxn  1

Calculation:

f(x)=|x3sinxcosx610pp2p3|

f(x) = x3(- 1 × p3 -  0 × p2) - sinx(6 × p3 - 0 × p) + cosx(6p2 + p)

⇒ f(x) =  - p3x3 - sinx.6p3 + cosx(6p2 + p)

⇒ f'(x) = -3p3x2 - cosx.6p3 - sinx(6p2 + p)

⇒ f''(x) = -6p3x + sinx.6p3 - cosx.(6p2 + p)

⇒ f'''(x) = -6p3 + cosx.6p3 + sinx.(6p2 + p)

Put x = 0 in above equation, we will get

 f'''(0) = -6p3 + cos0.6p3 + sin0.(6p2 + p)

⇒ f'''(0) = -6p+ 6p3

⇒ f'''(0) = 0

We can see that d3dx3(f(x)) at x = 0 is independent of p.

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