QBManager

Limits, Continuity and Differentiability

80272 If $x = \sin^{- 1} (\cos θ)$ and $y = \tan^{- 1} θ$ , then $\frac{d y}{d x} =$

1

\frac{1}{1 + θ^{2}}

2

1 + θ^{2}

3

\frac{- 1}{1 + θ^{2}}

4

- (1 + θ^{2})

Explanation:

(C) : Given, $x = \sin^{- 1} (\cos θ)$
$\frac{dx}{d θ} = \frac{1}{\sqrt{1 - \cos^{2} θ}} \frac{d}{d θ} (\cos θ) \Rightarrow \frac{dx}{d θ} = \frac{- \sin θ}{\sin θ} \Rightarrow \frac{dx}{d θ} = - 1$
And, $y = \tan^{- 1} θ$
$\frac{dy}{d θ} = \frac{1}{1 + θ^{2}}$
Therefore,
$\frac{d y}{d x} = \frac{\frac{d y}{d θ}}{\frac{d x}{d θ}}, \frac{d y}{d x} = \frac{\frac{1}{1 + θ^{2}}}{- 1} \Rightarrow \frac{d y}{d x} = - (\frac{1}{1 + θ^{2}})$

MHT CET-2019

Limits, Continuity and Differentiability

80273 If $y = \log [\sec (e^{x})]$ then $\frac{d y}{d x} =$

1

\frac{x \tan (e^{x})}{e^{x}}

2

- e^{x} \tan (e^{x})

3

e^{x} \tan (e^{x})

4

\frac{- x \tan (e^{x})}{e^{x}}

Explanation:

(C) : Given,
$y = \log [\sec (e^{x})]$
$\frac{d y}{d x} = \frac{1}{\sec (e^{x})} \cdot \frac{d}{d x} \sec (e^{x})$
$\frac{d y}{d x} = \frac{1}{\sec (e^{x})} \cdot \sec (e^{x}) \cdot \tan (e^{x}) \cdot \frac{d}{d x} \cdot e^{x}$
$\frac{d y}{d x} = e^{x} \tan (e^{x})$

MHT CET-2019

Limits, Continuity and Differentiability

80274 If $y = \tan^{- 1} (\sec x + \tan x)$ , then $\frac{d y}{d x} =$

1

\frac{- 1}{2}

2 1

3

\frac{1}{2}

4 -1

Explanation:

(C) : Given,
$y = \tan^{- 1} (see x + \tan x)$
$y = \tan^{- 1} (\frac{1}{\cos x} + \frac{\sin x}{\cos x}) = y = \tan^{- 1} (\frac{1 + \sin x}{\cos x})$
$y = \tan^{- 1} (\frac{\cos^{2} \frac{x}{2} + \sin^{2} \frac{x}{2} + 2 \sin \frac{x}{2} \cdot \cos \frac{x}{2}}{\cos^{2} \frac{x}{2} - \sin^{2} \frac{x}{2}})$
$y = \tan^{- 1} [\frac{{(\cos \frac{x}{2} + \sin \frac{x}{2})}^{2}}{(\cos \frac{x}{2} + \sin \frac{x}{2}) (\cos \frac{x}{2} - \sin \frac{x}{2})}]$
$y = \tan^{- 1} [\frac{\cos \frac{x}{2} + \sin \frac{x}{2}}{\cos \frac{x}{2} - \sin \frac{x}{2}}] = \tan^{- 1} [\frac{1 + \tan \frac{x}{2}}{1 - \tan \frac{x}{2}}]$
$y = \tan^{- 1} [\frac{\tan \frac{π}{4} + \tan \frac{x}{2}}{1 - \tan \frac{π}{4} \tan \frac{x}{2}}] = \tan^{- 1} [\tan (\frac{π}{4} + \frac{x}{2})]$
$y = \frac{π}{4} + \frac{x}{2}$
$∴ \frac{dy}{dx} = 0 + \frac{1}{2} \Rightarrow \frac{dy}{dx} = \frac{1}{2}$

MHT CET-2020

Limits, Continuity and Differentiability

80275 If $y = 3 e^{5 x} + 5 e^{3 x}$ , then $\frac{d^{2} y}{d x^{2}} - 8 \frac{d y}{d x} =$

1

- 10 y

2

- 15 y

3

15 y

4

10 y

Explanation:

(B) : Given,
$y = 3 e^{5 x} + 5 e^{3 x}$
$\frac{d y}{d x} = 3 e^{5 x} \frac{d}{d x} 5 x + 5 e^{3 x} \frac{d}{d x} \cdot 3 x$
$\frac{d y}{d x} = 15 e^{5 x} + 15 e^{3 x}$
$∴ \frac{d^{2} y}{{dx}^{2}} = 15 e^{5 x} \cdot \frac{d}{dx} 5 x + 15 e^{3 x} \frac{d}{dx} 3 x$
$\frac{d^{2} y}{d x^{2}} = 75 e^{5 x} + 45 e^{3 x}$
Therefore,
$\frac{d^{2} y}{d x^{2}} - 8 \frac{d y}{d x} = 75 e^{5 x} + 45 e^{3 x} - 8 (15 e^{5 x} + 15 e^{3 x})$
$\frac{d^{2} y}{d x^{2}} - 8 \frac{d y}{d x} = 75 e^{5 x} + 45 e^{3 x} - 120 e^{5 x} - 120 e^{3 x}$
$= - 45 e^{5 x} - 75 e^{3 x} = - 15 (3 e^{5 x} + 5 e^{3 x}) = - 15 y$

MHT CET-2020

Limits, Continuity and Differentiability

80272 If $x = \sin^{- 1} (\cos θ)$ and $y = \tan^{- 1} θ$ , then $\frac{d y}{d x} =$

1

\frac{1}{1 + θ^{2}}

2

1 + θ^{2}

3

\frac{- 1}{1 + θ^{2}}

4

- (1 + θ^{2})

Explanation:

(C) : Given, $x = \sin^{- 1} (\cos θ)$
$\frac{dx}{d θ} = \frac{1}{\sqrt{1 - \cos^{2} θ}} \frac{d}{d θ} (\cos θ) \Rightarrow \frac{dx}{d θ} = \frac{- \sin θ}{\sin θ} \Rightarrow \frac{dx}{d θ} = - 1$
And, $y = \tan^{- 1} θ$
$\frac{dy}{d θ} = \frac{1}{1 + θ^{2}}$
Therefore,
$\frac{d y}{d x} = \frac{\frac{d y}{d θ}}{\frac{d x}{d θ}}, \frac{d y}{d x} = \frac{\frac{1}{1 + θ^{2}}}{- 1} \Rightarrow \frac{d y}{d x} = - (\frac{1}{1 + θ^{2}})$

MHT CET-2019

Limits, Continuity and Differentiability

80273 If $y = \log [\sec (e^{x})]$ then $\frac{d y}{d x} =$

1

\frac{x \tan (e^{x})}{e^{x}}

2

- e^{x} \tan (e^{x})

3

e^{x} \tan (e^{x})

4

\frac{- x \tan (e^{x})}{e^{x}}

Explanation:

(C) : Given,
$y = \log [\sec (e^{x})]$
$\frac{d y}{d x} = \frac{1}{\sec (e^{x})} \cdot \frac{d}{d x} \sec (e^{x})$
$\frac{d y}{d x} = \frac{1}{\sec (e^{x})} \cdot \sec (e^{x}) \cdot \tan (e^{x}) \cdot \frac{d}{d x} \cdot e^{x}$
$\frac{d y}{d x} = e^{x} \tan (e^{x})$

MHT CET-2019

Limits, Continuity and Differentiability

80274 If $y = \tan^{- 1} (\sec x + \tan x)$ , then $\frac{d y}{d x} =$

1

\frac{- 1}{2}

2 1

3

\frac{1}{2}

4 -1

Explanation:

(C) : Given,
$y = \tan^{- 1} (see x + \tan x)$
$y = \tan^{- 1} (\frac{1}{\cos x} + \frac{\sin x}{\cos x}) = y = \tan^{- 1} (\frac{1 + \sin x}{\cos x})$
$y = \tan^{- 1} (\frac{\cos^{2} \frac{x}{2} + \sin^{2} \frac{x}{2} + 2 \sin \frac{x}{2} \cdot \cos \frac{x}{2}}{\cos^{2} \frac{x}{2} - \sin^{2} \frac{x}{2}})$
$y = \tan^{- 1} [\frac{{(\cos \frac{x}{2} + \sin \frac{x}{2})}^{2}}{(\cos \frac{x}{2} + \sin \frac{x}{2}) (\cos \frac{x}{2} - \sin \frac{x}{2})}]$
$y = \tan^{- 1} [\frac{\cos \frac{x}{2} + \sin \frac{x}{2}}{\cos \frac{x}{2} - \sin \frac{x}{2}}] = \tan^{- 1} [\frac{1 + \tan \frac{x}{2}}{1 - \tan \frac{x}{2}}]$
$y = \tan^{- 1} [\frac{\tan \frac{π}{4} + \tan \frac{x}{2}}{1 - \tan \frac{π}{4} \tan \frac{x}{2}}] = \tan^{- 1} [\tan (\frac{π}{4} + \frac{x}{2})]$
$y = \frac{π}{4} + \frac{x}{2}$
$∴ \frac{dy}{dx} = 0 + \frac{1}{2} \Rightarrow \frac{dy}{dx} = \frac{1}{2}$

MHT CET-2020

Limits, Continuity and Differentiability

80275 If $y = 3 e^{5 x} + 5 e^{3 x}$ , then $\frac{d^{2} y}{d x^{2}} - 8 \frac{d y}{d x} =$

1

- 10 y

2

- 15 y

3

15 y

4

10 y

Explanation:

(B) : Given,
$y = 3 e^{5 x} + 5 e^{3 x}$
$\frac{d y}{d x} = 3 e^{5 x} \frac{d}{d x} 5 x + 5 e^{3 x} \frac{d}{d x} \cdot 3 x$
$\frac{d y}{d x} = 15 e^{5 x} + 15 e^{3 x}$
$∴ \frac{d^{2} y}{{dx}^{2}} = 15 e^{5 x} \cdot \frac{d}{dx} 5 x + 15 e^{3 x} \frac{d}{dx} 3 x$
$\frac{d^{2} y}{d x^{2}} = 75 e^{5 x} + 45 e^{3 x}$
Therefore,
$\frac{d^{2} y}{d x^{2}} - 8 \frac{d y}{d x} = 75 e^{5 x} + 45 e^{3 x} - 8 (15 e^{5 x} + 15 e^{3 x})$
$\frac{d^{2} y}{d x^{2}} - 8 \frac{d y}{d x} = 75 e^{5 x} + 45 e^{3 x} - 120 e^{5 x} - 120 e^{3 x}$
$= - 45 e^{5 x} - 75 e^{3 x} = - 15 (3 e^{5 x} + 5 e^{3 x}) = - 15 y$

MHT CET-2020

Limits, Continuity and Differentiability

80272 If $x = \sin^{- 1} (\cos θ)$ and $y = \tan^{- 1} θ$ , then $\frac{d y}{d x} =$

1

\frac{1}{1 + θ^{2}}

2

1 + θ^{2}

3

\frac{- 1}{1 + θ^{2}}

4

- (1 + θ^{2})

Explanation:

(C) : Given, $x = \sin^{- 1} (\cos θ)$
$\frac{dx}{d θ} = \frac{1}{\sqrt{1 - \cos^{2} θ}} \frac{d}{d θ} (\cos θ) \Rightarrow \frac{dx}{d θ} = \frac{- \sin θ}{\sin θ} \Rightarrow \frac{dx}{d θ} = - 1$
And, $y = \tan^{- 1} θ$
$\frac{dy}{d θ} = \frac{1}{1 + θ^{2}}$
Therefore,
$\frac{d y}{d x} = \frac{\frac{d y}{d θ}}{\frac{d x}{d θ}}, \frac{d y}{d x} = \frac{\frac{1}{1 + θ^{2}}}{- 1} \Rightarrow \frac{d y}{d x} = - (\frac{1}{1 + θ^{2}})$

MHT CET-2019

Limits, Continuity and Differentiability

80273 If $y = \log [\sec (e^{x})]$ then $\frac{d y}{d x} =$

1

\frac{x \tan (e^{x})}{e^{x}}

2

- e^{x} \tan (e^{x})

3

e^{x} \tan (e^{x})

4

\frac{- x \tan (e^{x})}{e^{x}}

Explanation:

(C) : Given,
$y = \log [\sec (e^{x})]$
$\frac{d y}{d x} = \frac{1}{\sec (e^{x})} \cdot \frac{d}{d x} \sec (e^{x})$
$\frac{d y}{d x} = \frac{1}{\sec (e^{x})} \cdot \sec (e^{x}) \cdot \tan (e^{x}) \cdot \frac{d}{d x} \cdot e^{x}$
$\frac{d y}{d x} = e^{x} \tan (e^{x})$

MHT CET-2019

Limits, Continuity and Differentiability

80274 If $y = \tan^{- 1} (\sec x + \tan x)$ , then $\frac{d y}{d x} =$

1

\frac{- 1}{2}

2 1

3

\frac{1}{2}

4 -1

Explanation:

(C) : Given,
$y = \tan^{- 1} (see x + \tan x)$
$y = \tan^{- 1} (\frac{1}{\cos x} + \frac{\sin x}{\cos x}) = y = \tan^{- 1} (\frac{1 + \sin x}{\cos x})$
$y = \tan^{- 1} (\frac{\cos^{2} \frac{x}{2} + \sin^{2} \frac{x}{2} + 2 \sin \frac{x}{2} \cdot \cos \frac{x}{2}}{\cos^{2} \frac{x}{2} - \sin^{2} \frac{x}{2}})$
$y = \tan^{- 1} [\frac{{(\cos \frac{x}{2} + \sin \frac{x}{2})}^{2}}{(\cos \frac{x}{2} + \sin \frac{x}{2}) (\cos \frac{x}{2} - \sin \frac{x}{2})}]$
$y = \tan^{- 1} [\frac{\cos \frac{x}{2} + \sin \frac{x}{2}}{\cos \frac{x}{2} - \sin \frac{x}{2}}] = \tan^{- 1} [\frac{1 + \tan \frac{x}{2}}{1 - \tan \frac{x}{2}}]$
$y = \tan^{- 1} [\frac{\tan \frac{π}{4} + \tan \frac{x}{2}}{1 - \tan \frac{π}{4} \tan \frac{x}{2}}] = \tan^{- 1} [\tan (\frac{π}{4} + \frac{x}{2})]$
$y = \frac{π}{4} + \frac{x}{2}$
$∴ \frac{dy}{dx} = 0 + \frac{1}{2} \Rightarrow \frac{dy}{dx} = \frac{1}{2}$

MHT CET-2020

Limits, Continuity and Differentiability

80275 If $y = 3 e^{5 x} + 5 e^{3 x}$ , then $\frac{d^{2} y}{d x^{2}} - 8 \frac{d y}{d x} =$

1

- 10 y

2

- 15 y

3

15 y

4

10 y

Explanation:

(B) : Given,
$y = 3 e^{5 x} + 5 e^{3 x}$
$\frac{d y}{d x} = 3 e^{5 x} \frac{d}{d x} 5 x + 5 e^{3 x} \frac{d}{d x} \cdot 3 x$
$\frac{d y}{d x} = 15 e^{5 x} + 15 e^{3 x}$
$∴ \frac{d^{2} y}{{dx}^{2}} = 15 e^{5 x} \cdot \frac{d}{dx} 5 x + 15 e^{3 x} \frac{d}{dx} 3 x$
$\frac{d^{2} y}{d x^{2}} = 75 e^{5 x} + 45 e^{3 x}$
Therefore,
$\frac{d^{2} y}{d x^{2}} - 8 \frac{d y}{d x} = 75 e^{5 x} + 45 e^{3 x} - 8 (15 e^{5 x} + 15 e^{3 x})$
$\frac{d^{2} y}{d x^{2}} - 8 \frac{d y}{d x} = 75 e^{5 x} + 45 e^{3 x} - 120 e^{5 x} - 120 e^{3 x}$
$= - 45 e^{5 x} - 75 e^{3 x} = - 15 (3 e^{5 x} + 5 e^{3 x}) = - 15 y$

MHT CET-2020

Limits, Continuity and Differentiability

80272 If $x = \sin^{- 1} (\cos θ)$ and $y = \tan^{- 1} θ$ , then $\frac{d y}{d x} =$

1

\frac{1}{1 + θ^{2}}

2

1 + θ^{2}

3

\frac{- 1}{1 + θ^{2}}

4

- (1 + θ^{2})

Explanation:

(C) : Given, $x = \sin^{- 1} (\cos θ)$
$\frac{dx}{d θ} = \frac{1}{\sqrt{1 - \cos^{2} θ}} \frac{d}{d θ} (\cos θ) \Rightarrow \frac{dx}{d θ} = \frac{- \sin θ}{\sin θ} \Rightarrow \frac{dx}{d θ} = - 1$
And, $y = \tan^{- 1} θ$
$\frac{dy}{d θ} = \frac{1}{1 + θ^{2}}$
Therefore,
$\frac{d y}{d x} = \frac{\frac{d y}{d θ}}{\frac{d x}{d θ}}, \frac{d y}{d x} = \frac{\frac{1}{1 + θ^{2}}}{- 1} \Rightarrow \frac{d y}{d x} = - (\frac{1}{1 + θ^{2}})$

MHT CET-2019

Limits, Continuity and Differentiability

80273 If $y = \log [\sec (e^{x})]$ then $\frac{d y}{d x} =$

1

\frac{x \tan (e^{x})}{e^{x}}

2

- e^{x} \tan (e^{x})

3

e^{x} \tan (e^{x})

4

\frac{- x \tan (e^{x})}{e^{x}}

Explanation:

(C) : Given,
$y = \log [\sec (e^{x})]$
$\frac{d y}{d x} = \frac{1}{\sec (e^{x})} \cdot \frac{d}{d x} \sec (e^{x})$
$\frac{d y}{d x} = \frac{1}{\sec (e^{x})} \cdot \sec (e^{x}) \cdot \tan (e^{x}) \cdot \frac{d}{d x} \cdot e^{x}$
$\frac{d y}{d x} = e^{x} \tan (e^{x})$

MHT CET-2019

Limits, Continuity and Differentiability

80274 If $y = \tan^{- 1} (\sec x + \tan x)$ , then $\frac{d y}{d x} =$

1

\frac{- 1}{2}

2 1

3

\frac{1}{2}

4 -1

Explanation:

(C) : Given,
$y = \tan^{- 1} (see x + \tan x)$
$y = \tan^{- 1} (\frac{1}{\cos x} + \frac{\sin x}{\cos x}) = y = \tan^{- 1} (\frac{1 + \sin x}{\cos x})$
$y = \tan^{- 1} (\frac{\cos^{2} \frac{x}{2} + \sin^{2} \frac{x}{2} + 2 \sin \frac{x}{2} \cdot \cos \frac{x}{2}}{\cos^{2} \frac{x}{2} - \sin^{2} \frac{x}{2}})$
$y = \tan^{- 1} [\frac{{(\cos \frac{x}{2} + \sin \frac{x}{2})}^{2}}{(\cos \frac{x}{2} + \sin \frac{x}{2}) (\cos \frac{x}{2} - \sin \frac{x}{2})}]$
$y = \tan^{- 1} [\frac{\cos \frac{x}{2} + \sin \frac{x}{2}}{\cos \frac{x}{2} - \sin \frac{x}{2}}] = \tan^{- 1} [\frac{1 + \tan \frac{x}{2}}{1 - \tan \frac{x}{2}}]$
$y = \tan^{- 1} [\frac{\tan \frac{π}{4} + \tan \frac{x}{2}}{1 - \tan \frac{π}{4} \tan \frac{x}{2}}] = \tan^{- 1} [\tan (\frac{π}{4} + \frac{x}{2})]$
$y = \frac{π}{4} + \frac{x}{2}$
$∴ \frac{dy}{dx} = 0 + \frac{1}{2} \Rightarrow \frac{dy}{dx} = \frac{1}{2}$

MHT CET-2020

Limits, Continuity and Differentiability

80275 If $y = 3 e^{5 x} + 5 e^{3 x}$ , then $\frac{d^{2} y}{d x^{2}} - 8 \frac{d y}{d x} =$

1

- 10 y

2

- 15 y

3

15 y

4

10 y

Explanation:

(B) : Given,
$y = 3 e^{5 x} + 5 e^{3 x}$
$\frac{d y}{d x} = 3 e^{5 x} \frac{d}{d x} 5 x + 5 e^{3 x} \frac{d}{d x} \cdot 3 x$
$\frac{d y}{d x} = 15 e^{5 x} + 15 e^{3 x}$
$∴ \frac{d^{2} y}{{dx}^{2}} = 15 e^{5 x} \cdot \frac{d}{dx} 5 x + 15 e^{3 x} \frac{d}{dx} 3 x$
$\frac{d^{2} y}{d x^{2}} = 75 e^{5 x} + 45 e^{3 x}$
Therefore,
$\frac{d^{2} y}{d x^{2}} - 8 \frac{d y}{d x} = 75 e^{5 x} + 45 e^{3 x} - 8 (15 e^{5 x} + 15 e^{3 x})$
$\frac{d^{2} y}{d x^{2}} - 8 \frac{d y}{d x} = 75 e^{5 x} + 45 e^{3 x} - 120 e^{5 x} - 120 e^{3 x}$
$= - 45 e^{5 x} - 75 e^{3 x} = - 15 (3 e^{5 x} + 5 e^{3 x}) = - 15 y$

MHT CET-2020

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Question Bank Series

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation: