Maths M.C.Q (1 Marks)

2. 0

Solution:

$\begin{vmatrix}5^2&5^3&5^4\\5^3&5^4&5^5\\5^4&5^5&5^6\end{vmatrix}$

$=5^2\times5^3\times5^4\begin{vmatrix}1&5&5^2\\1&5&5^2\\1&5&5^2\end{vmatrix}$ [Taking out common factors from R₁, R₂, R₃]

$=5^2\times5^3\times5^4\times5\begin{vmatrix}1&1&5^2\\1&1&5^2\\1&1&5^2\end{vmatrix}$

$=5^2\times5^3\times5^4\times0$

$=0$

1. 4

Solution:

$\text{D}_\text{k}=\begin{vmatrix}1&\text{n}&\text{n}\\2\text{k}&\text{n}^2+\text{n}+2&\text{n}^2+\text{n}\\2\text{k}-1&\text{n}^2&\text{n}^2+\text{n}+2\end{vmatrix} $

$\text{D}_\text{k}=\begin{vmatrix}1&\text{n}&\text{n}\\1&\text{n}+2&-2\\2\text{k}-1&\text{n}^2&\text{n}^2+\text{n}+2\end{vmatrix}\ \text{Applying}\ \text{R}_2\rightarrow\text{R}_2-\text{R}_3$

$\text{D}_\text{k}=\begin{vmatrix}1&\text{n}&\text{n}\\0&2&-2-\text{n}\\2\text{k}-1&\text{n}^2&\text{n}^2+\text{n}+2\end{vmatrix}\ \text{Applying}\ \text{R}_2\rightarrow\text{R}_2-\text{R}_1$

Now,

$\sum\limits_{\text{k}=1}^\text{n}\text{D}_\text{k}=\begin{vmatrix}1&\text{n}&\text{n}\\0&2&-2-\text{n}\\1&\text{n}^2&\text{n}^2+\text{n}+2\end{vmatrix}+\begin{vmatrix}1&\text{n}&\text{n}\\0&2&-2-\text{n}\\3&\text{n}^2&\text{n}^2+\text{n}+2\end{vmatrix}+.......+\begin{vmatrix}1&\text{n}&\text{n}\\0&2&-2-\text{n}\\2\text{n}-1&\text{n}^2&\text{n}^2+\text{n}+2\end{vmatrix}$

$\sum\limits_{\text{k}=1}^\text{n}\text{D}_\text{k}=\Big[1\Big(2(\text{n}^2+\text{n}+2)+\text{n}^2(2+\text{n})\Big)+\text{n}\Big(0+1(-2-\text{n})\Big)+\text{n}\Big(0+2\Big)\Big]\\+\Big[1\Big(2(\text{n}^2+\text{n}+2)+\text{n}^2(2+\text{n})\Big)+\text{n}\Big(0+3(-2-\text{n})\Big)+\text{n}\Big(0+6\Big)\Big]+......\\+\Big[1\Big(2(\text{n}^2+\text{n}+2)+\text{n}^2(2+\text{n})\Big)+\text{n}\Big(0+(2\text{n}-1)(-2-\text{n})\Big)+\text{n}\Big(0+2(2\text{n}-1)\Big)\Big]$

$\sum\limits_{\text{k}=1}^\text{n}\text{D}_\text{k}=\Big[1\Big(2(\text{n}^2+\text{n}+2)+\text{n}^2(2+\text{n})\Big)+\text{n}\Big(-2-\text{n}\Big)-2\text{n}\Big]\Big(1+3+5+.....+\text{n}\Big)$

$\sum\limits_{\text{k}=1}^\text{n}\text{D}_\text{k}=\Big[1\Big(2(\text{n}^2+\text{n}+2)+\text{n}^2(2+\text{n})\Big)+\text{n}\Big(-2-\text{n}\Big)-2\text{n}\Big]\Big(\text{n}^2\Big)$

$\Rightarrow\ 2\text{n}^2+4\text{n}=48$

$\Rightarrow\ (\text{n}+6)(\text{n}-4)=0$

$\Rightarrow\ \text{n}=4$

2. 0

Solution:

Let $\begin{vmatrix}\text{x}^2+3\text{x}&\text{x}-1&\text{x}+ 3\\\text{x}+1&-2\text{x}&\text{x}-4\\\text{x}-3&\text{x}+4&3\text{x}\end{vmatrix}$

$=(\text{x}^2+3\text{x})\begin{vmatrix}-2\text{x}&\text{x}-4\\\text{x}+4&3\text{x}\end{vmatrix}-(\text{x}-1)\begin{vmatrix}\text{x}+1&\text{x}-4\\\text{x}-3&3\text{x}\end{vmatrix}+(\text{x}+3)\begin{vmatrix}\text{x}+1&-2\text{x}\\\text{x}-3&\text{x}+4\end{vmatrix}$

= (x² + 3x)(-6x - x² + 16) - (x - 1)(3x² + 3x - x² + 7x - 12) + (x + 3)(x² + 5x + 4 + 2x² - 6x)

= -7x⁴ + 16x² + 48x + 21x³ + 8x² - 22x - 2x³ - 12 + 8x² + x + 3x³ + 12

= -7x⁴ + 22x³ + 32x² + 27x + 0

But x is a root of ax⁴ + bx³ + cx² + dx + e

e = 0

4. 32

Solution:

$\begin{vmatrix}\text{p}+\text{x}&\text{a}+\text{x}&\text{a}+\text{p}\\\text{q}+\text{y}&\text{b}+\text{y}&\text{b}+\text{q}\\\text{r}+\text{z}&\text{c}+\text{z}&\text{c}+\text{r}\end{vmatrix}=\begin{vmatrix}\text{p}&\text{a}&\text{a}\\\text{q}&\text{b}&\text{b}\\\text{r}&\text{c}&\text{c}\end{vmatrix}+​​​​\begin{vmatrix}\text{p}&\text{a}&\text{p}\\\text{q}&\text{b}&\text{q}\\\text{r}&\text{c}&\text{r}\end{vmatrix}\begin{vmatrix}\text{p}&\text{x}&\text{a}\\\text{q}&\text{y}&\text{b}\\\text{r}&\text{z}&\text{c}\end{vmatrix}\\+​​​​\begin{vmatrix}\text{p}&\text{x}&\text{p}\\\text{q}&\text{y}&\text{q}\\\text{r}&\text{z}&\text{r}\end{vmatrix}+​​​​\begin{vmatrix}\text{x}&\text{a}&\text{a}\\\text{y}&\text{b}&\text{b}\\\text{r}&\text{c}&\text{c}\end{vmatrix}+​​​​\begin{vmatrix}\text{x}&\text{a}&\text{p}\\\text{y}&\text{b}&\text{q}\\\text{r}&\text{c}&\text{r}\end{vmatrix}+​​​​\begin{vmatrix}\text{x}&\text{x}&\text{a}\\\text{y}&\text{y}&\text{b}\\\text{z}&\text{z}&\text{c}\end{vmatrix}+​​​​\begin{vmatrix}\text{x}&\text{x}&\text{p}\\\text{y}&\text{y}&\text{q}\\\text{z}&\text{z}&\text{r}\end{vmatrix}$

$=0+0+\begin{vmatrix}\text{p}&\text{x}&\text{a}\\\text{q}&\text{y}&\text{b}\\\text{r}&\text{z}&\text{c}\end{vmatrix}+0+0+\begin{vmatrix}\text{x}&\text{a}&\text{p}\\\text{y}&\text{b}&\text{q}\\\text{z}&\text{c}&\text{r}\end{vmatrix}+0+0$

$=\begin{vmatrix}\text{p}&\text{x}&\text{a}\\\text{q}&\text{y}&\text{b}\\\text{r}&\text{z}&\text{c}\end{vmatrix}+\begin{vmatrix}\text{x}&\text{a}&\text{p}\\\text{y}&\text{b}&\text{q}\\\text{z}&\text{c}&\text{r}\end{vmatrix}$

$=2\begin{vmatrix}\text{a}&\text{p}&\text{x}\\\text{b}&\text{q}&\text{y}\\\text{c}&\text{r}&\text{z}\end{vmatrix}$

$=2\times16=32$

Hence, the correct option is (b)

3. f(0) = 0

Solution:

Let $\text{f(x)}=\begin{vmatrix}0&\text{x}-\text{a}&\text{x}-\text{b}\\\text{x}+\text{a}&0&\text{x}-\text{c}\\\text{x}+\text{b}&\text{x}+\text{c}&0\end{vmatrix}$

Now, $\text{f(a)}=\begin{vmatrix}0&\text{a}-\text{a}&\text{a}-\text{b}\\\text{a}+\text{a}&0&\text{a}-\text{c}\\\text{a}+\text{b}&\text{a}+\text{c}&0\end{vmatrix}$

$=\begin{vmatrix}0&0&\text{a}-\text{b}\\2\text{a}&0&\text{a}-\text{c}\\\text{a}+\text{b}&\text{a}+\text{c}&0\end{vmatrix}$

$=(\text{a}-\text{b})(2\text{a}^2+2\text{ac})\neq0$

$\text{f(b)}=\begin{vmatrix}0&\text{b}-\text{a}&\text{b}-\text{b}\\\text{b}+\text{a}&0&\text{b}-\text{c}\\\text{b}+\text{b}&\text{b}+\text{c}&0\end{vmatrix}$

$=\begin{vmatrix}0&\text{b}-\text{a}&0\\\text{b}+\text{a}&0&\text{b}-\text{c}\\2\text{a}&\text{b}+\text{c}&0\end{vmatrix}$

$=(\text{b}-\text{a})(2\text{ab}-2\text{ac})\neq0$

$\text{f(0)}=\begin{vmatrix}0&\text{0}-\text{a}&\text{0}-\text{b}\\\text{0}+\text{a}&0&\text{0}-\text{c}\\\text{0}+\text{b}&\text{0}+\text{c}&0\end{vmatrix}$

$=\begin{vmatrix}0&-\text{a}&-\text{b}\\\text{a}&0&\text{c}\\\text{b}&\text{c}&0\end{vmatrix}$

$=\text{a}(\text{bc})-\text{b}(\text{ac})=0$

Hence, the correct option is (c)

2. 10

Solution:

$\begin{vmatrix}\log_3512&\log_43\\\log_38&\log_49\end{vmatrix}\times\begin{vmatrix}\log_23&\log_83\\\log_34&\log_34\end{vmatrix}$

$=\begin{vmatrix}\log_32^9&\log_{2^{2}}3\\\log_32^3&\log_{2^2}3^3\end{vmatrix}\times\begin{vmatrix}\log_23&\log_{2^{3}}3\\\log_32^3&\log_32^2\end{vmatrix}$

$=\begin{vmatrix}9\log_32&\frac{1}{2}\log_23\\3\log_32&\frac{1}{2}\times2\log_23\end{vmatrix}\times\begin{vmatrix}\log_23&\frac{1}{3}\log_23\\2\log_32&2\log_32\end{vmatrix}$

$=\Big(\big(9\log_32\times\log_23\big)-\big(3\log_32\times\frac{1}{2}\log_23\big)\Big)\times\Big(\big(\log_23\times2\log_32\big)\\-\Big(\frac{1}{3}\log_23\times2\log_32\Big)\Big)$

$=\Big(9-\frac{3}{2}\Big)\times\Big(2-\frac{2}{3}\Big)$

$=\frac{15}{2}\times\frac{4}{3}$

$=10$

1. 0

Solution:

$\triangle=\begin{vmatrix}1&\omega^{\text{n}}&\omega^{2\text{n}}\\\omega^{2\text{n}}&1&\omega^{\text{n}}\\\omega^{\text{n}}&\omega^{2\text{n}}&1\end{vmatrix}$

$=\begin{vmatrix}1+\omega^{\text{n}}+\omega^{2\text{n}}&\omega^{\text{n}}&\omega^{2\text{n}}\\\omega^{2\text{n}}+1+\omega^{\text{n}}&1&\omega^{\text{n}}\\\omega^{\text{n}}+\omega^{2\text{n}}+1&\omega^{2\text{n}}&1\end{vmatrix}$ [Applying C₁ → C₁ + C₂ + C₃]

Now, $1+\omega+\omega^2=0$ $[\because$ is a complex cube root of unity$]$ 

$1+\omega^{\text{n}}+\omega^{2\text{n}}=0$ $[\because$ n is not a multiple of 3$]$

$\triangle=\begin{vmatrix}1&\omega^{\text{n}}&\omega^{2\text{n}}\\\omega^{2\text{n}}&1&\omega^{\text{n}}\\\omega^{\text{n}}&\omega^{2\text{n}}&1\end{vmatrix}=0$ 

4. None of these

Solution:

$\begin{vmatrix}\text{b}^2-\text{ab}&\text{b}-\text{c}&\text{bc}-\text{ac}\\\text{ab}-\text{a}^2&\text{a}-\text{b}&\text{b}^2-\text{ab}\\\text{bc}-\text{ac}&\text{c}-\text{a}&\text{ab}-\text{a}^2\end{vmatrix}$

$=\begin{vmatrix}\text{b}(\text{b}-\text{a})&\text{b}-\text{c}&\text{c}(\text{b}-\text{a})\\\text{a}(\text{b}-\text{a})&\text{a}-\text{b}&\text{b}(\text{b}-\text{a})\\\text{c}(\text{b}-\text{a})&\text{c}-\text{a}&\text{a}(\text{b}-\text{a})\end{vmatrix}$

$=(\text{b}-\text{a})^2\begin{vmatrix}\text{b}&\text{b}-\text{c}&\text{c}\\\text{a}&\text{a}-\text{b}&\text{b}\\\text{c}&\text{c}-\text{a}&\text{a}\end{vmatrix}$ [Taking (b - a) common from C₁ and C₃]

$=(\text{b}-\text{a})^2\begin{vmatrix}0&\text{b}-\text{c}&\text{c}\\0&\text{a}-\text{b}&\text{b}\\0&\text{c}-\text{a}&\text{a}\end{vmatrix}$ [Applying C₁ → C₁ - C₂ - C₃]

$=0$

Hence, the correct option is (d)

4. A + B = 0

Solution:

Let $\text{A}=[\text{a}_{\text{ij}}]$ and $\text{B}=[\text{b}_{\text{ij}}]$ be a square matrix of order 2

As their orders are same, A + B is defined as

$\text{A}+\text{B}=[\text{a}_{\text{ij}}+\text{b}_\text{ij}]$

$\Rightarrow|\text{A}+\text{B}|=|\text{a}_{\text{ij}}+\text{b}_\text{ij}|$

Now,

$|\text{A}+\text{B}|=0$

$\Rightarrow|\text{a}_{\text{ij}}+\text{b}_\text{ij}|=0$

$\Rightarrow[\text{a}_{\text{ij}}+\text{b}_\text{ij}]=0$

[corrsponding term is 0]

$\Rightarrow\text{A}+\text{B}=0$

1. 4

Solution:

$\triangle=\begin{vmatrix}-2\text{a}&\text{a}+\text{b}&\text{a}+\text{c}\\\text{b}+\text{a}&-2\text{b}&\text{b}+\text{c}\\\text{c}+\text{a}&\text{c}+\text{b}&-2\text{c}\end{vmatrix}$

Let a + b = 2C, b + c = 2A and c + a = 2B

⇒ a + b + b + c + c + a = 2A + 2B + 2C

⇒ 2(a + b + c) = (A + B + C)

Also, a = (a + b + c) - (b + c) = (A + B + C) - 2A = B + C - A

Similarly, b = C + A - B, c = A + B - C

Hence, 4 is the order factor of the determinant.

1. $\triangle_1+\triangle_2=0$

Solution:

$\triangle_2=\begin{vmatrix}1&\text{bc}&\text{a}\\1&\text{ca}&\text{b}\\1&\text{ab}&\text{c}\end{vmatrix}$

$=\frac{1}{\text{abc}}\begin{vmatrix}1&\text{abc}&\text{a}^2\\1&\text{bca}&\text{b}^2\\1&\text{cab}&\text{c}^2\end{vmatrix}$ [R₁, R₂, R₃ are multiplies by a, b and c respectively, therefore we divide by abc]

$=\frac{\text{abc}}{\text{abc}}\begin{vmatrix}1&1&\text{a}^2\\1&1&\text{b}^2\\1&1&\text{c}^2\end{vmatrix}$ [Taking abc common from C₂]

$=-\begin{vmatrix}1&\text{a}&\text{a}^2\\1&\text{b}&\text{b}^2\\1&\text{c}&\text{c}^2\end{vmatrix}$ $\text{C}_1\leftrightarrow\text{C}_2$

We know that the value of a determinant remains unchanged if its rows and columns are interchanged. so,

$\triangle_2=-\begin{vmatrix}1&1&1\\\text{a}&\text{b}&\text{c}\\\text{a}^2&\text{b}^2&\text{c}^2\end{vmatrix} $

$=-\triangle_1$

$\triangle_1+\triangle_2=0$

3. Negative

Solution:

Discriminant D of ax² + 2bx + c = (2b)² - 4ac < 0 [Given]

⇒ 4b² - 4ac < 0

⇒ b² - ac < 0, where a > 0 .....(i)

$\Rightarrow\triangle=\begin{vmatrix}\text{a}&\text{b}&\text{ax}+\text{b}\\\text{b}&\text{c}&\text{bx}+\text{c}\\\text{ax}+\text{b}&\text{bx}+\text{c}&0\end{vmatrix}$

$\Rightarrow\triangle=\begin{vmatrix}\text{ax}&\text{bx}&\text{ax}^2+\text{bx}\\\text{b}&\text{c}&\text{bx}+\text{c}\\\text{ax}+\text{b}&\text{bx}+\text{c}&0\end{vmatrix}$ [Applying R₁ → xR₁]

$\Rightarrow\triangle=\frac{1}{\text{x}}\begin{vmatrix}\text{ax}+\text{b}&\text{bx}+\text{c}&\text{ax}^2+\text{bx}+\text{bx}+\text{c}\\\text{b}&\text{c}&\text{bx}+\text{c}\\\text{ax}+\text{b}&\text{bx}+\text{c}&0\end{vmatrix}$ [Applying R₁ → R₁ + R₂] 

$\Rightarrow\triangle=\frac{1}{\text{x}}\begin{vmatrix}0&0&\text{ax}^2+2\text{bx}+\text{c}\\\text{b}&\text{c}&\text{bx}+\text{c}\\\text{ax}+\text{b}&\text{bx}+\text{c}&0\end{vmatrix}$ [Applying R₁ → R₁ - R₃]

$\Rightarrow\triangle=\frac{1}{\text{x}}\begin{Bmatrix}\text{ax}^2+2\text{bx}+\text{c}\begin{vmatrix}\text{b}&\text{c}\\\text{ax}+\text{b}&\text{bx}+\text{c}\end{vmatrix}\end{Bmatrix}$ [Expanding along R₁]

$\Rightarrow\triangle=\frac{1}{\text{x}}(\text{ax}^2+2\text{bx}+\text{c})(\text{b}^2\text{x}+\text{bc}-\text{acx}-\text{bc})$

$\Rightarrow\triangle=\frac{1}{\text{x}}(\text{ax}^2+2\text{bx}+\text{c})\text{ x }(\text{b}^2-\text{ac})$

$\Rightarrow\triangle=(\text{ax}^2+2\text{bx}+\text{c})(\text{b}^2\text{ac})<0$ [From eq. (i)]

$\Rightarrow\triangle<0$

2. 2

Solution:

Let $\triangle=\begin{vmatrix}\text{cosec x}&\sec\text{x}&\sec\text{x}\\\sec\text{x}&\text{cosec}\text{x}&\sec\text{x}\\\sec\text{x}&\sec\text{x}&\text{cosec x}\end{vmatrix}$

$=(\text{cosec x})^3\begin{vmatrix}1&\frac{\sec\text{x}}{\text{cosecx}}&\frac{\sec\text{x}}{\text{cosecx}}\\\frac{\sec\text{x}}{\text{cosecx}}&1&\frac{\sec\text{x}}{\text{cosecx}}\\\frac{\sec\text{x}}{\text{cosecx}}&\frac{\sec\text{x}}{\text{cosecx}}&1\end{vmatrix}$

$=(\text{cosec x})^3\begin{vmatrix}1&\tan\text{x}&\tan\text{x}\\\tan\text{x}&1&\tan\text{x}\\\tan\text{x}&\tan\text{x}&1\end{vmatrix}$

 $=(\text{cosec x})^3\begin{vmatrix}1-\tan\text{x}&\tan\text{x}-1&0\\0&1-\tan\text{x}&\tan\text{x}-1\\\tan\text{x}&\tan\text{x}&1\end{vmatrix}$ [Applying R₁ → R₁ - R₂, R₂ → R₂ - R₃]

$=(\text{cosec x})^3(1-\tan\text{x})^2\begin{vmatrix}1&-1&0\\0&1&-1\\\tan\text{x}&\tan\text{x}&1\end{vmatrix}$ [Taking out $(1-\tan\text{x})$ common from R₁ and R₂]

 $=(\text{cosec x})^3(1-\tan\text{x})^2\begin{Bmatrix}1\begin{vmatrix}1&-1\\\tan\text{x}&1\end{vmatrix}+\tan\text{x}\begin{vmatrix}-1&0\\1&-1\end{vmatrix}\end{Bmatrix}$ [Expanding along C₁]

 $=(\text{cosec x})^3(1-\tan\text{x})^2\{1+\tan\text{x}+\tan\text{x}\}$

$=(\text{cosec x})^3(1-\tan\text{x})^2\{1+2\tan\text{x}\}$

$\triangle=0$

$=(\text{cosec x})^3(1-\tan\text{x})^2(1+2\tan\text{x})=0$

$(1-\tan\text{x})=0,(\text{coses x})^3=0$ and $(1+2\tan\text{x})=0$

Or $\tan\text{x}=1,\text{cosec x}=0$ and $\tan\text{x}=\frac{-1}{2}$

$\Rightarrow-\frac{\pi}{4}\leq\text{x}\leq\frac{\pi}{4}$ $\Big[\tan\text{x}=1,\text{x}=\frac{-1}{2}$ are 2 real roots as $\text{cosec x}=0$ has no solution$\Big]$

Thus, these are 2 solutions.

3. a³ + b³ + c³ - 3abc

Solution:

$\begin{vmatrix}\text{a}-\text{b}&\text{b}+\text{c}&\text{a}\\\text{b}-\text{c}&\text{c}+\text{b}&\text{b}\\\text{c}-\text{a}&\text{a}+\text{b}&\text{c}\end{vmatrix}$

$=\begin{vmatrix}-\text{b}&\text{b}+\text{c}+\text{a}&\text{a}\\-\text{c}&\text{c}+\text{a}+\text{b}&\text{b}\\-\text{a}&\text{a}+\text{b}+\text{c}&\text{c}\end{vmatrix}$ [Applying C₁ → C₁ - C₃ and C₂ → C₂ + C₃]

$=(-1)(\text{a}+\text{b}+\text{c})\begin{vmatrix}\text{b}&1&\text{a}\\\text{c}&1&\text{b}\\\text{a}&1&\text{c}\end{vmatrix}$ [Taking (-1) common from C₁ and (a + b + c) common from C₂]

$=(-1)(\text{a}+\text{b}+\text{c})\begin{vmatrix}\text{b}&1&\text{a}\\\text{c}-\text{b}&0&\text{b}-\text{a}\\\text{a}-\text{b}&0&\text{c}-\text{a}\end{vmatrix}$ [Applying R₂ → R₂ - R₁ and R₃ → R₃ - R₁]

= (-1)(a + b + c)[-(c - b)(c - a) + (b - a)(a - b)]

= (-1)(a + b + c)[-c² + ac + bc - ab + ba - b² - a² + ab]

= (-1)(a + b + c)(-a² - b² - c² + ab + bc + ac)

= (a + b + c)(a² + b² + c² - ab - bc - ac)

= a³ + ab² + ac² - a²b - abc - a²c + ba² + b³ + bc² - ab²-b²c - abc + ca² + cb² + c³ - acb - bc² - ac²

= a³ + b³ + c³ - 3abc

Hence, the correct option is (c)

3. -2n³

Solution:

$\text{A}_\text{r}=\begin{vmatrix}1&\text{r}&2^{\text{r}}\\2&\text{n}&\text{n}^{2}\\\text{n}&\frac{\text{n}(\text{n}+1)}{2}&2^{\text{n}+1} \end{vmatrix}$

$\Rightarrow\sum\limits_{\text{r}=1}^\text{n}\text{A}_\text{r}=\begin{vmatrix}\sum\limits_{\text{r}=1}^\text{n}1&\sum\limits_{\text{r}=1}^\text{n}\text{r}&\sum\limits_{\text{r}=1}^\text{n}2\text{r}\\\sum\limits_{\text{r}=1}^\text{n}2&\text{n}&\text{n}^{2}\\\text{n}&\frac{\text{n}(\text{n}+1)}{2}&2^{\text{n}+1} \end{vmatrix}$

As $\sum\limits_{\text{r}=1}^\text{r}1=1+1+1\ ......+1(\text{n times})=\text{n}$

$\Rightarrow\sum\limits_{\text{r}=1}^\text{r}\text{r}=1+2+3+\ .....+\text{n}=\frac{\text{n}(\text{n}+1)}{2}$

Let $\text{S}=\sum\limits_{\text{r}=1}^\text{r}2^\text{r}=2+2^2+2^3=\ .....+2^{\text{n}}$

$\Rightarrow2\text{S}=2^2+3^2=\ ....+2^{\text{n}}+2^{\text{n}+1}$

$\Rightarrow2\text{S}-\text{S}$

$\Rightarrow\text{S}=\sum\limits_{\text{r}=1}^\text{n}2^{\text{r}}=2^{\text{n+1}}-2$

$\Rightarrow\sum\limits_{\text{r}=1}^\text{n}\text{A}_\text{r}=\begin{vmatrix}\text{n}&\frac{\text{n}(\text{n}+1)}{2}&2^{\text{n}+1}-2\\2\text{n}&\text{n}&\text{n}^{2}\\\text{n}&\frac{\text{n}(\text{n}+1)}{2}&2^{\text{n}+1} \end{vmatrix}$

[Applying R₁ → R₁ - R₂]

$\Rightarrow\sum\limits_{\text{r}=1}^\text{n}\text{A}_\text{r}=\begin{vmatrix}\text{n}-\text{n}&\frac{\text{n}(\text{n}+1)}{2}-\frac{\text{n}(\text{n}+1)}{2}&2^{\text{n}+1}-2-2^{\text{n}+1}\\2\text{n}&\text{n}&\text{n}^{2}\\\text{n}&\frac{\text{n}(\text{n}+1)}{2}&2^{\text{n}+1} \end{vmatrix}$

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

$=-2\times\begin{vmatrix}2\text{n}&\text{n}\\\text{n}&\frac{\text{n}(\text{n}+1)}{2}\end{vmatrix}$

$=-2\big[\text{n}^{3}+\text{n}^2-\text{n}^2\big]$

$=-2\text{n}^3$

3. $\pm6$

Solution:

$\begin{vmatrix}2\text{x}&5\\8&\text{x}\end{vmatrix}=\begin{vmatrix}6&-2\\7&3\end{vmatrix}$

$\Rightarrow2\text{x}^2-40=18+14$

$\Rightarrow2\text{x}^2-40=32$

$\Rightarrow2\text{x}^2=72$

$\Rightarrow\text{x}^2=36$

$\Rightarrow\text{x}=\pm6$

Hence, the correct option is (C)

1. 0

Solution:

$\begin{vmatrix}\text{x}+2&\text{x}+3&\text{x}+2\text{a}\\\text{x}+3&\text{x}+4&\text{x}+2\text{b}\\\text{x}+4&\text{x}+5&\text{x}+2\text{c}\end{vmatrix}$

$=\begin{vmatrix}0&0&2(\text{a}+\text{c}-2\text{b})\\\text{x}+3&\text{x}+4&\text{x}+2\text{b}\\\text{x}+4&\text{x}+5&\text{x}+2\text{c}\end{vmatrix}$

[Applying R₁ → R₁ + R₃ - R₂, R₁ → R₁ - R₂]

$=\begin{vmatrix}0&0&0\\\text{x}+3&\text{x}+4&\text{x}+2\text{b}\\\text{x}+4&\text{x}+5&\text{x}+2\text{c}\end{vmatrix}$ $[\because$ a, b, c are in A.P.$]$

$=0$

4. $\text{Det (A)}\in[2,4]$

Solution:

$\begin{vmatrix}1&\sin\theta&1\\-\sin\theta&1&\sin\theta\\-1&-\sin\theta&1\end{vmatrix}$

$=\begin{vmatrix}1&\sin\theta&2\\-\sin\theta&1&\sin\theta\\-1&-\sin\theta&1\end{vmatrix}$ [Applying C₃ → C₃ + C₁]

$=2\times\begin{vmatrix}-\sin\theta&1\\-1&-\sin\theta\end{vmatrix}$ [Expanding along C₃]

$=2(\sin^2\theta+1)$

Given, $0\leq\theta\leq2\pi$

$-1\leq\sin\theta\leq1$

$0\leq\sin^2\theta\leq1$

$|\text{A}|=2(\sin^2\theta+1)$

$|\text{A}|=2\times1=2$ $[\theta=0]$

$|\text{A}|=2\times2=4$ $[\theta=2\pi]$

$\text{Det (A)}\in[2,4]$

2. $\alpha$ is a root of 4ax² + 12bx + 9c = 0 or a, b, c are in G.P.

Solution:

Let $\triangle=\begin{vmatrix}\text{a}&\text{b}&2\text{a}\alpha+3\text{b}\\\text{b}&\text{c}&2\text{b}\alpha+3\text{c}\\2\text{a}\alpha+3\text{b}&2\text{b}\alpha+3\text{c}&0\end{vmatrix}$

$=\begin{vmatrix}\text{a}-\text{b}&\text{b}&2\text{a}\alpha+3\text{b}\\\text{b}-\text{c}& \text{c}&2\text{b}\alpha+3\text{c}\\2\text{a}\alpha+ 3\text{b}-2\text{b}\alpha-3&2\text{b}\alpha+3\text{c}&0\end{vmatrix}$ [Applying C₁ → C₁ - C₂]

$=\begin{vmatrix}\text{a}-\text{b}&\text{b}&2\text{a}\alpha+3\text{b}\\\text{b}-\text{c}& \text{c}&2\text{b}\alpha+3\text{c}\\2(\text{a}-\text{b})\alpha+3(\text{b}-\text{c})&2\text{b}\alpha+3\text{b}&0\end{vmatrix}$

$=2\alpha(2\text{a}\alpha+3\text{b})-3(2\text{b}\alpha+3\text{c})\begin{vmatrix}\text{a}-\text{b}&\text{b}\\\text{b}-\text{c}&\text{c}\end{vmatrix}$ [Expanding along R₃]

$=-(4\text{a}\alpha^2+12\text{b}\alpha+9\text{c})(\text{ac}-\text{b}^2)$

But $\triangle=0$ [Given]

$\Rightarrow-(4\text{a}\alpha^2+12\text{b}\alpha+9\text{c})(\text{ac}-\text{b}^2)=0$

$\Rightarrow(4\text{a}\alpha^2+12\text{b}\alpha+9\text{c})=0$

Or $(\text{ac}-\text{b}^2)=0$

$\Rightarrow\alpha$ is a root of 4ax² + 12bx + 9c = 0

Or ac = b², i.e. a, b, c are in G.P.

1. 0

solution:

$\text{A}+\text{B}+\text{C}=\pi$

$\text{A}+\text{C}=\pi-\text{B},\text{A}+\text{B}=\pi-\text{C}$ and $\text{B}+\text{C}=\pi-\text{A}$

Thus the determinant becomes

$\begin{vmatrix}\sin\pi&\sin(\pi-\text{B})&\cos\text{C}\\-\sin\text{B}&0&\tan\text{A}\\\cos(\pi-\text{C})&\tan(\pi-\text{A})&0\end{vmatrix}$

$=\begin{vmatrix}0&\sin\text{B}&\cos\text{C}\\-\sin\text{B}&0&\tan\text{A}\\-\cos\text{C}&-\tan\text{A}&0\end{vmatrix}$

$[\sin\pi=0,\sin(\pi-\text{B}),\cos(\pi-\text{C})=-\cos\text{C},\tan(\pi-\text{A})=-\tan\text{A}]$

It is a skew symmetric matrix of the odd order 3. Thus by property of determinants, we get

$|\triangle|=0$

$\begin{vmatrix}0&\sin\text{B}&\cos\text{C}\\-\sin\text{B}&0&\tan\text{A}\\-\cos\text{C}&-\tan\text{A}&0\end{vmatrix}$

1. 0

Solution:

$\text{f(x)}=\begin{vmatrix}\cos\text{x}&\text{x}&1\\2\sin\text{x}&\text{x}&2\text{x}\\\sin\text{x}&\text{x}&\text{x}\end{vmatrix}$

$=\begin{vmatrix}\cos\text{x}&\text{x}&1\\\sin\text{x}&0&\text{x}\\\sin\text{x}&\text{x}&\text{x}\end{vmatrix}$ [Applying R₂ → R₂ - R₃]

$=\begin{vmatrix}\cos\text{x}&\text{x}&1\\\sin\text{x}&0&\text{x}\\\sin\text{x}-\cos\text{x}&0&\text{x}-1\end{vmatrix}$ [Applying R₃ → R₃ - R₁]

$=-\text{x}[\text{x}\sin\text{x}-\sin\text{x}-\text{x}\sin\text{x}+\text{x}\cos\text{x}]$

$=-\text{x}(\text{x}\cos\text{x}-\sin\text{x})$

$\therefore\ \lim_\limits{\text{x}\rightarrow0}\frac{\text{f(x)}}{\text{x}^2}=\lim_\limits{\text{x}\rightarrow0}\frac{\text{x}(\sin\text{x}-\text{x}\cos\text{x})}{\text{x}^2}$

$=\lim_\limits{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}^2}-\lim_\limits{\text{x}\rightarrow0}\cos\text{x}$

$=1-1=0$

Hence, the correct option is (a)

4. -1

Solution:

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

$\Rightarrow\begin{vmatrix}\text{x}&0&-\text{z}\\0&\text{y}&-\text{z}\\1&1&1+\text{z}\end{vmatrix}=0$ [Applying R₂ → R₂ - R₃ and R₁ → R₁ - R₃]

$\Rightarrow\text{x}\big[\text{y}(1+\text{z})+\text{z}\big]+1(\text{yz})=0$ [Expanding along first column]

$\Rightarrow\text{x}[\text{y}+\text{yz}+\text{z}]+\text{yz}=0$

$\Rightarrow\text{xy}+\text{xyz}+\text{xz}+\text{yz}=0$

$\Rightarrow\text{xy}+\text{yz}+\text{zx}=-\text{xyz}$

$\Rightarrow\frac{\text{xy}}{\text{xyz}}+\frac{\text{yz}}{\text{xyz}}+\frac{\text{zx}}{\text{xyz}}=-\frac{\text{xyz}}{\text{xyz}}$

$\Rightarrow\frac{1}{\text{z}}+\frac{1}{\text{x}}+\frac{1}{\text{y}}=-1$

$\Rightarrow\text{x}^{-1}+\text{y}^{-1}+\text{z}^{-1}=-1$

Hence, the correct option is (d)

3. 8

Solution:

$\begin{vmatrix}1&1&1\\^\text{n}\text{C}_1&^{\text{n}+2}\text{C}_1&^{\text{n}+4}\text{C}_1\\^\text{n}\text{C}_2&^{\text{n}+2}\text{C}_2&^{\text{n}+4}\text{C}_2\end{vmatrix}$

$=\begin{vmatrix}1&1&1\\\text{n}&\text{n}+2&\text{n}+3\\\frac{\text{n}(\text{n}-1)}{2}&\frac{(\text{n}+2)(\text{n}+1)}{2}&\frac{(\text{n}+4)(\text{n}+3)}{2}\end{vmatrix}$

$=\begin{vmatrix}1&0&0\\\text{n}&2&4\\\frac{\text{n}(\text{n}+1)}{2}&\frac{4\text{n}+2}{2}&\frac{8\text{n}+12}{2}\end{vmatrix}$ [Applying C₂ → C₂ - C₁ and C₃ → C₃ - C₁]

$=\begin{vmatrix}1&0&0\\\text{n}&2&4\\\frac{\text{n}(\text{n}+1)}{2}&(2\text{n}+1)&(4\text{n}+6)\end{vmatrix}$

$=8\text{n}+12-8\text{n}-4$

$=8$

Hence, the correct option is (c)

1. $\Big[-\sqrt{2},\sqrt{2}\Big]$

Solution:

$\triangle=\begin{vmatrix}\cos\text{x}&-\sin\text{x}&1\\\sin\text{x}&\cos\text{x}&1\\\cos(\text{x}+\text{y})&-\sin(\text{x}+\text{y})&0\end{vmatrix}$

$=\begin{vmatrix}\cos\text{x}&-\sin\text{x}&1\\\sin\text{x}&\cos\text{x}&1\\0&0&\sin\text{y}-\cos\text{y}\end{vmatrix}$ [Applying R₃ → R₃ - cosy R₁ + siny R₂]

$=(\sin\text{y}-\cos\text{y})(\cos^2\text{x}+\sin^2\text{x})$

$=\sin\text{y}-\cos\text{y}$

$=\sqrt{2}\Big(\frac{1}{\sqrt{2}}\sin\text{y}-\frac{1}{\sqrt{2}}\cos\text{y}\Big)$

$=\sqrt{2}\Big(\cos\frac{\pi}{4}\sin\text{y}-\sin\frac{\pi}{4}\cos\text{y}\Big)$

$=\sqrt{2}\sin\Big(\text{y}-\frac{\pi}{4}\Big)$

Therefore, $-\sqrt{2}\leq\triangle\leq\sqrt{2}$

Hence, the correct option is (a)

2. 9y²(x + y)

Solution:

$\begin{vmatrix}\text{x}&\text{x}+\text{y}&\text{x}+2\text{y}\\\text{x}+2\text{y}&\text{x}&\text{x}+\text{y}\\\text{x}+\text{y}&\text{x}+2\text{y}&\text{x}\end{vmatrix}$

$=\begin{vmatrix}-2\text{y}&\text{y}&\text{y}\\\text{x}+2\text{y}&\text{x}&\text{x}+\text{y}\\-\text{y}&2\text{y}&-\text{y}\end{vmatrix}$ [Applying R₁ → R₁ - R₂ and R₃ → R₃ - R₂]

$=\text{y}^2\begin{vmatrix}-2&1&1\\\text{x}+2\text{y}&\text{x}&\text{x}+\text{y}\\-1&2&-1\end{vmatrix}$ [Taking (y) common from R₁ and from R₃]

$=\text{y}^2\begin{vmatrix}-2&-3&3\\\text{x}+2\text{y}&3\text{x}+4\text{y}&-\text{y}\\-1&0&0\end{vmatrix}$ [Applying C₂ → C₂ + 2C₁ and C₃ → C₃ - C₁]

$=\text{y}^2\big[-1(3\text{y}-9\text{x}-12\text{y})\big]$

$=\text{y}^2[9\text{y}+9\text{x}]$

$=9\text{y}^2(\text{y}+\text{x})$

Hence, the correct option is (b)

1. n

Solution:

Let A = nx, B = (n - 1)x, C = (n + 2)x

⇒ C - B = x, B - A = x, C - A = 2x

Thus, the given determinant is

$\begin{vmatrix}\text{a}^2&\text{a}&1\\\cos\text{A}&\cos\text{B}&\cos\text{C}\\\sin\text{A}&\sin\text{B}&\sin\text{C}\end{vmatrix}$

$=\text{a}^2(\cos\text{B}\sin\text{C}-\cos\text{C}\sin\text{B})-\text{a}\times(\cos\text{A}\sin\text{C}-\cos\text{C}\sin\text{A})\\+1\times(\cos\text{A}\sin\text{B}-\sin\text{A}\cos\text{B})$

$=\text{a}^2\sin(\text{C}-\text{B})-\text{a}\sin(\text{C}-\text{A})+\sin(\text{B}-\text{A})$

$=\text{a}^2\sin{\text{x}}-\text{a}\sin2\text{x}+\sin\text{x}$ [Independent of n] 

4. None of these.

Solution:

$\triangle=\begin{vmatrix}\text{x}&2&\text{x}\\\text{x}^2&\text{x}&6\\\text{x}&\text{x}&6\end{vmatrix}$

$=\text{x}\begin{vmatrix}\text{x}&6\\\text{x}&6\end{vmatrix}-\text{x}^2\begin{vmatrix}2&\text{x}\\\text{x}&6\end{vmatrix}+\text{x}\begin{vmatrix}2&\text{x}\\\text{x}&6\end{vmatrix}$

= 0 - x²(12 - x²) + x(12 - x²)

= x⁴ - 12x² + 12x - x³

= ax⁴ + bx³ + cx² + dx + e

⇒ x⁴ - 12x² + 12x - x³ = ax⁴ + bx³ + cx² + dx + e

⇒ a = 1, b = -1, c = -12, d = 12, e = 0

Thus,

5a + 4b + 3c + 2d + e = 5 - 4 - 36 + 24 + 0 = -11

3. -4

Solution:

$\triangle=\begin{vmatrix}1&-2&5\\2&\text{a}&-1\\0&4&2\text{a}\end{vmatrix}=86$

⇒ 1(2a² + 4) - 2(-4a - 20) = 86

⇒ 2a² + 4 + 8a + 40 = 86

⇒ 2a² + 8a - 42 = 0

⇒ a² + 4a - 21 = 0

⇒ a² + 7a - 3a - 21 = 0

⇒ a(a + 7) - 3(a + 7) = 0

⇒ a = -7, 3

Sum of the two values of a = -7 + 3 = -4

Hence, the correct option is (c)

1. $\frac{1}{2}$

Solution:

$\triangle=\begin{vmatrix}1&1&1\\1&1+\sin\theta&1\\1+\cos\theta&1&1\end{vmatrix}$

$=\begin{vmatrix}1&1&1\\0&\sin\theta&0\\\cos\theta&0&0\end{vmatrix}$ [Applying R₂ → R₂ - R₁ and R₃ → R₃ - R₁]

$=-\sin\theta\cos\theta$

$=-\frac{\sin2\theta}{2}$

Now, maximum and minimum value of $\sin\theta$ is 1 and -1.

So, the maximum value of $-\sin\theta$ is 1.

So, the maximum value of $-\sin2\theta$ is 1.

Therefore, the maximum value of $-\frac{\sin2\theta}{2}$ is $\frac{1}{2}$

Hence, the correct option is (a)

4. $\begin{vmatrix}\text{a}&\text{c}\\\text{e}&\text{g} \end{vmatrix}+\begin{vmatrix}\text{b}&\text{c}\\\text{f}&\text{g} \end{vmatrix}+\begin{vmatrix}\text{a}&\text{d}\\\text{e}&\text{h}\end{vmatrix}+\begin{vmatrix}\text{b}&\text{d}\\\text{f}&\text{h}\end{vmatrix}$

Solution:

$\begin{vmatrix}\text{a}+\text{b}&\text{c}+\text{d}\\\text{e}+\text{f}&\text{g}+\text{h} \end{vmatrix}=\begin{vmatrix}\text{a}+\text{b}&\text{c}\\\text{e}+\text{f}&\text{h} \end{vmatrix}+\begin{vmatrix}\text{a}+\text{b}&\text{d}\\\text{e}+\text{f}&\text{h}\end{vmatrix}$

$=\begin{vmatrix}\text{a}&\text{c}\\\text{e}&\text{g} \end{vmatrix}+\begin{vmatrix}\text{b}&\text{c}\\\text{f}&\text{g} \end{vmatrix}+\begin{vmatrix}\text{a}&\text{d}\\\text{e}&\text{h}\end{vmatrix}+\begin{vmatrix}\text{b}&\text{d}\\\text{f}&\text{h}\end{vmatrix}$