Advertisements
Advertisements
प्रश्न
The number of distinct real roots of \[\begin{vmatrix}cosec x & \sec x & \sec x \\ \sec x & cosec x & \sec x \\ \sec x & \sec x & cosec x\end{vmatrix} = 0\] lies in the interval
\[- \frac{\pi}{4} \leq x \leq \frac{\pi}{4}\]
विकल्प
1
2
3
0
Advertisements
उत्तर
\[\text{ Let }∆ = \begin{vmatrix} cosec x & \sec x & \sec x\\\sec x & cosec x & \sec x\\\sec x & \sec x & cosec x \end{vmatrix}\]
\[ = \left( cosec x \right)^3 \begin{vmatrix} 1 &\frac{\sec x}{cosec x} & \frac{\sec x}{cosec x}\\\frac{\sec x}{cosec x} & 1 & \frac{\sec x}{cosec x}\\\frac{\sec x}{cosec x} &\frac{\sec x}{cosec x} & 1 \end{vmatrix}\]
\[ = \left( cosec x \right)^3 \begin{vmatrix} 1 & \tan x & \tan x \\\tan x & 1 & \tan x\\\tan x & \tan x & 1 \end{vmatrix}\]
\[ = \left( cosec x \right)^3 \begin{vmatrix} 1 - \tan x & \tan x - 1 & 0 \\ 0 & 1 - \tan x & \tan x - 1\\\tan x & \tan x & 1 \end{vmatrix} \left[\text{ Applying }R_1 \to R_1 - R_2 , R_2 \to R_2 - R_3 \right]\]
\[ = \left( cosec x \right)^3 \left( 1 - \tan x \right)^2 \begin{vmatrix} 1 & - 1 & 0 \\ 0 & 1 & - 1\\\tan x & \tan x & 1 \end{vmatrix} \left[\text{ Taking out }\left( 1 - \tan x \right)\text{ common from }R_1\text{ and }R_2 \right]\]
\[ = \left( cosec x \right)^3 \left( 1 - \tan x \right)^2 \left\{ 1\begin{vmatrix}1 & - 1 \\ \tan x & 1\end{vmatrix} + \tan x\begin{vmatrix}- 1 & 0 \\ 1 & - 1\end{vmatrix} \right\} \left[ \text{ Expanding along }C_1 \right]\]
\[ = \left( cosec x \right)^3 \left( 1 - \tan x \right)^2 \left\{ 1 + \tan x + \tan x \right\}\]
\[ = \left( cosec x \right)^3 \left( 1 - \tan x \right)^2 \left\{ 1 + 2 \tan x \right\}\]
\[ ∆ = 0\]
\[ \left( cosec x \right)^3 \left( 1 - \tan x \right)^2 \left( 1 + 2 \tan x \right) = 0\]
\[ \Rightarrow \left( 1 - \tan x \right) = 0, \left( cosec x \right)^3 = 0\text{ and }\left( 1 + 2 \tan x \right) = 0\]
or
\[\tan x = 1, cosec x = 0\text{ and }\tan x = \frac{- 1}{2}\]
\[ \Rightarrow - \frac{\pi}{4} \leq x \leq \frac{\pi}{4} \left[ \tan x = 1, \tan x = \frac{- 1}{2}\text{ are 2 real roots as cosec x = 0 has no solution }\right]\]
Thus, there are 2 solutions .
APPEARS IN
संबंधित प्रश्न
If `|[x+1,x-1],[x-3,x+2]|=|[4,-1],[1,3]|`, then write the value of x.
Evaluate the following determinant:
\[\begin{vmatrix}\cos \theta & - \sin \theta \\ \sin \theta & \cos \theta\end{vmatrix}\]
If \[A = \begin{bmatrix}2 & 5 \\ 2 & 1\end{bmatrix} \text{ and } B = \begin{bmatrix}4 & - 3 \\ 2 & 5\end{bmatrix}\] , verify that |AB| = |A| |B|.
If A \[\begin{bmatrix}1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4\end{bmatrix}\] , then show that |3 A| = 27 |A|.
Find the value of x, if
\[\begin{vmatrix}2 & 3 \\ 4 & 5\end{vmatrix} = \begin{vmatrix}x & 3 \\ 2x & 5\end{vmatrix}\]
Evaluate the following determinant:
\[\begin{vmatrix}a & h & g \\ h & b & f \\ g & f & c\end{vmatrix}\]
Without expanding, show that the value of the following determinant is zero:
\[\begin{vmatrix}a + b & 2a + b & 3a + b \\ 2a + b & 3a + b & 4a + b \\ 4a + b & 5a + b & 6a + b\end{vmatrix}\]
Without expanding, show that the value of the following determinant is zero:
\[\begin{vmatrix}0 & x & y \\ - x & 0 & z \\ - y & - z & 0\end{vmatrix}\]
Evaluate the following:
\[\begin{vmatrix}a + x & y & z \\ x & a + y & z \\ x & y & a + z\end{vmatrix}\]
Show that
Solve the following determinant equation:
Using determinants show that the following points are collinear:
(3, −2), (8, 8) and (5, 2)
Using determinants, find the area of the triangle with vertices (−3, 5), (3, −6), (7, 2).
Find values of k, if area of triangle is 4 square units whose vertices are
(k, 0), (4, 0), (0, 2)
Prove that :
\[\begin{vmatrix}\left( b + c \right)^2 & a^2 & bc \\ \left( c + a \right)^2 & b^2 & ca \\ \left( a + b \right)^2 & c^2 & ab\end{vmatrix} = \left( a - b \right) \left( b - c \right) \left( c - a \right) \left( a + b + c \right) \left( a^2 + b^2 + c^2 \right)\]
Prove that :
2x − y = 17
3x + 5y = 6
2x + 3y = 10
x + 6y = 4
x+ y = 5
y + z = 3
x + z = 4
3x + y = 5
− 6x − 2y = 9
If A is a singular matrix, then write the value of |A|.
If \[A = \begin{bmatrix}1 & 2 \\ 3 & - 1\end{bmatrix}\text{ and }B = \begin{bmatrix}1 & 0 \\ - 1 & 0\end{bmatrix}\] , find |AB|.
If A = [aij] is a 3 × 3 scalar matrix such that a11 = 2, then write the value of |A|.
Write the value of
If \[∆_1 = \begin{vmatrix}1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2\end{vmatrix}, ∆_2 = \begin{vmatrix}1 & bc & a \\ 1 & ca & b \\ 1 & ab & c\end{vmatrix},\text{ then }\]}
Let \[\begin{vmatrix}x^2 + 3x & x - 1 & x + 3 \\ x + 1 & - 2x & x - 4 \\ x - 3 & x + 4 & 3x\end{vmatrix} = a x^4 + b x^3 + c x^2 + dx + e\]
be an identity in x, where a, b, c, d, e are independent of x. Then the value of e is
Let \[A = \begin{bmatrix}1 & \sin \theta & 1 \\ - \sin \theta & 1 & \sin \theta \\ - 1 & - \sin \theta & 1\end{bmatrix},\text{ where 0 }\leq \theta \leq 2\pi . \text{ Then,}\]
Solve the following system of equations by matrix method:
3x + y = 19
3x − y = 23
Solve the following system of equations by matrix method:
x − y + 2z = 7
3x + 4y − 5z = −5
2x − y + 3z = 12
Show that the following systems of linear equations is consistent and also find their solutions:
6x + 4y = 2
9x + 6y = 3
If A = `[(1, 2, 0), (-2, -1, -2), (0, -1, 1)]`, find A−1. Using A−1, solve the system of linear equations x − 2y = 10, 2x − y − z = 8, −2y + z = 7.
Two schools P and Q want to award their selected students on the values of Discipline, Politeness and Punctuality. The school P wants to award ₹x each, ₹y each and ₹z each the three respectively values to its 3, 2 and 1 students with a total award money of ₹1,000. School Q wants to spend ₹1,500 to award its 4, 1 and 3 students on the respective values (by giving the same award money for three values as before). If the total amount of awards for one prize on each value is ₹600, using matrices, find the award money for each value. Apart from the above three values, suggest one more value for awards.
2x − y + 2z = 0
5x + 3y − z = 0
x + 5y − 5z = 0
3x − y + 2z = 0
4x + 3y + 3z = 0
5x + 7y + 4z = 0
If `|(2x, 5),(8, x)| = |(6, -2),(7, 3)|`, then value of x is ______.
In system of equations, if inverse of matrix of coefficients A is multiplied by right side constant B vector then resultant will be?
The number of real values λ, such that the system of linear equations 2x – 3y + 5z = 9, x + 3y – z = –18 and 3x – y + (λ2 – |λ|z) = 16 has no solution, is ______.
