Category Archives: KVPY

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<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


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


So, we now we get the following two equations:



so, now we have 5f(x) = 3x + \frac{2}{x} so that we get f(x) = \frac{3}{5}x+\frac{2}{5x} andf(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} = \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}}


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+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


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


\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



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:


\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}


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,


Nalin Pithwa

Derivatives: Part 10: IITJEE maths tutorial problems for practice

Problem 1: If x=3\cos{\theta}-\cos^{3}{\theta}, and y=3\sin{\theta}-\sin^{3}{\theta}, then \frac{dy}{dx} is equal to:

(a) -\cot^{3}{\theta} (b) -\tan^{3}{\theta} (c) \cot^{3}{\theta} (d) \tan^{3}{\theta}

Problem 2: If x = \tan{\theta} + \cot{\theta}, and y=2 \log{(\cot{\theta})}, then \frac{dy}{dx} is equal to:

(a) \tan{(2\theta)} (b) \cot{(2\theta)} (c) \tan{\theta} (d) \sec^{2}{2\theta}

Problem 3: \frac{d}{dx}\log{\sqrt{\frac{1-\cos{x}}{1+\cos{x}}}} is equal to:

(a) \tan{\frac{x}{2}} (b) \sin{x} (c) cosec(x) (d) \tan{x}

Problem 4: y=2^{2(\log_{2}{(x+2)}-\log_{2}{(x+1)})}, then \frac{dy}{dx} is:

(a) \frac{-2(x+2)}{(x+1)^{3}} (b) \frac{4(x+2)}{(x+1)^{3}} (c) \frac{2(x+2)}{(x+1)^{3}} (d) \frac{-6(x+2)}{(x+1)^{3}}

Problem 5: \frac{d}{dx}(\arctan{\sqrt{\frac{e^{x}-1}{e^{x}+1}}}) is equal to:

(a) \frac{1}{1+e^{2x}} (b) \frac{1}{2\sqrt{e^{2x}-1}} (c) \frac{e^{x}}{2\sqrt{1+e^{2x}}} (d) \frac{1}{2\sqrt{1-e^{2x}}}

Problem 6: y=e^{m\arcsin{x}} then (1-x^{2}) (y^{'})^{2} is equal to :

(a) y^{2} (b) m^{2}(1-y^{2}) (c) -m^{2}y^{2} (d) m^{2}y^{2}

Problem 7: If y = \sin(m \arcsin{x}) then (1-x^{2})(\frac{dy}{dx})^{2} is

(a) m^{2}y^{2} (b) m^{2}(1-y^{2}) (c) -m^{2}y^{2} (d) m^{2}(1+y^{2})

Problem 8: \frac{d}{dx}(\cos{\arctan{x}}) is:

(a) \frac{1}{2\sqrt{1+x^{2}}} (b) \frac{-x}{(1+x^{2})^{\frac{3}{2}}}

(c) \frac{-x}{\sqrt{1+x^{2}}} (d) \frac{2x}{\sqrt{1+x^{2}}}

Problem 9: If y = \arcsin{\frac{a\cos{x}+b\sin{x}}{\sqrt{a^{2}+b^{2}}}} then \frac{d^{2}y}{dx^{2}} is:

(a) -1 (b) 0 (c) 1 (d) \arctan{\frac{b}{a}}

Problem 10: \arctan{\frac{4x}{4-x^{2}}} is:

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


Nalin Pithwa.

Derivatives: part 8: IITJEE mains tutorial problems practice

Problem 1: If y=b(\arctan{(\frac{x}{y})})+ \arctan{(\frac{y}{x})}, then \frac{dy}{dx} is equal to:

(a) \frac{x}{2} (b) -1 (c) 0 (d) b

Problem 2: If r=a(1+\cos{\theta}), then \sqrt{r^{2}+(\frac{dr}{d\theta})^{2}} is:

(a) 2a\cos{\theta} (b) 2a \sin{(\frac{\theta}{2})} (c) 2a \cos{(\frac{\theta}{2})} (d) 2a \sin{\theta}

Problem 3: \frac{d}{dx}\arctan{\log_{10}{x}} is equal to:

(a) \frac{1}{1 + (\log_{10}{x})^{2}} (b) \frac{1}{x \log_{10}{(1+ (\log_{10}{x})^{2})}} (c) \frac{1}{x(1+(\log_{10}{x})^{2})} (d) \frac{1}{10 \log{x}(1+(\log_{10}{x})^{2})}

Problem 4: If \sin^{2}(mx) + \cos^{2}(ny)=a^{2}, then \frac{dy}{dx} is equal to:

(a) \frac{m \sin{(2mx)}}{n \sin{(2ny)}} (b) \frac{n\sin{(2mx)}}{m\sin{(2ny)}} (c) \frac{n\sin{(2ny)}}{m\sin{(2mx)}} (d) \frac{-m\sin{(2mx)}}{n\sin{(2ny)}}

Problem 5: \frac{d}{dx}(\frac{\tan{x}-\cot{x}}{\tan{x}+\cot{x}}) is equal to:

(a) 2\sin{(2x)} (b) \sin{(2x)} (c) -2 \sin{(2x)} (d) 2\cos{(2x)}

Problem 6: If y=\log_{5}{(\log_{5}{x})} then the value of \frac{dy}{dx} is

(a) \frac{1}{x \log_{5}{x}} (b) \frac{1}{x \log_{5}{x}. (\log{5})^{2}} (c) \frac{1}{\log{5}.x\log{x}} (d) \frac{1}{x(\log_{5}{x})^{2}}

Problem 7: \frac{d}{dx}(ax+b)^{cx+d} is equal to:

(a) (ax+b)^{cx+d}(\frac{cx+d}{ax+b} + \log{(ax+b)}) (b) (ax+b)^{cx+d}(\frac{cx+d}{ax+b} + c \log{(ax+b)}) (c) a(ax+b)^{cx+d} (d) none

Problem 8: \frac{d}{dx}(\log{(\frac{\sin{(x-b)}}{\sin{(x-a)}})}) is equal to:

(a) \frac{\cos{(a-b)}}{\sin{(x-a)}\sin{(x-b)}} (b) \frac{\sin{(b-a)}}{\sin{(x-a)}\sin{(x-b)}}

(c) \frac{\sin{(a-b)}}{\sin{(x-a)}\sin{(x-b)}} (d) \frac{\cos{(b-a)}}{\sin{(x-a)}\sin{(x-b)}}

Problem 9: If y = \sqrt{\frac{\sec{x}+\tan{x}}{\sec{x}-\tan{x}}} and 0<x<\frac{\pi}{2}, then \frac{dy}{dx} is :

(a) \sec{x}(\sec{x}-\tan{x}) (b) \sec{x}(\sec{x}+\tan{x}) (c) \tan{x}(\sec{x}+\tan{x}) (d) \tan{x}(\sec{x}-\tan{x})

Problem 10: \frac{d}{dx}e^{ax}(a\sin{(bx)}-b\cos{(bx)}) is equal to:

(a) e^{ax}(\sin{(bx)}) (b) (a^{2}+b^{2})e^{ax}\sin{(bx)} (c) e^{ax}\cos{(bx)} (d) (a^{2}+b^{2})e^{ax}\cos{(bx)}


Nalin Pithwa.

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}}


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.


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,


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. This is a very useful famous classic result. 


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). 

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.


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).


Nalin Pithwa.