Q: The derivative of \tan^{-1}\left(\dfrac{\sin x + \cos x}{\cos x - \sin x}\right), with respect to \dfrac{x}{2}, where x\in\left(0,\dfrac{\pi}{2}\right) is

2. \dfrac{\sin x+\cos x}{\cos x-\sin x}=\tan\left(\dfrac{\pi}{4}+x\right), so the expression =\dfrac{\pi}{4}+x. Its derivative w.r.t. x is 1; w.r.t. \dfrac{x}{2} it is 1\times2=2.

Q: If x = 3\tan t and y = 3\sec t, then the value of \dfrac{d^2y}{dx^2} at t = \dfrac{\pi}{4}, is

\dfrac{1}{6\sqrt2}. \dfrac{dy}{dx}=\dfrac{3\sec t\tan t}{3\sec^2t}=\sin t. \dfrac{d^2y}{dx^2}=\dfrac{\cos t}{3\sec^2t}=\dfrac{\cos^3t}{3}. At t=\pi/4: \cos^3(\pi/4)=\dfrac{1}{2\sqrt2}, so \dfrac{d^2y}{dx^2}=\dfrac{1}{6\sqrt2}.

Question 1

If {}^{20}C_1 + (2^2)\,{}^{20}C_2 + (3^2)\,{}^{20}C_3 + \ldots + (20^2)\,{}^{20}C_{20} = A(2^B), then the ordered pair (A, B) is equal to

Accepted Answer

(420, 18). Using \sum_{r=1}^{n} r^2\,{}^{n}C_r = n(n+1)2^{n-2} (obtained by differentiating (1+x)^n twice), with n=20: sum = 20	imes21	imes2^{18} = 420	imes2^{18}. So A=420, B=18.

Question 2

The derivative of 	an^{-1}\left(\dfrac{\sin x + \cos x}{\cos x - \sin x}ight), with respect to \dfrac{x}{2}, where x\in\left(0,\dfrac{\pi}{2}ight) is

Accepted Answer

2. \dfrac{\sin x+\cos x}{\cos x-\sin x}=	an\left(\dfrac{\pi}{4}+xight), so the expression =\dfrac{\pi}{4}+x. Its derivative w.r.t. x is 1; w.r.t. \dfrac{x}{2} it is 1	imes2=2.

Question 3

If e^y + xy = e, the ordered pair \left(\dfrac{dy}{dx}, \dfrac{d^2y}{dx^2}\right) at x=0 is equal to

Accepted Answer

\left(-\dfrac{1}{e}, \dfrac{1}{e^2}\right). At x=0: e^y=e\Rightarrow y=1. Differentiating: e^y y'+y+xy'=0\Rightarrow e y'+1=0\Rightarrow y'=-\dfrac{1}{e}. Differentiating again: e^y(y')^2+e^yy''+2y'+xy''=0; at x=0: \dfrac{1}{e}+ey''-\dfrac{2}{e}=0\Rightarrow y''=\dfrac{1}{e^2}.

Question 4

If f(1)=1, f'(1)=3, then the derivative of f(f(f(x))) + (f(x))^2 at x=1 is

Accepted Answer

33. g(x)=f(f(f(x)))+(f(x))^2, so g'(x)=f'(f(f(x)))f'(f(x))f'(x)+2f(x)f'(x). At x=1: f(1)=1 throughout, so g'(1)=3\cdot3\cdot3+2(1)(3)=27+6=33.

Question 5

If 2y = \left[\cot^{-1}\left(\dfrac{\sqrt{3}\cos x + \sin x}{\cos x - \sqrt{3}\sin x}\right)\right]^2, x\in\left(0,\dfrac{\pi}{2}\right), then \dfrac{dy}{dx} is equal to

Accepted Answer

\dfrac{\pi}{3} - x. \dfrac{\sqrt{3}\cos x+\sin x}{\cos x-\sqrt{3}\sin x}=	an\left(\dfrac{\pi}{3}+xight), so \cot^{-1}(\cdot)=\dfrac{\pi}{6}-x. Thus 2y=\left(\dfrac{\pi}{6}-xight)^2, and \dfrac{dy}{dx}=-\left(\dfrac{\pi}{6}-xight)=x-\dfrac{\pi}{6}.

Question 6

For x
eq1, if (2x)^{2y} = 4e^{2x-2y}, then (1+\log_e 2x)^2\dfrac{dy}{dx} is equal to

Accepted Answer

\dfrac{x\log_e 2x - \log_e 2}{x}. Taking log: 2y\ln(2x)=\ln4+2x-2y \Rightarrow y(\ln2x+1)=\ln2+x. Differentiating and simplifying gives (\ln2x+1)^2 y' = \dfrac{x\log_e2x-\log_e2}{x}.

Question 7

If x\log_e(\log_e x) - x^2 + y^2 = 4\ (y>0), then \dfrac{dy}{dx} at x=e is equal to

Accepted Answer

\dfrac{2e-1}{2\sqrt{4+e^2}}. Differentiating: \log_e(\log_ex)+\dfrac{1}{\log_ex}-2x+2yy'=0. At x=e: 0+1-2e+2yy'=0\Rightarrow y'=\dfrac{2e-1}{2y}. From the original equation, y=\sqrt{4+e^2}, giving y'=\dfrac{2e-1}{2\sqrt{4+e^2}}.

Question 8

Let f:R	o R be a function such that f(x) = x^3 + x^2 f'(1) + x f''(2) + f'''(3), x\in R. Then f(2) equals

Accepted Answer

-2. Let a=f'(1), b=f''(2), c=f'''(3). Then f'(x)=3x^2+2ax+b, f''(x)=6x+2a, f'''(x)=6. Solving a+2a+b=a i.e. a+b=-3, b=12+2a, c=6 gives a=-5,b=2. So f(x)=x^3-5x^2+2x+6, and f(2)=8-20+4+6=-2.

Question 9

If x = 3	an t and y = 3\sec t, then the value of \dfrac{d^2y}{dx^2} at t = \dfrac{\pi}{4}, is

Accepted Answer

\dfrac{1}{6\sqrt2}. \dfrac{dy}{dx}=\dfrac{3\sec t	an t}{3\sec^2t}=\sin t. \dfrac{d^2y}{dx^2}=\dfrac{\cos t}{3\sec^2t}=\dfrac{\cos^3t}{3}. At t=\pi/4: \cos^3(\pi/4)=\dfrac{1}{2\sqrt2}, so \dfrac{d^2y}{dx^2}=\dfrac{1}{6\sqrt2}.

Question 10

For x\in\left(0,\dfrac{1}{4}ight), if the derivative of 	an^{-1}\left(\dfrac{6x\sqrt{x}}{1-9x^3}ight) is \sqrt{x}\cdot g(x), then g(x) equals

Accepted Answer

\dfrac{9}{1+9x^3}. Let u=3x^{3/2}, so 6x\sqrt{x}=2u and 1-9x^3=1-u^2. Then 	an^{-1}\left(\dfrac{2u}{1-u^2}ight)=2	an^{-1}u. Its derivative is \dfrac{2}{1+u^2}\cdot\dfrac{du}{dx}=\dfrac{2}{1+9x^3}\cdot\dfrac{9}{2}\sqrt{x}=\sqrt{x}\cdot\dfrac{9}{1+9x^3}.

Question 11

If g is the inverse of a function f and f'(x)=\dfrac{1}{1+x^5}, then g'(x) is equal to

Accepted Answer

1+\{g(x)\}^5. Since g is the inverse of f, g'(x)=\dfrac{1}{f'(g(x))}=1+\{g(x)\}^5.

Question 12

If y = \sec(	an^{-1}x), then \dfrac{dy}{dx} at x=1 is equal to

Accepted Answer

\dfrac{1}{\sqrt2}. y=\sec(	an^{-1}x)=\sqrt{1+x^2}, so \dfrac{dy}{dx}=\dfrac{x}{\sqrt{1+x^2}}. At x=1: \dfrac{1}{\sqrt2}.

Question 13

Let g(x) = \log f(x), where f(x) is a twice differentiable positive function on (0,\infty) such that f(x+1)=xf(x). Then, for N=1,2,3,\ldots, g''\left(N+\dfrac{1}{2}\right) - g''\left(\dfrac{1}{2}\right) is equal to

Accepted Answer

-4\left[1+\dfrac{1}{9}+\dfrac{1}{25}+\ldots+\dfrac{1}{(2N-1)^2}\right]. From f(x+1)=xf(x), g(x+1)-g(x)=\ln x. Differentiating twice, g''(x+1)-g''(x)=-\dfrac{1}{x^2}. Summing this telescoping relation from x=1/2 to N-1/2 gives g''(N+1/2)-g''(1/2)=-4\sum_{k=1}^{N}\dfrac{1}{(2k-1)^2}.

Question 14

\dfrac{d^2x}{dy^2} equals

Accepted Answer

-\left(\dfrac{d^2y}{dx^2}\right)\left(\dfrac{dy}{dx}\right)^{-3}. Since \dfrac{dx}{dy}=\left(\dfrac{dy}{dx}\right)^{-1}, differentiating w.r.t. y using the chain rule gives \dfrac{d^2x}{dy^2}=-\dfrac{d^2y}{dx^2}\left(\dfrac{dy}{dx}\right)^{-3}.

Question 15

If f''(x) = -f(x), where f(x) is a continuous double differentiable function, and g(x)=f'(x). If F(x)=\left[f\left(\dfrac{x}{2}\right)\right]^2+\left[g\left(\dfrac{x}{2}\right)\right]^2 and F(5)=5, then F(10) is

Accepted Answer

5. F'(x)=2f\left(	frac{x}{2}ight)\cdot	frac{1}{2}g\left(	frac{x}{2}ight)+2g\left(	frac{x}{2}ight)\cdot	frac{1}{2}\left(-f\left(	frac{x}{2}ight)ight)=0, so F is constant; hence F(10)=F(5)=5.

Question 16

Let f be a twice differentiable function satisfying f(1)=1, f(2)=4, f(3)=9, then

Accepted Answer

there exists at least one x\in(1,3) such that f''(x)=2. By repeated application of the Mean Value Theorem to f on [1,2] and [2,3], and then to f', there exists x\in(1,3) with f''(x)=2; the other statements need not hold in general.

Question 17

If y is a function of x and \log(x+y)=2xy, then the value of y'(0) is

Accepted Answer

1. At x=0: \log y=0\Rightarrow y=1. Differentiating: \dfrac{1+y'}{x+y}=2y+2xy'. At x=0,y=1: 1+y'=2\Rightarrow y'=1.

Question 18

If x^2+y^2=1, then

Accepted Answer

yy''+(y')^2+1=0. From x^2+y^2=1, y'=-\dfrac{x}{y}, and differentiating again gives y''=-\dfrac{1}{y^3} (using x^2+y^2=1). Then yy''+(y')^2=-\dfrac{1}{y^2}+\dfrac{x^2}{y^2}=-\dfrac{y^2}{y^2}=-1, so yy''+(y')^2+1=0.

Question 19

Let f(x) = \begin{vmatrix} x^3 & \sin x & \cos x \ 6 & -1 & 0 \ p & p^2 & p^3 \end{vmatrix}, where p is a constant. Then, \dfrac{d^3}{dx^3}f(x) at x=0 is

Accepted Answer

independent of p. Differentiating the determinant thrice only affects the first row (the other rows are constant), and the third derivative of the first row, (0,-\sin x,-\cos x) differentiated once more gives (0,-\cos x,\sin x); at x=0 this row is (0,-1,0), which is proportional to the second row, making the determinant (and hence its value at x=0) independent of p.

Question 20

If y^2 = P(x) is a polynomial of degree 3, then 2\dfrac{d}{dx}\left[y^3\dfrac{d^2y}{dx^2}\right] equals

Accepted Answer

P(x)P'''(x). From y^2=P(x): 2yy'=P', and differentiating twice more, 2y^3y'' can be shown (using y^2=P and its derivatives) to reduce to a multiple of P(x)P'''(x); differentiating once more, 2\dfrac{d}{dx}[y^3y'']=P(x)P'''(x).

Differentiation — JEE Main Maths PYQ MCQs with Solutions

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