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Commerce (English Medium) Class 12 - CBSE Question Bank Solutions

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< prev  11821 to 11840 of 18433  next > 
\[\int\frac{2x}{x^3 - 1} dx\]
[7] Integrals
Chapter: [7] Integrals
Concept: undefined >> undefined
\[\int\frac{1}{\left( x^2 - 1 \right) \sqrt{x^2 + 1}} \text{ dx }\]
[7] Integrals
Chapter: [7] Integrals
Concept: undefined >> undefined

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Integration of \[\frac{1}{1 + \left( \log_e x \right)^2}\] with respect to loge x is

[7] Integrals
Chapter: [7] Integrals
Concept: undefined >> undefined
\[\int \left| x \right|^3 dx\] is equal to
[7] Integrals
Chapter: [7] Integrals
Concept: undefined >> undefined

\[\int\frac{8x + 13}{\sqrt{4x + 7}} \text{ dx }\]

[7] Integrals
Chapter: [7] Integrals
Concept: undefined >> undefined

\[\int\frac{1 + x + x^2}{x^2 \left( 1 + x \right)} \text{ dx}\]

[7] Integrals
Chapter: [7] Integrals
Concept: undefined >> undefined

Find \[\frac{dy}{dx}\] at \[t = \frac{2\pi}{3}\] when x = 10 (t – sin t) and y = 12 (1 – cos t).

[5] Continuity and Differentiability
Chapter: [5] Continuity and Differentiability
Concept: undefined >> undefined

If xy = e(x – y), then show that `dy/dx = (y(x-1))/(x(y+1)) .`

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

If logy = tan–1 x, then show that `(1+x^2) (d^2y)/(dx^2) + (2x - 1) dy/dx = 0 .`

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

Differentiate \[\tan^{- 1} \left( \frac{\sqrt{1 + x^2} - 1}{x} \right) w . r . t . \sin^{- 1} \frac{2x}{1 + x^2},\]tan-11+x2-1x w.r.t. sin-12x1+x2, if x ∈ (–1, 1) .

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

If x = sin t and y = sin pt, prove that \[\left( 1 - x^2 \right)\frac{d^2 y}{d x^2} - x\frac{dy}{dx} + p^2 y = 0\] .

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

Find : \[\int\left( 2x + 5 \right)\sqrt{10 - 4x - 3 x^2}dx\] .

[7] Integrals
Chapter: [7] Integrals
Concept: undefined >> undefined

\[\text { If } y = \left( x + \sqrt{1 + x^2} \right)^n , \text { then show that }\]

\[\left( 1 + x^2 \right)\frac{d^2 y}{d x^2} + x\frac{dy}{dx} = n^2 y .\]

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

Find the minimum value of (ax + by), where xy = c2.

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

Let f : N → ℝ be a function defined as f(x) = 4x2 + 12x + 15. Show that f : N → S, where S is the range of f, is invertible. Also find the inverse of f.

[1] Relations and Functions
Chapter: [1] Relations and Functions
Concept: undefined >> undefined

If y = xx, prove that \[\frac{d^2 y}{d x^2} - \frac{1}{y} \left( \frac{dy}{dx} \right)^2 - \frac{y}{x} = 0 .\]

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

Find the shortest distance between the lines

\[\frac{x - 2}{- 1} = \frac{y - 5}{2} = \frac{z - 0}{3} \text{ and }  \frac{x - 0}{2} = \frac{y + 1}{- 1} = \frac{z - 1}{2} .\]
 
[11] Three - Dimensional Geometry
Chapter: [11] Three - Dimensional Geometry
Concept: undefined >> undefined

Find the shortest distance between the lines 

\[\frac{x + 1}{7} = \frac{y + 1}{- 6} = \frac{z + 1}{1} \text{ and } \frac{x - 3}{1} = \frac{y - 5}{- 2} = \frac{z - 7}{1} .\]
 
[11] Three - Dimensional Geometry
Chapter: [11] Three - Dimensional Geometry
Concept: undefined >> undefined

Find the shortest distance between the lines

\[\frac{x - 1}{2} = \frac{y - 3}{4} = \frac{z + 2}{1}\] and
\[3x - y - 2z + 4 = 0 = 2x + y + z + 1\]
 
[11] Three - Dimensional Geometry
Chapter: [11] Three - Dimensional Geometry
Concept: undefined >> undefined

Differentiate sin(log sin x) ?

[6] Applications of Derivatives
Chapter: [6] Applications of Derivatives
Concept: undefined >> undefined
< prev  11821 to 11840 of 18433  next > 
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