Mathematics Hothouse

Sir Andrew Wiles on being bright at math:

May 6, 2021 – 12:46 am

Yes, some people are brighter than others in math. But I really believe that most people can really reach a good level in math if they are prepared to handle pyschological issues arising out of situations when being stuck.

By Nalin Pithwa | Posted in mathematicians, memory power concentration retention, miscellaneous, motivational stuff, time management | Comments (0)

Derivatives part 14: IITJEE maths tutorial problems for practice

April 11, 2021 – 12:36 pm

This is part 14 of the series

Question 1:

Let $f(x)= \sqrt{x-1} + \sqrt{x+24-10\sqrt{x-1}}$ for $x<26$ be a real valued function. Then, find $f^{'}(x)$ for $1<x<26$ :

Answer 1:

Consider $(\sqrt{x-1}-5)^{2} = x-1+25-10\sqrt{x-1} = x+24 -10\sqrt{x-1}$ so that we have

$\sqrt{x+24-10\sqrt{x-1}}=\sqrt{x-1}-5$

Hence, $f(x) = \sqrt{x-1} + \sqrt{x-1}-5 = 2\sqrt{x-1}-5$ when $1<x<26$

Hence, $f^{'}{x} = \frac{-2}{2\sqrt{x-1}} = -\frac{1}{\sqrt{x-1}}$

Question 2:

Let $3f(x) - 2 f(\frac{1}{x})=x$ , then find $f^{'}(2)$ .

Answer 2:

Given that $3f(x) - 2 f(\frac{1}{x})=x$ ….call this I.

Also, from above, we get $3f(\frac{1}{x}) - 2 f(x)= \frac{1}{x}$ …call this II.

so we get $6f(x)-4f(\frac{1}{x})=2x$ ….call this I’

and $9f(\frac{1}{x})-6f(x) = 9/x$ …call this II’.

$5f(\frac{1}{x})=2x+ \frac{9}{x}$ and hence, $f^{'}(1/x) = \frac{2x}{5} + \frac{9}{5x}$

Also, again $3f(x)-2f(1/x)=x$ ….A

$3f(1/x)-2f(x)=1/x$ …B

So, we now we get the following two equations:

$9f(x)-6f(1/x)=3x$ …..A’

$6f(1/x)-4f(x)=2/x$ ….B’

so, now we have $5f(x) = 3x + \frac{2}{x}$ so that we get $f(x) = \frac{3}{5}x+\frac{2}{5x}$ and $f(1/x) = \frac{2}{5}x+\frac{9}{5x}$

so $f^{'}(x) = \frac{3}{5} + \frac{2}{5}\frac{-1}{x^{2}} = \frac{3}{5} - \frac{1}{10}=\frac{1}{2}$

Question 3:

If $x= \frac{a(1-t^{2})}{1+t^{2}}$ and $y = \frac{2bt}{1+t^{2}}$ , then find $\frac{dy}{dx}$ .

Answer 3:

Given that $x = \frac{a(1-t^{2})}{1+t^{2}}$ where a is a parameter (constant) and t is a variable.

Let $t=\tan{\theta}$ so that $x = \frac{a(1-\tan^{2}{\theta})}{1+\tan^{2}{\theta}} = a \cos{2\theta}$

so that $y = \frac{2bt}{1+t^{2}}=b \sin{2\theta}$

$\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{2b\cos{2\theta}}{-2a\sin{2\theta}}=- \frac{b}{a\tan{2\theta}}$

so that we have

$\frac{dy}{dx} = - \frac{b(1-t^{2})}{a}$

Question 4:

If $y = \arccos{\frac{x-x^{-1}}{x+x^{-1}}}$ then find $\frac{dy}{dx}$

Answer 4:

Given that $\arccos{\frac{x^{2}-1}{x^{2}+1}} = \arccos{\frac{1-x^{2}}{1+x^{2}}}$ and put $x=\tan{\theta}$

$\frac{1-\tan^{2}{\theta}}{1+\tan^{2}{\theta}}=\cos{2\theta}$ so that $y = \arccos {\cos{2\theta}}=2\theta$

$\frac{dy}{dx} = 2\frac{d}{dx}(\arctan{x})=\frac{2}{1+x^{2}}$ which is the required answer.

Question 5:

If $\arcsin{(\frac{x^{2}-y^{2}}{x^{2}+y^{2}})}=a$ , where a is a parameter, then find $\frac{dy}{dx}$ .

Answer 5:

Given that $a=\arcsin{(\frac{x^{2}-y^{2}}{x^{2}+y^{2}})}$ so that $\sin{a} = \frac{x^{2}-y^{2}}{x^{2}+y^{2}}$

$(x^{2}+y^{2})\sin{a} = x^{2}-y^{2}$

Differentiating both sides w.r.t. x, we get

$2x\sin{a} + \sin{a} ,2y.\frac{dy}{dx}=2x-2y\frac{dy}{dx}$

$\frac{dy}{dx}(2y\sin{a}+2y)=2x-2x\sin{2a}$

$\frac{dy}{dx} = \frac{2x(1-\sin{a})}{1+\sin{a}}=\frac{x}{y}\times \frac{2y^{2}}{2x^{2}}=\frac{y}{x}$

Question 6:

If $y = cot^{-1}{(\sqrt{\frac{1+x}{1-x}})}$ then find $\frac{dy}{d(\arccos{x})}$ .

Answer 6:

Given that $y = cot^{-1}(\sqrt{(\frac{1+x}{1-x})})$ so that $y = \arctan{(\sqrt{(\frac{1-x}{1+x})})} = \arctan{(cot {(2\theta)})}$ where $x=\tan^{2}{\theta}$ so that $\frac{d\theta}{dx} = \frac{1}{1+x^{2}}$

and $\sec^{2}{y}.\frac{dy}{dx} = - cosec^{2}{(2\theta)}.2\frac{d\theta}{dx}$

Let $f=\arccos{x}$ so that $\frac{df}{dx} = - \frac{1}{\sqrt{1-x^{2}}}$

Now, note that $\sec^{2}{y} = cosec^{2}{2\theta}$ so we get the following simplification:

$\frac{dy}{dx} = - \frac{2}{1+x^{2}}$

Now, $\frac{dy}{df} = -\frac{\frac{dy}{dx}}{\frac{df}{dx}}= \frac{2\sqrt{1-x^{2}}}{1+x^{2}}$

Cheers,

Nalin Pithwa

By Nalin Pithwa | Posted in calculus, IITJEE Mains, ISI Kolkatta Entrance Exam, KVPY | Comments (0)

Derivatives: part 13: IITJEE Math tutorial problems for practice

April 11, 2021 – 2:34 am

Question 1:

Let $f(x)$ be a differentiable function w.r.t. x at $x=1$ and $\lim_{h \rightarrow 0} \frac{1}{h}f(1+h)=5$ , then evaluate $f^{'}(1)$

Solution 1:

By definition, derivative of a function $f(x)$ is $f^{'}(x) = \lim_{t \rightarrow x}\frac{f(t)-f(x)}{t-x}$ , where let us substitute $t-x=h$ , $t=x+h$ , as $t \rightarrow x$ , then $h \rightarrow 0$

So that above expression is equal to $\lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$

$f^{'}(1) = \lim_{h \rightarrow 0} \frac{f(1+h)-f(1)}{h}$ exists and can be evaluated if we know the value of the function $f(x)$ at $x=1$ .

Question 2:

If $x\sqrt{1+y} + y \sqrt{1+x}=0$ , then find $\frac{dy}{dx}$ .

Answer 2:

Given that $x\sqrt{1+y} + y\sqrt{1+x}=0$

Taking derivative of both sides w.r.t. x, we get the following equation:

$\sqrt{1+y} \times 1 + \frac{x}{2\sqrt{1+y}}.\frac{dy}{dx} + \sqrt{1+x}\frac{dy}{dx}+\frac{y}{2\sqrt{1+x}}.1=0$

$\sqrt{1+y}+ \frac{y}{2\sqrt{1+x}}+\frac{dy}{dx} \times (\frac{x}{2\sqrt{1+y}}+\sqrt{1+x})=0$

This further simplifies to :

$\frac{dy}{dx}. (\frac{x+2\sqrt{(1+x)(1+y)}}{2\sqrt{1+y}}) = \frac{2\sqrt{(1+x)(1+y)}+y}{2\sqrt{1+x}}$

$\frac{}{} = 2 \times \sqrt{\frac{1+y}{1+x}} \times \frac{2\sqrt{(1+x)(1+y)+y}}{x+2\sqrt{(1+x)(1+y)}}$

But, we already know that $x\sqrt{1+y}=-y\sqrt{1+x}$ so that $\sqrt{\frac{1+y}{1+x}} = - \frac{y}{x}$

$\frac{dy}{dx} = - \frac{2y}{x} \times \frac{2\sqrt{(1+x)(1+y)}+y}{x+2\sqrt{(1+x)(1+y)}}$

$\frac{dy}{dx}= -2 \times \frac{2y\sqrt{(1+x)(1+y)}+y^{2}}{x^{2}+2x\sqrt{(1+x)(1+y)}}$

$\frac{dy}{dx}=-2. \frac{2x(1+y)+y^{2}}{x^{2}-2y(1+y)}$ is the desired answer.

You can see how ugly it looks. Is there any way to simplify above? Let us give it one more shot. As follows:

Given that $x\sqrt{1+y} + y\sqrt{1+x}=0$ Hence, $x^{2}(1+y)=y^{2}(1+x)$ so that

$x^{2}-y^{2}=y^{2}x-x^{2}y$

$(x+y)(x-y) = y^{2}x-x^{2}y=xy(y-x)$ . If $x \neq y$ , then

$x+y= -xy$ . Taking derivative of both sides w.r.t. x, we get:

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

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

$\frac{dy}{dx} = - \frac{1+y}{1+x}= - \frac{y^{2}}{x^{2}}$ which is such an elegant answer 🙂

Question 3:

If $x^{y}.y^{x}=c$ , where c is a parameter constant, then find $\frac{dy}{dx}$ at $(e,e)$ .

Solution 3:

Let $u=x^{y}$ and $v=y^{x}$ .

Taking logarithm of both sides:

$\log {u} = y \log {x}$ and $\log {v} = x \log{y}$ .

Consider the LHS equation:

Taking derivative of both sides w.r.t.x, we get:

$\frac{1}{u}\frac{du}{dx} = \frac{y}{x} + \frac{dy}{dx}. \log{x}$

$\frac{1}{u}. \frac{du}{dx} = \frac{y}{x} + \frac{dy}{dx} (\log{x})$

$\frac{du}{dx} = x^{y} \times (\frac{y}{x}+(\log{y}).\frac{dy}{dx})$

$\log {v} = x \log{y}$

$\frac{1}{v}\frac{dv}{dx} = \frac{x}{y}\frac{dy}{dx} + (\log{y})$

$\frac{dv}{dx} = y^{x} \times (\frac{x}{y}\frac{dy}{dx}+ \log{y})$ .

Also, $x^{y}\frac{dv}{dx} + y^{x}\frac{du}{dx}=0$

$x^{y}y^{x} \times (\frac{x}{y}\frac{dy}{dx}+\log{y}) + y^{x}.x^{y}. (\frac{y}{x}+(\log{x}).\frac{dy}{dx}) =0$

$x^{y}y^{x} \times (\frac{dy}{dx}(\frac{x}{y}+\log{x})+\log{y}+\frac{y}{x})=0$ Now substitute $(x,y)=(e,e)$ and get the required answer.

$x^{y}y^{x} \times (\frac{dy}{dx}(\frac{x}{y})+\log{x}) = - x^{y}y^{x}(\frac{y}{x}+\log{y})$

$\frac{dy}{dx} (\frac{x}{y}+\log{x})=-(\frac{y}{x}+\log{y})$

Substituting $(x,y) = (e,e)$

Hence, then, $(\frac{dy}{dx})_(e,e) = - \frac{e/e + log e}{e/e + \log {e}}=-1$ is the desired answer.

Question 4:

Find $\frac{d}{dx}(\tan(\arctan{x} + cot^{-1}(x+1))$

Answer 4:

Consider $\tan(A+B) = \frac{\tan{A}+\tan{B}}{1-\tan{A}\tan{B}}$

Subsituting $A= \arctan{x}$ and $B=cot^{-1}(x+1)$ , we get the following:

$\tan{(A+B)} = \frac{tan(\arctan{x}+tan(cot^{-1}(x+1)))}{1-x. tan(cot^{-1}(x+1))} = \frac{x+\frac{1}{(x+1)}}{1-\frac{x}{x+1}}$

which in turn equals $\frac{x+\frac{1}{x+1}}{1-\frac{x}{x+1}}=\frac{x^{2}+x+1}{1}$ noting that $\arctan{(\frac{1}{x+1})}=cot^{-1}(x+1)$

Hence, the answer is $\frac{d}{dx}(x^{2}+x+1)=2x+1$

Question 5:

If $y = \arctan{\sqrt{\frac{1+\sin{x}}{1-\sin{x}}}}$ , find $\frac{dy}{dx}$

Solution 5:

Given that $\tan{y} = \sqrt{\frac{1+\sin{x}}{1-\sin{x}}}$

$\tan^{2}{y} = \frac{1+\sin{x}}{1-\sin{x}}$ . Taking derivative of both sides w.r.t. x,

$2(\tan{y})/ \sec^{2}{y}\frac{dy}{dx} = \frac{(1-\sin{x})(\cos{x})}{(1-\sin{x})^{2}}$

$2\tan{y}\sec^{2}{y}\frac{dy}{dx} = \frac{\cos{x}-\sin{x}\cos{x}+\cos{x}+\sin{x}\cos{x}}{(1-\sin{x})^{2}}$ which in turn equals

$\frac{2\cos{x}}{(1-\sin{x})^{2}}$

But, $\tan^{2}{y} = \frac{1+\sin{x}}{1-\sin{x}}$

so that $\sec^{2}{y}=1+\tan^{2}{y}=1+\frac{1+\sin{x}}{1-\sin{x}} = \frac{2}{1-\sin{x}}$

Hence, $2. \sqrt{\frac{1+\sin{x}}{1-\sin{x}}} \times \frac{2}{1-\sin{x}} \times \frac{dy}{dx} = \frac{2\cos{x}}{(1-\sin{x})^{2}}$

Hence, $\frac{dy}{dx} = \frac{1}{2} \times \frac{\cos{x}}{\sqrt{(1+\sin{x})(1-\sin{x})}} = \frac{\cos{x}}{2\sqrt{1-\sin^{2}{x}}} = \frac{1}{2}$

Question 6:

Find $\frac{d}{dx}cot^{-1}(\frac{1+\sqrt{1+x^{2}}}{x})$

Solution 6:

Let $y = cot^{-1} (\frac{1+\sqrt{1+x^{2}}}{x})$

Put $x = \sin{\theta}$ so that $\sqrt{1-x^{2}}=\sqrt{1-\sin^{2}{\theta}}=\cos{\theta}$

$\frac{1+\cos{\theta}}{\sin{\theta}} = \frac{2\cos^{2}(\theta/2)}{2\sin{\theta/2}\cos{\theta/2}} = cot (\theta/2)$

$cot^{-1}(cot{(\theta/2)}) = \theta/2$

$y=\theta/2$

$\frac{dy}{dx} = \frac{1}{2}\frac{d}{dx}( \arcsin {x} )=\frac{1}{2\sqrt{1-x^{2}}}$ where $|x|<1$

Question 7:

If $y = \arcsin{(\frac{2x}{1+x^{2}})}+sec^{-1}(\frac{1+x^{2}}{1-x^{2}})$ . Find $\frac{dy}{dx}$ .

Solution 7:

Let $x = \tan{\theta}$ so that $\frac{2x}{1+x^{2}} = \frac{2\tan{\theta}}{1+\tan^{2}{\theta}} = \sin{2\theta}$

so that $\arcsin{(\frac{2x}{1+x^{2}})} = \arcsin{\sin{2\theta}}=2\theta$

We now have $\frac{1-x^{2}}{1+x^{2}} = \frac{1-\tan^{2}{\theta}}{1+\tan^{2}{\theta}} = \cos{2\theta}$

so that $sec^{-1}{(\frac{1+x^{2}}{1-x^{2}})} = sec^{-1}(sec {(2\theta)}) =2 \theta$

so the desired answer is $4\frac{d\theta}{dx}=\frac{4}{1+x^{2}}$

Question 8:

If $y= \arcsin{(\frac{\sqrt{1+x}+\sqrt{1-x}}{2})}$ then find $\frac{dy}{dx}$

Solution 8:

Given that $\sin{y} = \frac{\sqrt{1+x}+\sqrt{1-x}}{2}$

$2\cos{y}\frac{dy}{dx} = \frac{1}{2\sqrt{1+x}} \times 1 + \frac{1}{2\sqrt{1-x}} \times (-1)$

$2\cos{y} \frac{dy}{dx} = \frac{\sqrt{1-x}-\sqrt{1+x}}{2\sqrt{1-x^{2}}}$

$2\cos{y} \frac{dy}{dx} = \frac{1-x-(1+x)}{(2\sqrt{1-x^{2}})(\sqrt{1-x}+\sqrt{1+x})} = - \frac{2x}{(2\sqrt{1-x^{2}})(\sqrt{1-x}+\sqrt{1+x})}$

$2\cos{y} \frac{dy}{dx} = - \frac{x}{(\sqrt{1-x^{2}})(2\sin{y})}$

$\frac{dy}{dx} = - \frac{x}{2\sqrt{1-x^{2}} (\sin{2y})}$ but $\sin{2y} = \frac{\sqrt{1+x}+\sqrt{1-x}}{2}$ and $\cos{y} = \frac{\sqrt{1+x}-\sqrt{1-x}}{2}$

so now we have $\sin{2y} = \frac{2}{4} (1+x-(1-x)) = x$

Hence, we get $\frac{dy}{dx} = - \frac{x}{2\sqrt{1-x^{2}}(x)}=-\frac{1}{2\sqrt{1-x^{2}}}$ .

Question 9:

If $g(x) = x^{2}+2x+3f(x)$ and $f(0)=5$ and $\lim_{x \rightarrow 0} \frac{f(x)-5}{x}=4$ , then evaluate $g^{'}(0)$ .

Solution 9:

We have $g^{'}{(x)} = 2x+2+3f^{'}(x)$ and hence, $g^{'}{(0)}=2+3f^{'}{(0)}$

By definition of derivative, we have $f^{'}{(x)}= \lim_{t \rightarrow x} \frac{f(t)-f(x)}{t-x}$ where let us say $t-x=h$ so that $t \rightarrow x$ , and $h \rightarrow 0$

$f^{'}(x) = \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$

$\lim_{h \rightarrow 0} \frac{f(h)-5}{h}=4$ and $f(0)=5$ and hence, $f^{'}(0)=4$ .

Hence, $g^{'}(0)=2+3 \times 4=14$

Question 10:

If $\tan{y} = \frac{2t}{1-t^{2}}$ , and $\sin{x}=\frac{2t}{1+t^{2}}$ , then find $\frac{d^{2}y}{dx^{2}}$ .

Answer 10:

Let $t = \tan{\theta}$ and hence, $\frac{2t}{1-t^{2}} = \frac{2\tan{\theta}}{1-\tan^{2}{\theta}} = \tan{2\theta}$

Hence, $\tan{y} = \tan{2\theta}$

so that $\sec^{2}{y} \frac{dy}{dx} = \sec^{2}{(2\theta)}.2.\frac{d\theta}{dx}$

Let $t=\tan{\theta}$ so that $\theta = \arctan{t}$ and $\frac{d\theta}{dx}=\frac{1}{1+t^{2}} \times \frac{dt}{dx}$

Hence, we get $(\sec^{2}{y})\frac{dy}{dx} = \sec^{2}(2\theta) \times \frac{2}{(1+t^{2})}.\frac{dt}{dx}$ so that

$\sec^{2}{(2\theta)}\frac{dy}{dx} = \sec^{2}{(2\theta)} \times \frac{2}{1+t^{2}} \times \frac{dt}{dx}$

Hence, $\frac{dy}{dx} = \frac{2}{(1+t^{2})}\frac{dt}{dx}$

Hence, $\sin{x}=\frac{2t}{1+t^{2}} = \frac{2\tan{\theta}}{1+\tan^{2}{(\theta)}}=2 \frac{\sin{\theta}}{\cos{\theta}} \frac{\cos^{2}{\theta}}{1}$

Hence, $\sin{x} = \sin{2\theta}$ and hence $x=2\theta$ and so $t=\tan{\theta}$ hence, $\theta=\arctan{t}$

$x =\arctan{t}$ so that $t=\tan{x}$

$\frac{dt}{dx} = \frac{1}{1+x^{2}}$ ….call this A.

$\frac{dy}{dx} = \frac{2}{1+t^{2}} \frac{dt}{dx}$ …call this B.

$\frac{dy}{dx} = \frac{2}{(1+t^{2})(1+x^{2})}$

$\frac{d^{2}y}{dx^{2}} = \frac{d}{dx}(\frac{2}{(1+t^{2})(1+x^{2})}) = \frac{2}{(1+t^{2})} \frac{d}{dx}(\frac{1}{1+x^{2}}) + \frac{2}{(1+x^{2})} \frac{d}{dx}(\frac{1}{(1+t^{2})})$

$\frac{d^{2}y}{dx^{2}} = \frac{2}{(1+t^{2})}. \frac{-2x}{(1+x^{2})^{2}}+\frac{2}{(1+x^{2})}.\frac{-2t}{(1+t^{2})^{2}}.\frac{dt}{dx}$

$=\frac{-4x}{(1+t^{2})(1+x^{2})^{2}} - \frac{4t}{(1+x^{2})^{2}(1+t^{2})^{2}}$

$\frac{d^{2}y}{dx^{2}} = \frac{-4x(1+t^{2})-4t}{(1+x^{2})^{2}(1+t^{2})^{2}} = \frac{-4(t+x(1+t^{2}))}{(1+x^{2})^{2}(1+t^{2})^{2}}$ which in turn equals

$\frac{-4(\tan{x}+x (1+\tan^{2}{x}))}{(1+x^{2})^{2}(1+\tan^{2}{x})^{2}}$ so where did we go wrong….quite clearly, practice alone can help us develop foresight…below is a cute proof:

$\frac{dy}{dx}= \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{\frac{d}{dt}(\arctan{(\frac{2t}{1-t^{2}})})}{\frac{d}{dt}\arcsin(\frac{2t}{1+t^{2}})}$ and put $t=\tan{\theta}$

so that $\frac{dy}{dx} = \frac{2\frac{d\theta}{dt}}{2\frac{d\theta}{dt}}=1$ so we have bingo 🙂 an elegant answer

$\frac{d^{2}y}{dx^{2}}=1$

Cheers,

Nalin Pithwa.

By Nalin Pithwa | Posted in calculus, IITJEE Mains, ISI Kolkatta Entrance Exam, KVPY | Comments (0)

Derivatives: part 12:IITJEE maths tutorial problems for practice

March 20, 2021 – 11:03 pm

1, $x = a(t+\frac{1}{t})$ , $y=a(t-\frac{1}{t})$ , then find $\frac{dy}{dx}$ .

Option (A) $\frac{t^{2}-1}{t^{2}+1}$

Option (B) $\frac{t^{2}+1}{t^{2}-1}$

Option (C) $\frac{t^{2}+1}{1-t^{2}}$

Option (D) $\frac{1}{t}$

Solution 1: Given that $x=at+\frac{a}{t}$ so that $\frac{dx}{dt} = a +\frac{a(-1)}{t^{2}} = a(1-\frac{1}{t^{2}})$

and given that $y = at - \frac{a}{t}$ so that $\frac{dy}{dt} = x + \frac{a}{t^{2}} = a(1+ \frac{1}{t^{2}})$

and so we get $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{a(1 - \frac{1}{t^{2}})}{a(1+\frac{1}{t^{2}})} = \frac{t^{2}-1}{t^{2}+1}$ so that correct choice is option A.

2. If $x = a \sin{3t} + b \cos{3t}$ and $y= b \cos {t} + a \sin{t}$ then find $\frac{dy}{dx}$ when $t = \frac{\pi}{4}$

Option (A) 0

Option (B) $\frac{b-a}{3(a+b)}$

Option (C) $\frac{a-b}{3(a+b)}$

Option (D) $\frac{b-a}{b+a}$

Solution 2:

$\frac{dy}{dt} = -b \sin{t} + a\cos{t}$

$\frac{dx}{dt} = 3a \cos{3t} + -3b \sin{3t}$

$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = - \frac{-b\sin{t}+a \cos{t}}{3a \cos{3t}-3b\sin{3t}}$ which is equal to the following at $t = \frac{\pi}{4}$

$\frac{dy}{dx} = \frac{- \frac{b}{\sqrt{2}} + \frac{a}{\sqrt{2}}}{-\frac{3a}{\sqrt{2}} - \frac{3b}{\sqrt{2}}}=-\frac{1}{3} \times \frac{b-a}{a+b}$ so that the correct choice is C.

3. If $y = \frac{x \arcsin{x}}{\sqrt{1-x^{2}}} + \log{(\sqrt{1-x^{2}})}$ , then find $\frac{dy}{dx}$

Option A: $\frac{\arcsin{x}}{(1-x^{2})^{\frac{3}{2}}}$

Option B: $\frac{\arcsin{x}}{\sqrt{1-x^{2}}}$

Option C: $\frac{\arcsin{x}}{1-x^{2}}$

Option D: $(1-x^{2})^{\frac{3}{2}}\arcsin{x}$

Solution 3:

Let $y = f(x) + g(x)$ where we put $f(x) = \frac{x \arcsin{x}}{\sqrt{1-x^{2}}}$ so now let $x=\sin{\theta}$

So, we get $\frac{dx}{d\theta} = \cos{\theta}$ and $1-x^{2}= \cos^{2}{\theta}$ and $\sqrt{1-x^{2}} = \cos{\theta}$

So we get $f(\theta) = \frac{\theta \times \sin{\theta}}{\cos{\theta}} = \theta \times \tan{\theta}$

So now $\frac{df}{d\theta}= \tan{\theta}+ \theta \times \sin^{2}{\theta}$

And, $g(x) = \log{\sqrt{1-x^{2}}}$

$\frac{dg}{dx} = \frac{1}{\sqrt{1-x^{2}}} \times \frac{d}{dx} (\sqrt{1-x^{2}}) = \frac{1}{2(1-x^{2})} \times (-2x)= \frac{-x}{1-x^{2}}$

Hence, we get the following:

$\frac{dy}{dx} = \frac{x}{\sqrt{1-x^{2}}} + \frac{\arcsin{x}}{\frac{1}{1-x^{2}}} + \frac{-x}{1-x^{2}}$

Question 4: Find the following: $\frac{d}{dx}(sec^{-1}{(\frac{1}{\sqrt{1-x^{2}}})} + cot^{-1}(\frac{\sqrt{1-x^{2}}}{x}))$

Option a: $\frac{2}{\sqrt{1-x^{2}}}$

Option b: $\frac{1}{\sqrt{1-x^{2}}}$

Option c: $\frac{\sqrt{1-x^{2}}}{x}$

Option d: $\sqrt{1-x^{2}}$

Solution 4:

Let $f(x)=y_{1}=sec^{-1}(\frac{1}{\sqrt{1-x^{2}}})$

Let $x=\sin{\theta}$ , $1-x^{2}=\cos^{2}(\theta)$ , $\sqrt{1-x^{2}}=\cos{\theta}$ , and $\frac{1}{\sqrt{1-x^{2}}} = \sec{\theta}$

so $y_{1}=\theta=\arcsin{x}$

$\frac{dy_{1}}{dx} = \frac{1}{\sqrt{1-x^{2}}}$

Let $y_{2}=cot^{-1}{\frac{\sqrt{1-x^{2}}}{x}}$

Let $x=\sin{\theta}$ , $\sqrt{1-x^{2}}=\cos{\theta}$ and $\frac{\sqrt{1-x^{2}}}{x}=\cot{\theta}$

Let $y_{2}= \cot^{-1}{\cot{\theta}}=\theta$

so that $\frac{dy_{2}}{dx} = \frac{d}{dx}(\arcsin{x})=\frac{1}{1-x^{2}}$

so that $\frac{dy}{dx}=\frac{2}{\sqrt{1-x^{2}}}$ so the option is a.

Question 5:

If $y=(x+\sqrt{1+x^{2}})^{n}$ then find $(x^{2}+1)(\frac{dy}{dx})^{2}$ .

Solution 5:

$y=(x+\sqrt{1+x^{2}})^{n}$

$\frac{dy}{dx} = x(x+\sqrt{1+x^{2}})^{n-1}\frac{d}{dx}(x+\sqrt{1+x^{2}}) = n(x+\sqrt{1+x^{2}})^{n-1}(1+\frac{2x}{2\sqrt{1+x^{2}}})$

$(\frac{dy}{dx})^{2}.(x^{2}+1) = (x^{2}+1).n^{2}.(x+\sqrt{1+x^{2}})^{2n-2} \times (1+\frac{x}{\sqrt{1+x^{2}}})^{2}$

$=(x^{2}+1).n^{2}. (x+\sqrt{1+x^{2}})^{2n-2} \times (\frac{x+\sqrt{1+x^{2}}}{\sqrt{1+x^{2}}})^{2}$

$=(x^{2}+1).n^{2}.(x+\sqrt{1+x^{2}})^{2n}.\frac{1}{(1+x^{2})}=n^{2}y^{2}$

Question 6:

If $f(x)= \sqrt{\frac{(x+1)(x+2)}{(x+3)(x+6)}}$ , then $f^{'}(0)$ is equal to

(a) 1/2 (b) 1/3 (c) 1/6 (d) 0

Solution 6:

Given that $y = \sqrt{\frac{(x+1)(x+2)}{(x+3)(x+6)}}$

Hence, we have $(x+3)(x+6) y^{2}=(x+1)(x+2)$

$(x+3)\frac{d}{dx}(y^{2}(x+6))+y^{2}(x+6) \times 1 = (x+1) \times 1 + (x+2) \times 1=2x+3$

$(x+3)(x+6)2y\frac{dy}{dx} + (x+3)y^{2} \times 1 =2x+3$

$(x+3)(x+6)2y\frac{dy}{dx} + (x+3) \times \frac{(x+1)(x+2)}{(x+3)(x+6)} = 2x+3$

$(x+3)(x+6)^{2}.2. \sqrt{\frac{(x+1)(x+2)}{(x+3)(x+6)}}+(x+1)(x+2) = (2x+3)(x+6)$

$2\sqrt{(x+1)(x+2)(x+3)(x+6)} \times (x+6)\frac{dy}{dx} + (x+1)(x+2) = (2x+3)(x+6)$

So, at x=0, on substitution we get $f^{'}(0)$ .

Question 7:

If $y = \frac{1-t^{2}}{1+t^{2}}$ , $x = \frac{2t}{1+t^{2}}$ , then find $\frac{dy}{dx}$ .

Solution 7:

Given $y= \frac{1-t^{2}}{1+t^{2}}$ , let $t= \tan{\theta}$ so that $\frac{dt}{d\theta}= \sec^{2}(\theta)$

so that $y = \frac{1-t^{2}}{1+t^{2}} = \frac{1-\tan^{2}{\theta}}{1+\tan^{2}{\theta}} = \frac{\cos^{2}{\theta}-\sin^{2}{\theta}}{1} = 2 \cos^{2}{\theta}-1= \cos{2\theta}$

Now, $x = \frac{2t}{1+t^{2}}=\frac{2\tan{\theta}}{1+\tan^{2}{\theta}}$ so that $x = \sin{2\theta}$

so now $\frac{dx}{d\theta}=2 \cos{2\theta}$

$y = \cos{2\theta}$

$\frac{dy}{d\theta} = -2 \sin{2\theta}$

$\frac{dy}{dx} = - \frac{2\sin{(2\theta)}}{2(\cos{2\theta})}= - \tan{(2\theta)} = - \frac{2t}{1-t^{2}}= - \frac{x}{y}$ .

Question 8:

Find $\frac{d}{dx}(\arctan{x} + \arcsin{(\frac{x}{\sqrt{1+x^{2}}})})$

Solution 8:

Let it be given that $y = \arctan{x} + \arcsin{(\frac{x}{\sqrt{1+x^{2}}})}$

Now, let us simplify this as $y=y_{1}+y_{2}$ where $y_{1} = \arctan{x}$ and $y_{2} = \arcsin{(\frac{x}{\sqrt{1+x^{2}}})}$

Now, first consider $y_{1} = \arctan{x}$ . Taking derivative of both sides w.r.t. x, we get

$\frac{dy_{1}}{dx} = \frac{d}{dx}(\arctan{x}) = \frac{1}{1+x^{2}}$ ….A

Now, next consider $y_{2} = \arcsin{(\frac{x}{\sqrt{1+x^{2}}})}$ . Takind derivative of both sides w.r.t. x, we get

$\frac{dy_{2}}{dx} = \frac{d}{dx}(\arcsin{(\frac{x}{\sqrt{1+x^{2}}})}) = \frac{1}{1- \frac{x^{2}}{1+x^{2}}} \frac{d}{dx}(\frac{x}{\sqrt{1+x^{2}}}) = \frac{1}{1+x^{2}}(\frac{1}{\sqrt{1+x^{2}}+\frac{x}{2}(1+x^{2})^{-3/2}})$ ….B

So that we get $\frac{dy}{dx} = \frac{dy_{1}}{dx} + \frac{dy_{2}}{dx}$ using A and B.

Question 9:

If $x^{y} = y^{x}$ , then find $\frac{dy}{dx}$

Solution 9:

Given that $x^{y} = y^{x}$

$y \log{x}= x \log{y}$ . Taking derivative of both sides w.r.t. x, we get

$(\log{x}).\frac{dy}{dx} + \frac{y}{x} = \frac{x}{y}. \frac{dy}{dx} + (\log{y}) \times 1$

$(\log{x}- \frac{x}{y}).\frac{dy}{dx} = (\log{y}) - \frac{y}{x}$

$\frac{dy}{dx} = \frac{\frac{x(\log{y}-y)}{x}}{\frac{y\log{x}-x}{y}}= \frac{y}{x} \times \frac{x(\log{y})-y}{y(\log{x})-x}$ which is the required answer.

Question 10:

If $(x+y)^{m+n} = x^{m}y^{n}$ , then find $\frac{dy}{dx}$ .

Solution 10:

Given that $(x+y)^{m+n} = x^{m}y^{n}$

Taking logarithm of both sides w.r.t. any arbitrary valid base,

$(m+n) \times \log{(x+y)} = \log{(x^{m}y^{n})} = \log(x^{m}) + \log{y^{n}}$ so that $(m+n).\log{(x+y)}=m \log{x} + n \log{y}$

Taking derivative of both sides w.r.t. x, we get the following:

$\frac{m+n}{x+y}. \times (1+\frac{dy}{dx}) = \frac{m}{x} + \frac{n}{y}.\times \frac{dy}{dx}$

$\frac{m+n}{x+y} \times \frac{dy}{dx} - \frac{x}{y}. \frac{dy}{dx} = \frac{m}{x} - \frac{m+n}{x+y}$

$\frac{(m+n)y-n(x+y)}{y(x+y)}. \frac{}{} = \frac{mx+my-mx-nx}{x(x+y)}$

$\frac{my+ny-nx-ny}{} = \frac{my-nx}{x(x+y)}$

$\frac{my-nx}{y(n+y)}. \frac{dy}{dx} = \frac{my-nx}{x(x+y)}$ , so that finally we get the desired answer:

$\frac{dy}{dx} = \frac{y}{x}$

More later,

Cheers,

Nalin Pithwa

By Nalin Pithwa | Posted in calculus, IITJEE Mains, ISI Kolkatta Entrance Exam, KVPY | Comments (0)

Annual Essay Contest: Mentoris Project: Aug 2021

March 20, 2021 – 9:29 pm

https://mentorisproject.org/essay-contest

By Nalin Pithwa | Posted in memory power concentration retention, miscellaneous, motivational stuff | Comments (0)

Match making, STEM, math, algorithm by 18 year old

March 19, 2021 – 7:32 am

https://www.npr.org/2021/03/18/978721494/matchmaker-matchmaker-make-me-an-algorithm-stem-contest-winner-pairs-data

By Nalin Pithwa | Posted in applications of maths | Comments (0)

Derivatives: part 11: IITJEE maths tutorial problems for practice

March 11, 2021 – 2:25 am

Problem 1: Find $\frac{d}{dx}\arctan{\frac{4x}{4-x^{2}}}$ .

Choose (a) $\frac{1}{4-x^{2}}$ (b) $\frac{1}{4+x^{2}}$ (c) $\frac{4}{4+x^{2}}$ (d) $\frac{4}{4-x^{2}}$

Solution 1:

Let $y = \arctan{\frac{4x}{4-x^{2}}}$ . Hence, $\tan{y} = \frac{4x}{4-x^{2}}$ . Differentiating both sides w.r.t. x, we get the following:

$\sec^{2}{y} \times \frac{dy}{dx}= \frac{d}{dx} (\frac{4x}{4-x^{2}})$

$\sec^{2}{y} \times \frac{dy}{dx} = \frac{(4-x^{2}) \times 4 - 4x \times (-2x)}{(4-x^{2})^{2}} = \frac{16+4x^{2}}{(4-x^{2})}$

But, $\sec^{2}{y}=\tan^{2}{y}+1=\frac{(x^{2}+4)^{2}}{(4-x^{2})^{2}}$

Hence, the answer is $\frac{dy}{dx}= \frac{4}{4+x^{2}}$ . Option c.

Problem 2: Find $\frac{dy}{dx}$ if $\sqrt{\frac{x}{y}} + \sqrt{\frac{y}{x}}=1$

Choose (a) $\frac{-(2x+y)}{x+y}$ (b) $\frac{-(2x+y)}{x+2y}$ (c) $\frac{x+2y}{x+y}$ (d) $-\frac{2x+y}{x+2y}$

Solution 2:

The given equation is $x+y = \sqrt{xy}$ . Differentiating both sides wrt x,

$1+ \frac{dy}{dx} = \sqrt{y} \times \frac{d}{dx} (\sqrt{x})+ \sqrt{x}\frac{d}{dx}(y^{\frac{1}{2}})$

$1+\frac{dy}{dx} = \frac{\sqrt{y}}{2\sqrt{x}} + \frac{\sqrt{x}}{2\sqrt{y}} \times \frac{dy}{dx}$

$(1- \frac{1}{2}\sqrt{\frac{x}{y}}) \times \frac{dy}{dx} =\frac{\sqrt{y}}{2\sqrt{x}} -1$

$\frac{dy}{dx} = \frac{\sqrt{y}-2\sqrt{x}}{2\sqrt{y}-\sqrt{x}} \times \frac{2\sqrt{y}}{2\sqrt{x}}$

$\frac{}{} = \frac{2y - 4\sqrt{xy}}{4\sqrt{xy}-2x} = \frac{2y-4(x+y)}{4(x+y)-2x} = - \frac{2x+y}{x+2y}$ is the answer. Option D.

Problem 3: If $y=\arctan{\frac{\log{(\frac{e}{x^{2}})}}{\log{(ex^{2})}}}$ then $\frac{dy}{dx}$ is

choose (a) $e$ (b) $\frac{2}{x(1+4(\log{x})^{2})}$ (c) $\frac{-2}{x(1+4(\log{x})^{2})}$ (d) $\frac{2}{1+x^{2}}$

Solution 3:

Given that $y = \arctan{(\frac{\log(\frac{e}{x^{2}})}{\log(ex^{2})})}$ so that we have

$\tan{y} = \frac{\log{\frac{e}{x^{2}}}}{\log{ex^{2}}}$ so now differentiating both sides w.r.t. x,

$\sec^{2}{y}\frac{dy}{dx} = \frac{\frac{\log{(ex^{2})}}{\frac{e}{x^{2}}} \frac{d}{dx}(\frac{e}{x^{2}}) - \log{(\frac{e}{x^{2}})} \times \frac{1}{ex^{2}} \times \frac{d}{dx}(ex^{2})}{(\log{(ex^{2})})}$

$\sec^{2}{y}\frac{dy}{dx} = \frac{-\frac{2}{x}(\log{(ex^{2})})- \frac{2}{x}\log{(\frac{e}{x^{2}})}}{(\log{(ex^{2})})^{2}}$

$\sec^{y}(\frac{dy}{dx}) = \frac{-\frac{2}{x}(\log{(ex^{2}) \times (\frac{e}{x^{2}})})}{(\log{ex^{2}})^{2}}$

$\sec^{y}\frac{dy}{dx} = \frac{-\frac{4}{x}}{(\log{(ex^{2})^{2}})} = \frac{-4}{x(\log{(ex^{2})})^{2}}$

Now, we also know that $\sec^{2}{y} = 1 + \tan^{2}{y} = \frac{(\log{(\frac{e}{x^{2}})})^{2}}{(\log{(ex^{2})})^{2}} + 1 = \frac{(\log(\frac{e}{x^{2}}))^{2}+(\log{(ex^{2})})^{2}}{(\log{(ex^{2})})^{2}}$

But, note that by laws of logarithms, on simplification, we get

$\log{(\frac{e}{x^{2}})} = 1 - 2\log{x}$ and $\log{(ex^{2})} = 1 + 2 \log{x}$ so that on squaring, we get

$(\log{(e/x^{2})})^{2} = 1-4\log{x} + 4 (\log{x})^{2}$

$(\log{(ex^{2})})^{2}=1+4\log{x} + 4 (\log{x})^{2}$ so that now we get

$(\log{(\frac{e}{x^{2}})})^{2} + (\log{(ex^{2})})^{2} = 2 + 8 (\log{x})^{2}$ , which all put together simplifies to

$\frac{dy}{dx} = \frac{1}{\sec^{2}{y}} \times \frac{-(\frac{4}{x})}{(\log{(ex^{2})})^{2}}$

$\frac{dy}{dx} = - \frac{(\frac{2}{x})}{1+4(\log{x})^{2}}$ so that the answer is option C.

Problem 4: Find $\frac{d}{dx}(\arcsin{(3x-4x^{3})}+\arccos{(2x(\sqrt{(1-x^{2})}))})$

Choose option (a) $\frac{1}{\sqrt{1-x^{2}}}$ (b) $\frac{-1}{\sqrt{1-x^{2}}}$ (c) $\frac{5}{\sqrt{1-x^{2}}}$ (d) $\frac{-2}{\sqrt{1-x^{2}}}$

Solution 4:

Let us consider the first differential. Let us substitute $x = \sin{\theta}$ . Hence,

$3x-4x^{3}=3\sin{\theta} - 4\sin^{3}{\theta}= \sin{3\theta}$ and so we $\arcsin{3x-4x^{3}} = \arcsin{\sin{3\theta}} = 3 \theta$ , and so also, we get $\arccos{2x\sqrt{1-x^{2}}}=\arccos{2\sin{\theta}\cos{\theta}} = \arccos{\cos{2\theta}}=2\theta$ so we get

required derivative

$\frac{dy}{dx} = \frac{d}{dx}(3\theta) + \frac{d}{dx}(2\theta) = \frac{d}{dx}(5\theta) = 5 \frac{d\theta}{dx} = 5 \frac{d}{dx}(\arcsin{x})= 5 \frac{1}{\sqrt{1-x^{2}}}$ . Answer is option C.

Problem 5: Find $\frac{d}{dx}(x-a)(x-b)(x-c)\ldots (x-z)$

Choose option (a) zero (b) 26 (c) 26! (d) does not exist

Solution 5: the expression also includes a term $0 = (x-x)$ so that the final answer is zero only.

Problem 6: Find $\frac{d}{dx}(x^{x})^{x}$ .

Solution 6: Let $y= (x^{x})^{x}$

so $\log{y} = \log{(x^{x})^{x}}$

so $\log{y} = x^{2} \times \log{x}$ so that differentiating both sides w.r.t. x, we get

we get $\frac{1}{y} \times \frac{dy}{dx} = \frac{x^{2}}{x} + \log{x} \times 2x$

we get $\frac{1}{y} \times \frac{dy}{dx} = x + 2x \log{x} = x(1+2\log{x})$

we get $\frac{dy}{dx} = yx (1+2 \log{x}) = (x^{x})^{x} \times x \times (1+2\log{x})$

so the answer is option B.

Choose option (a): $x.x^{x}(1+2\log{x})$ (b) $x^{x^{2}+1} \times (1+2\log{x})$ (c) ${x^{{x}^{2}}}(1+\log{x})$ (d) none of these

Problem 7:

Find $\frac{d}{dx}(e^{x^{x}})$

Choose option (a) $e^{x^{x}}.x^{x}.(1+\log{x})$ (b) $e^{x^{x}}. x^{x}.\log{(\frac{x}{e})}$ (c) $e^{x^{x}}.x^{x}$ (d) $e^{x^{x}}. (\log{(e^{x})})$

Solution 7: Let $y = (e^{(x^{x})})$ so that taking logarithm of both sides

$\log{y} = \log{(e^{(x^{x})})}$ so that $\log{y} = x^{x} \log{e} = x^{x}$

$\log {(\log{y})}= x \times (\log{x})$ . Differentiating both sides w.r.t.x we get:

$\frac{1}{\log{y}} \times \frac{d}{dx} \times (\log{y})= \frac{x}{x} + \log{x}$ so that we get now

$latex\frac{1}{y(\log{y})} \times \frac{dy}{dx} = 1 + \log{x} $

$\frac{dy}{dx} = e^{x^{x}} \times x^{x} \times (1+\log{x})$ so we get option a as the answer.

Problem 8:

Find $\frac{d}{dx}(x^{x^{x}})$

Choose option (a): $x^{x^{x}} \times (1+\log{x})$ (b) $x^{x^{x}} \times (x^{x}\log{x})(1+\log{x}+\frac{1}{x})$ (c) $x^{x^{x}} \times (x^{x}\log{x}) \times (1+\log{x}+\frac{1}{x\log{x}})$ (d) none of these.

Solution 8:

let $y=x^{x^{x}}$ taking logarithm of both sides we get

$\log{y} = x^{x} \times \log{x}$ and now differentiating both sides w.r.t.x, we get

$\frac{1}{y} \times \frac{dy}{dx} = \frac{x^{x}}{x} + (\log{x}) \times \frac{d}{dx} (x^{x})$ and now let $t=x^{x}$ and again take logarithm of both sides so that we get (this is quite a classic example…worth memorizing and applying wherever it arises):

$\log{t}= x\log{x}$

$\frac{1}{t} \frac{dt}{dx} = \frac{x}{x} + \log{x}=1+\log{x}$

$\frac{dt}{dx} = x^{x}(1+\log{x})$

$\frac{dy}{dx} \times \frac{1}{y} = x^{x-1} + (\log{x}).x^{x}.(1+\log{x})$

$\frac{dy}{dx} = x^{x}(x^{x-1}+x^{x} \times \log{x} \times (1+\log{x}))$

$\frac{dy}{dx} = x^{x^{x}} (x^{x} \times (\log{x})) \times (1+ \log{x}+ \frac{1}{x \log{x}})$

The answer is option C.

Problem 9:

Find $\frac{d}{dx}(x+a)(x^{2}+a^{2})(x^{4}+a^{4})(x^{8}+a^{8})$ .

Choose option (a): $\frac{15x^{16}-16x^{15}a+a^{16}}{(x-a)^{2}}$ (b) $\frac{x^{16}-a^{16}}{x-a}$ (c) $\frac{x^{16}-x^{15}a+a^{16}}{(x-a)^{2}}$ (d) none of these

Solution 9:

Given that $y = (x+a)(x^{2}+a^{2})(x^{4}+a^{4})(x^{8}+a^{8})$

Remark: Simply multpilying out thinking the symmetry will simplify itself is going to lead to a mess…because there will be no cancellation of terms …:-) The way out is a simple algebra observation…this is why we should never ever forget the fundamentals of our foundation math:-)

note that the above can be re written as follows:

$y = \frac{(x^{2}-a^{2})}{(x-a)} \times \frac{(x^{4}-a^{4})}{(x^{2}-a^{2})} \times \frac{x^{8}-a^{8}}{(x^{4}-a^{4})} \times \frac{(x^{16}-a^{16})}{(x^{8}+a^{8})}$

Now, we are happy like little children because many terms cancel out 🙂 hahaha…lol 🙂

$y = \frac{(x^{16}-a^{16})}{(x-a)}$ and now differentiating both sides w.r.t.x we get

$\frac{dy}{dx} = \frac{(x-a)(16x^{15})- (x^{16}-a^{16})(1)}{(x-a)^{2}}$

$\frac{dy}{dx} = \frac{15x^{16}-16x^{15}a+a^{16}}{(x-a)^{2}}$

The answer is option A.

Problem 10:

If $x= \theta {\cos{\theta}}+\sin{\theta}$ and $y = \cos{\theta}-\theta \times \sin{\theta}$ then find the value of $\frac{dy}{dx}$ at $\theta = \frac{\pi}{2}$

Choose option (a): $-\frac{\pi}{2}$ (b) $\frac{2}{\pi}$ (c) $\frac{\pi}{4}$ (d) $\frac{4}{\pi}$

Solution 10:

$\frac{dx}{d\theta} = \cos{\theta} - \theta \times \sin{\theta} + \cos{\theta}$

$\frac{dy}{d\theta} = -\sin{\theta} - (\sin{\theta} + \theta \times \cos{\theta})$

$\frac{dy}{d\theta} = \theta \times \cos{\theta} - 2\sin{\theta}$

$\frac{dx}{d\theta} = 2 \cos{\theta} - \theta \times \sin{\theta}$

$\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dy}{d\theta}} = \frac{\theta \times \cos{\theta}-2\sin{\theta}}{2\cos{\theta}-\theta \times \sin{\theta}} = \frac{-2}{-\pi/2}=\frac{4}{\pi}$

Answer is option D.

Regards,

Nalin Pithwa.

By Nalin Pithwa | Posted in calculus, IITJEE Mains | Comments (0)

A problem of log, GP and HP…

January 27, 2021 – 3:48 am

Question: If $a^{x}=b^{y}=c^{z}$ and $b^{2}=ac$ , pyrove that: $y = \frac{2xz}{x+z}$

Solution: This is same as proving: y is Harmonic Mean (HM) of x and z;

That is, to prove that $y=\frac{2xz}{x+z}$ is the same as the proof for : $\frac{1}{y} - \frac{1}{x} = \frac{1}{z} - \frac{1}{y}$

Now, it is given that $a^{x} = b^{y} = c^{z}$ —– I

and $b^{2}=ac$ —– II
Let $a^{x} = b^{y}=c^{z}=N$ say. By definition of logarithm,

$x = \log_{a}{N}$ ; $y=\log_{b}{N}$ ; $z=\log_{c}{N}$

$\frac{1}{x} = \frac{1}{\log_{a}{N}}$ ; $\frac{1}{y} = \frac{1}{\log_{b}{N}}$ ; $\frac{1}{z} = \frac{1}{\log_{a}{N}}$ .

Now let us see what happens to the following two algebraic entities, namely, $\frac{1}{y} - \frac{1}{x}$ and $\frac{1}{z} - \frac{1}{y}$ ;

Now, $\frac{1}{y} - \frac{1}{x} = \frac{1}{\log_{b}{N}} - \frac{1}{\log_{a}{N}} = \frac{\log_{b}{b}}{\log_{b}{N}} - \frac{\log_{a}{a}}{\log_{a}{N}} = \log_{N}{b} - \log_{N}{a} = \log_{N}{(\frac{b}{a})}$ …call this III

Now, $\frac{1}{z} - \frac{1}{y} = \frac{1}{\log_{c}{N}} - \frac{1}{\log_{b}{N}} = \frac{\log_{c}{c}}{\log_{c}{}N} -\frac{\log_{b}{b}}{\log_{b}{N}}= \log_{N}{c}-\log_{N}{b}$

Hence, $\frac{1}{z} - \frac{1}{y}=\log_{N}{c/b}$ ….equation IV

but it is also given that $b^{2}=ac$ …see equation II

Hence, $\frac{b}{a} = \frac{c}{b}$

Take log of above both sides w.r.t. base N:

So, above is equivalent to $\log_{N}{b/a} = \log_{N}{c/b}$

But now see relations III and IV:

Hence, $\frac{1}{y} -\frac{1}{x} = \frac{1}{x} - \frac{1}{y}$

Hence, $\frac{2}{y} = \frac{1}{x} + \frac{1}{z} = \frac{x+z}{xz}$

Hence, $y= \frac{2xz}{x+z}$ as desired.

Regards,

Nalin Pithwa

By Nalin Pithwa | Posted in algebra, IITJEE Foundation mathematics, IITJEE Mains, Pre-RMO | Comments (0)

Express a given integral number in any scale (radix)

January 19, 2021 – 6:06 pm

Several scales (radix) have been used by mathematicians. Binary (2), Ternary (3), Quaternary (4), Quinary (5), Senary (6), Septenary (7), Octenary(8), Nonary (9), Denary (10/Decimal), Undenary(11), Duodenary (12) and of course, hexadecimal (16). Note that in any scale the base radix is “10”. Thus, “10” stands for 2 in “binary”, “ten” in “decimal”, 8 for “octal” radix respectively, etc.

Let the digits used in a proposed scale(radix r) be $a_{0}, a_{1}, a_{2}, \ldots, a_{n}$ . Let us express an integer in this scale. Let $a_{0}$ be unit’s digits. Analagous to the place value system (in decimal):

$N=a_{0} + a_{1} \times r^{1} + a_{2} \times r^{2} + \ldots + a_{n} \times r^{n}$

Now, let us say we want to express this number N in terms of these digits $a_{i}$ s.

Dividing N by $r$ , we get the unit’s digit $a_{0}$ as the remainder; and the quotient is:

$a_{1} + a_{2} \times r^{1} + a_{3} \times r^{2} + \ldots + a_{n} \times r^{n-1}$ .

Dividing the above quotient by r, we get $a_{1}$ as the remainder and the quotient as:

$a_{2} \times r^{1} + a_{3} \times r^{2} + a_{4} \times r^{3} + \ldots + a_{n} \times r^{n-2}$ , and so on.

Example: Express the denary number 5213 in the scale of seven.

Solution: $(5213)_{10} \div 7$ gives 5 as remainder and $(744)_{10}$ as quotient.

$(744)_{10} \div 7$ gives 2 as remainder and $(106)_{10}$ as remainder.

Continuing this way, we are able to express:

$(5213)_{10} = 2 \times 7^{4} + 1 \times 7^{3} + 1 \times 7^{2} + 2 \times 7 +5$ . That is $(21125)_{7}$ . You can check the equivalence by converting both to decimal values.