Differentiation (Part-38) | S N De

Question 94

If $a^2 + b^2 + c^2 = -2$ and

$$f(x) = \begin{vmatrix} 1 + a^2x & (1+b^2)x & (1+c^2)x \\ (1+a^2)x & 1+b^2x & (1+c^2)x \\ (1+a^2)x & (1+b^2)x & 1+c^2x \end{vmatrix}$$

then the degree of the polynomial $f(x)$ is—

(a) 0

(b) 1

(c) 2

(d) 3

Solution:

Let's simplify the determinant using the column operation $C_1 \to C_1 + C_2 + C_3$. The first column becomes:

Row 1: $$1 + a^2x + x + b^2x + x + c^2x = 1 + x(a^2 + b^2 + c^2 + 2)$$ Row 2: $$x + a^2x + 1 + b^2x + x + c^2x = 1 + x(a^2 + b^2 + c^2 + 2)$$ Row 3: $$x + a^2x + x + b^2x + 1 + c^2x = 1 + x(a^2 + b^2 + c^2 + 2)$$

Given $a^2 + b^2 + c^2 = -2$, we have $a^2 + b^2 + c^2 + 2 = 0$. Therefore, all elements in $C_1$ reduce to 1.

$$f(x) = \begin{vmatrix} 1 & (1+b^2)x & (1+c^2)x \\ 1 & 1+b^2x & (1+c^2)x \\ 1 & (1+b^2)x & 1+c^2x \end{vmatrix}$$

Applying row operations $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_1$:

$$f(x) = \begin{vmatrix} 1 & (1+b^2)x & (1+c^2)x \\ 0 & 1-x & 0 \\ 0 & 0 & 1-x \end{vmatrix}$$

Expanding the determinant along $C_1$:

$$f(x) = 1 \cdot (1-x) \cdot (1-x) = (1-x)^2 = x^2 - 2x + 1$$

The highest power of $x$ is 2, so the degree of the polynomial is 2. Correct Option: (c) 2

Question 95

If $A = \begin{pmatrix} 5 & 5\alpha & \alpha \\ 0 & \alpha & 5\alpha \\ 0 & 0 & 5 \end{pmatrix}$ and $|A^2| = 25$, then the value of $|\alpha|$ is—

(a) $\frac{1}{5}$

(b) 5

(c) $5^2$

(d) 1

Solution:

The determinant of an upper triangular matrix is the product of its principal diagonal elements.

$$|A| = 5 \cdot \alpha \cdot 5 = 25\alpha$$

We are given $|A^2| = 25$. Using the property $|A^n| = |A|^n$, we get:

$$|A|^2 = 25 \implies (25\alpha)^2 = 25$$ $$625\alpha^2 = 25 \implies \alpha^2 = \frac{25}{625} = \frac{1}{25}$$

Taking the square root on both sides:

$$|\alpha| = \frac{1}{5}$$

Correct Option: (a) $\frac{1}{5}$

Question 96

For all $\alpha, \beta, \gamma \in \mathbb{R}$, the value of

$$\begin{vmatrix} (e^{i\alpha} + e^{-i\alpha})^2 & (e^{i\alpha} - e^{-i\alpha})^2 & 1 \\ (e^{i\beta} + e^{-i\beta})^2 & (e^{i\beta} - e^{-i\beta})^2 & 1 \\ (e^{i\gamma} + e^{-i\gamma})^2 & (e^{i\gamma} - e^{-i\gamma})^2 & 1 \end{vmatrix}$$

is—

(a) $a^{i(\alpha+\beta+\gamma)}$

(b) $e^{-i(\alpha+\beta+\gamma)}$

(c) $e^{i\alpha\beta\gamma}$

(d) 0

Solution:

Let's apply the column operation $C_1 \to C_1 - C_2$. Recall the algebraic identity: $(a+b)^2 - (a-b)^2 = 4ab$. For the first row:

$$(e^{i\alpha} + e^{-i\alpha})^2 - (e^{i\alpha} - e^{-i\alpha})^2 = 4(e^{i\alpha})(e^{-i\alpha}) = 4(1) = 4$$

Similarly, for the second and third rows, the values will also be 4. The determinant becomes:

$$\begin{vmatrix} 4 & (e^{i\alpha} - e^{-i\alpha})^2 & 1 \\ 4 & (e^{i\beta} - e^{-i\beta})^2 & 1 \\ 4 & (e^{i\gamma} - e^{-i\gamma})^2 & 1 \end{vmatrix}$$

Now, take 4 common from $C_1$:

$$4 \begin{vmatrix} 1 & (e^{i\alpha} - e^{-i\alpha})^2 & 1 \\ 1 & (e^{i\beta} - e^{-i\beta})^2 & 1 \\ 1 & (e^{i\gamma} - e^{-i\gamma})^2 & 1 \end{vmatrix}$$

Since column 1 ($C_1$) and column 3 ($C_3$) are identical, the determinant evaluates to 0. Correct Option: (d) 0

Question 97

If $i^2 = -1, \forall n \in \mathbb{Z}^+$, the value of

$$\begin{vmatrix} i^n & i^{n+1} & i^{n+2} \\ i^{n+5} & i^{n+4} & i^{n+3} \\ i^{n+6} & i^{n+7} & i^{n+8} \end{vmatrix}$$

is—

(a) 1 if $n$ is a multiple of 4

(b) $i$ if $n$ is a multiple of 4

(c) $(-i)$ for all values of $n$

(d) 0 for all values of $n$

Solution:

Let's extract common factors from each row: Take $i^n$ common from $R_1$, $i^{n+3}$ common from $R_2$, and $i^{n+6}$ common from $R_3$. The determinant becomes:

$$i^n \cdot i^{n+3} \cdot i^{n+6} \begin{vmatrix} 1 & i & i^2 \\ i^2 & i & 1 \\ 1 & i & i^2 \end{vmatrix}$$

Looking at the simplified matrix, Row 1 ($R_1 = 1, i, i^2$) and Row 3 ($R_3 = 1, i, i^2$) are completely identical. Because two rows are identical, the value of the determinant is 0 regardless of the value of $n$. Correct Option: (d) 0 for all values of n

Question 98

The value of $\Delta = \begin{vmatrix} 1 & 21 & 28 \\ 0 & \sec\theta & \tan\theta \\ 0 & \tan\theta & \sec\theta \end{vmatrix}$ is—

(a) 0

(b) 1

(c) 525

(d) -1

Solution:

We can evaluate the determinant by expanding along the first column ($C_1$), as it contains two zeros.

$$\Delta = 1 \cdot (\sec\theta \cdot \sec\theta - \tan\theta \cdot \tan\theta) - 0 + 0$$ $$\Delta = \sec^2\theta - \tan^2\theta$$

By the standard trigonometric identity, we know $\sec^2\theta - \tan^2\theta = 1$. Correct Option: (b) 1

Group-B: Fill in the Blanks

Question 99

If $\Delta = \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix}$ and $\Delta' = \begin{vmatrix} a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \\ a_1 & b_1 & c_1 \end{vmatrix}$, then $\Delta =$ ____

(a) $-\Delta'$

(b) $\Delta'$

(c) 0

(d) $2\Delta'$

Solution:

We can transform $\Delta'$ into $\Delta$ by performing row exchanges. Remember that swapping two adjacent rows of a determinant changes its sign. Start with:

$$\Delta' = \begin{vmatrix} a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \\ a_1 & b_1 & c_1 \end{vmatrix}$$

Step 1: Swap $R_2$ and $R_3$. The sign becomes negative.

$$-\begin{vmatrix} a_2 & b_2 & c_2 \\ a_1 & b_1 & c_1 \\ a_3 & b_3 & c_3 \end{vmatrix}$$

Step 2: Swap $R_1$ and $R_2$. The sign flips back to positive.

$$+\begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} = \Delta$$

Since we did two row swaps, the net sign change is $(-1) \times (-1) = +1$. Therefore, $\Delta = \Delta'$. Correct Option: (b) $\Delta'$

Question 100

The value of the determinant $\begin{vmatrix} \sin\theta & \cos\theta \\ -\cos\theta & \sin\theta \end{vmatrix}$ is ____

(a) 1

(b) -1

(c) 0

(d) $\sin\theta\cos\theta$

Solution:

Expanding the $2 \times 2$ determinant:

$$= (\sin\theta \cdot \sin\theta) - (\cos\theta \cdot -\cos\theta)$$ $$= \sin^2\theta - (-\cos^2\theta)$$ $$= \sin^2\theta + \cos^2\theta$$

Using the fundamental trigonometric identity, $\sin^2\theta + \cos^2\theta = 1$. Correct Option: (a) 1

Question 101

If A and B are two square matrices of the same order, then $\det(AB) =$ ____

(a) $\det A + \det B$

(b) $\det A - \det B$

(c) $(\det A).(\det B)$

(d) $(\det A)(\det B) + \det A + \det B$

Solution:

According to the multiplicativity property of determinants for square matrices of the same order, the determinant of a product of matrices equals the product of their individual determinants.

$$\det(AB) = \det(A) \cdot \det(B)$$

Correct Option: (c) $(\det A).(\det B)$

Question 102

If $\begin{vmatrix} 1+x & 1 & 1 \\ 1 & 1+y & 1 \\ 1 & 1 & 1+z \end{vmatrix} = 0$ and $xyz \neq 0$, then $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} =$ ____

(a) 0

(b) 1

(c) -1

(d) $\frac{1}{2}$

Solution:

Take $x, y, z$ common from $R_1, R_2, R_3$ respectively:

$$xyz \begin{vmatrix} 1/x + 1 & 1/x & 1/x \\ 1/y & 1/y + 1 & 1/y \\ 1/z & 1/z & 1/z + 1 \end{vmatrix} = 0$$

Apply row operation $R_1 \to R_1 + R_2 + R_3$: The first row becomes $(1 + 1/x + 1/y + 1/z)$ for all elements. Take this term common from $R_1$:

$$xyz \left(1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{z}\right) \begin{vmatrix} 1 & 1 & 1 \\ 1/y & 1/y + 1 & 1/y \\ 1/z & 1/z & 1/z + 1 \end{vmatrix} = 0$$

Evaluate the remaining determinant by $C_2 \to C_2 - C_1$ and $C_3 \to C_3 - C_1$:

$$\begin{vmatrix} 1 & 0 & 0 \\ 1/y & 1 & 0 \\ 1/z & 0 & 1 \end{vmatrix} = 1 \cdot (1 - 0) = 1$$

Thus, $xyz \left(1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{z}\right) = 0$. Since $xyz \neq 0$:

$$1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0 \implies \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = -1$$

Correct Option: (c) -1

Question 103

$\begin{vmatrix} 1 & a & bc \\ 1 & b & ca \\ 1 & c & ab \end{vmatrix} =$ ____

(a) $\begin{vmatrix} 1 & bc & a \\ 1 & ca & b \\ 1 & ab & c \end{vmatrix}$

(b) $\begin{vmatrix} 1 & a^2 & a \\ 1 & b^2 & b \\ 1 & c^2 & c \end{vmatrix}$

(c) $\begin{vmatrix} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{vmatrix}$

(d) $\begin{vmatrix} a & 1 & a^2 \\ b & 1 & b^2 \\ c & 1 & c^2 \end{vmatrix}$

Solution:

Let $\Delta = \begin{vmatrix} 1 & a & bc \\ 1 & b & ca \\ 1 & c & ab \end{vmatrix}$. Multiply $R_1$ by $a$, $R_2$ by $b$, $R_3$ by $c$, and balance it by dividing the determinant by $abc$:

$$\Delta = \frac{1}{abc} \begin{vmatrix} a & a^2 & abc \\ b & b^2 & abc \\ c & c^2 & abc \end{vmatrix}$$

Take $abc$ common from $C_3$:

$$\Delta = \frac{abc}{abc} \begin{vmatrix} a & a^2 & 1 \\ b & b^2 & 1 \\ c & c^2 & 1 \end{vmatrix} = \begin{vmatrix} a & a^2 & 1 \\ b & b^2 & 1 \\ c & c^2 & 1 \end{vmatrix}$$

Swap $C_2$ and $C_3$ (changes sign):

$$-\begin{vmatrix} a & 1 & a^2 \\ b & 1 & b^2 \\ c & 1 & c^2 \end{vmatrix}$$

Swap $C_1$ and $C_2$ (changes sign back to positive):

$$+\begin{vmatrix} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{vmatrix}$$

Correct Option: (c) $\begin{vmatrix} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{vmatrix}$

Question 104

$\begin{vmatrix} 1 & 1 & 1 \\ ^nC_1 & ^{n+1}C_1 & ^{n+2}C_1 \\ ^{n+1}C_2 & ^{n+2}C_2 & ^{n+3}C_2 \end{vmatrix} =$ ____

(a) 0

(b) 1

(c) -1

(d) n

Solution:

Expand the combinatorial terms:

Row 2: $$^nC_1 = n, \ ^{n+1}C_1 = n+1, \ ^{n+2}C_1 = n+2$$ Row 3: $$^{n+1}C_2 = \frac{n(n+1)}{2}, \ ^{n+2}C_2 = \frac{(n+1)(n+2)}{2}, \ ^{n+3}C_2 = \frac{(n+2)(n+3)}{2}$$

The determinant becomes:

$$\begin{vmatrix} 1 & 1 & 1 \\ n & n+1 & n+2 \\ \frac{n(n+1)}{2} & \frac{(n+1)(n+2)}{2} & \frac{(n+2)(n+3)}{2} \end{vmatrix}$$

Apply column operations $C_3 \to C_3 - C_2$ and $C_2 \to C_2 - C_1$:

$C_2$ middle term: $$(n+1) - n = 1$$ $C_3$ middle term: $$(n+2) - (n+1) = 1$$ $C_2$ bottom term: $$\frac{n^2+3n+2 - (n^2+n)}{2} = \frac{2n+2}{2} = n+1$$ $C_3$ bottom term: $$\frac{n^2+5n+6 - (n^2+3n+2)}{2} = \frac{2n+4}{2} = n+2$$

Matrix is now:

$$\begin{vmatrix} 1 & 0 & 0 \\ n & 1 & 1 \\ \frac{n(n+1)}{2} & n+1 & n+2 \end{vmatrix}$$

Expanding along $R_1$:

$$= 1 \cdot (1(n+2) - 1(n+1)) = n+2 - n - 1 = 1$$

Correct Option: (b) 1

Question 105

If the coordinates of A, B, and C are $(\alpha, \frac{1}{\alpha})$, $(\beta, \frac{1}{\beta})$, and $(\gamma, \frac{1}{\gamma})$ respectively, then the area of $\Delta ABC$ = ____

(a) $(\alpha-\beta)(\beta-\gamma)(\gamma-\alpha)$ sq. units

(b) $\frac{1}{2}\left(\alpha - \frac{1}{\alpha}\right)\left(\beta - \frac{1}{\beta}\right)\left(\gamma - \frac{1}{\gamma}\right)$ sq. units

(c) $\frac{1}{2} \left[ \alpha\left(\frac{1}{\beta} - \frac{1}{\gamma}\right) + \beta\left(\frac{1}{\gamma} - \frac{1}{\alpha}\right) + \gamma\left(\frac{1}{\alpha} - \frac{1}{\beta}\right) \right]$ sq. units

(d) $\left[ \frac{1}{\alpha}(\beta - \gamma) + \frac{1}{\beta}(\gamma - \alpha) + \frac{1}{\gamma}(\alpha - \beta) \right]$ sq. units

Solution:

The area of a triangle formed by coordinates $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is given by the determinant formula:

$$\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$$

Substitute the given coordinates:

$$x_1 = \alpha, \ y_1 = \frac{1}{\alpha}$$ $$x_2 = \beta, \ y_2 = \frac{1}{\beta}$$ $$x_3 = \gamma, \ y_3 = \frac{1}{\gamma}$$

Plugging these values into the formula yields:

$$\text{Area} = \frac{1}{2} \left| \alpha\left(\frac{1}{\beta} - \frac{1}{\gamma}\right) + \beta\left(\frac{1}{\gamma} - \frac{1}{\alpha}\right) + \gamma\left(\frac{1}{\alpha} - \frac{1}{\beta}\right) \right|$$

This expression precisely matches option (c). Correct Option: (c) $\frac{1}{2} \left[ \alpha\left(\frac{1}{\beta} - \frac{1}{\gamma}\right) + \beta\left(\frac{1}{\gamma} - \frac{1}{\alpha}\right) + \gamma\left(\frac{1}{\alpha} - \frac{1}{\beta}\right) \right]$ sq. units