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The order of the differential equation whose general solution is given by y = c1 cos (2x + c2) − (c3 + c4) ax + c5 + c6 sin (x − c7) is

[9] Differential Equations
Chapter: [9] Differential Equations
Concept: undefined >> undefined

The order of the differential equation satisfying
\[\sqrt{1 - x^4} + \sqrt{1 - y^4} = a\left( x^2 - y^2 \right)\] is

[9] Differential Equations
Chapter: [9] Differential Equations
Concept: undefined >> undefined

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If p and q are the order and degree of the differential equation \[y\frac{dy}{dx} + x^3 \frac{d^2 y}{d x^2} + xy\] = cos x, then

[9] Differential Equations
Chapter: [9] Differential Equations
Concept: undefined >> undefined

The degree of the differential equation \[\left( \frac{d^2 y}{d x^2} \right)^3 + \left( \frac{dy}{dx} \right)^2 + \sin\left( \frac{dy}{dx} \right) + 1 = 0\], is

[9] Differential Equations
Chapter: [9] Differential Equations
Concept: undefined >> undefined

The order of the differential equation \[2 x^2 \frac{d^2 y}{d x^2} - 3\frac{dy}{dx} + y = 0\], is

[9] Differential Equations
Chapter: [9] Differential Equations
Concept: undefined >> undefined

Find the vector and cartesian equations of the line through the point (5, 2, −4) and which is parallel to the vector  \[3 \hat{i} + 2 \hat{j} - 8 \hat{k} .\]

[11] Three - Dimensional Geometry
Chapter: [11] Three - Dimensional Geometry
Concept: undefined >> undefined

Find the vector equation of the line passing through the points (−1, 0, 2) and (3, 4, 6).

[11] Three - Dimensional Geometry
Chapter: [11] Three - Dimensional Geometry
Concept: undefined >> undefined

Find the vector equation of a line which is parallel to the vector \[2 \hat{i} - \hat{j} + 3 \hat{k}\]  and which passes through the point (5, −2, 4). Also, reduce it to cartesian form.

[11] Three - Dimensional Geometry
Chapter: [11] Three - Dimensional Geometry
Concept: undefined >> undefined

A line passes through the point with position vector \[2 \hat{i} - 3 \hat{j} + 4 \hat{k} \] and is in the direction of  \[3 \hat{i} + 4 \hat{j} - 5 \hat{k} .\] Find equations of the line in vector and cartesian form. 

[11] Three - Dimensional Geometry
Chapter: [11] Three - Dimensional Geometry
Concept: undefined >> undefined

ABCD is a parallelogram. The position vectors of the points AB and C are respectively, \[4 \hat{ i} + 5 \hat{j} -10 \hat{k} , 2 \hat{i} - 3 \hat{j} + 4 \hat{k}  \text{ and } - \hat{i} + 2 \hat{j} + \hat{k} .\]  Find the vector equation of the line BD. Also, reduce it to cartesian form.

[11] Three - Dimensional Geometry
Chapter: [11] Three - Dimensional Geometry
Concept: undefined >> undefined

Find in vector form as well as in cartesian form, the equation of the line passing through the points A (1, 2, −1) and B (2, 1, 1).

[11] Three - Dimensional Geometry
Chapter: [11] Three - Dimensional Geometry
Concept: undefined >> undefined

Prove that the function f(x) = loge x is increasing on (0, ∞) ?

[6] Applications of Derivatives
Chapter: [6] Applications of Derivatives
Concept: undefined >> undefined

Prove that the function f(x) = loga x is increasing on (0, ∞) if a > 1 and decreasing on (0, ∞), if 0 < a < 1 ?

[6] Applications of Derivatives
Chapter: [6] Applications of Derivatives
Concept: undefined >> undefined

Prove that f(x) = ax + b, where a, b are constants and a > 0 is an increasing function on R ?

[6] Applications of Derivatives
Chapter: [6] Applications of Derivatives
Concept: undefined >> undefined

Prove that f(x) = ax + b, where a, b are constants and a < 0 is a decreasing function on R ?

[6] Applications of Derivatives
Chapter: [6] Applications of Derivatives
Concept: undefined >> undefined

Find the vector equation for the line which passes through the point (1, 2, 3) and parallel to the vector \[\hat{i} - 2 \hat{j} + 3 \hat{k} .\]  Reduce the corresponding equation in cartesian from.

[11] Three - Dimensional Geometry
Chapter: [11] Three - Dimensional Geometry
Concept: undefined >> undefined

Show that f(x) = \[\frac{1}{x}\] is a decreasing function on (0, ∞) ?

[6] Applications of Derivatives
Chapter: [6] Applications of Derivatives
Concept: undefined >> undefined

Find the vector equation of a line passing through (2, −1, 1) and parallel to the line whose equations are \[\frac{x - 3}{2} = \frac{y + 1}{7} = \frac{z - 2}{- 3} .\]

[11] Three - Dimensional Geometry
Chapter: [11] Three - Dimensional Geometry
Concept: undefined >> undefined

Show that f(x) = \[\frac{1}{1 + x^2}\] decreases in the interval [0, ∞) and increases in the interval (−∞, 0] ?

[6] Applications of Derivatives
Chapter: [6] Applications of Derivatives
Concept: undefined >> undefined

The cartesian equations of a line are \[\frac{x - 5}{3} = \frac{y + 4}{7} = \frac{z - 6}{2} .\]  Find a vector equation for the line.

[11] Three - Dimensional Geometry
Chapter: [11] Three - Dimensional Geometry
Concept: undefined >> undefined
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CBSE Science (English Medium) कक्षा १२ Question Bank Solutions
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Question Bank Solutions for CBSE Science (English Medium) कक्षा १२ English Elective - NCERT
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