Differentiation (Part-38) | S N De

Question 80

If \( p, q, r \) are not in Geometric Progression (G.P.), and

\[ \begin{vmatrix} 1 & \frac{q}{p} & \alpha + \frac{q}{p} \\ 1 & \frac{r}{q} & \alpha + \frac{r}{q} \\ p\alpha+q & q\alpha+r & 0 \end{vmatrix} = 0 \]

then which of the following is true?

(a) \( p\alpha^2 + 2q\alpha + r = 0 \)
(b) \( p\alpha^2 - 2q\alpha + r = 0 \)
(c) \( p\alpha^2 - 2q\alpha - r = 0 \)
(d) \( p\alpha^2 + 2q\alpha + r^2 = 0 \)

Solution:

Let \(\Delta = \begin{vmatrix} 1 & \frac{q}{p} & \alpha + \frac{q}{p} \\ 1 & \frac{r}{q} & \alpha + \frac{r}{q} \\ p\alpha+q & q\alpha+r & 0 \end{vmatrix} = 0 \).

Multiply \(R_1\) by \(p\) and \(R_2\) by \(q\): \[ \frac{1}{pq} \begin{vmatrix} p & q & p\alpha + q \\ q & r & q\alpha + r \\ p\alpha+q & q\alpha+r & 0 \end{vmatrix} = 0 \] Apply column operation \( C_3 \to C_3 - (\alpha C_1 + C_2) \): \[ \begin{vmatrix} p & q & 0 \\ q & r & 0 \\ p\alpha+q & q\alpha+r & -(p\alpha^2 + 2q\alpha + r) \end{vmatrix} = 0 \] Expanding along \( C_3 \): \[ -(p\alpha^2 + 2q\alpha + r) (pr - q^2) = 0 \] Since \( p, q, r \) are not in G.P., \( q^2 \neq pr \), which means \( pr - q^2 \neq 0 \).
Therefore, \( p\alpha^2 + 2q\alpha + r = 0 \).

Correct Option: (a)

Question 81

If \( a, b, c \) are positive and \( a \neq b \neq c \), then which relation is correct for \( \Delta = \begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix} \)?

(a) \( \Delta = 0 \)
(b) \( \Delta > 0 \)
(c) \( \Delta < 0 \)
(d) None of these

Solution:

Expanding the determinant: \[ \Delta = a(bc - a^2) - b(b^2 - ac) + c(ab - c^2) \] \[ \Delta = 3abc - a^3 - b^3 - c^3 = -(a^3 + b^3 + c^3 - 3abc) \] Using the algebraic identity: \[ a^3 + b^3 + c^3 - 3abc = \frac{1}{2}(a+b+c)[(a-b)^2 + (b-c)^2 + (c-a)^2] \] Since \( a, b, c \) are positive, \( (a+b+c) > 0 \).
Since \( a, b, c \) are distinct (\( a \neq b \neq c \)), the sum of squares is positive.
Thus, \( a^3 + b^3 + c^3 - 3abc > 0 \).
Therefore, \( \Delta = -(positive \ value) < 0 \).

Correct Option: (c)

Question 82

If \( x + 3y = 4 \), \( y + 3z = 7 \), \( 4x + z = 6 \), then \( (x, y, z) = \)

(a) \( (1, 1, 2) \)
(b) \( (1, 2, 1) \)
(c) \( (2, 1, 1) \)
(d) \( (1, -1, 2) \)

Solution:

We can verify the options by substituting them into the equations.

Let's check Option (a) \( x=1, y=1, z=2 \):
Eq 1: \( 1 + 3(1) = 4 \) (True)
Eq 2: \( 1 + 3(2) = 7 \) (True)
Eq 3: \( 4(1) + 2 = 6 \) (True)

Since all three equations are satisfied, this is the correct solution set.

Correct Option: (a)

Question 83

The determinant \( \begin{vmatrix} a^2+10 & ab & ac \\ ab & b^2+10 & bc \\ ac & bc & c^2+10 \end{vmatrix} \) is divisible by—

(a) 100
(b) 15
(c) 30
(d) 45

Solution:

Let \( \Delta = \begin{vmatrix} a^2+10 & ab & ac \\ ab & b^2+10 & bc \\ ac & bc & c^2+10 \end{vmatrix} \)

Multiply \(R_1\) by \(a\), \(R_2\) by \(b\), and \(R_3\) by \(c\), and divide by \(abc\): \[ \Delta = \frac{1}{abc} \begin{vmatrix} a(a^2+10) & a^2b & a^2c \\ ab^2 & b(b^2+10) & b^2c \\ ac^2 & bc^2 & c(c^2+10) \end{vmatrix} \] Now take \(a\) common from \(C_1\), \(b\) from \(C_2\), and \(c\) from \(C_3\): \[ \Delta = \begin{vmatrix} a^2+10 & a^2 & a^2 \\ b^2 & b^2+10 & b^2 \\ c^2 & c^2 & c^2+10 \end{vmatrix} \] Apply \( C_1 \to C_1 - C_2 \) and \( C_2 \to C_2 - C_3 \): \[ \Delta = \begin{vmatrix} 10 & 0 & a^2 \\ -10 & 10 & b^2 \\ 0 & -10 & c^2+10 \end{vmatrix} \] Expanding along \( R_1 \): \[ = 10[10(c^2+10) - (-10)b^2] + a^2(100 - 0) \] \[ = 10[10c^2 + 100 + 10b^2] + 100a^2 \] \[ = 100c^2 + 1000 + 100b^2 + 100a^2 = 100(a^2 + b^2 + c^2 + 10) \] Clearly, the expression has a multiple of 100, meaning it is perfectly divisible by 100.

Correct Option: (a)

Question 84

The value of \( \begin{vmatrix} 1 & 1 & 1+3x \\ 1+3y & 1 & 1 \\ 1 & 1+3z & 1 \end{vmatrix} \) is—

(a) \( 3xyz + xy + yz + zx \)
(b) \( 9(3xyz + xy + yz + zx) \)
(c) \( 9(xy + yz + zx) \)
(d) \( 9xyz \)

Solution:

Let \( \Delta = \begin{vmatrix} 1 & 1 & 1+3x \\ 1+3y & 1 & 1 \\ 1 & 1+3z & 1 \end{vmatrix} \)

Applying \( C_1 \to C_1 - C_2 \) and \( C_2 \to C_2 - C_3 \): \[ \Delta = \begin{vmatrix} 0 & -3x & 1+3x \\ 3y & 0 & 1 \\ -3z & 3z & 1 \end{vmatrix} \] Expanding along \( R_1 \): \[ = -(-3x)[3y(1) - (-3z)(1)] + (1+3x)[3y(3z) - 0] \] \[ = 3x(3y + 3z) + (1+3x)(9yz) \] \[ = 9xy + 9xz + 9yz + 27xyz \] Taking 9 common: \[ = 9(xy + xz + yz + 3xyz) \] This matches Option (b).

Correct Option: (b)

Question 85

The value of \( \begin{vmatrix} 0 & \sin\alpha & -\cos\alpha \\ -\sin\alpha & 0 & \sin\beta \\ \cos\alpha & -\sin\beta & 0 \end{vmatrix} \) is—

(a) 0    (b) -1    (c) 1    (d) 2

Solution:

The given determinant represents a skew-symmetric matrix of odd order (3x3). A fundamental property of skew-symmetric matrices is that the determinant of an odd-ordered skew-symmetric matrix is always zero.

You can also verify by standard expansion: \[ = 0 - \sin\alpha(0 - \sin\beta\cos\alpha) - \cos\alpha(\sin\alpha\sin\beta - 0) \] \[ = \sin\alpha\sin\beta\cos\alpha - \sin\alpha\sin\beta\cos\alpha = 0 \]

Correct Option: (a)

Question 86

If \( \begin{vmatrix} x & \sin\theta & \cos\theta \\ -\sin\theta & -x & 1 \\ \cos\theta & 1 & x \end{vmatrix} = 8 \), then the value of \( x \) is—

(a) -2    (b) 2    (c) 1    (d) -1

Solution:

Expanding the determinant along \( R_1 \): \[ x(-x^2 - 1) - \sin\theta(-x\sin\theta - \cos\theta) + \cos\theta(-\sin\theta + x\cos\theta) = 8 \] \[ -x^3 - x + x\sin^2\theta + \sin\theta\cos\theta - \sin\theta\cos\theta + x\cos^2\theta = 8 \] \[ -x^3 - x + x(\sin^2\theta + \cos^2\theta) = 8 \] Since \( \sin^2\theta + \cos^2\theta = 1 \): \[ -x^3 - x + x(1) = 8 \] \[ -x^3 = 8 \implies x^3 = -8 \implies x = -2 \]

Correct Option: (a)

Question 87

If \( \begin{vmatrix} 2 & 3 & 2 \\ x & x & x \\ 4 & 9 & 1 \end{vmatrix} + 3 = 0 \), then the value of \( x \) is—

(a) 3    (b) 0    (c) -1    (d) 1

Solution:

Taking \( x \) common from \( R_2 \): \[ x \begin{vmatrix} 2 & 3 & 2 \\ 1 & 1 & 1 \\ 4 & 9 & 1 \end{vmatrix} + 3 = 0 \] Evaluating the determinant: \[ x \left[ 2(1 - 9) - 3(1 - 4) + 2(9 - 4) \right] + 3 = 0 \] \[ x \left[ 2(-8) - 3(-3) + 2(5) \right] + 3 = 0 \] \[ x \left[ -16 + 9 + 10 \right] + 3 = 0 \] \[ x(3) + 3 = 0 \implies 3x = -3 \implies x = -1 \]

Correct Option: (c)

Question 88

If the area of a triangle with vertices \( (2, -6), (5, 4) \) and \( (K, 4) \) is 35 sq. units, then the value of \( K \) is—

(a) 12
(b) -2
(c) -12, -2
(d) 12, -2

Solution:

The area of a triangle given by vertices is: \[ \text{Area} = \frac{1}{2} \left| \begin{vmatrix} 2 & -6 & 1 \\ 5 & 4 & 1 \\ K & 4 & 1 \end{vmatrix} \right| = 35 \] \[ \begin{vmatrix} 2 & -6 & 1 \\ 5 & 4 & 1 \\ K & 4 & 1 \end{vmatrix} = \pm 70 \] Expanding the determinant: \[ 2(4 - 4) - (-6)(5 - K) + 1(20 - 4K) = \pm 70 \] \[ 0 + 6(5 - K) + 20 - 4K = \pm 70 \] \[ 30 - 6K + 20 - 4K = \pm 70 \implies 50 - 10K = \pm 70 \] Case 1: \( 50 - 10K = 70 \implies -10K = 20 \implies K = -2 \)
Case 2: \( 50 - 10K = -70 \implies -10K = -120 \implies K = 12 \)
Thus, \( K = 12, -2 \).

Correct Option: (d)

Question 89

If the points \( (K, -2), (5, 2) \) and \( (8, 8) \) are collinear, then the value of \( K \) is—

(a) 1    (b) 2    (c) 3    (d) 4

Solution:

For three points to be collinear, the area of the triangle formed by them must be zero. \[ \begin{vmatrix} K & -2 & 1 \\ 5 & 2 & 1 \\ 8 & 8 & 1 \end{vmatrix} = 0 \] Expanding the determinant: \[ K(2 - 8) - (-2)(5 - 8) + 1(40 - 16) = 0 \] \[ -6K + 2(-3) + 24 = 0 \] \[ -6K - 6 + 24 = 0 \implies -6K + 18 = 0 \implies 6K = 18 \implies K = 3 \]

Correct Option: (c)

Question 90

If \( (a, 0), (0, b) \) and \( (x, y) \) are collinear, then the value of \( \frac{x}{a} + \frac{y}{b} - 1 \) is—

(a) 0    (b) 1    (c) -1    (d) 2

Solution:

The equation of the line passing through the points \( (a, 0) \) and \( (0, b) \) in intercept form is: \[ \frac{x}{a} + \frac{y}{b} = 1 \] Since the point \( (x, y) \) is collinear with these two points, it must lie on this straight line. Thus, it satisfies the line's equation: \[ \frac{x}{a} + \frac{y}{b} = 1 \implies \frac{x}{a} + \frac{y}{b} - 1 = 0 \]

Correct Option: (a)

Question 91

For what value of \( m \) will the system of equations \( x + y - 2z = 0 \), \( 2x - 3y + z = 0 \) and \( x - 5y + 4z = m \) have a solution?

(a) 1    (b) -1    (c) 0    (d) None of these

Solution:

Check the determinant of the coefficient matrix \( \Delta \): \[ \Delta = \begin{vmatrix} 1 & 1 & -2 \\ 2 & -3 & 1 \\ 1 & -5 & 4 \end{vmatrix} \] \[ = 1(-12 + 5) - 1(8 - 1) - 2(-10 + 3) = -7 - 7 - 2(-7) = -14 + 14 = 0 \] Since \( \Delta = 0 \), for the system to be consistent (have a solution), we must also have \( \Delta_x = \Delta_y = \Delta_z = 0 \) (for infinitely many solutions). Let's evaluate \( \Delta_z \): \[ \Delta_z = \begin{vmatrix} 1 & 1 & 0 \\ 2 & -3 & 0 \\ 1 & -5 & m \end{vmatrix} \] Expanding along \( C_3 \): \[ = m(-3 - 2) = -5m \] For the system to be solvable, \( -5m = 0 \implies m = 0 \).

Correct Option: (c)

Question 92

If \( (at_1^2, 2at_1), (at_2^2, 2at_2) \) and \( (at_3^2, 2at_3) \) are the vertices of a triangle, then the area of the triangle is—

(a) \( a^2(t_1 - t_2)(t_2 - t_3)(t_3 - t_1) \) sq. units
(b) \( a(t_1 - t_2)(t_2 - t_3)(t_3 - t_1) \) sq. units
(c) \( a^3(t_1 - t_2)(t_2 - t_3)(t_3 - t_1) \) sq. units
(d) \( a^4(t_1 - t_2)(t_2 - t_3)(t_3 - t_1) \) sq. units

Solution:

The area of a triangle with the given coordinates is: \[ \text{Area} = \frac{1}{2} \left| \begin{vmatrix} at_1^2 & 2at_1 & 1 \\ at_2^2 & 2at_2 & 1 \\ at_3^2 & 2at_3 & 1 \end{vmatrix} \right| \] Take \( a \) common from \( C_1 \) and \( 2a \) common from \( C_2 \): \[ = \frac{1}{2} (a)(2a) \left| \begin{vmatrix} t_1^2 & t_1 & 1 \\ t_2^2 & t_2 & 1 \\ t_3^2 & t_3 & 1 \end{vmatrix} \right| = a^2 \left| \begin{vmatrix} t_1^2 & t_1 & 1 \\ t_2^2 & t_2 & 1 \\ t_3^2 & t_3 & 1 \end{vmatrix} \right| \] This represents the absolute value of a standard Vandermonde determinant. Evaluating it yields the product of the differences of variables: \[ = a^2 |(t_1 - t_2)(t_2 - t_3)(t_3 - t_1)| \] Represented generally as a positive area expression.

Correct Option: (a)

Question 93

The equivalent determinant of \( \begin{vmatrix} a & b & c \\ x & y & z \\ yz & zx & xy \end{vmatrix} \) is—

(a) \( \begin{vmatrix} ax & by & cz \\ x^2 & y^2 & z^2 \\ 1 & 1 & 1 \end{vmatrix} \)
(b) \( \begin{vmatrix} 1 & 1 & 1 \\ x & y & z \\ xy & zx & xy \end{vmatrix} \)
(c) \( \begin{vmatrix} x^2 & y^2 & z^2 \\ x & y & z \\ xy & yz & zx \end{vmatrix} \)
(d) \( abc \begin{vmatrix} 1 & 1 & 1 \\ x & y & z \\ x^2 & y^2 & z^2 \end{vmatrix} \)

Solution:

Let \( \Delta = \begin{vmatrix} a & b & c \\ x & y & z \\ yz & zx & xy \end{vmatrix} \)

Multiply \( C_1 \) by \( x \), \( C_2 \) by \( y \), and \( C_3 \) by \( z \) and balance by dividing by \( xyz \): \[ \Delta = \frac{1}{xyz} \begin{vmatrix} ax & by & cz \\ x^2 & y^2 & z^2 \\ xyz & xyz & xyz \end{vmatrix} \] Take \( xyz \) common from the third row (\( R_3 \)): \[ \Delta = \frac{xyz}{xyz} \begin{vmatrix} ax & by & cz \\ x^2 & y^2 & z^2 \\ 1 & 1 & 1 \end{vmatrix} = \begin{vmatrix} ax & by & cz \\ x^2 & y^2 & z^2 \\ 1 & 1 & 1 \end{vmatrix} \]

Correct Option: (a)