JEE Mathematics Differentiation MCQs Set 02

Answer: (c) 0

Explanation:
1. \( f^{(k)}(1)=n(n-1)\cdots(n-k+1) \), so \( \dfrac{f^{(k)}(1)}{k!}={}^nC_k \).
2. The sum is \( {}^nC_0-{}^nC_1+{}^nC_2-\cdots+(-1)^n\,{}^nC_n \).
3. This equals \( (1-1)^n=0 \).

Teacher's Note:
1. Recognise the binomial coefficients in the derivatives of \( x^n \).
2. The alternating sum of binomial coefficients is zero.

Answer: (b) 0

Explanation:
1. \( \dfrac{\log e-2\log x}{\log e+2\log x}=\dfrac{1-2\log x}{1+2\log x} \), and \( \tan^{-1}\dfrac{1-t}{1+t}=\tan^{-1}1-\tan^{-1}t \) with \( t=2\log x \).
2. \( \tan^{-1}\dfrac{3+t}{1-3t}=\tan^{-1}3+\tan^{-1}t \).
3. The sum is \( \tan^{-1}1+\tan^{-1}3 \), a constant. So every derivative is 0.

Teacher's Note:
1. Use the identity \( \tan^{-1}\frac{a\pm b}{1\mp ab}=\tan^{-1}a\pm\tan^{-1}b \).
2. A constant function has all derivatives equal to zero.

Answer: (b) \( -n^4f(x) \)

Explanation:
1. By De Moivre, \( f(x)=\cos(1+3+5+\cdots+(2n-1))x+i\sin(\cdots)x=\cos n^2x+i\sin n^2x=e^{in^2x} \).
2. \( f'(x)=in^2f(x) \), \( f''(x)=(in^2)^2f(x)=-n^4f(x) \).

Teacher's Note:
1. The sum of the first n odd numbers is \( n^2 \).
2. Write the product as \( e^{in^2x} \) so that differentiation is easy.

Answer: (a) 0

Explanation:
1. \( F'(x) \) is the sum of three determinants, each with one row differentiated.
2. At \( x=a \) each determinant has two equal rows because \( f_r(a)=g_r(a)=h_r(a) \).
3. Each determinant is 0, so \( F'(a)=0 \).

Teacher's Note:
1. A determinant with two identical rows is zero.
2. Differentiate one row at a time to differentiate a determinant.

Answer: (d) independent of p

Explanation:
1. Only the first row depends on x, so \( f'''(x) \) is the determinant with the first row replaced by its third derivatives \( 6,\ \sin\left(x+\tfrac{3\pi}2\right),\ \cos\left(x+\tfrac{3\pi}2\right) \).
2. At \( x=0 \) the first row becomes \( (6,-1,0) \), identical to the second row.
3. The determinant is 0 for every p, so it is independent of p.

Teacher's Note:
1. Differentiating a determinant row by row is easy when other rows are constant.
2. Two equal rows make the determinant vanish.

Answer: (a) \( (9x^2-4x)\log x \)

Explanation:
1. By Leibniz's rule: \( F'(x)=\log(x^3)\cdot3x^2-\log(x^2)\cdot2x \).
2. \( =9x^2\log x-4x\log x \).
3. \( =(9x^2-4x)\log x \).

Teacher's Note:
1. Differentiating an integral with variable limits: upper-limit integrand times its derivative minus the same for the lower limit.
2. Use \( \log x^k=k\log x \).

Answer: (b) \( \{x^2+2nx+n(n-1)\}e^x \)

Explanation:
1. Use Leibniz's theorem with \( u=e^x \) and \( v=x^2 \): \( D^n(uv)=u_nv+nu_{n-1}v'+\dfrac{n(n-1)}{2}u_{n-2}v''+\cdots \).
2. \( =e^x\left[x^2+n\cdot2x+\dfrac{n(n-1)}{2}\cdot2\right] \).
3. \( =\{x^2+2nx+n(n-1)\}e^x \).

Teacher's Note:
1. Choose the factor that becomes zero after a few differentiations (here \( x^2 \)) as the second function.
2. Only three terms of the Leibniz sum survive.

Answer: (b) \( \dfrac{(-1)^{n-1}(n-3)!}{x^{n-2}}\cdot2 \)

Explanation:
1. Take \( u=\log x \), \( v=x^2 \): \( D^n(uv)=u_nx^2+nu_{n-1}\cdot2x+\dfrac{n(n-1)}2u_{n-2}\cdot2 \), with \( u_k=\dfrac{(-1)^{k-1}(k-1)!}{x^k} \).
2. Every term has the factor \( \dfrac{(-1)^{n-1}(n-3)!}{x^{n-2}} \): \( (n-1)(n-2)-2n(n-2)+n(n-1)=2 \).
3. So \( y_n=\dfrac{2(-1)^{n-1}(n-3)!}{x^{n-2}} \) (for \( n\ge3 \)).

Teacher's Note:
1. Check the formula with \( n=4 \): \( y_4=-\dfrac2{x^2} \), which matches direct differentiation.
2. Only three Leibniz terms are needed because the third derivative of \( x^2 \) is zero.

Answer: (c) \( a\in\mathbb R,\ b=0 \)

Explanation:
1. Continuity at 0 needs \( b=0 \).
2. Left derivative at 0: \( 2ax|_{x=0}=0 \). Right derivative: \( 2x|_{x=0}=0 \).
3. They agree for every a, so \( a\in\mathbb R \) and \( b=0 \).

Teacher's Note:
1. Differentiability needs continuity first, then equal left and right derivatives.
2. The coefficient a does not affect the slope at 0.

Answer: (b) \( -3 \)

Explanation:
1. Near \( x=-3 \), \( 2+x<0 \), so \( f(x)=-3(2+x) \).
2. \( f'(x)=-3 \).

Teacher's Note:
1. Remove the modulus using the sign of the expression near the point.
2. The corner of \( |2+x| \) is at \( x=-2 \), not \( -3 \).

Answer: (b) 0

Explanation:
1. For \( 1<x<5 \): \( |x-1|=x-1 \) and \( |x-5|=5-x \).
2. The sum is \( 4 \), a constant.
3. Its derivative is 0.

Teacher's Note:
1. Between the two corners the sum of the two moduli is constant.
2. Always test which interval the point lies in.

Answer: (b) 0

Explanation:
1. For \( 1<x<3 \): \( |x-1|+|x-3|=(x-1)+(3-x)=2 \).
2. This is constant near \( x=2 \).
3. So the derivative is 0.

Teacher's Note:
1. The corners of the function are at \( x=1 \) and \( x=3 \).
2. The point \( x=2 \) lies between them, where the function is smooth.

Answer: (a) \( (-\infty,\infty) \)

Explanation:
1. For \( x\ge0 \): \( f=\dfrac x{1+x} \), \( f'=\dfrac1{(1+x)^2} \). For \( x<0 \): \( f=\dfrac x{1-x} \), \( f'=\dfrac1{(1-x)^2} \).
2. At \( x=0 \) both one-sided derivatives equal 1.
3. So f is differentiable for all real x.

Teacher's Note:
1. Check the one-sided derivatives at the point where the modulus changes.
2. The corner of the modulus can be smoothed out by the quotient.

Answer: (d) none of these

Explanation:
1. \( h'(x)=2ff'+2gg'=2fg+2g(-f)=0 \) (since \( g'=f''=-f \)).
2. So h is constant: \( h(10)=h(5)=11 \).
3. 11 is not among (a), (b), (c), so the answer is none of these.

Teacher's Note:
1. Show that a function is constant by proving its derivative is zero.
2. This is the identity \( \sin^2x+\cos^2x=1 \) in disguise.

Answer: (b) 33

Explanation:
1. \( f'(3)=\displaystyle\lim_{h\to0}\dfrac{f(3+h)-f(3)}h=f(3)\lim_{h\to0}\dfrac{f(h)-1}h \).
2. Since \( f(0)=1 \), the limit is \( f'(0)=11 \).
3. \( f'(3)=3\times11=33 \).

Teacher's Note:
1. For the exponential functional equation, \( f'(x)=f(x)f'(0) \).
2. Always confirm \( f(0)=1 \) first.

Answer: (d) \( \dfrac1v \)

Explanation:
1. Differentiate with respect to t: \( 2=2v\dfrac{dv}{dt} \).
2. \( \dfrac{dv}{dt}=\dfrac1v \).

Teacher's Note:
1. Differentiate implicitly with respect to the stated variable.
2. Here \( v \) is a function of t.

Answer: (c) \( \dfrac{\sqrt{1-y^2}}{1-2y^2} \)

Explanation:
1. \( \dfrac{dx}{dy}=\sqrt{1-y^2}-\dfrac{y^2}{\sqrt{1-y^2}}=\dfrac{1-2y^2}{\sqrt{1-y^2}} \).
2. \( \dfrac{dy}{dx}=\dfrac1{dx/dy} \).
3. \( =\dfrac{\sqrt{1-y^2}}{1-2y^2} \).

Teacher's Note:
1. When x is given as a function of y, take the reciprocal of \( dx/dy \).
2. Combine the two terms over a common denominator first.

Answer: (b) \( -1 \)

Explanation:
1. \( p=\dfrac{81}v \), so \( \dfrac{dp}{dv}=-\dfrac{81}{v^2} \).
2. At \( v=9 \): \( -\dfrac{81}{81}=-1 \).

Teacher's Note:
1. Write one variable explicitly in terms of the other.
2. The derivative of \( 1/v \) is \( -1/v^2 \).

Answer: (d) 2|x|

Explanation:
1. For \( x\ge0 \): \( f=x^2 \), \( f'=2x=2|x| \).
2. For \( x<0 \): \( f=-x^2 \), \( f'=-2x=2|x| \).
3. So \( f'(x)=2|x| \) everywhere.

Teacher's Note:
1. Write the modulus function piecewise and differentiate each part.
2. The two pieces join smoothly at 0.

Answer: (d) \( F'[f\{\phi(x)\}]\,f'\{\phi(x)\}\,\phi'(x) \)

Explanation:
1. Apply the chain rule layer by layer.
2. Outer: \( F' \) of the inside. Middle: \( f' \) of \( \phi(x) \).
3. Inner: \( \phi'(x) \). The product is option (d).

Teacher's Note:
1. A chain of three functions gives a product of three derivatives.
2. Each derivative is evaluated at the function inside it.

Answer: (a) \( 4x\cos(2x^2) \)

Explanation:
1. By the chain rule: \( \cos(2x^2)\cdot\dfrac d{dx}(2x^2) \).
2. \( =\cos(2x^2)\cdot4x \).

Teacher's Note:
1. \( \sin2x^2 \) means \( \sin(2x^2) \), not \( \sin^2 \).
2. Differentiate the inner function \( 2x^2 \) last.

Answer: (c) \( \dfrac\pi{180}\sec x^\circ\tan x^\circ \)

Explanation:
1. \( x^\circ=\dfrac{\pi x}{180} \) radians, so \( y=\sec\dfrac{\pi x}{180} \).
2. \( \dfrac{dy}{dx}=\sec\dfrac{\pi x}{180}\tan\dfrac{\pi x}{180}\cdot\dfrac\pi{180} \).
3. \( =\dfrac\pi{180}\sec x^\circ\tan x^\circ \).

Teacher's Note:
1. Derivative formulas for trigonometric functions hold only when the angle is in radians.
2. Converting degrees to radians gives the factor \( \pi/180 \).

Answer: (c) e

Explanation:
1. Differentiate: \( \cos y\,y'+e^{-x\cos y}\left(-\cos y+x\sin y\,y'\right)=0 \).
2. At \( (1,\pi) \): \( \cos\pi=-1,\ \sin\pi=0,\ e^{-x\cos y}=e \).
3. \( -y'+e(1+0)=0 \), so \( y'=e \).

Teacher's Note:
1. Substitute the point only after differentiating.
2. Check that the point satisfies the given equation: \( 0+e=e \).

Answer: (d) constant

Explanation:
1. \( y'=a\cos x-b\sin x \).
2. \( y^2+y'^2=(a\sin x+b\cos x)^2+(a\cos x-b\sin x)^2 \).
3. \( =a^2+b^2 \), a constant.

Teacher's Note:
1. Expand and use \( \sin^2x+\cos^2x=1 \).
2. The cross terms cancel.

Answer: (c) \( 4x(1-x^2)\sin(1-x^2)^2 \)

Explanation:
1. \( \dfrac d{dx}\cos u=-\sin u\cdot u' \) with \( u=(1-x^2)^2 \).
2. \( u'=2(1-x^2)(-2x)=-4x(1-x^2) \).
3. The derivative is \( 4x(1-x^2)\sin(1-x^2)^2 \).

Teacher's Note:
1. Handle two nested powers carefully: \( (1-x^2)^2 \) is the angle.
2. The signs from \( -\sin \) and \( -2x \) combine to give a plus.

Answer: (d) 0

Explanation:
1. \( y'=-\sin(\sin x^2)\cdot\cos x^2\cdot2x \).
2. At \( x^2=\dfrac\pi2 \): \( \cos x^2=0 \).
3. So \( y'=0 \).

Teacher's Note:
1. One factor of the chain-rule product is zero at the given point.
2. Check such factors before computing the rest.