JEE Mathematics Differentiation MCQs Set 09

Answer: (b) \( xy_n=(n-1)! \)

Explanation:
1. Check small n: for \( n=1 \), \( y=\log x \), \( y_1=\dfrac1x \), so \( xy_1=1=0! \). For \( n=2 \), \( y=x\log x \), \( y_2=\dfrac1x \), so \( xy_2=1=1! \).
2. In general \( y_n=\dfrac{(n-1)!}{x} \), so \( xy_n=(n-1)! \).

Teacher's Note:
1. Differentiate two or three times to see the pattern and then generalise.
2. Use the standard nth derivatives of sin(ax+b), cos(ax+b), e^(ax) and 1/(ax+b).

Answer: (c) \( p(x)\cdot p'''(x) \)

Explanation:
1. From \( y^2=p \): \( yy'=\dfrac{p'}2 \) and \( y'^2+yy''=\dfrac{p''}2 \).
2. So \( y^3y''=y^2\left(\dfrac{p''}2-y'^2\right)=\dfrac{pp''}2-\dfrac{p'^2}4 \).
3. Differentiate: \( \dfrac{p'p''+pp'''}2-\dfrac{p'p''}2=\dfrac{pp'''}2 \). Twice this is \( p\,p''' \).

Teacher's Note:
1. Find dy/dx first and then differentiate again, using the product or chain rule.
2. Substituting the original expression back into the differential equation is a good check.

Answer: (c) \( -\dfrac{3b}{a^3}\csc^4\theta\cot\theta \)

Explanation:
1. \( \dfrac{dy}{dx}=-\dfrac ba\cot\theta \) and \( \dfrac{d^2y}{dx^2}=-\dfrac b{a^2}\csc^3\theta \).
2. \( \dfrac{d^3y}{dx^3}=\dfrac{3b}{a^2}\csc^3\theta\cot\theta\cdot\dfrac{1}{-a\sin\theta}=-\dfrac{3b}{a^3}\csc^4\theta\cot\theta \).

Teacher's Note:
1. For parametric equations, dy/dx equals (dy/dt) divided by (dx/dt).
2. Differentiate x and y separately with respect to the parameter and take the ratio.

Answer: (c) \( \dfrac{\sin x-e^x}{(\cos x+e^x)^3} \)

Explanation:
1. \( \dfrac{d^2x}{dy^2}=-\dfrac{y''}{y'^3} \) with \( y'=\cos x+e^x \) and \( y''=-\sin x+e^x \).
2. So \( \dfrac{d^2x}{dy^2}=\dfrac{\sin x-e^x}{(\cos x+e^x)^3} \).

Teacher's Note:
1. Differentiate the given relation twice and eliminate the constants.
2. Find dy/dx first and then differentiate again, using the product or chain rule.

Answer: (d) \( -\dfrac1{2at^3} \)

Explanation:
1. \( \dfrac{dy}{dx}=\dfrac{2a}{2at}=\dfrac1t \).
2. \( \dfrac{d^2y}{dx^2}=-\dfrac1{t^2}\cdot\dfrac1{2at}=-\dfrac1{2at^3} \).

Teacher's Note:
1. Differentiate x and y separately with respect to the parameter and take the ratio.
2. Simplify the ratio with trigonometric identities before substituting the given parameter value.

Answer: (d) \( (n-1)! \)

Explanation:
1. Test small cases: \( I_1=\log x+1 \), \( I_0=\log x \), so \( I_1-I_0=1=0! \).
2. Next \( I_2=2\log x+3 \), and \( I_2-2I_1=3-2=1=1! \). Then \( I_3=6\log x+11 \), and \( I_3-3I_2=11-9=2=2! \).
3. The pattern is \( I_n-nI_{n-1}=(n-1)! \).

Teacher's Note:
1. Convert products of trigonometric functions to sums before taking the nth derivative.
2. Differentiate two or three times to see the pattern and then generalise.

Answer: (a) \( (1+x^2)y_2+(2x-1)y_1=0 \)

Explanation:
1. \( y_1=\dfrac{e^{\tan^{-1}x}}{1+x^2} \), so \( (1+x^2)y_1=y \).
2. Differentiate: \( 2xy_1+(1+x^2)y_2=y_1 \), i.e. \( (1+x^2)y_2+(2x-1)y_1=0 \).

Teacher's Note:
1. Substituting the original expression back into the differential equation is a good check.
2. Differentiate the given relation twice and eliminate the constants.

Answer: (b) \( \dfrac{d^2u}{dx^2}=2v \)

Explanation:
1. \( \dfrac{du}{dx}=e^x(\sin x+\cos x)=u+v \) and \( \dfrac{dv}{dx}=e^x(\cos x-\sin x)=v-u \).
2. \( \dfrac{d^2u}{dx^2}=\dfrac{du}{dx}+\dfrac{dv}{dx}=(u+v)+(v-u)=2v \), so option (b) holds.
3. Also \( \dfrac{d^2v}{dx^2}=(v-u)-(u+v)=-2u \), so (c) is false, and \( v\dfrac{du}{dx}-u\dfrac{dv}{dx}=u^2+v^2 \), so (a) is false.

Teacher's Note:
1. Differentiate the given relation twice and eliminate the constants.
2. Find dy/dx first and then differentiate again, using the product or chain rule.

Answer: (b) \( \dfrac{|y''|}{\sqrt{(1+y'^2)^3}} \)

Explanation:
1. \( y'=-\dfrac xy \) and \( y''=-\dfrac{y-xy'}{y^2}=-\dfrac{y^2+x^2}{y^3}=-\dfrac{a^2}{y^3} \).
2. \( 1+y'^2=\dfrac{x^2+y^2}{y^2}=\dfrac{a^2}{y^2} \), so \( (1+y'^2)^{3/2}=\dfrac{a^3}{|y|^3} \).
3. \( \dfrac{|y''|}{(1+y'^2)^{3/2}}=\dfrac{a^2/|y|^3}{a^3/|y|^3}=\dfrac1a=k \).

Teacher's Note:
1. Find dy/dx first and then differentiate again, using the product or chain rule.
2. Substituting the original expression back into the differential equation is a good check.

Answer: (a) \( (1+x^2)\dfrac{d^2y}{dx^2}+x\dfrac{dy}{dx}=2 \)

Explanation:
1. \( \dfrac{d}{dx}\log(x+\sqrt{x^2+1})=\dfrac1{\sqrt{x^2+1}} \), so \( y'=\dfrac{2\log(x+\sqrt{x^2+1})}{\sqrt{x^2+1}} \).
2. \( \sqrt{x^2+1}\,y'=2\log(x+\sqrt{x^2+1}) \). Differentiate: \( \sqrt{x^2+1}\,y''+\dfrac{x}{\sqrt{x^2+1}}y'=\dfrac2{\sqrt{x^2+1}} \).
3. Multiply by \( \sqrt{x^2+1} \): \( (1+x^2)y''+xy'=2 \).

Teacher's Note:
1. Differentiate the given relation twice and eliminate the constants.
2. Find dy/dx first and then differentiate again, using the product or chain rule.

Answer: (a) \( 4 \)

Explanation:
1. Since \( \cos^{-1}x=\dfrac\pi2-\sin^{-1}x \), \( y=2(\sin^{-1}x)^2-\pi\sin^{-1}x+\dfrac{\pi^2}4 \).
2. \( \sqrt{1-x^2}\,y'=4\sin^{-1}x-\pi \). Differentiating: \( (1-x^2)y''-xy'=4 \).

Teacher's Note:
1. Differentiate the given relation twice and eliminate the constants.
2. Find dy/dx first and then differentiate again, using the product or chain rule.

Answer: (b) \( 0 \)

Explanation:
1. \( \dfrac{d^6}{dx^6}\sin2x=2^6\sin\left(2x+3\pi\right)=-64\sin2x \).
2. At \( x=\dfrac\pi2 \): \( \sin\pi=0 \), so the value is 0.

Teacher's Note:
1. Differentiate two or three times to see the pattern and then generalise.
2. Use the standard nth derivatives of sin(ax+b), cos(ax+b), e^(ax) and 1/(ax+b).

Answer: (c) \( 2^{n-1}\cos\left(2x+\dfrac{n\pi}2\right) \)

Explanation:
1. \( \cos^2x=\dfrac{1+\cos2x}2 \).
2. For \( n\ge1 \), \( y_n=\dfrac12\cdot2^n\cos\left(2x+\dfrac{n\pi}2\right)=2^{n-1}\cos\left(2x+\dfrac{n\pi}2\right) \).

Teacher's Note:
1. Use the standard nth derivatives of sin(ax+b), cos(ax+b), e^(ax) and 1/(ax+b).
2. Convert products of trigonometric functions to sums before taking the nth derivative.

Answer: (c) \( (-1)^nn!\left[\dfrac1{(x+2)^{n+1}}-\dfrac1{(x+3)^{n+1}}\right] \)

Explanation:
1. \( \dfrac1{(x+2)(x+3)}=\dfrac1{x+2}-\dfrac1{x+3} \).
2. Use \( \dfrac{d^n}{dx^n}\dfrac1{x+k}=\dfrac{(-1)^nn!}{(x+k)^{n+1}} \) on each fraction to get option (c).

Teacher's Note:
1. Convert products of trigonometric functions to sums before taking the nth derivative.
2. Differentiate two or three times to see the pattern and then generalise.

Answer: (c) \( 6 x^{2} \)

Explanation:
1. Differentiate one row at a time. Differentiating row 1 gives a determinant with rows 1 and 2 equal (value 0); differentiating row 2 gives rows 2 and 3 equal (value 0).
2. Differentiating row 3 gives \( \begin{vmatrix}x&x^2&x^3\\1&2x&3x^2\\0&0&6\end{vmatrix}=6(2x^2-x^2)=6x^2 \).

Teacher's Note:
1. Differentiate a determinant one row (or column) at a time and add the resulting determinants.
2. A determinant with two identical or proportional rows is zero, which kills many terms.

Answer: (b) \( \begin{vmatrix}f'&g'\\h&k\end{vmatrix}+\begin{vmatrix}f&g\\h'&k'\end{vmatrix} \)

Explanation:
1. \( F=fk-gh \), so \( F'=f'k+fk'-g'h-gh' \).
2. Group as \( (f'k-g'h)+(fk'-gh') \), which equals \( \begin{vmatrix}f'&g'\\h&k\end{vmatrix}+\begin{vmatrix}f&g\\h'&k'\end{vmatrix} \).

Teacher's Note:
1. Expand the determinant first when it is small and the entries are simple.
2. Differentiate a determinant one row (or column) at a time and add the resulting determinants.

Answer: (d) A constant

Explanation:
1. Expand along row 1: \( f(x)=x^3(-6p^3-4p^2)-x^2(p^3-4p)+3x^2(p^2+6p) \).
2. This is a cubic in x, so \( f'''(x) \) is a constant, equal to \( 6(-6p^3-4p^2) \).

Teacher's Note:
1. Differentiate a determinant one row (or column) at a time and add the resulting determinants.
2. A determinant with two identical or proportional rows is zero, which kills many terms.

Answer: (a) \( x \sin{\left(x \right)} \)

Explanation:
1. By the fundamental theorem of calculus, \( \dfrac{d}{dx}\int_0^xg(t)\,dt=g(x) \).
2. So \( f'(x)=x\sin x \).

Teacher's Note:
1. Differentiate under the integral sign carefully and substitute the given point at the end.
2. Convert the integral equation into a differential equation by differentiating both sides.

Answer: (a) \( \frac{32}{9} \)

Explanation:
1. \( x^2F(x)=\int_4^x(4t^2-2F'(t))\,dt \), so \( F(4)=0 \).
2. Differentiate: \( 2xF(x)+x^2F'(x)=4x^2-2F'(x) \).
3. At \( x=4 \): \( 0+16F'(4)=64-2F'(4) \), so \( F'(4)=\dfrac{64}{18}=\dfrac{32}{9} \).

Teacher's Note:
1. Convert the integral equation into a differential equation by differentiating both sides.
2. Use Leibnitz's rule: the derivative of an integral with variable limits is the integrand at the upper limit times its derivative minus the same for the lower limit.

Answer: (a) \( 0 \)

Explanation:
1. By Leibnitz's rule, \( y_n=x\sin\left(x+\dfrac{n\pi}2\right)+n\sin\left(x+\dfrac{(n-1)\pi}2\right) \).
2. At \( x=0 \): \( y_{15}(0)=15\sin\dfrac{14\pi}2=15\sin7\pi=0 \).

Teacher's Note:
1. Convert products of trigonometric functions to sums before taking the nth derivative.
2. Differentiate two or three times to see the pattern and then generalise.

Answer: (c) \( (x+n)e^x \)

Explanation:
1. By Leibnitz's rule, \( y_n=xe^x+ne^x \).
2. So \( y_n=(x+n)e^x \).

Teacher's Note:
1. Differentiate two or three times to see the pattern and then generalise.
2. Use the standard nth derivatives of sin(ax+b), cos(ax+b), e^(ax) and 1/(ax+b).

Answer: (b) \( x^2 \)

Explanation:
1. We need \( f' \) to be additive, i.e. \( f'(x)=kx \) (linear through the origin).
2. For \( f=x^2 \): \( f'(x)=2x \) and \( 2(a+b)=2a+2b \). For \( f=x \), \( f'=1 \) and \( 1\ne2 \); for \( x^3 \), \( x^4 \), \( f' \) is not additive.

Teacher's Note:
1. Treat f as an unknown function and apply the chain rule symbolically.
2. Substitute the point only after you have found the derivative expression.

Answer: (c) \( \dfrac1x\left(\dfrac a{a-x}+\dfrac b{b-x}+\dfrac c{c-x}\right) \)

Explanation:
1. Combine the terms stage by stage: \( \dfrac{c}{x-c}+1=\dfrac x{x-c} \). Then \( \dfrac{bx}{(x-b)(x-c)}+\dfrac x{x-c}=\dfrac{x\cdot x}{(x-b)(x-c)}=\dfrac{x^2}{(x-b)(x-c)} \).
2. Next \( \dfrac{ax^2}{(x-a)(x-b)(x-c)}+\dfrac{x^2}{(x-b)(x-c)}=\dfrac{x^3}{(x-a)(x-b)(x-c)} \). So \( y=\dfrac{x^3}{(x-a)(x-b)(x-c)} \).
3. \( \dfrac{y'}y=\dfrac3x-\dfrac1{x-a}-\dfrac1{x-b}-\dfrac1{x-c} \). Since \( \dfrac1x-\dfrac1{x-a}=\dfrac{a}{x(a-x)} \), this equals \( \dfrac1x\left[\dfrac a{a-x}+\dfrac b{b-x}+\dfrac c{c-x}\right] \).

Teacher's Note:
1. For a quotient, either use the quotient rule or rewrite it as a product with a negative power.
2. Take logarithms first when the expression has many factors in numerator and denominator.

Answer: (a) A.P.

Explanation:
1. Let \( f(x)=px^2+qx+r \). \( f(1)=f(-1) \) gives \( q=0 \), so \( f'(x)=2px \).
2. \( f'(a_i)=2pa_i \) are constant multiples of an A.P., hence they are in A.P.

Teacher's Note:
1. Use the given values of the function and its derivative at the point after applying the rule.
2. Treat f as an unknown function and apply the chain rule symbolically.