Mathematics Hothouse

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.

By Nalin Pithwa | Posted in miscellaneous | Comments (0)

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.

By Nalin Pithwa | Posted in careers in mathematics, mathematicians, miscellaneous, motivational stuff | Comments (0)

STEM for Britain

January 19, 2021 – 2:30 pm

Home

By Nalin Pithwa | Posted in careers in mathematics, miscellaneous, motivational stuff | 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

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

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

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) $(a^{2}+b^{2})e^{ax}\cos{(bx)}$

Cheers,

Nalin Pithwa.

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

Derivatives: part 7: IITJEE tutorial problems practice

January 7, 2021 – 8:39 pm

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) $\sin{(\sqrt{x})} \times \sin{(\sqrt{a})}$

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) $\frac{1}{x\log{x}} - \frac{\log{a}}{x(\log{x})^{2}}$

Cheers,

Nalin Pithwa.

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

Derivatives: part 6: IITJEE tutorial practice problems

January 6, 2021 – 1:41 am

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

Regards,

Nalin Pithwa

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

Derivatives: part 5: IITJEE maths tutorial problems for practice

November 14, 2020 – 5:09 pm

Problem 1:

The derivative of $arcsec (\frac{1}{1-2x^{2}})$ w.r.t. $\sqrt{1-x^{2}}$ at $x=\frac{1}{2}$ is

(a) 2 (b) -4 (c) 1 (d) -2

Problem 2:

If $y = \sin{\sin{x}}$ and $\frac{d^{2}y}{dx^{2}} + \frac{dy}{dx} \tan{x} + f(x)=0$ , then $f(x) =$

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

Problem 3:

If $f(x) = \log_{a}{\log_{a}{x}}$ , then $f^{'}(x)$ is

(a) $\frac{\log_{a}{e}}{x \log_{e}{x}}$ (b) $\frac{\log_{e}{a}}{x}$ (c) $\frac{\log_{e}{a}}{x\log_{a}{x}}$ (d) $\frac{x}{\log_{e}{a}}$

Problem 4:

If $y=\log {\tan{\frac{x}{2}}} + \arcsin{\cos{x}}$ , then $\frac{dy}{dx}$ is

(a) $cosec (x) -1$ (b) $cosec (x) +1$ (c) $cosec (x)$ (d) x

Problem 5:

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

(a) $\frac{y}{x}$ (b) $\frac{x}{y}$ (c) $\frac{y(x\log{y}-y)}{x(y\log{x}-x)}$ (d) $\frac{x \log{y}}{y \log{x}}$

Problem 6:

Let f, g, h and k be differentiable in $(a,b)$ , if F is defined as $F(x) = \left | \begin{array}{cc} f(x) & g(x) \\ h(x) & k(x) \end{array} \right |$ for all a, b, then $F^{'}$ is given by:

(i) $\left | \begin{array}{cc} f & g \\ h & k \end{array} \right| + \left | \begin{array}{cc}f & g \\ h^{'} & k \end{array} \right |$

(ii) $\left | \begin{array}{cc}f & g^{'} \\ h & k^{'} \end{array}\right | + \left | \begin{array}{cc} f^{'} & g \\ h & k^{'} \end{array} \right |$

(iii) $\left | \begin{array}{cc}f^{'} & g^{'} \\ h & k \end{array} \right | + \left | \begin{array}{cc}f & g \\ h^{'} & h^{'} \end{array} \right |$

(iv) $\left | \begin{array}{cc}f & g \\ h^{'} & k^{'} \end{array} \right | + \left | \begin{array}{cc}f^{'} & g \\h & k \end{array} \right |$

Problem 7:

If $pv=81$ , then $\frac{dp}{dv}$ at $v=9$ is equal to:

(i) 1 (ii) -1 (iii) 2 (iv) 3

Problem 8:

If $x^{2}+y^{2}=1$ , then

(i) $yy^{''}-2(y^{'})^{2}+1=0$ (ii) $yy^{''} - (y^{'})^{2}-1=0$ (iii) $yy^{''} + (y^{'})^{2} + 1 = 0$ (iv) $yy^{''} - 2(y^{'})^{2}-1=0$

Problem 9:

If $y = \arctan{\frac{\sqrt{x}-1}{\sqrt{x}+1}} + \arctan{\frac{\sqrt{x}+1}{\sqrt{x}-1}}$ , then the value of $\frac{dy}{dx}$ will be

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

Problem 10:

Let $f(x) = \left | \begin{array}{ccc} x^{3} & \sin{x} & \cos{x} \\ 0 & -1 & 0 \\ p & p^{2} & p^{3} \end{array} \right |$ , where p is a constant, then $\frac{d^{3}}{dx^{3}}(f(x))$ at $x=0$ is

(a) p (b) $p+p^{2}$ (c) $p+p^{3}$ (d) independent of p

Regards,

Nalin Pithwa

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

Derivatives: part 4: IITJEE maths tutorial problems for practice

November 13, 2020 – 11:16 pm

Problem 1:

Given $x=x(t)$ , $y=y(t)$ , then $\frac{d^{2}y}{dx^{2}}$ is equal to

(a) $\frac{\frac{d^{2}y}{dt^{2}}}{\frac{d^{2}x}{dt^{2}}}$

(b) $\frac{\frac{d^{2}y}{dt^{2}} \times \frac{dx}{dt} - \frac{dy}{dt} \times \frac{d^{2}x}{dt^{2}}}{(\frac{dx}{dt})^{3}}$

(c) $\frac{\frac{dx}{dt} \times \frac{d^{2}y}{dt^{2}} - \frac{d^{2}x}{dt^{2}} \times \frac{dy}{dt}}{(\frac{dx}{dt})^{2}}$

(d) $\frac{1}{\frac{d^{2}x}{dy^{2}}}$

Problem 2:

$\frac{d}{dx}(\arctan{\sec{x}+ \tan{x}})$ is equal to

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

Problem 3:

If $y= \sqrt{x + \sqrt{x + \sqrt{x} + \ldots}}$ , then $\frac{dy}{dx}$ is equal to :

(a) 1 (b) \ $\frac{1}{xy}$ (c) $\frac{1}{2y-x}$ (d) $\frac{1}{2y-1}$

Problem 4:

If $f(x) = \left| \begin{array}{ccc} x & x^{2} & x^{3} \\ 1 & 2x & 3x^{2} \\ 0 & 2 & 6x \end{array} \right|$ , then $f^{'}(x) =$

(a) 12 (b) $6x^{2}$ (c) $6x$ (d) $12x^{2}$

Problem 5:

If $y = (\frac{x^{a}}{x^{b}}) ^{a+b} \times (\frac{x^{b}}{x^{c}})^{b+c} \times (\frac{x^{c}}{x^{a}})^{c+a}$ , then $\frac{dy}{dx}=$

(a) 0 (b) 1 (c) $a+b+c$ (d) abc

Problem 6:

If $y = \arctan{\frac{x-\sqrt{1-x^{2}}}{x+\sqrt{1-x^{2}}}}$ , then $\frac{dy}{dx}$ is equal to

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

Problem 7:

If $x=at^{2}$ , $y=2at$ , then $\frac{d^{2}y}{dx^{2}}=$

(a) $\frac{1}{t^{2}}$ (b) $\frac{1}{2at^{3}}$ (c) $\frac{1}{t^{3}}$ (d) $\frac{-1}{2at^{3}}$

Problem 8:

If $y=ax^{n+1} +bx^{-n}$ , then $x^{2}\frac{d^{2}y}{dx^{2}}=$

(a) $n(n-1)y$ (b) $ny$ (c) $n(n+1)y$ (d) $n^{2}y$

Problem 9:

If $x=t^{2}$ , $y=t^{3}$ , then $\frac{d^{2}y}{dx^{2}}=$

(a) $\frac{3}{2}$ (b) $\frac{3}{4t}$ (c) $\frac{3}{2t}$ (d) 0

Problem 10:

If $y=a+bx^{2}$ , a, b arbitrary constants, then

(a) $\frac{d^{2}}{dx^{2}} = 2xy$ (b) $x \frac{d^{2}y}{dx^{2}} - \frac{dy}{dx} + y=0$ (c) $x \frac{d^{2}y}{dx^{2}} = \frac{dy}{dx}$ (d) $x \frac{d^{2}y}{dx^{2}} = 2xy$