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**Objective Type Questions (M.C.Q.)****Q.24. Ifthen value of x is****(a) 3 ****(b) ± 3 ****(c) ± 6 ****(d) 6****Ans.** (c)**Solution.**

Given that

⇒

⇒ 2x^{2} – 40 = 18 + 14 ⇒ 2x^{2} = 32 + 40

⇒ 2x^{2} = 72 ⇒ x^{2} = 36

∴ x = ± 6

Hence, the correct option is (c).**Q.25. The value of determinant****(a) a ^{3} + b^{3} + c^{3} **

Here, we have

C

(Taking a + b + c common from C

R

Expanding along C

⇒ (a + b + c) (a - b) (b - c) (- 1)

⇒ (a + b + c) (a - b) (c - b)

Hence, the correct option is (d).

Area of triangle with vertices (x

⇒

⇒

⇒

3k = 9 ⇒ k = 3

Hence, the correct option is (b).

(b) (b – c) (c – a) (a – b)

(c) (a + b + c) (b – c) (c – a) (a – b)

(d) None of these

Let

(Taking (b – a) common from C

C

(C

= (a – b)

= 0

Hence, the correct option is (d).

Given that

C

⇒

Taking 2 cos x + sin x common from C

⇒

R

⇒ (2 cos x + sin x) (cos x – sin x)

Hence, the correct option is (c).

Let

C

Hence, the correct option is (a).

Solution.

We have f(t) =

Expanding along R

= cos t(t

= -t

∴

⇒

⇒

Hence, the correct option is (a).

(d) 2√3/4

Solution.

Given that:

C

Expanding along R

⇒

but maximum value of sin 2θ = 1

Hence, the correct option is (a).

Solution.

Given that:

Expanding along R

= (a – b) [2a(a + c)] = (a – b) × 2a × (a + c) ≠ 0

Expanding along R

= - (b - a) [( - 2b) (b - c)] = 2b(b – a) (b – c) ≠ 0

Expanding along R

= a(bc) – b(ac)= abc – abc = 0

Hence, the correct option is (c).

Solution.

We have,

Expanding along R

= 2(6 – 5) – λ(0 – 5) – 3(0 – 2)

= 2 + 5λ + 6 = 8 + 5λ

If A

∴ 8 + 5λ ≠ 0 so λ ≠

Hence, the correct option is (d).

Solution.

If A and B are two invertible matrices then

(a) adj A = |A| × A

(b) det (A)

(c) Also, (AB)

(d) (A + B)

∴ (A + B)

Hence, the correct option is (d).

Solution.

Given that

Taking x, y and z common from R

R

Takingcommon from R

C

Expanding along R

⇒

⇒+ 1 = 0 and xyz ≠ 0 (x ≠ y ≠ z ≠ 0)

∴ x^{–1} + y^{–1} + z^{–1} = – 1

Hence, the correct option is (d).**Q.36. The value of the determinant****(a) 9x ^{2}(x + y) **

Solution.

Let

C

[Taking (3x + 3y) common from C

R

Expanding along C

⇒ 3(x + y) (y

Hence, the correct option is (b).

Solution.

Given that,

Expanding along C

⇒ (2a^{2} + 4) – 2(– 4a – 20) = 86

⇒ 2a^{2} + 4 + 8a + 40 = 86

⇒ 2a^{2} + 8a + 4 + 40 – 86 = 0

⇒ 2a^{2} + 8a – 42 = 0

⇒ a^{2} + 4a – 21 = 0

⇒ a^{2} + 7a – 3a – 21 = 0

⇒ a(a + 7) – 3(a + 7) = 0

⇒ (a – 3) (a + 7) = 0

∴ a = 3, – 7

Required sum of the two numbers = 3 – 7 = – 4.

Hence, the correct option is (c).**Fill in the blanks****Q.38. If A is a matrix of order 3 × 3, then |3A| = _______ .****Ans.**

We know that for a matrix of order 3 × 3,

|KA| = K^{3} |A|

∴ **Q.39. If A is invertible matrix of order 3 × 3, then |A ^{–1} | _______ .**

We know that for an invertible matrix A of any order,

We have,

C

[applying (a + b)

(Taking 4 common from C

⇒ 4 × 0 = 0 (∵ C

Given that: cos 2θ = 0

⇒

∴

The determinant can be written as

Expanding along C

For any square matrix A, (A

The order of a matrix is 3 × 3

∴ Total number of elements = 3 × 3 = 9

Hence, the number of minors in the determinant is 9.

The sum of the products of elements of any row with the co-factors of corresponding elements is equal to the value of the determinant of the given matrix.

Let

Expanding along R

⇒ a

(where M

We have,

Expanding along R

⇒

⇒ x(x

⇒ x

x

The roots of the equation may be the factors of 126 i.e., 2 × 7 × 9

9 is given the root of the determinant put x = 2 in eq. (1)

(2)

Hence, x = 2 is the other root.

Now, put x = 7 in eq. (1)

(7)

Hence, x = 7 is also the other root of the determinant.

Let

C

Taking (z – x) common from C

R

Taking (y – z) common from R

= (z – x) (y – z) (xyz – x + y) = (y – z) (z – x) (y – x + xyz)

Given that

Taking (1 + x)

⇒ (1 + x)

∴ 0 = A + Bx + Cx

By comparing the like terms, we get A = 0.

Since (A

So, (A

If A is a non-singular square matrix, then for any non-zero

scalar ‘a‘, aA is invertible.

∴

So, (aA) is inverse of

is true.

False.

Since = for a non-singular matrix.

True.

True.

Since

If A is a square matrix of order n

then

∴

True.

Let

R

a, b, c are in A.P.

∴ b – a = c – b ⇒ 2b = a + c

False.

Sincewhere n is the order of the square matrix.

True.

Let

Splitting up C

[∵ C

[Taking cos A and cos B common from C

= cos A cos B (0) [∵ C

= 0

True.

Let

Splitting up C

Splitting up C

Similarly by splitting C

Given that:

L.H.S.

C

[Taking 2 common from C

C

C

Splitting up C

= 2 × 16 = 32

True.

Let

C

Expanding along C

= sin θ cos θ – 0 = sin θ cos θ

[Maximum value of sin 2θ = 1]

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