QBManager

Matrix and Determinant

79234 Let $Δ = | \begin{array}{lll} A x & x^{2} & 1 \\ B y & y^{2} & 1 \\ C z & z^{2} & 1 \end{array} |$ and $Δ_{1} = | \begin{array}{ccc} A & B & C \\ x & y & z \\ z y & z x & x y \end{array} |$ then

1

Δ_{1} = - Δ

2

Δ_{1} = Δ

3

Δ_{1} = 2 Δ

4

Δ_{1} \neq Δ

Explanation:

(B) : Given,
$Δ_{1} = | \begin{array}{ccc} A & B & C \\ x & y & z \\ 2 y & 2 x & xy \end{array} |$
According to gives questions,
$Δ = | \begin{array}{ccc} A x & x^{2} & 1 \\ B y & y^{2} & 1 \\ C z & z^{2} & 1 \end{array} | = x y z | \begin{array}{ccc} A & x & \frac{1}{x} \\ B & y & \frac{1}{y} \\ C & z & \frac{1}{z} \end{array} |$
Operating $C_{3} \to$ xyz $C_{3}$
$= | \begin{array}{ccc} A & x & yz \\ B & y & xz \\ C & z & xy \end{array} | = | \begin{array}{ccc} A & B & C \\ x & y & z \\ zy & zx & xy \end{array} | = Δ_{1}$
$∴ Δ = Δ_{1}$

Karnataka CET-2017

Matrix and Determinant

79235 If $f (x), g (x)$ and $h (x)$ are three polynomials of degree 2 and
$Δ (x) = | \begin{array}{ccc} f (x) & g (x) & h (x) \\ f^{'} (x) & g^{'} (x) & h^{'} (x) \\ f^{''} (x) & g^{''} (x) & h^{''} (x) \end{array} |$
then $Δ (x)$ is a polynomial of degree 2 and

1 2

2 3

3 0

4 atmost 3

Explanation:

(C) : It is given that
$f (x), g (x)$ and $h (x)$ are the polynomials of degree 2 ,
$∴ f^{m} (x) = g^{m} (x) = h^{m} (x) = 0$
$Δ (x) = | \begin{array}{ccc} f (x) & g (x) & h (x) \\ f^{'} (x) & g^{'} (x) & h^{'} (x) \\ f^{''} (x) & g^{''} (x) & h^{''} (x) \end{array} |$
$Δ^{'} (x) = | \begin{array}{ccc} f^{'} (x) & g^{'} (x) & h^{'} (x) \\ f^{'} (x) & g^{'} (x) & h^{'} (x) \\ f^{''} (x) & g^{''} (x) & h^{''} (x) \end{array} | + | \begin{array}{ccc} f (x) & g (x) & h (x) \\ f^{''} (x) & g^{''} (x) & h^{''} (x) \\ f^{''} (x) & g^{''} (x) & h^{''} (x) \end{array}$
$+ | \begin{array}{ccc} f (x) & g (x) & h (x) \\ f^{'} (x) & g^{'} (x) & h^{'} (x) \\ f^{'''} (x) & g^{'''} (x) & h^{'''} (x) \end{array} |$
$Δ^{'} (x) = 0 + 0 + 0 = 0$
$Δ (x) = constant$
Thus, $Δ (x)$ is the polynomial of degree zero.

VITEEE-2010

Matrix and Determinant

79236 If $C = 2 \cos θ$ , then the value of the determinant
$Δ = | \begin{array}{lll} C & 1 & 0 \\ 1 & C & 1 \\ 6 & 1 & C \end{array} |$ is

1

\frac{\sin 4 θ}{\sin θ}

2

\frac{2 \sin^{2} 2 θ}{\sin θ}

3

4 \cos^{2} θ (2 \cos θ - 1)

4 None of the above

Explanation:

(D) : Given, $C = 2 \cos θ$
According to given summation
Let, $Δ = | \begin{array}{lll} C & 1 & 0 \\ 1 & C & 1 \\ 6 & 1 & C \end{array} | = C (C^{2} - 1) - 1 (C - 6) = C^{3} - 2 C + 6$
Put $C = 2 \cos θ$ , we get
$Δ = (2 \cos θ)^{3} - 2 (2 \cos θ) + 6$
$Δ = 8 \cos^{3} θ - 4 \cos θ + 6$

BCECE-2012

Matrix and Determinant

79237 The only integral root of the equation
$| \begin{array}{ccc} 2 - y & 2 & 3 \\ 2 & 5 - y & 6 \\ 3 & 4 & 10 - y \end{array} | = 0$ is

1

y = 3

2

y = 2

3

y = 1

4 none of these

Explanation:

(C) : Given,
$| \begin{array}{ccc} 2 - y & 2 & 3 \\ 2 & 5 - y & 6 \\ 3 & 4 & 10 - y \end{array} | = 0$
$(2 - y) [(5 - y) (10 - y) - 24] - 2 [20 - 2 y - 18] + 3 (8 -$ $15 + 3 y) = 0$
$(2 - y) [50 - 5 y - 10 y + y^{2} - 24] - 40 + 4 y + 36 - 21 +$ $9 y = 0$
$(2 - y) [y^{2} - 15 y + 26] + 13 y - 25 = 0$
$2 y^{2} - 30 y + 52 - y^{3} + 15 y^{2} - 26 y + 13 y - 25 = 0$
$- y^{3} + 17 y^{2} - 43 y + 27 = 0$
Only $y = 1$ satisfy this equation.

AMU-2016

Matrix and Determinant

79238 If
$= | \begin{array}{ccc} 2 \cos x & 1 & 0 \\ x - \frac{π}{2} & 2 \cos x & 1 \end{array} | then f^{'} (π / 2) is$
equal to.

1 0

2 2

3

\frac{π}{2}

4

π - 6

Explanation:

(B) : Given,
$f (x) = | \begin{array}{ccc} 2 \cos x & 1 & 0 \\ π - π / 2 & 2 \cos x & 1 \\ 0 & 1 & 2 \cos x \end{array} |$
Expanding always first column.
$= 2 \cos x (4 \cos^{2} x - 1) - [(x - π / 2) 2 \cos x - 0] + 0$
$= 8 \cos^{3} x - 2 \cos x - 2 x \cos x + π \cos x$
$(x) = 24 \cos^{2} x (- \sin x) + 2 \sin x - 2 (\cos x - x \sin x)$
$- π \sin x$
$f^{'} (π / 2) = 0 + 2 + 0 = 2$

AP EAMCET-2010

Matrix and Determinant

79234 Let $Δ = | \begin{array}{lll} A x & x^{2} & 1 \\ B y & y^{2} & 1 \\ C z & z^{2} & 1 \end{array} |$ and $Δ_{1} = | \begin{array}{ccc} A & B & C \\ x & y & z \\ z y & z x & x y \end{array} |$ then

1

Δ_{1} = - Δ

2

Δ_{1} = Δ

3

Δ_{1} = 2 Δ

4

Δ_{1} \neq Δ

Explanation:

(B) : Given,
$Δ_{1} = | \begin{array}{ccc} A & B & C \\ x & y & z \\ 2 y & 2 x & xy \end{array} |$
According to gives questions,
$Δ = | \begin{array}{ccc} A x & x^{2} & 1 \\ B y & y^{2} & 1 \\ C z & z^{2} & 1 \end{array} | = x y z | \begin{array}{ccc} A & x & \frac{1}{x} \\ B & y & \frac{1}{y} \\ C & z & \frac{1}{z} \end{array} |$
Operating $C_{3} \to$ xyz $C_{3}$
$= | \begin{array}{ccc} A & x & yz \\ B & y & xz \\ C & z & xy \end{array} | = | \begin{array}{ccc} A & B & C \\ x & y & z \\ zy & zx & xy \end{array} | = Δ_{1}$
$∴ Δ = Δ_{1}$

Karnataka CET-2017

Matrix and Determinant

79235 If $f (x), g (x)$ and $h (x)$ are three polynomials of degree 2 and
$Δ (x) = | \begin{array}{ccc} f (x) & g (x) & h (x) \\ f^{'} (x) & g^{'} (x) & h^{'} (x) \\ f^{''} (x) & g^{''} (x) & h^{''} (x) \end{array} |$
then $Δ (x)$ is a polynomial of degree 2 and

1 2

2 3

3 0

4 atmost 3

Explanation:

(C) : It is given that
$f (x), g (x)$ and $h (x)$ are the polynomials of degree 2 ,
$∴ f^{m} (x) = g^{m} (x) = h^{m} (x) = 0$
$Δ (x) = | \begin{array}{ccc} f (x) & g (x) & h (x) \\ f^{'} (x) & g^{'} (x) & h^{'} (x) \\ f^{''} (x) & g^{''} (x) & h^{''} (x) \end{array} |$
$Δ^{'} (x) = | \begin{array}{ccc} f^{'} (x) & g^{'} (x) & h^{'} (x) \\ f^{'} (x) & g^{'} (x) & h^{'} (x) \\ f^{''} (x) & g^{''} (x) & h^{''} (x) \end{array} | + | \begin{array}{ccc} f (x) & g (x) & h (x) \\ f^{''} (x) & g^{''} (x) & h^{''} (x) \\ f^{''} (x) & g^{''} (x) & h^{''} (x) \end{array}$
$+ | \begin{array}{ccc} f (x) & g (x) & h (x) \\ f^{'} (x) & g^{'} (x) & h^{'} (x) \\ f^{'''} (x) & g^{'''} (x) & h^{'''} (x) \end{array} |$
$Δ^{'} (x) = 0 + 0 + 0 = 0$
$Δ (x) = constant$
Thus, $Δ (x)$ is the polynomial of degree zero.

VITEEE-2010

Matrix and Determinant

79236 If $C = 2 \cos θ$ , then the value of the determinant
$Δ = | \begin{array}{lll} C & 1 & 0 \\ 1 & C & 1 \\ 6 & 1 & C \end{array} |$ is

1

\frac{\sin 4 θ}{\sin θ}

2

\frac{2 \sin^{2} 2 θ}{\sin θ}

3

4 \cos^{2} θ (2 \cos θ - 1)

4 None of the above

Explanation:

(D) : Given, $C = 2 \cos θ$
According to given summation
Let, $Δ = | \begin{array}{lll} C & 1 & 0 \\ 1 & C & 1 \\ 6 & 1 & C \end{array} | = C (C^{2} - 1) - 1 (C - 6) = C^{3} - 2 C + 6$
Put $C = 2 \cos θ$ , we get
$Δ = (2 \cos θ)^{3} - 2 (2 \cos θ) + 6$
$Δ = 8 \cos^{3} θ - 4 \cos θ + 6$

BCECE-2012

Matrix and Determinant

79237 The only integral root of the equation
$| \begin{array}{ccc} 2 - y & 2 & 3 \\ 2 & 5 - y & 6 \\ 3 & 4 & 10 - y \end{array} | = 0$ is

1

y = 3

2

y = 2

3

y = 1

4 none of these

Explanation:

(C) : Given,
$| \begin{array}{ccc} 2 - y & 2 & 3 \\ 2 & 5 - y & 6 \\ 3 & 4 & 10 - y \end{array} | = 0$
$(2 - y) [(5 - y) (10 - y) - 24] - 2 [20 - 2 y - 18] + 3 (8 -$ $15 + 3 y) = 0$
$(2 - y) [50 - 5 y - 10 y + y^{2} - 24] - 40 + 4 y + 36 - 21 +$ $9 y = 0$
$(2 - y) [y^{2} - 15 y + 26] + 13 y - 25 = 0$
$2 y^{2} - 30 y + 52 - y^{3} + 15 y^{2} - 26 y + 13 y - 25 = 0$
$- y^{3} + 17 y^{2} - 43 y + 27 = 0$
Only $y = 1$ satisfy this equation.

AMU-2016

Matrix and Determinant

79238 If
$= | \begin{array}{ccc} 2 \cos x & 1 & 0 \\ x - \frac{π}{2} & 2 \cos x & 1 \end{array} | then f^{'} (π / 2) is$
equal to.

1 0

2 2

3

\frac{π}{2}

4

π - 6

Explanation:

(B) : Given,
$f (x) = | \begin{array}{ccc} 2 \cos x & 1 & 0 \\ π - π / 2 & 2 \cos x & 1 \\ 0 & 1 & 2 \cos x \end{array} |$
Expanding always first column.
$= 2 \cos x (4 \cos^{2} x - 1) - [(x - π / 2) 2 \cos x - 0] + 0$
$= 8 \cos^{3} x - 2 \cos x - 2 x \cos x + π \cos x$
$(x) = 24 \cos^{2} x (- \sin x) + 2 \sin x - 2 (\cos x - x \sin x)$
$- π \sin x$
$f^{'} (π / 2) = 0 + 2 + 0 = 2$

AP EAMCET-2010

Matrix and Determinant

79234 Let $Δ = | \begin{array}{lll} A x & x^{2} & 1 \\ B y & y^{2} & 1 \\ C z & z^{2} & 1 \end{array} |$ and $Δ_{1} = | \begin{array}{ccc} A & B & C \\ x & y & z \\ z y & z x & x y \end{array} |$ then

1

Δ_{1} = - Δ

2

Δ_{1} = Δ

3

Δ_{1} = 2 Δ

4

Δ_{1} \neq Δ

Explanation:

(B) : Given,
$Δ_{1} = | \begin{array}{ccc} A & B & C \\ x & y & z \\ 2 y & 2 x & xy \end{array} |$
According to gives questions,
$Δ = | \begin{array}{ccc} A x & x^{2} & 1 \\ B y & y^{2} & 1 \\ C z & z^{2} & 1 \end{array} | = x y z | \begin{array}{ccc} A & x & \frac{1}{x} \\ B & y & \frac{1}{y} \\ C & z & \frac{1}{z} \end{array} |$
Operating $C_{3} \to$ xyz $C_{3}$
$= | \begin{array}{ccc} A & x & yz \\ B & y & xz \\ C & z & xy \end{array} | = | \begin{array}{ccc} A & B & C \\ x & y & z \\ zy & zx & xy \end{array} | = Δ_{1}$
$∴ Δ = Δ_{1}$

Karnataka CET-2017

Matrix and Determinant

79235 If $f (x), g (x)$ and $h (x)$ are three polynomials of degree 2 and
$Δ (x) = | \begin{array}{ccc} f (x) & g (x) & h (x) \\ f^{'} (x) & g^{'} (x) & h^{'} (x) \\ f^{''} (x) & g^{''} (x) & h^{''} (x) \end{array} |$
then $Δ (x)$ is a polynomial of degree 2 and

1 2

2 3

3 0

4 atmost 3

Explanation:

(C) : It is given that
$f (x), g (x)$ and $h (x)$ are the polynomials of degree 2 ,
$∴ f^{m} (x) = g^{m} (x) = h^{m} (x) = 0$
$Δ (x) = | \begin{array}{ccc} f (x) & g (x) & h (x) \\ f^{'} (x) & g^{'} (x) & h^{'} (x) \\ f^{''} (x) & g^{''} (x) & h^{''} (x) \end{array} |$
$Δ^{'} (x) = | \begin{array}{ccc} f^{'} (x) & g^{'} (x) & h^{'} (x) \\ f^{'} (x) & g^{'} (x) & h^{'} (x) \\ f^{''} (x) & g^{''} (x) & h^{''} (x) \end{array} | + | \begin{array}{ccc} f (x) & g (x) & h (x) \\ f^{''} (x) & g^{''} (x) & h^{''} (x) \\ f^{''} (x) & g^{''} (x) & h^{''} (x) \end{array}$
$+ | \begin{array}{ccc} f (x) & g (x) & h (x) \\ f^{'} (x) & g^{'} (x) & h^{'} (x) \\ f^{'''} (x) & g^{'''} (x) & h^{'''} (x) \end{array} |$
$Δ^{'} (x) = 0 + 0 + 0 = 0$
$Δ (x) = constant$
Thus, $Δ (x)$ is the polynomial of degree zero.

VITEEE-2010

Matrix and Determinant

79236 If $C = 2 \cos θ$ , then the value of the determinant
$Δ = | \begin{array}{lll} C & 1 & 0 \\ 1 & C & 1 \\ 6 & 1 & C \end{array} |$ is

1

\frac{\sin 4 θ}{\sin θ}

2

\frac{2 \sin^{2} 2 θ}{\sin θ}

3

4 \cos^{2} θ (2 \cos θ - 1)

4 None of the above

Explanation:

(D) : Given, $C = 2 \cos θ$
According to given summation
Let, $Δ = | \begin{array}{lll} C & 1 & 0 \\ 1 & C & 1 \\ 6 & 1 & C \end{array} | = C (C^{2} - 1) - 1 (C - 6) = C^{3} - 2 C + 6$
Put $C = 2 \cos θ$ , we get
$Δ = (2 \cos θ)^{3} - 2 (2 \cos θ) + 6$
$Δ = 8 \cos^{3} θ - 4 \cos θ + 6$

BCECE-2012

Matrix and Determinant

79237 The only integral root of the equation
$| \begin{array}{ccc} 2 - y & 2 & 3 \\ 2 & 5 - y & 6 \\ 3 & 4 & 10 - y \end{array} | = 0$ is

1

y = 3

2

y = 2

3

y = 1

4 none of these

Explanation:

(C) : Given,
$| \begin{array}{ccc} 2 - y & 2 & 3 \\ 2 & 5 - y & 6 \\ 3 & 4 & 10 - y \end{array} | = 0$
$(2 - y) [(5 - y) (10 - y) - 24] - 2 [20 - 2 y - 18] + 3 (8 -$ $15 + 3 y) = 0$
$(2 - y) [50 - 5 y - 10 y + y^{2} - 24] - 40 + 4 y + 36 - 21 +$ $9 y = 0$
$(2 - y) [y^{2} - 15 y + 26] + 13 y - 25 = 0$
$2 y^{2} - 30 y + 52 - y^{3} + 15 y^{2} - 26 y + 13 y - 25 = 0$
$- y^{3} + 17 y^{2} - 43 y + 27 = 0$
Only $y = 1$ satisfy this equation.

AMU-2016

Matrix and Determinant

79238 If
$= | \begin{array}{ccc} 2 \cos x & 1 & 0 \\ x - \frac{π}{2} & 2 \cos x & 1 \end{array} | then f^{'} (π / 2) is$
equal to.

1 0

2 2

3

\frac{π}{2}

4

π - 6

Explanation:

(B) : Given,
$f (x) = | \begin{array}{ccc} 2 \cos x & 1 & 0 \\ π - π / 2 & 2 \cos x & 1 \\ 0 & 1 & 2 \cos x \end{array} |$
Expanding always first column.
$= 2 \cos x (4 \cos^{2} x - 1) - [(x - π / 2) 2 \cos x - 0] + 0$
$= 8 \cos^{3} x - 2 \cos x - 2 x \cos x + π \cos x$
$(x) = 24 \cos^{2} x (- \sin x) + 2 \sin x - 2 (\cos x - x \sin x)$
$- π \sin x$
$f^{'} (π / 2) = 0 + 2 + 0 = 2$

AP EAMCET-2010

Matrix and Determinant

79234 Let $Δ = | \begin{array}{lll} A x & x^{2} & 1 \\ B y & y^{2} & 1 \\ C z & z^{2} & 1 \end{array} |$ and $Δ_{1} = | \begin{array}{ccc} A & B & C \\ x & y & z \\ z y & z x & x y \end{array} |$ then

1

Δ_{1} = - Δ

2

Δ_{1} = Δ

3

Δ_{1} = 2 Δ

4

Δ_{1} \neq Δ

Explanation:

(B) : Given,
$Δ_{1} = | \begin{array}{ccc} A & B & C \\ x & y & z \\ 2 y & 2 x & xy \end{array} |$
According to gives questions,
$Δ = | \begin{array}{ccc} A x & x^{2} & 1 \\ B y & y^{2} & 1 \\ C z & z^{2} & 1 \end{array} | = x y z | \begin{array}{ccc} A & x & \frac{1}{x} \\ B & y & \frac{1}{y} \\ C & z & \frac{1}{z} \end{array} |$
Operating $C_{3} \to$ xyz $C_{3}$
$= | \begin{array}{ccc} A & x & yz \\ B & y & xz \\ C & z & xy \end{array} | = | \begin{array}{ccc} A & B & C \\ x & y & z \\ zy & zx & xy \end{array} | = Δ_{1}$
$∴ Δ = Δ_{1}$

Karnataka CET-2017

Matrix and Determinant

79235 If $f (x), g (x)$ and $h (x)$ are three polynomials of degree 2 and
$Δ (x) = | \begin{array}{ccc} f (x) & g (x) & h (x) \\ f^{'} (x) & g^{'} (x) & h^{'} (x) \\ f^{''} (x) & g^{''} (x) & h^{''} (x) \end{array} |$
then $Δ (x)$ is a polynomial of degree 2 and

1 2

2 3

3 0

4 atmost 3

Explanation:

(C) : It is given that
$f (x), g (x)$ and $h (x)$ are the polynomials of degree 2 ,
$∴ f^{m} (x) = g^{m} (x) = h^{m} (x) = 0$
$Δ (x) = | \begin{array}{ccc} f (x) & g (x) & h (x) \\ f^{'} (x) & g^{'} (x) & h^{'} (x) \\ f^{''} (x) & g^{''} (x) & h^{''} (x) \end{array} |$
$Δ^{'} (x) = | \begin{array}{ccc} f^{'} (x) & g^{'} (x) & h^{'} (x) \\ f^{'} (x) & g^{'} (x) & h^{'} (x) \\ f^{''} (x) & g^{''} (x) & h^{''} (x) \end{array} | + | \begin{array}{ccc} f (x) & g (x) & h (x) \\ f^{''} (x) & g^{''} (x) & h^{''} (x) \\ f^{''} (x) & g^{''} (x) & h^{''} (x) \end{array}$
$+ | \begin{array}{ccc} f (x) & g (x) & h (x) \\ f^{'} (x) & g^{'} (x) & h^{'} (x) \\ f^{'''} (x) & g^{'''} (x) & h^{'''} (x) \end{array} |$
$Δ^{'} (x) = 0 + 0 + 0 = 0$
$Δ (x) = constant$
Thus, $Δ (x)$ is the polynomial of degree zero.

VITEEE-2010

Matrix and Determinant

79236 If $C = 2 \cos θ$ , then the value of the determinant
$Δ = | \begin{array}{lll} C & 1 & 0 \\ 1 & C & 1 \\ 6 & 1 & C \end{array} |$ is

1

\frac{\sin 4 θ}{\sin θ}

2

\frac{2 \sin^{2} 2 θ}{\sin θ}

3

4 \cos^{2} θ (2 \cos θ - 1)

4 None of the above

Explanation:

(D) : Given, $C = 2 \cos θ$
According to given summation
Let, $Δ = | \begin{array}{lll} C & 1 & 0 \\ 1 & C & 1 \\ 6 & 1 & C \end{array} | = C (C^{2} - 1) - 1 (C - 6) = C^{3} - 2 C + 6$
Put $C = 2 \cos θ$ , we get
$Δ = (2 \cos θ)^{3} - 2 (2 \cos θ) + 6$
$Δ = 8 \cos^{3} θ - 4 \cos θ + 6$

BCECE-2012

Matrix and Determinant

79237 The only integral root of the equation
$| \begin{array}{ccc} 2 - y & 2 & 3 \\ 2 & 5 - y & 6 \\ 3 & 4 & 10 - y \end{array} | = 0$ is

1

y = 3

2

y = 2

3

y = 1

4 none of these

Explanation:

(C) : Given,
$| \begin{array}{ccc} 2 - y & 2 & 3 \\ 2 & 5 - y & 6 \\ 3 & 4 & 10 - y \end{array} | = 0$
$(2 - y) [(5 - y) (10 - y) - 24] - 2 [20 - 2 y - 18] + 3 (8 -$ $15 + 3 y) = 0$
$(2 - y) [50 - 5 y - 10 y + y^{2} - 24] - 40 + 4 y + 36 - 21 +$ $9 y = 0$
$(2 - y) [y^{2} - 15 y + 26] + 13 y - 25 = 0$
$2 y^{2} - 30 y + 52 - y^{3} + 15 y^{2} - 26 y + 13 y - 25 = 0$
$- y^{3} + 17 y^{2} - 43 y + 27 = 0$
Only $y = 1$ satisfy this equation.

AMU-2016

Matrix and Determinant

79238 If
$= | \begin{array}{ccc} 2 \cos x & 1 & 0 \\ x - \frac{π}{2} & 2 \cos x & 1 \end{array} | then f^{'} (π / 2) is$
equal to.

1 0

2 2

3

\frac{π}{2}

4

π - 6

Explanation:

(B) : Given,
$f (x) = | \begin{array}{ccc} 2 \cos x & 1 & 0 \\ π - π / 2 & 2 \cos x & 1 \\ 0 & 1 & 2 \cos x \end{array} |$
Expanding always first column.
$= 2 \cos x (4 \cos^{2} x - 1) - [(x - π / 2) 2 \cos x - 0] + 0$
$= 8 \cos^{3} x - 2 \cos x - 2 x \cos x + π \cos x$
$(x) = 24 \cos^{2} x (- \sin x) + 2 \sin x - 2 (\cos x - x \sin x)$
$- π \sin x$
$f^{'} (π / 2) = 0 + 2 + 0 = 2$

AP EAMCET-2010

Matrix and Determinant

79234 Let $Δ = | \begin{array}{lll} A x & x^{2} & 1 \\ B y & y^{2} & 1 \\ C z & z^{2} & 1 \end{array} |$ and $Δ_{1} = | \begin{array}{ccc} A & B & C \\ x & y & z \\ z y & z x & x y \end{array} |$ then

1

Δ_{1} = - Δ

2

Δ_{1} = Δ

3

Δ_{1} = 2 Δ

4

Δ_{1} \neq Δ

Explanation:

(B) : Given,
$Δ_{1} = | \begin{array}{ccc} A & B & C \\ x & y & z \\ 2 y & 2 x & xy \end{array} |$
According to gives questions,
$Δ = | \begin{array}{ccc} A x & x^{2} & 1 \\ B y & y^{2} & 1 \\ C z & z^{2} & 1 \end{array} | = x y z | \begin{array}{ccc} A & x & \frac{1}{x} \\ B & y & \frac{1}{y} \\ C & z & \frac{1}{z} \end{array} |$
Operating $C_{3} \to$ xyz $C_{3}$
$= | \begin{array}{ccc} A & x & yz \\ B & y & xz \\ C & z & xy \end{array} | = | \begin{array}{ccc} A & B & C \\ x & y & z \\ zy & zx & xy \end{array} | = Δ_{1}$
$∴ Δ = Δ_{1}$

Karnataka CET-2017

Matrix and Determinant

79235 If $f (x), g (x)$ and $h (x)$ are three polynomials of degree 2 and
$Δ (x) = | \begin{array}{ccc} f (x) & g (x) & h (x) \\ f^{'} (x) & g^{'} (x) & h^{'} (x) \\ f^{''} (x) & g^{''} (x) & h^{''} (x) \end{array} |$
then $Δ (x)$ is a polynomial of degree 2 and

1 2

2 3

3 0

4 atmost 3

Explanation:

(C) : It is given that
$f (x), g (x)$ and $h (x)$ are the polynomials of degree 2 ,
$∴ f^{m} (x) = g^{m} (x) = h^{m} (x) = 0$
$Δ (x) = | \begin{array}{ccc} f (x) & g (x) & h (x) \\ f^{'} (x) & g^{'} (x) & h^{'} (x) \\ f^{''} (x) & g^{''} (x) & h^{''} (x) \end{array} |$
$Δ^{'} (x) = | \begin{array}{ccc} f^{'} (x) & g^{'} (x) & h^{'} (x) \\ f^{'} (x) & g^{'} (x) & h^{'} (x) \\ f^{''} (x) & g^{''} (x) & h^{''} (x) \end{array} | + | \begin{array}{ccc} f (x) & g (x) & h (x) \\ f^{''} (x) & g^{''} (x) & h^{''} (x) \\ f^{''} (x) & g^{''} (x) & h^{''} (x) \end{array}$
$+ | \begin{array}{ccc} f (x) & g (x) & h (x) \\ f^{'} (x) & g^{'} (x) & h^{'} (x) \\ f^{'''} (x) & g^{'''} (x) & h^{'''} (x) \end{array} |$
$Δ^{'} (x) = 0 + 0 + 0 = 0$
$Δ (x) = constant$
Thus, $Δ (x)$ is the polynomial of degree zero.

VITEEE-2010

Matrix and Determinant

79236 If $C = 2 \cos θ$ , then the value of the determinant
$Δ = | \begin{array}{lll} C & 1 & 0 \\ 1 & C & 1 \\ 6 & 1 & C \end{array} |$ is

1

\frac{\sin 4 θ}{\sin θ}

2

\frac{2 \sin^{2} 2 θ}{\sin θ}

3

4 \cos^{2} θ (2 \cos θ - 1)

4 None of the above

Explanation:

(D) : Given, $C = 2 \cos θ$
According to given summation
Let, $Δ = | \begin{array}{lll} C & 1 & 0 \\ 1 & C & 1 \\ 6 & 1 & C \end{array} | = C (C^{2} - 1) - 1 (C - 6) = C^{3} - 2 C + 6$
Put $C = 2 \cos θ$ , we get
$Δ = (2 \cos θ)^{3} - 2 (2 \cos θ) + 6$
$Δ = 8 \cos^{3} θ - 4 \cos θ + 6$

BCECE-2012

Matrix and Determinant

79237 The only integral root of the equation
$| \begin{array}{ccc} 2 - y & 2 & 3 \\ 2 & 5 - y & 6 \\ 3 & 4 & 10 - y \end{array} | = 0$ is

1

y = 3

2

y = 2

3

y = 1

4 none of these

Explanation:

(C) : Given,
$| \begin{array}{ccc} 2 - y & 2 & 3 \\ 2 & 5 - y & 6 \\ 3 & 4 & 10 - y \end{array} | = 0$
$(2 - y) [(5 - y) (10 - y) - 24] - 2 [20 - 2 y - 18] + 3 (8 -$ $15 + 3 y) = 0$
$(2 - y) [50 - 5 y - 10 y + y^{2} - 24] - 40 + 4 y + 36 - 21 +$ $9 y = 0$
$(2 - y) [y^{2} - 15 y + 26] + 13 y - 25 = 0$
$2 y^{2} - 30 y + 52 - y^{3} + 15 y^{2} - 26 y + 13 y - 25 = 0$
$- y^{3} + 17 y^{2} - 43 y + 27 = 0$
Only $y = 1$ satisfy this equation.

AMU-2016

Matrix and Determinant

79238 If
$= | \begin{array}{ccc} 2 \cos x & 1 & 0 \\ x - \frac{π}{2} & 2 \cos x & 1 \end{array} | then f^{'} (π / 2) is$
equal to.

1 0

2 2

3

\frac{π}{2}

4

π - 6

Explanation:

(B) : Given,
$f (x) = | \begin{array}{ccc} 2 \cos x & 1 & 0 \\ π - π / 2 & 2 \cos x & 1 \\ 0 & 1 & 2 \cos x \end{array} |$
Expanding always first column.
$= 2 \cos x (4 \cos^{2} x - 1) - [(x - π / 2) 2 \cos x - 0] + 0$
$= 8 \cos^{3} x - 2 \cos x - 2 x \cos x + π \cos x$
$(x) = 24 \cos^{2} x (- \sin x) + 2 \sin x - 2 (\cos x - x \sin x)$
$- π \sin x$
$f^{'} (π / 2) = 0 + 2 + 0 = 2$

AP EAMCET-2010