If $${ y }^{ 2 } + { b }^{ 2 } = 2xy$$, then $$\displaystyle\frac { dy }{ dx } $$ equals

$$\displaystyle\frac { xy-{ b }^{ 2 } }{ { \left( y-x \right) }^{ 2 } } $$

$$\displaystyle\frac { 1 }{ xy-{ b }^{ 2 } } $$

$$\displaystyle\frac { y }{ y-x } $$

$$\displaystyle\frac { xy-{ b }^{ 2 } }{ y } $$

MCQ Questions for Class 11 Commerce Applied Mathematics Differentiation Quiz 6

If for a non-zero

$x,$ the function

$f(x)$ satisfies the equation

$\displaystyle af\left( x \right)+bf\left( \frac { 1 }{ x } \right) =\frac { 1 }{ x } -5\left( a\neq b \right)$ then

$f'(x)$ is equal to

0%
$\displaystyle \frac { 1 }{ { b }^{ 2 }-{ a }^{ 2 } } \left( \frac { a }{ { x }^{ 2 } } +b \right)$
0%
$\displaystyle \frac { 1 }{ { a }^{ 2 }-{ b }^{ 2 } } \left( \frac { a }{ { x }^{ 2 } } +b \right)$
0%
$\displaystyle \frac { 1 }{ { a }^{ 2 }-{ b }^{ 2 } } \left( \frac { a }{ { x }^{ 2 } } -b \right)$
0%
none of these

Explanation

We have,

$\displaystyle af\left( x \right)+bf\left( \frac { 1 }{ x } \right) =\frac { 1 }{ x } -5$

Substituting

$\displaystyle\frac{1}{x}$ in place of

$x,$ we have

$\displaystyle af\left( \frac { 1 }{ x } \right) +bf\left( x \right)=x-5$ ...(2)

Eliminating

$\displaystyle f\left( \frac { 1 }{ x } \right)$ from equation (1) and (2), we have

$\displaystyle \left( { a }^{ 2 }-{ b }^{ 2 } \right) f\left( x \right) =a\left( \frac { 1 }{ x } -5 \right) -b\left( x-5 \right)$

$\displaystyle \Rightarrow f\left( x \right) =\frac { 1 }{ { a }^{ 2 }-{ b }^{ 2 } } \left[ \frac { a }{ x } -bx+5\left( b-a \right) \right]$

$\displaystyle \therefore f'\left( x \right) =\frac { 1 }{ { b }^{ 2 }-{ a }^{ 2 } } \left[ \frac { a }{ { x }^{ 2 } } +b \right]$

If for all

$x,y$ the function

$f$ is defined by

$f\left( x \right)+f\left( y \right)+f\left( x \right).f\left( y \right)=1$ and

$f\left( x \right)>0$ , then

0%
$f'\left( x \right)$ does not exist
0%
$f'\left( x \right) =0$ for all $x$
0%
$f'\left( 0 \right) <f'\left( 1 \right)$
0%
None of these

Explanation

Putting

$x=0,y=0,$ we get

$2f(0)+{ \left\{ f\left( 0 \right) \right\} }^{ 2 }=1\Rightarrow f\left( 0 \right) =\sqrt { 2 } -1\quad \left( \because f\left( x \right) >0 \right)$

Putting

$y=x,2f(x)+{ \left\{ f\left( x \right) \right\} }^{ 2 }=1$

Differentiating w.r.t.

$x,$ we get

$2f'\left( x \right) +2f\left( x \right) .f'\left( x \right) =0\quad f'\left( x \right) \left\{ 1+f\left( x \right) \right\} =0$

$\Rightarrow f'\left( x \right) =0,$ because

$f\left( x \right) >0$

$\displaystyle {f}'\left ( x \right )=f\left ( x \right )$ for all x and

$\displaystyle {f}'\left ( 0\right )=4$ then

$\displaystyle f\left ( x \right )$ is equal to

0%
$\displaystyle 2e^{2x}$
0%
$\displaystyle e^{4x}$
0%
$\displaystyle x^{4}+4x^{2}+4x$
0%
$\displaystyle 4e^{x}$

Explanation

given

$f^{ 1 }\left( x \right) =f\left( x \right)$

$\frac { df }{ dx } =f$ which is equal to

$\frac { df }{ f } =dx$

now integrate on both sides , we get

$\ln { f(x) } =x$

which implies

$f(x)=k{ e }^{ x }$ where

$k$ is constant.

given

$f^{ 1 }\left( 0 \right) =4=k$

so the equation is

$f(x)=4{ e }^{ x }$

If the functions

$\displaystyle f\left ( x \right )=\sin \left ( x+a \right )$ and

$\displaystyle g\left ( x \right )=b\sin x+c\cos x$ satisfy

$\displaystyle f\left ( 0 \right )=g\left ( 0 \right )$ and

$\displaystyle {f}'\left ( 0 \right )={g}'\left ( 0 \right )$ then

0%
$\displaystyle b=\dfrac{\pi}2$
0%
$\displaystyle b=\cos a$
0%
$\displaystyle c=\sin a$
0%
$\displaystyle c=\cos a$

Explanation

Given,

$f(x)=\sin(x+a)\quad$

and

$g(x)=b\sin x+c\cos x\\ f(0)=g(0)\Rightarrow \sin a=c\\ f'(x)=\cos(x+a)\\ g'(x)=b\cos x-c\sin x\\ f'(0)=g'(0)\Rightarrow \cos a=b$

$\displaystyle y=\sec ^{ -1 }{ \left( \frac { x+1 }{ x-1 } \right) } +\sin ^{ -1 }{ \left( \frac { x-1 }{ x+1 } \right) }$ then

$\displaystyle \frac{dy}{dx}$ is equal to

0%
$0$
0%
$\displaystyle x+1$
0%
$1$
0%
$-1$

Explanation

$\displaystyle y=\sec ^{ -1 }{ \left( \frac { x+1 }{ x-1 } \right) } +\sin ^{ -1 }{ \left( \frac { x-1 }{ x+1 } \right) }$

$\displaystyle =\cos ^{ -1 }{ \left( \frac { x-1 }{ x+1 } \right) } +\sin ^{ -1 }{ \left( \frac { x-1 }{ x+1 } \right) } =\frac { \pi }{ 2 }$
However the above function is defined only for values of x, given by

$\displaystyle -1\le \frac { x-1 }{ x+1 } \le 1$
i.e.

$\displaystyle \frac { x-1 }{ x+1 } +1\ge 0$ and

$\displaystyle \frac { x-1 }{ x+1 } -1\le 0$
i.e.

$\displaystyle \frac { 2x }{ x+1 } \ge 0$ and

$\displaystyle \frac { 2 }{ x+1 } \ge 0$
i.e

$x<-1$ or

$\ge 0$ and

$x>-1$
i.e

$x\ge 0$
Hence we have

$\displaystyle y=\frac { \pi }{ 2 } ,\quad x\ge 0\Rightarrow \frac { dy }{ dx } =0,\quad x>0$

$\displaystyle \sqrt{1-x^{6}} +\sqrt{1-y^{6}}=a\left ( x^{3}-y^{3} \right )$ and

$\displaystyle \frac{dy}{dx}=f\left ( x,y \right )\sqrt{\frac{1-y^{6}}{1-x^{6}}}$ then

0%
$\displaystyle f\left ( x,y \right )=\dfrac{y}{x}$
0%
$\displaystyle f\left ( x,y \right )=\dfrac{x^{2}}{y^{2}}$
0%
$\displaystyle f\left ( x,y \right )=2\dfrac{y^{2}}{x^{2}}$
0%
$\displaystyle f\left ( x,y \right )=\dfrac{y^{2}}{x^{2}}$

Explanation

$\sqrt { 1-{ x }^{ 6 } } +\sqrt { 1-{ y }^{ 6 } } =a\left( { x }^{ 3 }-{ y }^{ 3 } \right)$

$\displaystyle \Rightarrow -\frac { 6\times { x }^{ 5 } }{ 2\sqrt { { 1-x }^{ 6 } } } -\frac { 6\times { y }^{ 5 } }{ 2\sqrt { { 1- }y^{ 6 } } } \frac { dy }{ dx } =3a\left( { x }^{ 2 }-{ y }^{ 2 }\frac { dy }{ dx } \right)$

$\displaystyle \Rightarrow \left( -\dfrac { 3{ y }^{ 5 } }{ \sqrt { { 1- }y^{ 6 } } } +{ 3ay }^{ 2 } \right) \frac { dy }{ dx } =3ax^{ 2 }+\dfrac { { 3x }^{ 5 } }{ \sqrt { { 1-x }^{ 6 } } }$

$\displaystyle \Rightarrow \dfrac { dy }{ dx } =\dfrac { { a }x^{ 2 }+\dfrac { {x }^{ 5 } }{ \sqrt { { 1-x }^{ 6 } } } }{ { ay }^{ 2 }-\dfrac { { y }^{ 5 } }{ \sqrt { { 1-y }^{ 6 } } } }$

$\displaystyle \Rightarrow \dfrac { dy }{ dx } =\dfrac{x^2}{y^2}\left(\dfrac { { a }+\dfrac { {x }^{ 3 } }{ \sqrt { { 1-x }^{ 6 } } } }{ { a }-\dfrac { { y }^{ 3 } }{ \sqrt { { 1-y }^{ 6 } } } } \right)$

$\displaystyle \Rightarrow \dfrac { dy }{ dx } =\dfrac{x^2}{y^2}\left(\dfrac { \dfrac{\sqrt{1-x^6}+\sqrt{1-y^6}}{x^{3}-y^{3}}+\dfrac { {x }^{ 3 } }{ \sqrt { { 1-x }^{ 6 } } } }{ \dfrac{\sqrt{1-x^6}+\sqrt{1-y^6}}{x^3-y^3}-\dfrac { { y }^{ 3 } }{ \sqrt { { 1-y }^{ 6 } } } } \right)$

$\displaystyle \Rightarrow \dfrac { dy }{ dx } =\dfrac{x^2}{y^2}\left(\dfrac { \dfrac{{1-x^6}+\sqrt{(1-y^6)(1-x^6)}+x^{6}-x^{3}y^{3}}{ \sqrt { { 1-x }^{ 6 } } } }{ \dfrac{\sqrt{(1-x^6)(1-y^6)}+{1-y^6}-x^3 y^3+y^6}{ \sqrt { { 1-y }^{ 6 } } } } \right)$

$\Rightarrow \dfrac { dy }{ dx } =\dfrac { { x }^{ 2 } }{ { y }^{ 2 } } \sqrt { \dfrac { { 1-y }^{ 6 } }{ { 1-x }^{ 6 } } }$

Hence,

$f(x,y)=\dfrac{x^2}{y^2}$

$f(x+y)=f(x)f(y) \forall x,y$ and

$f(5)=2, f'(0)=3$ ; then

$f'(5)$ is equal to-

0%
$2$
0%
$4$
0%
$6$
0%
$8$

Explanation

We have

$f(x + y) = f(x)f(y)$

Also,

$f(5) = 2, f'(0) = 3$

$f(x + y) = f(x)f(y)$

Putting

$x,y=0,$ we get

$f(0) = 0$ or

$1.$

$f(0) = 1$ not possible.

So,

$f(0) = 0$

Now, differentiating both sides w.r.t. x, we have

$f'(x + y) = f'(x)f(y) + f(x)f'(y)$

Thus, substituting

$x = 0$ and

$y = 5, f'(5) = f'(0) f(5)+f(0)f'(5)$

$\Rightarrow f'(5) = 3 \times 2 = 6$

Consider the function:

$f(-\infty,\infty)\rightarrow (-\infty,\infty)$ defined by

$\displaystyle f(x)=\frac{x^{2}-ax+1}{x^{2}+ax+1},0<a<2$

Which of the following is true?

0%
$(2+a)^{2}f''(1)+(2-a)^{2}f'(-1)=0$
0%
$(2-a)^{2}f''(1)-(2+a)^{2}f'(-1)=0$
0%
$f(1)+f(-1)=(2-a)^{2}$
0%
$f(1)f(-1)=-(2-a)^{2}$

Explanation

$(x^{2}+ax+1)f(x)=x^{2}-ax+1$
Differentiating we obtain

$\displaystyle (2x+a)f(x)+(x^{2}+ax+1) f'(x)=2x-a............. (i)$
Putting

$x=1$ and noting that

$\displaystyle f(1)=\frac{2-a}{2+a},$ we obtain

$(2+a)f(1)+(2+a)f'(1)=2-a\Rightarrow f'(1)=0$
Differentiating (i) again (ii)

$2f(x)+(2x+a)f'(x)+(2x+a)f'(x) +(x^{2}+ax+1)f''(x)=2$
Putting

$x=1$ , we have

$\displaystyle 2\cdot \frac{2-a}{2+a}+(2+a) f''(1)=2$

$\displaystyle \Rightarrow f''(1)=\frac{4a}{(2+a)^{2}}$
Putting

$x=-1$ in (i) and noting that

$\displaystyle f(-1)=\frac{2+a}{2-a} (-2+a)f(-1)+(2-a) f'(-1)=-(2+a)\Rightarrow f'(-1)=0$
Putting

$x=-1$ in (ii), we obtain

$\displaystyle f''(-1)=\frac{-4a}{(2-a)^{2}}$ so

$(2+a)^{2} f''(1)+(2-a)^{2}f'(-1)=0$

Differentiation of

$x \sin x$ with respect to

$x$ is

0%
$\displaystyle x\cos x+\sin x$
0%
$\displaystyle x\sin x + \cos x$
0%
$\displaystyle x\cos x$
0%
$\displaystyle x\cos x - \sin x$

Explanation

Given,

$y= x \sin x$

$\therefore \displaystyle \frac{d}{dx}\left ( x\sin x \right )=x\frac{d}{dx}\left ( \sin x \right )+\sin x.\frac{d}{dx}\left ( x \right )$

$\displaystyle =x\cos x+\sin x (1)$

$\displaystyle =x\cos x+\sin x$

Let

$y = \sqrt { x + \sqrt { x + \sqrt { x + ......\infty } } }$ then

$\displaystyle\frac { dy }{ dx }$

0%
$\displaystyle\frac { 1 }{ 2y-1 }$
0%
$\displaystyle\frac { x }{ x-2y }$
0%
$\displaystyle\frac { 1 }{ \sqrt { 1+4x } }$
0%
$\displaystyle\frac { y }{ 2x+y }$

Explanation

$y=\sqrt { x+\sqrt { x+\sqrt { x+......\infty } } } \\ \therefore y=\sqrt { x+y } \\ Squaring\quad on\quad both\ sides,\\ y^{ 2 }=x+y\\ Differentiating\ on\ both\ sides,\\ 2y\dfrac { dy }{ dx }=1+\dfrac { dy }{ dx } \\ \therefore \dfrac { dy }{ dx }=\dfrac { 1 }{ 2y-1 } \\ \Rightarrow \dfrac { 1 }{ \sqrt { (2y-1)^{ 2 } } }=\dfrac { 1 }{ \sqrt { 4y^{ 2 }-4y+1 } }=\dfrac { 1 }{ \sqrt { 1+4x } } \\ \Rightarrow \dfrac { \sqrt { x+y } }{ \sqrt { (1 + 4x)(x+y) } } = \dfrac { \sqrt { y^{ 2 } } }{ \sqrt { 4x^{ 2 } + 4xy + x + y } } \\\Rightarrow \dfrac { \sqrt { y^{ 2 } } }{ \sqrt { 4x^{ 2 } + 4xy + y^{ 2 } } } = \dfrac { \sqrt { y^{ 2 } } }{ \sqrt { (2x+y)^{ 2 } } } = \dfrac { y }{ 2x + y }$

$x=\sqrt { { a }^{ \sin ^{ -1 }{ t } } }$ and

$y=\sqrt { { a }^{ \cos ^{ -1 }{ t } } }$ then

$\displaystyle\frac{dy}{dx}=$

0%
$\displaystyle\frac{y}{x}$
0%
$\displaystyle\frac{x}{y}$
0%
$\displaystyle-\frac{x}{y}$
0%
$\displaystyle-\frac{y}{x}$

Explanation

Let

$\displaystyle x=\sqrt { { a }^{ \sin ^{ -1 }{ t } } }$ ...(1)

and

$\displaystyle y=\sqrt { { a }^{ \cos ^{ -1 }{ t } } }$ ...(2)

Multiplying (1) and (2), we get

$\displaystyle xy=\sqrt { { a }^{ \sin ^{ -1 }{ t } } } \times \sqrt { { a }^{ \cos ^{ -1 }{ t } } } =\sqrt { { a }^{ \sin ^{ -1 }{ t } }\times { a }^{ \cos ^{ -1 }{ t } } }$

$\displaystyle =\sqrt { { a }^{ \sin ^{ -1 }{ t } +\cos ^{ -1 }{ t } } } =\sqrt { { a }^{ \pi /2 } } \quad \quad \quad \left[ \because \sin ^{ -1 }{ t } +\cos ^{ -1 }{ t } =\frac { \pi }{ 2 } \right]$

Differentiating w.r.t

$x$ , we get

$\displaystyle x\frac { dy }{ dx } +y=0\Rightarrow \frac { dy }{ dx } =-\frac { y }{ x }$

Let

$f(x+y)=f(x)f(y)$ for all

$x$ and

$y$ . If

$f(7)=2$ and

${f}'\left ( 0 \right )=3$ , then

${f}'\left ( 7 \right )$ is equal to

0%
$5$
0%
$6$
0%
$0$
0%
none of these

Explanation

Given,

$f(7)=2, f'(0)=3$ Also,

$f(x+y)=f(x)f(y)$

$\therefore \displaystyle {f}'\left ( 7 \right )=\lim_{h\rightarrow 0}\frac{f\left ( 7+h \right )-f\left (7 \right )}{h}$

$\displaystyle =\lim_{h\rightarrow 0}\frac{f\left ( 7 \right )f\left ( h \right )-f\left ( 7 \right )f\left ( 0 \right )}{h}$ , Since

$f(0)=1$

$\displaystyle =f\left ( 7 \right )\lim_{h\rightarrow 0}\frac{f\left ( h \right )-f\left ( 0 \right )}{h}=f\left ( 7 \right ){f}'\left ( 0 \right )=2\cdot3=6$

$\displaystyle f^{ ' }\left( x \right) =g\left( x \right)$ and $\displaystyle g^{ ' }\left( x \right) =-f\left( x \right)$ for all real x and $\displaystyle f\left( 5 \right) =2=f^{ ' }\left( 5 \right)$ then $\displaystyle f^{ 2 }\left( 10 \right) +g^{ 2 }\left( 10 \right)$ is -

0%
$2$
0%
$4$
0%
$8$
0%
None of these

Explanation

Given, $\displaystyle f^{ ' }\left( x \right) =g\left( x \right)$ and $g^{ ' }\left( x \right) =-f\left( x \right)$

Now $\displaystyle \frac { d }{ dx } \left[ f^{ 2 }\left( x \right) +g^{ 2 }\left( x \right) \right] =2f\left( x \right) f^{ ' }\left( x \right) +2g\left( x \right) g^{ ' }\left( x \right)$

$\displaystyle =2f\left( x \right) g\left( x \right) -2g\left( x \right) f\left( x \right) =0$

$\displaystyle \therefore \quad f^{ 2 }\left( x \right) +g^{ 2 }\left( x \right)$ =constant

$\displaystyle f^{ 2 }\left( 5 \right) +g^{ 2 }\left( 5 \right) = 4 + 4 = 8$

$\displaystyle \therefore \quad f^{ 2 }\left( 10 \right) +g^{ 2 }\left( 10 \right)$ = 8

$\sqrt { y + x } + \sqrt { y - x } = c$ , then

$\displaystyle\frac { dy }{ dx }$ is equal to

0%
$\displaystyle\frac { 2x }{ { c }^{ 2 } }$
0%
$\displaystyle\frac { x }{ y+\sqrt { { y }^{ 2 }-{ x }^{ 2 } } }$
0%
$\displaystyle\frac { y-\sqrt { { y }^{ 2 }-{ x }^{ 2 } } }{ x }$
0%
$\displaystyle\frac { { c }^{ 2 } }{ 2y }$

Explanation

Given,

$\sqrt { y + x } + \sqrt { y - x } = c$

$\displaystyle\frac { 1 }{ 2\sqrt { y + x } } \left( \displaystyle\frac { dy }{ dx } + 1 \right) + \displaystyle\frac { 1 }{ 2\sqrt { y - x } } \left( \displaystyle\frac { dy }{ dx } - 1 \right) = 0$

$\displaystyle\frac { dy }{ dx } \left( \displaystyle\frac { 1 }{ \sqrt { y + x } } + \displaystyle\frac { 1 }{ \sqrt { y - x } } \right) = \displaystyle\frac { 1 }{ \sqrt { y - x } } - \displaystyle\frac { 1 }{ \sqrt { y + x } }$

$\displaystyle\frac { dy }{ dx } = \displaystyle\frac { \sqrt { y + x } - \sqrt { y - x } }{ \sqrt { y + x } + \sqrt { y - x } }$

(By rationalizing Numerator or Denominator)

$\displaystyle\frac { dy }{ dx } = \displaystyle\frac { \sqrt { y + x } - \sqrt { y - x } }{ \sqrt { y + x } + \sqrt { y - x } } \times \displaystyle\frac { \sqrt { y + x } - \sqrt { y - x } }{ \sqrt { y + x } - \sqrt { y - x } } \,or\, \displaystyle\frac { \sqrt { y + x } - \sqrt { y - x } }{ \sqrt { y + x } + \sqrt { y - x } } \times \displaystyle\frac { \sqrt { y + x } + \sqrt { y - x } }{ \sqrt { y + x } + \sqrt { y - x } }$

$= \displaystyle\frac { y - \sqrt { { y }^{ 2 } - { x }^{ 2 } } }{ x } \, or \, \displaystyle\frac { x }{ y + \sqrt { { y }^{ 2 } - { x }^{ 2 } } }$

$g$ is the inverse of

$f$ and

$f'(x) = \displaystyle \frac{1}{1+x^{3}}$ then

$g'(x)$ is equal to-

0%
$\displaystyle 1+\left [ g\left ( x \right ) \right ]^{3}$
0%
$\displaystyle \frac{-1}{2\left ( 1+x^{2} \right )}$
0%
$\displaystyle \frac{1}{2\left ( 1+x^{2} \right )}$
0%
None of these

Explanation

Here

$g$ is inverse of

$f\left( x \right)$

$\Rightarrow fog\left( x \right) =x$
on differentiating w.r.t x, we get

$f'\left( g\left( x \right) \right) \times g'\left( x \right) =1$

$\displaystyle \Rightarrow g'\left( x \right) =\frac { 1 }{ f'\left( g\left( x \right) \right) } =\frac {1 }{ \dfrac{1}{1+g{ \left( x \right) }^{ 3 } }}$

$\displaystyle \left[ \because f'\left( x \right) =\frac { 1 }{ 1+{ x }^{ 3 } } \right]$

$\therefore g'\left( x \right) =1+g{ \left( x \right) }^{ 3 }$

If $\displaystyle x\sqrt { \left( 1+y \right) } +y\sqrt { \left( 1+x \right) } =0$ . then $\displaystyle \frac{dy}{dx}$ equals -

0%
$\displaystyle \frac { 1 }{ \left( 1+x \right) ^{ 2 } }$
0%
- $\displaystyle \frac { 1 }{ \left( 1+x \right) ^{ 2 } }$
0%
$\displaystyle -\frac { 1 }{ \left( 1+x \right) } +\frac { 1 }{ \left( 1+x \right) ^{ 2 } }$
0%
None of these

Explanation

$\displaystyle x\sqrt { \left( 1+y \right) } +y\sqrt { \left( 1+x \right) } =0$ .

$\Rightarrow \displaystyle x\sqrt { \left( 1+y \right) } =-y\sqrt { \left( 1+x \right) }$ .

Squaring both sides, $\displaystyle { x }^{ 2 }\left( 1+y \right)={ y }^{ 2 }\left( 1+x \right) \Rightarrow { x }^{ 2 }-{ y }^{ 2 }+xy\left( x-y \right) = 0$

$\displaystyle \Rightarrow \left( x-y \right) \left( x+y+xy \right) = 0$

Now $\displaystyle x\neq y$ [does not satisfy the given equation]

$\displaystyle \therefore x+y+xy=0 \Rightarrow y=\frac { -x }{ 1+x }$

$\displaystyle \therefore \quad \frac { dy }{ dx } =\frac { -\left( 1+x \right) +x }{ \left( 1+x \right) ^{ 2 } } =-\frac { 1 }{ \left( 1+x \right) ^{ 2 } }$

If $\displaystyle y=\dfrac { x }{ a+\dfrac { x }{ b+y } }$ , then $\displaystyle \frac{dy}{dx}$ is

0%
$\displaystyle \frac{a}{ab+2ay}$
0%
$\displaystyle \frac{b}{ab+2by}$
0%
$\displaystyle \frac{a}{ab+2by}$
0%
$\displaystyle \frac{b}{ab+2ay}$

Explanation

$\displaystyle y=\dfrac { x }{ a+\dfrac { x }{ b+y } }$

$\displaystyle y=\frac { x(b+y) }{ ab+ay+x }$

$\displaystyle \Rightarrow$ $aby+ay^{2}+xy = xb+xy$

$\displaystyle \Rightarrow$ $aby+ay^{2} = xb$

Differentiating both side w.r.t $x$ we get,

$\displaystyle \Rightarrow ab\frac { dy }{ dx } +2ay\frac { dy }{ dx }=b \Rightarrow \frac { dy }{ dx } =\frac { b }{ ab+2ay }$

${ S }_{ n }$ denotes the sum of

$n$ terms of

$g.p$ . whose common ratio is

$r$ , then

$\displaystyle \left( r-1 \right) \frac { d{ S }_{ n } }{ dr }$ is equal to

0%
$\left( n-1 \right) { S }_{ n }+n{ S }_{ n-1 }$
0%
$\left( n-1 \right) { S }_{ n }-n{ S }_{ n-1 }$
0%
$\left( n-1 \right) { S }_{ n }$
0%
None of these

Explanation

We have,

$\displaystyle {S}_{n}=\frac { a\left( { r }^{ n }+1 \right) }{ r-1 }$

$\Rightarrow \left( r-1 \right) { S }_{ n }={ ar }^{ n }-a$

Differentiating both sides with respect to

$r,$ we get

$\displaystyle \left( r-1 \right) \frac { { dS }_{ n } }{ dr } +{ S }_{ n }=na{ r }^{ n-1 }-0$

$\displaystyle \Rightarrow \left( r-1 \right) \frac { { dS }_{ n } }{ dr } =na{ r }^{ n-1 }-{ S }_{ n }$

$=n[n$ th term from of

$G.P.]$

$-{S}_{n}$

$=n\left( { S }_{ n }-{ S }_{ n-1 } \right) -{ S }_{ n }$

$=\left( n-1 \right) { S }_{ n }-n{ S }_{ n-1 }.$

The values of

$f'(1)$ is

0%
$0$
0%
$1$
0%
$2$
0%
$3$

Explanation

$\displaystyle f\left ( x \right )=x^{2}f\left ( 1 \right )-xf'(2)+f"(3)$
given

$f(0)=2\Rightarrow f''(3)=2$
Now

$f'(x)=2xf(1)-f'(2)$

$\Rightarrow f''(x)=2f(1)\Rightarrow f''(3)=2=2f(1)\Rightarrow f(1)=1$
and using this we get

$f'(2)=2$
Thus our function is,

$f(x)=x^2-2x+2$

$f'(x)=2x-2\Rightarrow f'(1)=0$

${ y }^{ 2 } + { b }^{ 2 } = 2xy$ , then

$\displaystyle\frac { dy }{ dx }$ equals

0%
$\displaystyle\frac { 1 }{ xy-{ b }^{ 2 } }$
0%
$\displaystyle\frac { y }{ y-x }$
0%
$\displaystyle\frac { xy-{ b }^{ 2 } }{ { \left( y-x \right) }^{ 2 } }$
0%
$\displaystyle\frac { xy-{ b }^{ 2 } }{ y }$

Explanation

$y^2+b^2 =2xy\Rightarrow$
differentiating both sides w.r.t.

$x$

$\Rightarrow 2y\dfrac{dy}{dx} =2\left(y+x\dfrac{dy}{dx}\right)$

$\Rightarrow \dfrac{dy}{dx} =\dfrac{y}{y-x}= \dfrac{y(y-x)}{(y-x)^2}=\dfrac{xy-b^2}{(y-x)^2}$ , using (1)

Find

$\displaystyle \frac{dy}{dx}$ if

$y=x^{x}$

0%
$\displaystyle x^{x}\left ( lnx+1 \right )$
0%
$\displaystyle x^{x}\left ( lnx-1 \right )$
0%
$\displaystyle x .x^{x-1}$
0%
$\displaystyle x^{x-1}\left ( lnx+1 \right )$

Explanation

$y = x^x$
Take log both sides,

$\log y = x\log x$
Now differentiate both side w.r.t

$x$

$\Rightarrow \cfrac{1}{y}\cfrac{dy}{dx} = 1+\log x$

$\Rightarrow \cfrac{dy}{dx} =x^x(1+\log x)$

Find $\displaystyle \frac { dy }{ dx }$ , if $x + y =\displaystyle \sin { \left( x-y \right) }$

0%
$\displaystyle \frac { \cos { \left( x-y \right) -1 } }{ \cos { \left( x-y \right) +1 } }$
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$\displaystyle \frac { \cos { \left( x-y \right) +1 } }{ \cos { \left( x-y \right) -1 } }$
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$\displaystyle \frac { \cos { \left( x+y \right) +1 } }{ \cos { \left( x-y \right) -1 } }$
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$\displaystyle \frac { \cos { \left( x+y \right) -1 } }{ \cos { \left( x-y \right)+1 } }$

Explanation

Given,

$x + y = \displaystyle \sin { \left( x-y \right) }$

Differentiate both side w.r.t

$x$

$1+\cfrac{dy}{dx} = \cos(x-y)(1-\cfrac{dy}{dx})$

$\Rightarrow \cfrac{dy}{dx} = \cfrac{\cos(x-y)-1}{\cos(x-y)+1}$

A balloon which always remains spherical, has a variable diameter

$\displaystyle \frac {3} {2}(2x+3)$ . The rate of change of volume with respect to

$x$ will be

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$\displaystyle \frac {27 \pi}{8} (2x-3)^2$
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$\displaystyle \frac {27 \pi}{8} (2x+3)^2$
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$\displaystyle \frac {27 \pi}{8} (3x-2)^2$
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$\displaystyle \frac {8} {27 \pi} (2x+3)^2$

Explanation

Given diameter

$2r =\dfrac {3}{2}(2x + 3)$

$\Rightarrow r =\dfrac {3}{4}(2x + 3)$

$\Rightarrow \dfrac {dr}{dx}=\dfrac {3}{4}\times 2 =\dfrac {3}{2}$

$V=\dfrac {4}{3}\pi r^3$

$\Rightarrow \dfrac {dv}{dx}=\dfrac {4}{3}\pi \times 3r^2 \times \dfrac {dr}{dx}$

$\Rightarrow \dfrac{dV}{dx}=4\pi \times \dfrac {9}{16} (2x + 3)^2 \times \dfrac {3}{2}$

$\dfrac {dV}{dx}= \dfrac {27\pi }{8} (2x + 3)^2$

The surface area of a cube is increasing at the rate of

$2 cm^2/sec$ . When its edge is 90 cm, the volume is increasing at the rate of.

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$1620 cm^3/sec$
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$810 cm^3/sec$
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$405 cm^3/sec$
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$45 cm^3/sec$

Explanation

Given

$\dfrac{dA}{dt}=2 cm^2/sec$ ,

$x=90 cm$
Surface area of cube is

$A=6x^2$

$\dfrac {dA}{dt}=12 \times x \times \dfrac {dx}{dt}$

$2 = 12 \times 90 \times \dfrac {dx}{dt}$

$\Rightarrow \dfrac {1}{6 \times 90}=\dfrac {dx}{dt}$

Now,

$V=x^3$

$\dfrac {dV}{dx}=3x^2 \dfrac {dx}{dt}$

$\Rightarrow \dfrac{dV}{dt}=3 \times 90 \times 90 \times \dfrac {1}{6 \times 90}=45 cm^3/sec$

$f(x) = \displaystyle \left | \cos x-\sin x \right |$ then

$\displaystyle f'\left ( \dfrac{\pi}4 \right )$ is equal to-

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$\displaystyle \sqrt{2}$
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$\displaystyle -\sqrt{2}$
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$0$
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$Does\ not\ exist$

Explanation

$f\left( x \right) =\left| \cos { x } -\sin { x } \right|$

$\displaystyle =\begin{cases} \cos { x-\sin { x } \quad \quad x<\dfrac { \pi }{ 4 } } \\ -\cos { x+\sin { x } } \quad x>\dfrac { \pi }{ 4 } \end{cases}$

$\displaystyle f'\left( x \right) =\begin{cases} -\sin { x } -\cos { x } \quad x<\dfrac { \pi }{ 4 } \\ +\sin { x+\cos { x } \quad x>\dfrac { \pi }{ 4 } } \end{cases}$
Hence

$\displaystyle f'\left( \dfrac { \pi }{ 4 } \right)$ does not exists.

The surface area of a spherical bubble is increasing at the rate of

$2 cm^2/s$ . When the radius of the bubble is 6 cm, then the rate by which the volume of the bubble increasing is.

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$6 cm^3/sec$
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$9 cm^3/sec$
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$3 cm^3/sec$
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$13 cm^3/sec$

Explanation

Given

$\dfrac {dA}{dt}= 2 cm^2/sec$ ,

$r = 6 cm$
Surface area of sphere

$A = 4\pi r^2$

$\dfrac {dA}{dt}= 8\pi r \dfrac {dr}{dt}$

$\Rightarrow 2 = 8 \times \pi \times 6 \times \dfrac {dr}{dt}$

$\dfrac {dr}{dt}= \dfrac {1}{24\pi }$ cm/sec

Now

$V =\dfrac {4}{3}\pi r^3$

$\Rightarrow \dfrac {dV}{dt}= \dfrac {4}{3} \pi 3r^2 \dfrac {dr}{dt }$

$\Rightarrow \dfrac {dV}{dt}=\dfrac {4.\pi .(6)^2}{24\pi }$

$\Rightarrow \dfrac {dV}{dt}=6 cm^3/sec.$

The radius of a sphere is changing at the rate of 0.1 cm/sec. The rate of change of its surface area when the radius is 200 cm, is.

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$8\pi cm^2/sec$
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$12\pi cm^2/sec$
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$160\pi cm^2/sec$
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$200\pi cm^2/sec$

Explanation

Given

$r = 200 cm \,and\, \dfrac {dr}{dt}=0.1 cm/sec$

Surface area of sphere

$A=4\pi r^2$

$\Rightarrow \dfrac {dA}{dt}=8\pi r \dfrac {dr}{dt}$

$\Rightarrow \dfrac {dA}{dt}=8\pi \times 200 \times 0.1$

$\dfrac {dA}{dt}=160 \pi cm^2/sec$

$\displaystyle y=\left | \cos x \right |+\left | \sin x \right |$ then

$\displaystyle \frac{dy}{dx}$ at

$x=\dfrac{2\pi }{3}$ is:

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$\displaystyle \frac{1-\sqrt{3}}{2}$
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$0$
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$\displaystyle \frac{\sqrt{3}-1}{2}$
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None of these

Explanation

In the neighborhood of

$\:x=\dfrac{2\pi}3$

$\Rightarrow y=-\cos\:x+\sin\:x$

$\Rightarrow \left ( \cfrac{dy}{dx} \right )=\sin\:x+\cos\:x$
Thus at

$\:x=\dfrac{2\pi}3$

$\cfrac{dy}{dx}=\dfrac{\sqrt{3}}{2}-\dfrac{1}{\sqrt{2}}=\dfrac{1}{2}\left ( \sqrt{3}-1 \right )$

The surface area of a sphere when its volume is increasing at the same rate as its radius, is.

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1 sq. units
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$\dfrac {1} {2 \sqrt {\pi}}$ sq. units
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$4 \pi$ sq. units
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$\dfrac {4 \pi} {3 }$ sq. units

Explanation

Given

$\displaystyle\frac{dV}{dt}=\frac{dr}{dt}$

$V=\displaystyle\frac{4}{3}\pi r^3$

$\displaystyle\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}$

$\Rightarrow 4\pi r^2=1$

$\therefore$ Surface Area of sphere is

$1$ sq.unit
Hence, option A.

$f(x) = \displaystyle \left | \left ( x-4 \right )\left ( x-5 \right ) \right |$ then

$f'(x)$ is equal to-

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$-2x + 9$ , for all $\displaystyle x\: \: \epsilon \: \: R$
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$2x - 9$ , if $4 < x < 5$
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$-2x + 9$ , if $4 < x < 5$
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None of these

Explanation

$f\left( x \right) =\left| \left( x-4 \right) \left( x-5 \right) \right|$

$=\begin{cases} \left( x-4 \right) \left( x-5 \right) \quad \quad x<4 \\ -\left( x-4 \right) \left( x-5 \right) \quad 4\le x<5 \\ \left( x-4 \right) \left( x-5 \right) \quad \quad x\ge 5 \end{cases}=\begin{cases} { x }^{ 2 }-9x+20\quad \quad x<4 \\ -{ x }^{ 2 }+9x-20\quad 4\le x<5 \\ { x }^{ 2 }-9x+20\quad \quad x\ge 5 \end{cases}\\ \therefore f'\left( x \right) =\begin{cases} 2x-9\quad \quad \quad x<0 \\ -2x+9\quad \quad 4\le x<5 \\ 2x-9\quad \quad \quad x\ge 5 \end{cases}$

CBSE Questions for Class 11 Commerce Applied Mathematics Differentiation Quiz 6 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 11 Commerce Applied Mathematics Differentiation Quiz 6 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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