Monthly Archives: November 2020

Derivatives: part 4: IITJEE maths tutorial problems for practice

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


Nalin Pithwa

How to find square root of a binomial quadratic surd

Assume \sqrt{a+ \sqrt{b} + \sqrt{c} + \sqrt{d}}=\sqrt{x} + \sqrt{y} + \sqrt{z};

Hence, a+\sqrt{b} + \sqrt{c} + \sqrt{d} = x+y+z+ 2\sqrt{xy} + 2\sqrt{yz}+ 2\sqrt{zx}

If then, 2\sqrt{xy}=\sqrt{b}, 2\sqrt{yz}=\sqrt{c}, 2\sqrt{zx}=\sqrt{d},

And, if simultaneously, the values of x, y, z thus found satisfy x+y+z=a, we shall have obtained the required root.


Find the square root of 21-4\sqrt{5}+5\sqrt{3}-4\sqrt{15}.


Clearly, we can’t have anything like

21--4\sqrt{5}+8\sqrt{3}-4\sqrt{15}=\sqrt{x} + \sqrt{y} +\sqrt{z}

We will have to try the following options:

21-4\sqrt{5}+8\sqrt{3}-4\sqrt{15}=\sqrt{x} - \sqrt{y} - \sqrt{z}



Only the last option will work as we now show:

So, once again, assume that \sqrt{21-4\sqrt{5}+8\sqrt{3}-4\sqrt{15}}=\sqrt{x}+\sqrt{y}-\sqrt{z}

Hence, 21-4\sqrt{5}+8\sqrt{3}-4\sqrt{15}=z+y+z+2\sqrt{xy}-2\sqrt{yz}+2\sqrt{zx}

Put 2\sqrt{xy}=8\sqrt{3}, 2\sqrt{xz}=4\sqrt{15}, 2\sqrt{yz}=4\sqrt{5};

by multiplication, xyz = 240; that is \sqrt{xyz}=4\sqrt{15}; so it follows that : \sqrt{x}=2\sqrt{3}, \sqrt{y}=2, \sqrt{z}=\sqrt{5}.

And, since, these values satisfy the equation x+y+z=21, the required root is 2\sqrt{3}+2-\sqrt{5}.

That is all, for now,


Nalin Pithwa

IITJEE Mains or Advanced Maths Doubt Solving Tutorials

Any maths questions from any where or any other branded class problems sets.

Contact Nalin Pithwa