Monthly Archives: January 2021

A problem of log, GP and HP…

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.


Nalin Pithwa

Express a given integral number in any scale (radix)

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.


Nalin Pithwa.

STEM for Australia,Engineering%20and%20Mathematics%20in%20Australia.

STEM for Britain

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 9: IITJEE maths tutorial problems practice

Problem 1: \frac{d}{dx}((\frac{1}{b}\arctan{\frac{x}{b}})-\frac{1}{a}\arctan{(\frac{x}{a})}) is equal to:

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

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

Problem 2: \frac{d}{dx}(\frac{x}{2} + \frac{1}{2}\log{(\sin{x}+\cos{x})}) is equal to:

(a) \frac{\tan{x}}{1+\tan{x}} (b) \frac{1}{1+\cot{x}} (c) \frac{1-\tan{x}}{1+\tan{x}} (d) \frac{1}{1+\tan{x}}

Problem 3: If y=\sqrt{\frac{cosec{x}-\cot{x}}{cosec{x}+\cot{x}}} where 0<x<\frac{\pi}{2}, then \frac{dy}{dx} is given by :

(a) cosec{x}(cosec{x}-\cot{x}) (b) cosec{x}(\cot{x}-cosec{x}) (c) cosec{x}(\cot{x}-cosec{x}) (d) \cot{x}(cosec{x}-\cot{x})

Problem 4: \frac{d}{dx}\log {|\sec{(x-\frac{\pi}{4})}+\tan{(x-\frac{\pi}{4})}|} is equal to:

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

Problem 5:

If r=a(1+\cos{\theta}), and \tan{\phi}=r\frac{d\theta}{dr}, then \phi is equal to:

(a) \frac{-2}{\theta} (b) \frac{\pi}{2} + \frac{\theta}{2} (c) -\frac{\theta}{2} (d) \frac{\pi}{2} - \frac{\theta}{2}

Problem 6: \frac{d}{dx}\log{(\sqrt{x+ \sqrt{x^{2}+a^{2}}})} is equal to:

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

Problem 7: \frac{d}{dx}(\log{(1+\sin{(2x)})} + 2 \log{\sec{(\frac{\pi}{4}-x)}}) is equal to

(a) 0 (b) \log{2} (c) \frac{4(\cos{x}-\tan{x})}{\sin{x}+\cos{x}} (d) \frac{2\cos{(2x)}}{1+\sin{(2x)}} + \tan{(\frac{\pi}{4}-x)}

Problem 8: If x^{2}+xy+y^{2}=1, then \frac{dy}{dx} is equal to:

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

Problem 9: \frac{d}{dx}(\arcsin{(\sqrt{\frac{1-x}{2}})}) is equal to:

(a) \frac{1}{\sqrt{1-x^{2}}} (b) \frac{-1}{\sqrt{1-x^{2}}} (c) \frac{1}{2\sqrt{1-x^{2}}} 9d) \frac{-1}{2\sqrt{1-x^{2}}}

Problem 10: If y = \arctan{(\frac{3a^{2}x-x^{3}}{x^{3}-3ax^{2}})} then \frac{dy}{dx} is equal to:

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

Derivatives: part 7: IITJEE tutorial problems practice

Problem 1: Differential coefficient of \sec{\arctan{x}} is

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

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

(a) 2 (b) -2 (c) 1 (d) -1

Problem 3: If y = \arcsin{\sqrt{x-ax}-\sqrt{a-ax}}, then \frac{dy}{dx} is equal to:

(a) \frac{1}{2\sqrt{x}\sqrt{1-x}} (b) \sin{(\sqrt{x})} \times \sin{(\sqrt{a})}

(c) \frac{1}{\sin{\sqrt{a-ax}}} (d) zero

Problem 4: For the differentiable function f, the value of : \lim_{h \rightarrow 0} \frac{(f(x+h))^{2}-(f(x))^{2}}{2h} is equal to:

(a) (f^{'}(x))^{2} (b) \frac{1}{2}(f(x))^{2} (c) f(x)f^{'}(x) (d) zero

Problem 5: The derivative of \arctan{\frac{\sqrt{1+x^{2}}-1}{x}} w.r.t. \arctan{(\frac{2x\sqrt{1-x^{2}}}{1-2x^{2}})} at x=0 is :

(a) \frac{1}{8} (b) \frac{1}{4} (c) \frac{1}{2} (d) 1

Problem 6: If x = e^{y+e^{y+e^{y+e^{y+ \ldots}}}} then \frac{dy}{dx} is

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

Problem 7: Consider the following statements:

(1) (\frac{f}{g})^{'} = \frac{f^{'}}{g^{'}} (2) \frac{(fg)^{'}}{fg} = \frac{f^{'}}{f} + \frac{g^{'}}{g}

(3) \frac{(f+g)^{'}}{f+g} = \frac{f^{'}}{f} + \frac{g^{'}}{g} (4) \frac{(f/g)^{'}}{f/g} = \frac{f^{'}}{f} + \frac{g^{'}}{g}

Which of the following statements are true?

(a) 1 and 2 (b) 2 and 3 (c) 2 and 4 (d) 3 and 4

Problem 8: If y=e^{x+3\log{x}} then \frac{dy}{dx} =

(a) e^{x+3\log{x}} (b) e^{x}.x^{2}(x+3) (c) e^{x}. e^{3\log{x}} (d) 3x^{2}e^{x}

Problem 9: If y=\sin^{2}(x \deg), then find the value of \frac{dy}{dx} is:

(a) \frac{\pi}{360}\sin{(2 x \deg)} (b) \frac{\pi}{2}\sin{(2x\deg)} (c) 180 \sin {(2x\deg)} (d) \frac{\pi}{180}\sin{(2x\deg)}

Problem 10: If y=\log_{a}{x} + \log_{x}{a} + \log_{x}{x}+ \log_{a}{a} then the value of \frac{dy}{dx} is:

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


Nalin Pithwa.

Derivatives: part 6: IITJEE tutorial practice problems

Problem 1:

If \sec {(\frac{x+y}{x-y})}=a, then \frac{dy}{dx} is (i) \frac{x}{y} (ii) \frac{y}{x} (iii) y (iv) x

Problem 2:

If f(x) = x+ 2, when -1<x<1;

f(x)=5, when x=3;

f(x) = 8-x, when x>3; then, at x=3, the value of f^{'}(x) is

(a) 1 (b) -1 (c) 0 (d) does not exist.

Problem 3:

If y = x \tan{y}, then \frac{dy}{dx} is equal to

(i) \frac{\tan{y}}{x-x^{2}-y^{2}} (ii) \frac{\tan{y}}{y-x}

(iii) \frac{y}{x-x^{2}-y^{2}} (iv) \frac{\tan{x}}{x-y^{2}}

Problem 4:

If g is the inverse function of f and f^{'}(x) = \frac{1}{1+x^{n}}, then g^{'}(x) is equal to

(i) 1 + (g(x))^{n} (ii) 1+g(x) (iii) 1-g(x) (iv) 1-(g(x))^{n}

Problem 5:

If f(x) = \log_{x^{2}}(\log{x}) then f(x) at x=c is :

(i) 0 (ii) 1 (iii) \frac{1}{e} (iv) \frac{1}{2e}

Problem 6:

If y = (\sin{x})^{\tan{x}} then \frac{dy}{dx} is equal to :

(i) (\sin{x})^{\tan{x}}(1+ \sec^{2}{x} \log{\sin{x}})

(ii) \tan{x}. (\sin{x})^{\tan{x}-1} \times \cos{x}

(iii) (\sin{x})^{\tan{x}}\sec^{2}{x} \times \log{\sin{x}}

(iv) \tan{x} (\sin{x})^{\tan{x}-1}

Problem 7:

If y = \sqrt{\sin{x}+y}, then \frac{dy}{dx} equals:

(i) \frac{\sin{x}}{2y-1} (ii) \frac{\sin{x}}{1-2y} (iii) \frac{\cos{x}}{1-2y}

(iv) \frac{\cos{x}}{2y-1}

Problem 8:

If x = \sqrt{\frac{1-t^{2}}{1+t^{2}}} and y = \sqrt{\frac{\sqrt{1+t^{2}}-sqrt{1-t^{2}}}{\sqrt{1+t^{2}}+\sqrt{1-t^{2}}}}

then the value of \frac{d^{2}y}{dx^{2}} at t=0 is given by:

(a) 0 (b) 1/2 (c) 1 (d) -1

Problem 9:

If x = a \cos^{3}{\theta}, y = a \sin^{3}{\theta}, then \sqrt{1 + (\frac{dy}{dx})^{2}} is equal to:

(i) \sec^{2}{\theta} (ii) \tan^{2}{\theta} (iii) \sec{\theta} (iv) |\sec{\theta}|

Problem 10:

If y = \arcsin{\sqrt{1-x}} + \arccos{\sqrt{x}}, then \frac{dy}{dx} equals:

(a) \frac{1}{\sqrt{x(1-x)}} (b) \frac{1}{x(1+x)} (c) \frac{-1}{\sqrt{x(1-x)}} (d) none


Nalin Pithwa