QBManager

Limits, Continuity and Differentiability

80254 For constant $a, \frac{d}{d x} (x^{x} + x^{a} + a^{x} + a^{a})$ is

1

x^{x} (1 + \log x) + a x^{a - 1}

2

x^{x} (1 + \log x) + a x^{a - 1} + a^{x} \log a

3

x^{x} (1 + \log x) + a^{a} (1 + \log x)

4

x^{x} (1 + \log x) + a^{a} (1 + \log a) + a^{a - 1}

Explanation:

(B) : Given,
$\frac{d}{d x} (x^{x} + x^{a} + a^{x} + a^{a})$
Let, $y = x^{x}$
Taking $\log$ both sides.
$\log y = x \log x$
Differentiating both sides w.r.t. $x$
$\frac{1}{y} \frac{dy}{dx} = x \frac{1}{x} + \log x .1$
$\frac{dy}{dx} = y [1 + \log x]$
$\frac{d y}{d x} = x^{x} [1 + \log x]$
Then,
$\frac{d}{d x} (x^{x} + x^{a} + a^{x} + a^{a}) = x^{x} [1 + \log x] + a x^{a - 1} + a^{x} \log a + 0$
$= x^{x} [1 + \log x] + a x^{a - 1} + a^{x} \log a$

Karnataka CET-2021

Limits, Continuity and Differentiability

80255 If $y = {(\cos x^{2})}^{2}$ , then $\frac{d y}{d x}$ is equal to

1

- 4 x \sin 2 x^{2}

2

- x \sin x^{2}

3

- 2 x \sin 2 x^{2}

4

- x \cos 2 x^{2}

Explanation:

(C) : Given,
$y = {(\cos x^{2})}^{2}$
Differentiating both sides w.r.t. $x$ , we get
$\frac{d y}{d x} = 2 \cos x^{2} (- \sin x^{2}) \cdot 2 x$
$\frac{d y}{d x} = - 2 x (2 \sin x^{2} \cdot \cos x^{2}) \Rightarrow \frac{d y}{d x} = - 2 x \sin 2 x^{2}$

Karnataka CET-2021

Limits, Continuity and Differentiability

80256 The value of $C$ in mean value theorem for the function $f (x) = x^{2}$ in $[2, 4]$ is

1 2

2 4

3

\frac{7}{2}

4 3

Explanation:

(D) : Given,
$f (x) = x^{2}$
$f^{'} (x) = \frac{d}{d x} x^{2}$
$f^{'} (x) = 2 x$
From mean value therefore,
$f^{'} (C) = \frac{f (b) - f (a)}{b - a}$
Where, $a = 2$ and $b = 4$
Now, $f (2) = 2^{2} = 4$
$f (4) = 4^{2} = 16$
Then,
$2 C = \frac{f (4) - f (2)}{4 - 2} \Rightarrow 2 C = \frac{16 - 4}{2} \Rightarrow C = 3$

Karnataka CET-2017

Limits, Continuity and Differentiability

80257 If $\log_{10} (\frac{x^{3} - y^{3}}{x^{3} + y^{3}}) = 2$ , then $\frac{d y}{d x} =$

1

\frac{x}{y}

2

- \frac{y}{x}

3

- \frac{x}{y}

4

\frac{y}{x}

Explanation:

(D) : Given,
$\log_{10} (\frac{x^{3} - y^{3}}{x^{3} + y^{3}}) = 2$
We know that,
$\log_{b} a = c$
$a = b^{c}$
Then,
$\frac{x^{3} - y^{3}}{x^{3} + y^{3}} = 10^{2}$
$x^{3} - y^{3} = 100 (x^{3} + y^{3})$
$x^{3} - y^{3} = 100 x^{3} + 100 y^{3}$
$- 99 x^{3} = 101 y^{3}$
Differentiating both sides w.r.t $x$ ,
$- 99 \times 3 x^{2} = 101 \times 3 y^{2} \frac{d y}{d x}$
$101 y^{2} \frac{d y}{d x} = - 99 x^{2}$
Multiplying by $x$ both sides,
$101 x y^{2} \frac{d y}{d x} = - 99 x^{3}$
$\frac{d y}{d x} = - \frac{99 x^{3}}{101 x y^{2}}$
On putting the value $(- 99 x^{3} = 101 y^{3})$ in above equation.
$\frac{d y}{d x} = \frac{101 y^{3}}{101 x y^{2}} \Rightarrow \frac{d y}{d x} = \frac{y}{x}$

MHT CET-2018

Limits, Continuity and Differentiability

80254 For constant $a, \frac{d}{d x} (x^{x} + x^{a} + a^{x} + a^{a})$ is

1

x^{x} (1 + \log x) + a x^{a - 1}

2

x^{x} (1 + \log x) + a x^{a - 1} + a^{x} \log a

3

x^{x} (1 + \log x) + a^{a} (1 + \log x)

4

x^{x} (1 + \log x) + a^{a} (1 + \log a) + a^{a - 1}

Explanation:

(B) : Given,
$\frac{d}{d x} (x^{x} + x^{a} + a^{x} + a^{a})$
Let, $y = x^{x}$
Taking $\log$ both sides.
$\log y = x \log x$
Differentiating both sides w.r.t. $x$
$\frac{1}{y} \frac{dy}{dx} = x \frac{1}{x} + \log x .1$
$\frac{dy}{dx} = y [1 + \log x]$
$\frac{d y}{d x} = x^{x} [1 + \log x]$
Then,
$\frac{d}{d x} (x^{x} + x^{a} + a^{x} + a^{a}) = x^{x} [1 + \log x] + a x^{a - 1} + a^{x} \log a + 0$
$= x^{x} [1 + \log x] + a x^{a - 1} + a^{x} \log a$

Karnataka CET-2021

Limits, Continuity and Differentiability

80255 If $y = {(\cos x^{2})}^{2}$ , then $\frac{d y}{d x}$ is equal to

1

- 4 x \sin 2 x^{2}

2

- x \sin x^{2}

3

- 2 x \sin 2 x^{2}

4

- x \cos 2 x^{2}

Explanation:

(C) : Given,
$y = {(\cos x^{2})}^{2}$
Differentiating both sides w.r.t. $x$ , we get
$\frac{d y}{d x} = 2 \cos x^{2} (- \sin x^{2}) \cdot 2 x$
$\frac{d y}{d x} = - 2 x (2 \sin x^{2} \cdot \cos x^{2}) \Rightarrow \frac{d y}{d x} = - 2 x \sin 2 x^{2}$

Karnataka CET-2021

Limits, Continuity and Differentiability

80256 The value of $C$ in mean value theorem for the function $f (x) = x^{2}$ in $[2, 4]$ is

1 2

2 4

3

\frac{7}{2}

4 3

Explanation:

(D) : Given,
$f (x) = x^{2}$
$f^{'} (x) = \frac{d}{d x} x^{2}$
$f^{'} (x) = 2 x$
From mean value therefore,
$f^{'} (C) = \frac{f (b) - f (a)}{b - a}$
Where, $a = 2$ and $b = 4$
Now, $f (2) = 2^{2} = 4$
$f (4) = 4^{2} = 16$
Then,
$2 C = \frac{f (4) - f (2)}{4 - 2} \Rightarrow 2 C = \frac{16 - 4}{2} \Rightarrow C = 3$

Karnataka CET-2017

Limits, Continuity and Differentiability

80257 If $\log_{10} (\frac{x^{3} - y^{3}}{x^{3} + y^{3}}) = 2$ , then $\frac{d y}{d x} =$

1

\frac{x}{y}

2

- \frac{y}{x}

3

- \frac{x}{y}

4

\frac{y}{x}

Explanation:

(D) : Given,
$\log_{10} (\frac{x^{3} - y^{3}}{x^{3} + y^{3}}) = 2$
We know that,
$\log_{b} a = c$
$a = b^{c}$
Then,
$\frac{x^{3} - y^{3}}{x^{3} + y^{3}} = 10^{2}$
$x^{3} - y^{3} = 100 (x^{3} + y^{3})$
$x^{3} - y^{3} = 100 x^{3} + 100 y^{3}$
$- 99 x^{3} = 101 y^{3}$
Differentiating both sides w.r.t $x$ ,
$- 99 \times 3 x^{2} = 101 \times 3 y^{2} \frac{d y}{d x}$
$101 y^{2} \frac{d y}{d x} = - 99 x^{2}$
Multiplying by $x$ both sides,
$101 x y^{2} \frac{d y}{d x} = - 99 x^{3}$
$\frac{d y}{d x} = - \frac{99 x^{3}}{101 x y^{2}}$
On putting the value $(- 99 x^{3} = 101 y^{3})$ in above equation.
$\frac{d y}{d x} = \frac{101 y^{3}}{101 x y^{2}} \Rightarrow \frac{d y}{d x} = \frac{y}{x}$

MHT CET-2018

Limits, Continuity and Differentiability

80254 For constant $a, \frac{d}{d x} (x^{x} + x^{a} + a^{x} + a^{a})$ is

1

x^{x} (1 + \log x) + a x^{a - 1}

2

x^{x} (1 + \log x) + a x^{a - 1} + a^{x} \log a

3

x^{x} (1 + \log x) + a^{a} (1 + \log x)

4

x^{x} (1 + \log x) + a^{a} (1 + \log a) + a^{a - 1}

Explanation:

(B) : Given,
$\frac{d}{d x} (x^{x} + x^{a} + a^{x} + a^{a})$
Let, $y = x^{x}$
Taking $\log$ both sides.
$\log y = x \log x$
Differentiating both sides w.r.t. $x$
$\frac{1}{y} \frac{dy}{dx} = x \frac{1}{x} + \log x .1$
$\frac{dy}{dx} = y [1 + \log x]$
$\frac{d y}{d x} = x^{x} [1 + \log x]$
Then,
$\frac{d}{d x} (x^{x} + x^{a} + a^{x} + a^{a}) = x^{x} [1 + \log x] + a x^{a - 1} + a^{x} \log a + 0$
$= x^{x} [1 + \log x] + a x^{a - 1} + a^{x} \log a$

Karnataka CET-2021

Limits, Continuity and Differentiability

80255 If $y = {(\cos x^{2})}^{2}$ , then $\frac{d y}{d x}$ is equal to

1

- 4 x \sin 2 x^{2}

2

- x \sin x^{2}

3

- 2 x \sin 2 x^{2}

4

- x \cos 2 x^{2}

Explanation:

(C) : Given,
$y = {(\cos x^{2})}^{2}$
Differentiating both sides w.r.t. $x$ , we get
$\frac{d y}{d x} = 2 \cos x^{2} (- \sin x^{2}) \cdot 2 x$
$\frac{d y}{d x} = - 2 x (2 \sin x^{2} \cdot \cos x^{2}) \Rightarrow \frac{d y}{d x} = - 2 x \sin 2 x^{2}$

Karnataka CET-2021

Limits, Continuity and Differentiability

80256 The value of $C$ in mean value theorem for the function $f (x) = x^{2}$ in $[2, 4]$ is

1 2

2 4

3

\frac{7}{2}

4 3

Explanation:

(D) : Given,
$f (x) = x^{2}$
$f^{'} (x) = \frac{d}{d x} x^{2}$
$f^{'} (x) = 2 x$
From mean value therefore,
$f^{'} (C) = \frac{f (b) - f (a)}{b - a}$
Where, $a = 2$ and $b = 4$
Now, $f (2) = 2^{2} = 4$
$f (4) = 4^{2} = 16$
Then,
$2 C = \frac{f (4) - f (2)}{4 - 2} \Rightarrow 2 C = \frac{16 - 4}{2} \Rightarrow C = 3$

Karnataka CET-2017

Limits, Continuity and Differentiability

80257 If $\log_{10} (\frac{x^{3} - y^{3}}{x^{3} + y^{3}}) = 2$ , then $\frac{d y}{d x} =$

1

\frac{x}{y}

2

- \frac{y}{x}

3

- \frac{x}{y}

4

\frac{y}{x}

Explanation:

(D) : Given,
$\log_{10} (\frac{x^{3} - y^{3}}{x^{3} + y^{3}}) = 2$
We know that,
$\log_{b} a = c$
$a = b^{c}$
Then,
$\frac{x^{3} - y^{3}}{x^{3} + y^{3}} = 10^{2}$
$x^{3} - y^{3} = 100 (x^{3} + y^{3})$
$x^{3} - y^{3} = 100 x^{3} + 100 y^{3}$
$- 99 x^{3} = 101 y^{3}$
Differentiating both sides w.r.t $x$ ,
$- 99 \times 3 x^{2} = 101 \times 3 y^{2} \frac{d y}{d x}$
$101 y^{2} \frac{d y}{d x} = - 99 x^{2}$
Multiplying by $x$ both sides,
$101 x y^{2} \frac{d y}{d x} = - 99 x^{3}$
$\frac{d y}{d x} = - \frac{99 x^{3}}{101 x y^{2}}$
On putting the value $(- 99 x^{3} = 101 y^{3})$ in above equation.
$\frac{d y}{d x} = \frac{101 y^{3}}{101 x y^{2}} \Rightarrow \frac{d y}{d x} = \frac{y}{x}$

MHT CET-2018

Limits, Continuity and Differentiability

80254 For constant $a, \frac{d}{d x} (x^{x} + x^{a} + a^{x} + a^{a})$ is

1

x^{x} (1 + \log x) + a x^{a - 1}

2

x^{x} (1 + \log x) + a x^{a - 1} + a^{x} \log a

3

x^{x} (1 + \log x) + a^{a} (1 + \log x)

4

x^{x} (1 + \log x) + a^{a} (1 + \log a) + a^{a - 1}

Explanation:

(B) : Given,
$\frac{d}{d x} (x^{x} + x^{a} + a^{x} + a^{a})$
Let, $y = x^{x}$
Taking $\log$ both sides.
$\log y = x \log x$
Differentiating both sides w.r.t. $x$
$\frac{1}{y} \frac{dy}{dx} = x \frac{1}{x} + \log x .1$
$\frac{dy}{dx} = y [1 + \log x]$
$\frac{d y}{d x} = x^{x} [1 + \log x]$
Then,
$\frac{d}{d x} (x^{x} + x^{a} + a^{x} + a^{a}) = x^{x} [1 + \log x] + a x^{a - 1} + a^{x} \log a + 0$
$= x^{x} [1 + \log x] + a x^{a - 1} + a^{x} \log a$

Karnataka CET-2021

Limits, Continuity and Differentiability

80255 If $y = {(\cos x^{2})}^{2}$ , then $\frac{d y}{d x}$ is equal to

1

- 4 x \sin 2 x^{2}

2

- x \sin x^{2}

3

- 2 x \sin 2 x^{2}

4

- x \cos 2 x^{2}

Explanation:

(C) : Given,
$y = {(\cos x^{2})}^{2}$
Differentiating both sides w.r.t. $x$ , we get
$\frac{d y}{d x} = 2 \cos x^{2} (- \sin x^{2}) \cdot 2 x$
$\frac{d y}{d x} = - 2 x (2 \sin x^{2} \cdot \cos x^{2}) \Rightarrow \frac{d y}{d x} = - 2 x \sin 2 x^{2}$

Karnataka CET-2021

Limits, Continuity and Differentiability

80256 The value of $C$ in mean value theorem for the function $f (x) = x^{2}$ in $[2, 4]$ is

1 2

2 4

3

\frac{7}{2}

4 3

Explanation:

(D) : Given,
$f (x) = x^{2}$
$f^{'} (x) = \frac{d}{d x} x^{2}$
$f^{'} (x) = 2 x$
From mean value therefore,
$f^{'} (C) = \frac{f (b) - f (a)}{b - a}$
Where, $a = 2$ and $b = 4$
Now, $f (2) = 2^{2} = 4$
$f (4) = 4^{2} = 16$
Then,
$2 C = \frac{f (4) - f (2)}{4 - 2} \Rightarrow 2 C = \frac{16 - 4}{2} \Rightarrow C = 3$

Karnataka CET-2017

Limits, Continuity and Differentiability

80257 If $\log_{10} (\frac{x^{3} - y^{3}}{x^{3} + y^{3}}) = 2$ , then $\frac{d y}{d x} =$

1

\frac{x}{y}

2

- \frac{y}{x}

3

- \frac{x}{y}

4

\frac{y}{x}

Explanation:

(D) : Given,
$\log_{10} (\frac{x^{3} - y^{3}}{x^{3} + y^{3}}) = 2$
We know that,
$\log_{b} a = c$
$a = b^{c}$
Then,
$\frac{x^{3} - y^{3}}{x^{3} + y^{3}} = 10^{2}$
$x^{3} - y^{3} = 100 (x^{3} + y^{3})$
$x^{3} - y^{3} = 100 x^{3} + 100 y^{3}$
$- 99 x^{3} = 101 y^{3}$
Differentiating both sides w.r.t $x$ ,
$- 99 \times 3 x^{2} = 101 \times 3 y^{2} \frac{d y}{d x}$
$101 y^{2} \frac{d y}{d x} = - 99 x^{2}$
Multiplying by $x$ both sides,
$101 x y^{2} \frac{d y}{d x} = - 99 x^{3}$
$\frac{d y}{d x} = - \frac{99 x^{3}}{101 x y^{2}}$
On putting the value $(- 99 x^{3} = 101 y^{3})$ in above equation.
$\frac{d y}{d x} = \frac{101 y^{3}}{101 x y^{2}} \Rightarrow \frac{d y}{d x} = \frac{y}{x}$

MHT CET-2018