## Category Archives: ISI Kolkatta Entrance Exam

### Derivatives part 14: IITJEE maths tutorial problems for practice

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:

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

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

### Derivatives: part 13: IITJEE Math tutorial problems for practice

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 ðŸ™‚

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 ðŸ™‚ an elegant answer

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

Cheers,

Nalin Pithwa.

### Derivatives: part 12:IITJEE maths tutorial problems for practice

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

### Set Theory Primer : Some basic thinking and problem solving

Reference: AMS, Student Mathematical Library: Basic Set Theory by A. Shen, et al. Chapter 1. Section 1.

Problem 1:

Consider the oldest mathematician amongst chess players and the oldest chess player amongst mathematicians. Could they be two different people?

Problem 2:

The same question for the best mathematician amongst chess players and the best chess player amongst mathematicians.

Problem 3:

One tenth of mathematicians are chess players, and one sixth of chess players are mathematicians. Which group (mathematicians or chess players) is bigger? What is the ratio of sizes of these two groups?

Problem 4:

Do there exist sets A, B and C such that $A \bigcap B \neq \phi$, $A \bigcap C = \phi$ and $(A\bigcap B)-C = \phi$ ?

Problem 5:

Which of the following formulas are true for arbitrary sets A, B and C:

i) $(A \bigcap B) \bigcup C = (A \bigcup C) \bigcap (B \bigcup C)$

ii) $(A \bigcup B) \bigcap C = (A \bigcap C) \bigcup (B \bigcap C)$

iii) $(A \bigcup B) - C = (A-C)\bigcup B$

iv) $(A \bigcap B) - C = (A - C) \bigcap B$

v) $A - (B \bigcup C) = (A-B) \bigcap (A-C)$

vi) $A - (B \bigcap C) = (A-B) \bigcup (A-C)$

Problem 6:

Give formal proofs of all valid formulas from the preceding problem. (Your proof should go like this : “We have to prove that the left hand side equals the right hand side. Let x be any element of the left hand side set. Then, ….Therefore, x belongs to the right hand side set. On the other hand, let…”)

Please give counterexamples to the formulas which are not true.

Problem 7:

Prove that the symmetric difference is associative:

$A \triangle (B \triangle C) = (A \triangle B) \triangle C$ for any sets A, B and C. Hint: Addition modulo two is associative.

Problem 8:

Prove that:

$(A_{1}\bigcap A_{2}\bigcap \ldots A_{n}) \triangle (B_{1} \bigcap B_{2} \bigcap \ldots B_{n}) = (A_{1} \triangle B_{1}) \bigcup (A_{2} \triangle B_{2}) \ldots (A_{n} \triangle B_{n})$ for arbitrary sets $A_{1}, A_{2}, \ldots, A_{n}, B_{1}, B_{2}, \ldots, B_{n}$.

Problem 9:

Consider an inequality whose left hand side and right hand side contain set variables and operations $\bigcap, \bigcup$ and -. Prove that if this equality is false for some sets, then it is false for some sets that contain at most one element.

Problem 10:

How many different expressions can be formed from set variables A and B by using union, intersection and set difference? (Variables and operations can be used more than once. Two expressions are considered identical if they assume the same value for each set of values of the variables involved.) Solve the same problem for three sets and for n sets. (Answer: In the general case, $2^{2^{n}-1}$)

Problem 11:

Solve the same problem if only $\bigcup$ and $\bigcap$ are allowed. For n=2 and n-3, this problem is easy to solve; however, no general formula for any n is known. This problem is also called “counting monotone Boolean functions in n variables”.)

Problem 12:

How many subsets does an n-element subset have?

Problem 13:

Assume that A consists of n elements and $B \subset A$ consists of k elements. Find the number of different sets C such that $B \subset C \subset A$.

Problem 14:

A set U contains 2n elements. We select k subsets of A in such a way that none of them is a subset of another one. What is the maximum possible value of k? (Hint: Maximal k is achieved when all subsets have n elements. Indeed, imagine the following process: We start with an empty set and add random elements one by one until we get U. At most one selected set can appear in this process. On the other hand, the expected number of selected sets that appear during this process can be computed using the linearity of expectation. Take into account that the probability to come across some set $Z \subset U$ is minimal when Z contains n elements, since all the sets of a given size are equiprobable.)

Your comments/solutions are welcome.

Regards,

Nalin Pithwa.

Purva building, 5A
Flat 06
Near Dimple Arcade, Thakur Complex
Mumbai, Maharastra 400101
India.

### Binomial Theorem Tutorial problems I: IITJEE mains practice

I. Expand up to 5 terms the following expressions:

1. $(1+x)^{\frac{1}{2}}$
2. $(1+x)^{\frac{7}{2}}$
3. $(1-x)^{\frac{2}{5}}$
4. $(1+x^{2})^{-2}$
5. $(1-3x)^{\frac{1}{3}}$
6. $(1-3x)^{\frac{-1}{2}}$
7. $(1+2x)^{-\frac{1}{2}}$
8. $(1+\frac{x}{3})^{-2}$
9. $(1+\frac{2x}{3})^{\frac{3}{2}}$
10. $(1+\frac{1}{2}a)^{-4}$
11. $(2+x)^{-2}$
12. $(9+2x)^{\frac{1}{2}}$
13. $(8+12a)^{\frac{3}{2}}$
14. $(9-6x)^{-\frac{3}{2}}$
15. $(4a-8x)^{-\frac{1}{2}}$

II. Write down and simplify:

1. The 8th term of $(1+2x)^{-\frac{1}{2}}$
2. The 11th term of $(1-2x^{3})^{\frac{11}{2}}$
3. The 16th term of $(1+3a^{2})^{\frac{16}{3}}$
4. The 6th term of $(3a-2b)^{-1}$
5. The $(r+1)^{th}$ term of $(1-x)^{-2}$
6. The $(r+1)^{th}$ term of $(1-x)^{-4}$
7. The $(r+1)^{th}$ term of $(1+x)^{\frac{1}{2}}$
8. The $(r+1)^{th}$ term of $(1+x)^{\frac{11}{3}}$
9. The 14th term of $(2^{10}-2^{7}x)^{\frac{13}{2}}$
10. The 7th term of $(3^{8}+6^{4}x)^{\frac{11}{4}}$

Regards,

Nalin Pithwa

### best explanation of epsilon delta definition

Refer any edition of (i) Calculus and Analytic Geometry by Thomas and Finney (ii) recent editions which go by the title “Thomas’ Calculus”. If you need, you will have to go through the previous stuff (given in the text) on “preliminaries” and/or functions also. For Sets, Functions and Relations, I have also presented a long series of articles on this blog.

Ref:

https://www.amazon.in/Thomas-Calculus-George-B/dp/9353060419/ref=sr_1_1?crid=3F1XO0L9KBT1F&keywords=thomas+calculus&qid=1581323971&s=books&sprefix=Thomas+%2Caps%2C265&sr=1-1

### Theory of Quadratic Equations: Part III: Tutorial practice problems: IITJEE Mains and preRMO

Problem 1:

Find the condition that a quadratic function of x and y may be resolved into two linear factors. For instance, a general form of such a function would be : $ax^{2}+2hxy+by^{2}+2gx+2fy+c$.

Problem 2:

Find the condition that the equations $ax^{2}+bx+c=0$ and $a^{'}x^{2}+b^{'}x+c^{'}=0$ may have a common root.

Using the above result, find the condition that the two quadratic functions $ax^{2}+bxy+cy^{2}$ and $a^{'}x^{2}+b^{'}xy+c^{'}y^{2}$ may have a common linear factor.

Problem 3:

For what values of m will the expression $y^{2}+2xy+2x+my-3$ be capable of resolution into two rational factors?

Problem 4:

Find the values of m which will make $2x^{2}+mxy+3y^{2}-5y-2$ equivalent to the product of two linear factors.

Problem 5:

Show that the expression $A(x^{2}-y^{2})-xy(B-C)$ always admits of two real linear factors.

Problem 6:

If the equations $x^{2}+px+q=0$ and $x^{2}+p^{'}x+q^{'}=0$ have a common root, show that it must be equal to $\frac{pq^{'}-p^{'}q}{q-q^{'}}$ or $\frac{q-q^{'}}{p^{'}-p}$.

Problem 7:

Find the condition that the expression $lx^{2}+mxy+ny^{2}$ and $l^{'}x^{2}+m^{'}xy+n^{'}y^{2}$ may have a common linear factor.

Problem 8:

If the expression $3x^{2}+2Pxy+2y^{2}+2ax-4y+1$ can be resolved into linear factors, prove that P must be be one of the roots of the equation $P^{2}+4aP+2a^{2}+6=0$.

Problem 9:

Find the condition that the expressions $ax^{2}+2hxy+by^{2}$ and $a^{'}x^{2}+2h^{'}xy+b^{'}y^{2}$ may be respectively divisible by factors of the form $y-mx$ and $my+x$.

Problem 10:

Prove that the equation $x^{2}-3xy+2y^{2}-2x-3y-35=0$ for every real value of x, there is a real value of y, and for every real value of y, there is a real value of x.

Problem 11:

If x and y are two real quantities connected by the equation $9x^{2}+2xy+y^{2}-92x-20y+244=0$, then will x lie between 3 and 6, and y between 1 and 10.

Problem 11:

If $(ax^{2}+bx+c)y+a^{'}x^{2}+b^{'}x+c^{'}=0$, find the condition that x may be a rational function of y.

More later,

Regards,

Nalin Pithwa.

### Theory of Quadratic Equations: part II: tutorial problems: IITJEE Mains, preRMO

Problem 1:

If x is a real number, prove that the rational function $\frac{x^{2}+2x-11}{2(x-3)}$ can have all numerical values except such as lie between 2 and 6. In other words, find the range of this rational function. (the domain of this rational function is all real numbers except $x=3$ quite obviously.

Problem 2:

For all real values of x, prove that the quadratic function $y=f(x)=ax^{2}+bx+c$ has the same sign as a, except when the roots of the quadratic equation $ax^{2}+bx+c=0$ are real and unequal, and x has a value lying between them.Â This is a very useful famous classic result.Â

Remarks:

a) From your proof, you can conclude the following also: The expression $ax^{2}+bx+c$ will always have the same sign, whatever real value x may have, provided that $b^{2}-4ac$ is negative or zero; and if this condition is satisfied, the expression is positive, or negative accordingly as a is positive or negative.

b) From your proof, and using the above conclusion, you can also conclude the following: Conversely, in order that the expression $ax^{2}+bx+c$ may be always positive, $b^{2}-4ac$ must be negative or zero; and, a must be positive; and, in order that $ax^{2}+bx+c$ may be always negative, $b^{2}-4ac$ must be negative or zero, and a must be negative.

Further Remarks:

Please note that the function $y=f(x)=ax^{2}+bx+c$, where $a, b, c \in \Re$ and $a \neq 0$ is a parabola. The roots of this $y=f(x)=0$ are the points where the parabola cuts the y axis. Can you find the vertex of this parabola? Compare the graph of the elementary parabola $y=x^{2}$, with the graph of $y=ax^{2}$ where $a \neq 0$ and further with the graph of the general parabola $y=ax^{2}+bx+c$. Note you will just have to convert the expression $ax^{2}+bx+c$ to a perfect square form.

Problem 3:

Find the limits between which a must lie in order that the rational function $\frac{ax^{2}-7x+5}{5x^{2}-7x+a}$ may be real, if x is real.

Problem 4:

Determine the limits between which n must lie in order that the equation $2ax(ax+nc)+(n^{2}-2)c^{2}=0$ may have real roots.

Problem 5:

If x be real, prove that $\frac{x}{x^{2}-5x+9}$ must lie between 1 and $-\frac{1}{11}$.

Problem 6:

Prove that the range of the rational function $y=f(x)=\frac{x^{2}-x+1}{x^{2}+x+1}$ lies between 3 and $\frac{1}{3}$ for all real values of x.

Problem 7:

If $x \in \Re$, Prove that the rational function $y=f(x)=\frac{x^{2}+34x-71}{x^{2}+2x-7}$ can have no value between 5 and 9. In other words, prove that the range of the function is $(x <5)\bigcup(x>9)$.

Problem 8:

Find the equation whose roots are $\frac{\sqrt{a}}{\sqrt{a} \pm \sqrt(a-b)}$.

Problem 9:

If $\alpha, \beta$ are roots of the quadratic equation $x^{2}-px+q=0$, find the value of (a) $\alpha^{2}(\alpha^{2}\beta^{-1}-\beta)+\beta^{2}(\beta^{2}\alpha^{-1}-\alpha)$ (b) $(\alpha-p)^{-4}+(\beta-p)^{-4}$.

Problem 10:

If the roots of $lx^{2}+mx+n=0$ be in the ratio p:q, prove that $\sqrt{\frac{p}{q}}+\sqrt{\frac{q}{p}}+\sqrt{\frac{n}{l}}=0$

Problem 11:

If x be real, the expression $\frac{(x+m)^{2}-4mn}{2(x-n)}$ admits of all values except such as those that lie between 2n and 2m.

Problem 12:

If the roots of the equation $ax^{2}+2bx+c=0$ are $\alpha$ and $\beta$, and those of the equation $Ax^{2}+2Bx+C=0$ be $\alpha+\delta$ and $\beta+\delta$, prove that $\frac{b^{2}-ac}{a^{2}} = \frac{B^{2}-AC}{A^{2}}$.

Problem 13:

Prove that the rational function $y=f(x)=\frac{px^{2}+3x-4}{p+3x-4x^{2}}$ will be capable of all values when x is real, provided that p has any real value between 1 and 7. That is, under the conditions on p, we have to show that the given rational function has as its range the full real numbers. (Of course, the domain is real except those values of x for which the denominator is zero).

Problem 14:

Find the greatest value of $\frac{x+2}{2x^{2}+3x+6}$ for any real value of x. (Remarks: this is maxima-minima problem which can be solved with algebra only, calculus is not needed).Â

Problem 15:

Show that if x is real, the expression $(x^{2}-bc)(2x-b-c)^{-1}$ has no real value between b and a.

Problem 16:

If the roots of $ax^{2}+bx+c=0$ be possible (real) and different, then the roots of $(a+c)(ax^{2}+2bx+c)=2(ac-b^{2})(x^{2}+1)$ will not be real, and vice-versa. Prove this.

Problem 17:

Prove that the rational function $y=f(x)=\frac{(ax-b)(dx-c)}{(bx-a)(cx-a)}$ will be capable of all real values when x is real, if $a^{2}-b^{2}$ and $c^{2}-a^{2}$ have the same sign.

Cheers,

Nalin Pithwa

### Theory of Quadratic Equations: Tutorial problems : Part I: IITJEE Mains, preRMO

I) Form the equations whose roots are:

a) $-\frac{4}{5}, \frac{3}{7}$ (b) $\frac{m}{n}, -\frac{n}{m}$ (c) $\frac{p-q}{p+q}, -\frac{p+q}{p-q}$ (d) $7 \pm 2\sqrt{5}$ (e) $-p \pm 2\sqrt{2q}$ (f) $-3 \pm 5i$ (g) $-a \pm ib$ (h) $\pm i(a-b)$ (i) $-3, \frac{2}{3}, \frac{1}{2}$ (j) $\frac{a}{2},0, -\frac{2}{a}$ (k) $2 \pm \sqrt{3}, 4$

II) Prove that the roots of the following equations are real:

i) $x^{2}-2ax+a^{2}-b^{2}-c^{2}=0$

ii) $(a-b+c)x^{2}+4(a-b)x+(a-b-c)=0$

III) If the equation $x^{2}-15-m(2x-8)=0$ has equal roots, find the values of m.

IV) For what values of m will the equation $x^{2}-2x(1+3m)+7(3+2m)=0$ have equal roots?

V) For what value of m will the equation $\frac{x^{2}-bx}{ax-c} = \frac{m-1}{m+1}$ have roots equal in magnitude but opposite in sign?

VI) Prove that the roots of the following equations are rational:

(i) $(a+c-b)x^{2}+2ax+(b+c-a)=0$

(ii) $abc^{2}x^{2}+3a^{2}cx+b^{2}ax-6a^{2}-ab+2b^{2}=0$

VII) If $\alpha, \beta$ are the roots of the equation $ax^{2}+bx+c=0$, find the values of

(i) $\frac{1}{\alpha^{2}} + \frac{1}{\beta^{2}}$

(ii) $\alpha^{4}\beta^{7}+\alpha^{7}\beta^{4}$

(iii) $(\frac{\alpha}{\beta}-\frac{\beta}{\alpha})^{2}$

VIII) Find the value of:

(a) $x^{3}+x^{2}-x+22$ when $x=1+2i$

(b) $x^{3}-3x^{2}-8x+16$ when $x=3+i$

(c) $x^{3}-ax^{2}+2a^{2}x+4a^{3}$ when $\frac{x}{a}=1-\sqrt{-3}$

IX) If $\alpha$ and $\beta$ are the roots of $x^{2}+px+q=0$ form the equation whose roots are $(\alpha-\beta)^{2}$ and $(\alpha+\beta)^{2}$/

X) Prove that the roots of $(x-a)(x-b)=k^{2}$ are always real.

XI) If $\alpha_{1}, \alpha_{2}$ are the roots of $ax^{2}+bx+c=0$, find the value of (i) $(ax_{1}+b)^{-2}+(ax_{2}+b)^{-2}$ (ii) $(ax_{1}+b)^{-3}+(ax_{2}+b)^{-3}$

XII) Find the condition that one root of $ax^{2}+bx+c=0$ shall be n times the other.

XIII) If $\alpha, \beta$ are the roots of $ax^{2}+bx+c=0$ form the equation whose roots are $\alpha^{2}+\beta^{2}$ and $\alpha^{-2}+\beta^{-2}$.

XIV) Form the equation whose roots are the squares of the sum and of the differences of the roots of $2x^{2}+2(m+n)x+m^{2}+n^{2}=0$.

XV) Discuss the signs of the roots of the equation $px^{2}+qx+r=0$

XVI) If a, b and c are odd integers, prove that the roots of the equation $ax^{2}+bx+c=0$ cannot be rational numbers.

XVII) Given that the equation $x^{4}+px^{3}+qx^{2}+rx+s=0$ has four real positive roots, prove that (a) $pr-16s \geq 0$ (b) $q^{2}-36s \geq 0$, where equality holds, in each case, if and only if the roots are equal.

XVIII) Let $p(x)=x^{2}+ax+b$ be a quadratic polynomial in which a and b are integers. Given any integer n, show that there is an integer M such that $p(n)p(n+1)=p(M)$.

Cheers,

Nalin Pithwa.

### Rules for Inequalities

If a, b and c are real numbers, then

1. $a < b \Longrightarrow a + c< b + c$
2. $a < b \Longrightarrow a - c < b - c$
3. $a < b \hspace{0.1in} and \hspace{0.1in}c > 0 \Longrightarrow ac < bc$
4. $a < b \hspace{0.1in} and \hspace{0.1in}c < 0 \Longrightarrow bc < ac$ special case: $a < b \Longrightarrow -b < -a$
5. $a > 0 \Longrightarrow \frac{1}{a} > 0$
6. If a and b are both positive or both negative, then $a < b \Longrightarrow \frac{1}{b} < \frac{1}{a}$.

Remarks:

Notice the rules for multiplying an inequality by a number: Multiplying by a positive number preserves the inequality; multiplying by a negative number reverses the inequality. Also, reciprocation reverses the inequality for numbers of the same sign.

Regards,

Nalin Pithwa.