QBManager

Limits, Continuity and Differentiability

80306 If $x = e^{y + e^{y + e^{y + e}}}$ then $\frac{d y}{d x} =$

1

\frac{1}{x}

2

\frac{x}{1 + x}

3

\frac{1 - x}{x}

4

\frac{1 + x}{x}

Explanation:

(C) : It is given that,
Here,
$x = e^{y + ey + ey + ey + \dots . . . \infty}$
$x = e^{y + x}$
Differentiating w.r.to $x$ .
$1 = e^{y + x} (\frac{d y}{d x} + 1) \Rightarrow 1 = e^{y + x} \frac{d y}{d x} + e^{y + x}$
$1 = x \frac{d y}{d x} + x, 1 - x = x \frac{d y}{d x} \Rightarrow \frac{d y}{d x} = \frac{1 - x}{x}$

MHT CET-2020

Limits, Continuity and Differentiability

80307 If $y = \cos^{2} (\frac{5 x}{2}) - \sin^{2} (\frac{5 x}{2})$ , then $(\frac{d^{2} y}{d x^{2}}) =$

1

5 \sqrt{1 - y^{2}}

2

25 y

3

- 25 y

4

- 5 \sqrt{1 - y^{2}}

[MНT CET-2020]

Explanation:

(C) : Given,
$y = \cos^{2} (\frac{5 x}{2}) - \sin^{2} (\frac{5 x}{2})$
We know that,
So,
$\cos^{2} x - \sin^{2} x = \cos 2 x$
$y = \cos 2 (\frac{5 x}{2})$
$y = \cos 5 x$
Now, differentiate with respect to $x$
$\frac{d y}{d x} = - 5 \sin 5 x$
$∴ \frac{d^{2} y}{{dx}^{2}} = - 25 \cos 5 x$
On putting the value $(\cos 5 x = y)$ in above equation. Then,
$\frac{d^{2} y}{d x^{2}} = - 25 y$

Limits, Continuity and Differentiability

80308 If $y = 2^{a x}$ and ${(\frac{d y}{d x})}_{x = 1} = \log 256$ , then $a =$

1 3

2 2

3 4

4 8

Explanation:

(B) : Given, $y = 2^{ax}$
On taking $\log$ both sides.
$\log y = \log 2^{a x} \Rightarrow \log y = a x \log 2$
Differentiating on both sides w.r.t.x
$\frac{1}{y} \frac{dy}{dx} = a \times 1 \times \log 2 \Rightarrow \frac{dy}{dx} = ay \log 2$
On putting the value $y$ in above equation,
$\frac{d y}{d x} = 2^{a x} a \log 2 \Rightarrow {(\frac{d y}{d x})}_{x = 1} = a 2^{a \times 1} \log 2$
${(\frac{d y}{d x})}_{x = 1} = a 2^{a} \log 2$
Given by,
${(\frac{d y}{d x})}_{x = 1} = \log 256 \Rightarrow {(\frac{d y}{d x})}_{x = 1} = \log 2^{8}$
${(\frac{d y}{d x})}_{x = 1} = 8 \log 2$
On comparing equation (i) and equation (ii)
$a 2^{a} \log 2 = 8 \log 2$
$a \cdot 2^{a} = 2 \times 2^{2}$
On comparing both sides, $a = 2$

MHT CET-2020

Limits, Continuity and Differentiability

80309 If $x^{p} y^{q} = (x + y)^{p + q}$ , then $\frac{d y}{d x}$ is equal to

1

\frac{y}{x}

2

\frac{p y}{q x}

3

\frac{x}{y}

4

\frac{qy}{px}

Explanation:

(A) : Given, $x^{p} y^{q} = (x + y)^{p + q}$
On taking $\log$ both sides.
$\log (x^{p} y^{q}) = \log (x + y)^{p + q}$
$p \log x + q \log y = (p + q) \log (x + y)$
Differentiating on both sides w.r.t.x
$\frac{p}{x} + \frac{q}{y} (\frac{d y}{d x}) = (\frac{p + q}{x + y}) \times (1 + \frac{d y}{d x})$
$\frac{p}{x} + \frac{q}{y} (\frac{d y}{d x}) = \frac{p + q}{x + y} + \frac{p + q}{x + y} (\frac{d y}{d x})$
$\frac{d y}{d x} [\frac{q}{y} - \frac{p + q}{x + y}] = \frac{p + q}{x + y} - \frac{p}{x}$
$\frac{d y}{d x} [\frac{q (x + y) - y (p + q)}{y (x + y)}] = \frac{x (p + q) - p (x + y)}{x (x + y)}$
$\frac{d y}{d x} [\frac{q x + q y - p y - q y}{y (x + y)}] = \frac{p x + q x - p x - p y}{x (x + y)}$
$\frac{d y}{d x} [\frac{q x - p y}{y (x + y)}] = \frac{q x - p y}{x (x + y)} \Rightarrow \frac{d y}{d x} = \frac{y (x + y)}{x (x + y)} \Rightarrow \frac{d y}{d x} = \frac{y}{x}$

MHT CET-2011

Limits, Continuity and Differentiability

80306 If $x = e^{y + e^{y + e^{y + e}}}$ then $\frac{d y}{d x} =$

1

\frac{1}{x}

2

\frac{x}{1 + x}

3

\frac{1 - x}{x}

4

\frac{1 + x}{x}

Explanation:

(C) : It is given that,
Here,
$x = e^{y + ey + ey + ey + \dots . . . \infty}$
$x = e^{y + x}$
Differentiating w.r.to $x$ .
$1 = e^{y + x} (\frac{d y}{d x} + 1) \Rightarrow 1 = e^{y + x} \frac{d y}{d x} + e^{y + x}$
$1 = x \frac{d y}{d x} + x, 1 - x = x \frac{d y}{d x} \Rightarrow \frac{d y}{d x} = \frac{1 - x}{x}$

MHT CET-2020

Limits, Continuity and Differentiability

80307 If $y = \cos^{2} (\frac{5 x}{2}) - \sin^{2} (\frac{5 x}{2})$ , then $(\frac{d^{2} y}{d x^{2}}) =$

1

5 \sqrt{1 - y^{2}}

2

25 y

3

- 25 y

4

- 5 \sqrt{1 - y^{2}}

[MНT CET-2020]

Explanation:

(C) : Given,
$y = \cos^{2} (\frac{5 x}{2}) - \sin^{2} (\frac{5 x}{2})$
We know that,
So,
$\cos^{2} x - \sin^{2} x = \cos 2 x$
$y = \cos 2 (\frac{5 x}{2})$
$y = \cos 5 x$
Now, differentiate with respect to $x$
$\frac{d y}{d x} = - 5 \sin 5 x$
$∴ \frac{d^{2} y}{{dx}^{2}} = - 25 \cos 5 x$
On putting the value $(\cos 5 x = y)$ in above equation. Then,
$\frac{d^{2} y}{d x^{2}} = - 25 y$

Limits, Continuity and Differentiability

80308 If $y = 2^{a x}$ and ${(\frac{d y}{d x})}_{x = 1} = \log 256$ , then $a =$

1 3

2 2

3 4

4 8

Explanation:

(B) : Given, $y = 2^{ax}$
On taking $\log$ both sides.
$\log y = \log 2^{a x} \Rightarrow \log y = a x \log 2$
Differentiating on both sides w.r.t.x
$\frac{1}{y} \frac{dy}{dx} = a \times 1 \times \log 2 \Rightarrow \frac{dy}{dx} = ay \log 2$
On putting the value $y$ in above equation,
$\frac{d y}{d x} = 2^{a x} a \log 2 \Rightarrow {(\frac{d y}{d x})}_{x = 1} = a 2^{a \times 1} \log 2$
${(\frac{d y}{d x})}_{x = 1} = a 2^{a} \log 2$
Given by,
${(\frac{d y}{d x})}_{x = 1} = \log 256 \Rightarrow {(\frac{d y}{d x})}_{x = 1} = \log 2^{8}$
${(\frac{d y}{d x})}_{x = 1} = 8 \log 2$
On comparing equation (i) and equation (ii)
$a 2^{a} \log 2 = 8 \log 2$
$a \cdot 2^{a} = 2 \times 2^{2}$
On comparing both sides, $a = 2$

MHT CET-2020

Limits, Continuity and Differentiability

80309 If $x^{p} y^{q} = (x + y)^{p + q}$ , then $\frac{d y}{d x}$ is equal to

1

\frac{y}{x}

2

\frac{p y}{q x}

3

\frac{x}{y}

4

\frac{qy}{px}

Explanation:

(A) : Given, $x^{p} y^{q} = (x + y)^{p + q}$
On taking $\log$ both sides.
$\log (x^{p} y^{q}) = \log (x + y)^{p + q}$
$p \log x + q \log y = (p + q) \log (x + y)$
Differentiating on both sides w.r.t.x
$\frac{p}{x} + \frac{q}{y} (\frac{d y}{d x}) = (\frac{p + q}{x + y}) \times (1 + \frac{d y}{d x})$
$\frac{p}{x} + \frac{q}{y} (\frac{d y}{d x}) = \frac{p + q}{x + y} + \frac{p + q}{x + y} (\frac{d y}{d x})$
$\frac{d y}{d x} [\frac{q}{y} - \frac{p + q}{x + y}] = \frac{p + q}{x + y} - \frac{p}{x}$
$\frac{d y}{d x} [\frac{q (x + y) - y (p + q)}{y (x + y)}] = \frac{x (p + q) - p (x + y)}{x (x + y)}$
$\frac{d y}{d x} [\frac{q x + q y - p y - q y}{y (x + y)}] = \frac{p x + q x - p x - p y}{x (x + y)}$
$\frac{d y}{d x} [\frac{q x - p y}{y (x + y)}] = \frac{q x - p y}{x (x + y)} \Rightarrow \frac{d y}{d x} = \frac{y (x + y)}{x (x + y)} \Rightarrow \frac{d y}{d x} = \frac{y}{x}$

MHT CET-2011

Limits, Continuity and Differentiability

80306 If $x = e^{y + e^{y + e^{y + e}}}$ then $\frac{d y}{d x} =$

1

\frac{1}{x}

2

\frac{x}{1 + x}

3

\frac{1 - x}{x}

4

\frac{1 + x}{x}

Explanation:

(C) : It is given that,
Here,
$x = e^{y + ey + ey + ey + \dots . . . \infty}$
$x = e^{y + x}$
Differentiating w.r.to $x$ .
$1 = e^{y + x} (\frac{d y}{d x} + 1) \Rightarrow 1 = e^{y + x} \frac{d y}{d x} + e^{y + x}$
$1 = x \frac{d y}{d x} + x, 1 - x = x \frac{d y}{d x} \Rightarrow \frac{d y}{d x} = \frac{1 - x}{x}$

MHT CET-2020

Limits, Continuity and Differentiability

80307 If $y = \cos^{2} (\frac{5 x}{2}) - \sin^{2} (\frac{5 x}{2})$ , then $(\frac{d^{2} y}{d x^{2}}) =$

1

5 \sqrt{1 - y^{2}}

2

25 y

3

- 25 y

4

- 5 \sqrt{1 - y^{2}}

[MНT CET-2020]

Explanation:

(C) : Given,
$y = \cos^{2} (\frac{5 x}{2}) - \sin^{2} (\frac{5 x}{2})$
We know that,
So,
$\cos^{2} x - \sin^{2} x = \cos 2 x$
$y = \cos 2 (\frac{5 x}{2})$
$y = \cos 5 x$
Now, differentiate with respect to $x$
$\frac{d y}{d x} = - 5 \sin 5 x$
$∴ \frac{d^{2} y}{{dx}^{2}} = - 25 \cos 5 x$
On putting the value $(\cos 5 x = y)$ in above equation. Then,
$\frac{d^{2} y}{d x^{2}} = - 25 y$

Limits, Continuity and Differentiability

80308 If $y = 2^{a x}$ and ${(\frac{d y}{d x})}_{x = 1} = \log 256$ , then $a =$

1 3

2 2

3 4

4 8

Explanation:

(B) : Given, $y = 2^{ax}$
On taking $\log$ both sides.
$\log y = \log 2^{a x} \Rightarrow \log y = a x \log 2$
Differentiating on both sides w.r.t.x
$\frac{1}{y} \frac{dy}{dx} = a \times 1 \times \log 2 \Rightarrow \frac{dy}{dx} = ay \log 2$
On putting the value $y$ in above equation,
$\frac{d y}{d x} = 2^{a x} a \log 2 \Rightarrow {(\frac{d y}{d x})}_{x = 1} = a 2^{a \times 1} \log 2$
${(\frac{d y}{d x})}_{x = 1} = a 2^{a} \log 2$
Given by,
${(\frac{d y}{d x})}_{x = 1} = \log 256 \Rightarrow {(\frac{d y}{d x})}_{x = 1} = \log 2^{8}$
${(\frac{d y}{d x})}_{x = 1} = 8 \log 2$
On comparing equation (i) and equation (ii)
$a 2^{a} \log 2 = 8 \log 2$
$a \cdot 2^{a} = 2 \times 2^{2}$
On comparing both sides, $a = 2$

MHT CET-2020

Limits, Continuity and Differentiability

80309 If $x^{p} y^{q} = (x + y)^{p + q}$ , then $\frac{d y}{d x}$ is equal to

1

\frac{y}{x}

2

\frac{p y}{q x}

3

\frac{x}{y}

4

\frac{qy}{px}

Explanation:

(A) : Given, $x^{p} y^{q} = (x + y)^{p + q}$
On taking $\log$ both sides.
$\log (x^{p} y^{q}) = \log (x + y)^{p + q}$
$p \log x + q \log y = (p + q) \log (x + y)$
Differentiating on both sides w.r.t.x
$\frac{p}{x} + \frac{q}{y} (\frac{d y}{d x}) = (\frac{p + q}{x + y}) \times (1 + \frac{d y}{d x})$
$\frac{p}{x} + \frac{q}{y} (\frac{d y}{d x}) = \frac{p + q}{x + y} + \frac{p + q}{x + y} (\frac{d y}{d x})$
$\frac{d y}{d x} [\frac{q}{y} - \frac{p + q}{x + y}] = \frac{p + q}{x + y} - \frac{p}{x}$
$\frac{d y}{d x} [\frac{q (x + y) - y (p + q)}{y (x + y)}] = \frac{x (p + q) - p (x + y)}{x (x + y)}$
$\frac{d y}{d x} [\frac{q x + q y - p y - q y}{y (x + y)}] = \frac{p x + q x - p x - p y}{x (x + y)}$
$\frac{d y}{d x} [\frac{q x - p y}{y (x + y)}] = \frac{q x - p y}{x (x + y)} \Rightarrow \frac{d y}{d x} = \frac{y (x + y)}{x (x + y)} \Rightarrow \frac{d y}{d x} = \frac{y}{x}$

MHT CET-2011

Limits, Continuity and Differentiability

80306 If $x = e^{y + e^{y + e^{y + e}}}$ then $\frac{d y}{d x} =$

1

\frac{1}{x}

2

\frac{x}{1 + x}

3

\frac{1 - x}{x}

4

\frac{1 + x}{x}

Explanation:

(C) : It is given that,
Here,
$x = e^{y + ey + ey + ey + \dots . . . \infty}$
$x = e^{y + x}$
Differentiating w.r.to $x$ .
$1 = e^{y + x} (\frac{d y}{d x} + 1) \Rightarrow 1 = e^{y + x} \frac{d y}{d x} + e^{y + x}$
$1 = x \frac{d y}{d x} + x, 1 - x = x \frac{d y}{d x} \Rightarrow \frac{d y}{d x} = \frac{1 - x}{x}$

MHT CET-2020

Limits, Continuity and Differentiability

80307 If $y = \cos^{2} (\frac{5 x}{2}) - \sin^{2} (\frac{5 x}{2})$ , then $(\frac{d^{2} y}{d x^{2}}) =$

1

5 \sqrt{1 - y^{2}}

2

25 y

3

- 25 y

4

- 5 \sqrt{1 - y^{2}}

[MНT CET-2020]

Explanation:

(C) : Given,
$y = \cos^{2} (\frac{5 x}{2}) - \sin^{2} (\frac{5 x}{2})$
We know that,
So,
$\cos^{2} x - \sin^{2} x = \cos 2 x$
$y = \cos 2 (\frac{5 x}{2})$
$y = \cos 5 x$
Now, differentiate with respect to $x$
$\frac{d y}{d x} = - 5 \sin 5 x$
$∴ \frac{d^{2} y}{{dx}^{2}} = - 25 \cos 5 x$
On putting the value $(\cos 5 x = y)$ in above equation. Then,
$\frac{d^{2} y}{d x^{2}} = - 25 y$

Limits, Continuity and Differentiability

80308 If $y = 2^{a x}$ and ${(\frac{d y}{d x})}_{x = 1} = \log 256$ , then $a =$

1 3

2 2

3 4

4 8

Explanation:

(B) : Given, $y = 2^{ax}$
On taking $\log$ both sides.
$\log y = \log 2^{a x} \Rightarrow \log y = a x \log 2$
Differentiating on both sides w.r.t.x
$\frac{1}{y} \frac{dy}{dx} = a \times 1 \times \log 2 \Rightarrow \frac{dy}{dx} = ay \log 2$
On putting the value $y$ in above equation,
$\frac{d y}{d x} = 2^{a x} a \log 2 \Rightarrow {(\frac{d y}{d x})}_{x = 1} = a 2^{a \times 1} \log 2$
${(\frac{d y}{d x})}_{x = 1} = a 2^{a} \log 2$
Given by,
${(\frac{d y}{d x})}_{x = 1} = \log 256 \Rightarrow {(\frac{d y}{d x})}_{x = 1} = \log 2^{8}$
${(\frac{d y}{d x})}_{x = 1} = 8 \log 2$
On comparing equation (i) and equation (ii)
$a 2^{a} \log 2 = 8 \log 2$
$a \cdot 2^{a} = 2 \times 2^{2}$
On comparing both sides, $a = 2$

MHT CET-2020

Limits, Continuity and Differentiability

80309 If $x^{p} y^{q} = (x + y)^{p + q}$ , then $\frac{d y}{d x}$ is equal to

1

\frac{y}{x}

2

\frac{p y}{q x}

3

\frac{x}{y}

4

\frac{qy}{px}

Explanation:

(A) : Given, $x^{p} y^{q} = (x + y)^{p + q}$
On taking $\log$ both sides.
$\log (x^{p} y^{q}) = \log (x + y)^{p + q}$
$p \log x + q \log y = (p + q) \log (x + y)$
Differentiating on both sides w.r.t.x
$\frac{p}{x} + \frac{q}{y} (\frac{d y}{d x}) = (\frac{p + q}{x + y}) \times (1 + \frac{d y}{d x})$
$\frac{p}{x} + \frac{q}{y} (\frac{d y}{d x}) = \frac{p + q}{x + y} + \frac{p + q}{x + y} (\frac{d y}{d x})$
$\frac{d y}{d x} [\frac{q}{y} - \frac{p + q}{x + y}] = \frac{p + q}{x + y} - \frac{p}{x}$
$\frac{d y}{d x} [\frac{q (x + y) - y (p + q)}{y (x + y)}] = \frac{x (p + q) - p (x + y)}{x (x + y)}$
$\frac{d y}{d x} [\frac{q x + q y - p y - q y}{y (x + y)}] = \frac{p x + q x - p x - p y}{x (x + y)}$
$\frac{d y}{d x} [\frac{q x - p y}{y (x + y)}] = \frac{q x - p y}{x (x + y)} \Rightarrow \frac{d y}{d x} = \frac{y (x + y)}{x (x + y)} \Rightarrow \frac{d y}{d x} = \frac{y}{x}$

MHT CET-2011