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

By 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.

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 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}}}

(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}}

Regards,

Nalin Pithwa.

By 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})}

(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}}

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)}}

(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)}

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})}

(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}}

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)