Advertisements
Advertisements
प्रश्न
Solve the equation for x: `sin^(-1) 5/x + sin^(-1) 12/x = π/2, x ≠ 0`
Solve the equation x: `sin^(-1) (5/x) + sin^(-1) (12/x) = π/2 (x ≠ 0)`
Advertisements
उत्तर १
`sin^(-1) (5/x) + sin^(-1) (12/x) = π/2`
`sin^(-1) + cos^(-1) sqrt(1 - 144/x^2) = π/2`
Let `sin^(-1) 12/x = β`
`12/x = sin β = "OPP"/"HYP"`
`sqrt(x^2 - 144)/x = cos β = "Adj"/"HYP"`
`β = cos^(-1) (sqrt(x^2 - 144)/x^2)`
∴ `5/x = sqrt(1 - 144/x^2)`
`25/x^2 = 1 - 144/x^2`
`169/(x^2) = 1`
x2 = 169
x = 13
उत्तर २
`sin^(-1) (5/x) + sin^(-1) (12/x) = π/2`
⇒ `sin^-1 (5/x) = π/2 - sin^-1 (12/x)`
⇒ `sin^-1 (5/x) = cos^-1 (12/x)` ...`[∵ sin^-1x + cos^-1x = π/2]`
⇒ `sin^-1 (5/x) = cos^-1 (12/x) = θ`
⇒ `sin θ = 5/x, cos θ = 12/x`
Since, sin2θ + cos2θ = 1
⇒ `(5/x)^2 + (12/x)^2 = 1`
⇒ `25/(x^2) + 144/(x^2) = 1`
⇒ `169/(x^2) = 1`
⇒ x = ±13
Since, x = –13 does not satisfy the given equation.
So, x = 13.
Notes
Students should refer to the answer according to their questions.
संबंधित प्रश्न
Show that the differential equation of which y = 2(x2 − 1) + \[c e^{- x^2}\] is a solution, is \[\frac{dy}{dx} + 2xy = 4 x^3\]
Form the differential equation representing the family of ellipses having centre at the origin and foci on x-axis.
For the following differential equation verify that the accompanying function is a solution:
| Differential equation | Function |
|
\[x\frac{dy}{dx} + y = y^2\]
|
\[y = \frac{a}{x + a}\]
|
(1 − x2) dy + xy dx = xy2 dx
(y + xy) dx + (x − xy2) dy = 0
Solve the following differential equation:
\[y\left( 1 - x^2 \right)\frac{dy}{dx} = x\left( 1 + y^2 \right)\]
In a bank principal increases at the rate of r% per year. Find the value of r if ₹100 double itself in 10 years (loge 2 = 0.6931).
(x + y) (dx − dy) = dx + dy
Solve the following initial value problem:
\[\frac{dy}{dx} + y \cot x = 4x\text{ cosec }x, y\left( \frac{\pi}{2} \right) = 0\]
Solve the following initial value problem:-
\[\tan x\left( \frac{dy}{dx} \right) = 2x\tan x + x^2 - y; \tan x \neq 0\] given that y = 0 when \[x = \frac{\pi}{2}\]
The decay rate of radium at any time t is proportional to its mass at that time. Find the time when the mass will be halved of its initial mass.
Experiments show that radium disintegrates at a rate proportional to the amount of radium present at the moment. Its half-life is 1590 years. What percentage will disappear in one year?
Write the differential equation obtained eliminating the arbitrary constant C in the equation xy = C2.
Solve the following differential equation : \[y^2 dx + \left( x^2 - xy + y^2 \right)dy = 0\] .
Solve the following differential equation.
`y^3 - dy/dx = x dy/dx`
Solve the following differential equation.
`dy/dx + y` = 3
The solution of `dy/dx + x^2/y^2 = 0` is ______
A solution of a differential equation which can be obtained from the general solution by giving particular values to the arbitrary constants is called ___________ solution.
The integrating factor of the differential equation `dy/dx - y = x` is e−x.
Solve the differential equation sec2y tan x dy + sec2x tan y dx = 0
Solve the following differential equation `("d"y)/("d"x)` = cos(x + y)
Solution: `("d"y)/("d"x)` = cos(x + y) ......(1)
Put `square`
∴ `1 + ("d"y)/("d"x) = "dv"/("d"x)`
∴ `("d"y)/("d"x) = "dv"/("d"x) - 1`
∴ (1) becomes `"dv"/("d"x) - 1` = cos v
∴ `"dv"/("d"x)` = 1 + cos v
∴ `square` dv = dx
Integrating, we get
`int 1/(1 + cos "v") "d"v = int "d"x`
∴ `int 1/(2cos^2 ("v"/2)) "dv" = int "d"x`
∴ `1/2 int square "dv" = int "d"x`
∴ `1/2* (tan("v"/2))/(1/2)` = x + c
∴ `square` = x + c
