Nalin Pithwa

Author Archives: Nalin Pithwa

I am a DSP Engineer and a mathematician working in closely related areas of DSP, Digital Control, Digital Comm and Error Control Coding. I have a passion for both Pure and Applied Mathematics.

Annual Essay Contest: Mentoris Project: Aug 2021

March 20, 2021 – 9:29 pm

https://mentorisproject.org/essay-contest

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

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.

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

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.

Cheers,

Nalin Pithwa.

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STEM for Australia

January 19, 2021 – 2:34 pm

http://www.stemaustralia.edu.au/#:~:text=STEM%20Australia%20is%20a%20combination,Engineering%20and%20Mathematics%20in%20Australia.

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STEM for Britain

January 19, 2021 – 2:30 pm

Home

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Derivatives: Part 10: IITJEE maths tutorial problems for practice

January 18, 2021 – 5:18 am

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) $\frac{-x}{(1+x^{2})^{\frac{3}{2}}}$

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) $\frac{4}{4-x^{2}}$

Regards,

Nalin Pithwa.

Posted in calculus, IITJEE Mains, KVPY | Comments (0)

Derivatives: Part 9: IITJEE maths tutorial problems practice

January 17, 2021 – 11:24 pm

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

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) $\frac{3a}{x^{2}+a^{2}}$

Cheers,

Nalin Pithwa

Posted in calculus, IITJEE Mains | Comments (0)

Derivatives: part 8: IITJEE mains tutorial problems practice

January 17, 2021 – 3:42 pm

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) $\frac{-m\sin{(2mx)}}{n\sin{(2ny)}}$