Mathematics Test 231

Calculation:

Option 1: (p' ∧ q')→q

It means that if both p and q are false, then q is true.

This is not equivalent to p ↔ q.

Option 2: (p' ∧ q') ∨ (p ∧ q)

It means that either both p and q are false, or both p and q are true.

This is exactly what p ↔ q represents.

Option 3: (p' ∨ q') → p

It means that if either p or q is false, then p is true.

This is not equivalent to p ↔ q.

Option 4: (p' ∧ q') ∧ (p ∧ q)

It means that both p and q are false and both p and q are true simultaneously, which is a contradiction.

This is not equivalent to p ↔ q.

∴ p ↔ q is equivalent to (p' ∧ q') ∨ (p ∧ q).

The correct answer is Option 2.

Calculation:

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

Hence, Option 1 is incorrect

If A and B are invertible, then it is not necessary that (A + B) is invertible.

Hence, Option 2 is incorrect.

AB = O

⇒ |AB| = |O|

⇒ |A||B| = 0

⇒ |A| = 0 or |B| = 0

Hecne, Option 3 is correct.

AB = O

⇒ |AB| = |O|

⇒ |A||B| = 0

⇒ |A| = 0 or |B| = 0

⇒ It is not necessary |A| = 0 and |B| = 0

Hecne, Option 4 is incorrect. 

∴ If AB = O, then |A| = 0 or |B| = 0.

The correct answer is Option 3.

Calculation:

Let the components of the line segment vector be a, b, c

Since, a < 0 as the line makes an obtuse angle with x -axis

⇒ t = − 9

⇒ a =  − 27, b = 18, c = − 54

∴ The required components are (–27, 18, –54).

The correct answer is 3.

Calculation:

In x ∈ (0, 1), f(x) = |x ln x| = - x ln x

For maxima, f '(x) = 0 and f ''(x) < 0

∴ f '(x) = - ln x - 1 = 0

⇒ ln x = - 1

⇒ x = 1/e

Now, f ''(x) = - 1/x

At x = 1/e, f ''(x) = - e < 0

⇒ x = 1/e is point of maxima.

⇒ Maximum value = f(1/e) = |- 1/e| = 1/e

∴ The maximum value of f(x) = |x ln x| in x ∈ (0, 1) is 1/e.

The correct answer is Option 1.

Calculation:

A normal to the parabola y² =4ax is y = mx − 2am − am³

∴ Normal to parabola y² =4x is y = mx − 2m − m³

Now, it passes through (5, 0)

∴ 0 = 5m − 2m − m³

⇒ m(5 − 2 − m²) = 0

⇒ m(3 − m²) = 0

⇒ m = 0, ± √3 ​

Now, feet of the normal is given by (am² ,− 2am) = (m², −2m)

Hence, feet of normal are (0,0), (3, −2√3) and (3, 2√3)

∴ Centroid of the triangle 

= (2, 0)

∴ Centroid of the triangle formed by the feet of the three normals is (2, 0).

The correct answer is Option 1.

Calculation:

Given, There are 3 number a, b and c where abc = 30.

Now, all divisors of 30 are 1, 2, 3, 5, 6, 10, 15 and 30.

Selecting any 3 numbers from all divisors so that we get multiplication as 30

(1) 1 × 1 × 30 = 30

Number of ways to arrange 1, 1 and 30 

(2) 1 × 3 × 10 = 30

Number of ways to arrange 1, 3 and 10 = 3! = 6

(3) 2 × 3 × 5 = 30

Number of ways to arrange 2, 3 and 5 = 3! = 6

(4) 1 × 5 × 6 = 30

Number of ways to arrange 1, 5 and 6 = 3! = 6

(5) 1 × 2 × 15 = 30

Number of ways to arrange 1, 2 and 15 = 3! = 6

∴ Total solutions = 6 + 6 + 6 + 6 + 3

= 27 ways

∴ Total number of integral solutions is 27.

The correct answer is Option 2.

Calculation:

Given, 3x² + 2x + p (p – 1) = 0 

Let the roots be α and β.

⇒ αβ = p(p – 1)

Since, the roots are of opposite signs, then the product of roots will be negative.

⇒ p(p – 1) < 0

⇒ p ∈ (0, 1)

∴ The set of values of p is (0, 1).

The correct answer is Option 2.

Now, the curve passes through (1, 1).

⇒ C = 1

⇒ xy = 1

∴ The equation of the curve is xy = 1

The correct answer is Option 3.

Calculation:

f (x) = |(x – 1) (x – 2) (x – 3)|

∴  f (x – 2) = 2

⇒ |(x – 3) (x – 4) (x – 5)| = 2

From the graph, we can observe there are 2 solutions.

∴ The number of solutions of the equation f (x – 2) = 2 are 2.

The correct answer is Option 2.