Practice Start Test

Differentiation

JEE Mathematics · 66 questions · Page 1 of 7 · Click an option or "Show Solution" to reveal answer

Q1
If y=(x+1+x2)n,y = {\left( {x + \sqrt {1 + {x^2}} } \right)^n}, then (1+x2)d2ydx2+xdydx\left( {1 + {x^2}} \right){{{d^2}y} \over {d{x^2}}} + x{{dy} \over {dx}} is
A n2y{n^2}y
B −n2y-{n^2}y
C −y-y
D 2x2y2{x^2}y
Correct Answer
Option A
Solution
y=(x+1+x2)ny = {\left( {x + \sqrt {1 + {x^2}} } \right)^n}
dydx=n(x+1+x2)n−1{{dy} \over {dx}} = n{\left( {x + \sqrt {1 + {x^2}} } \right)^{n - 1}}
                  (1+12(1+x2)−1/2.2x);\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {1 + {1 \over 2}{{\left( {1 + {x^2}} \right)}^{ - 1/2}}.2x} \right);
dydx=n(x+1+x2)n−1{{dy} \over {dx}} = n{\left( {x + \sqrt {1 + {x^2}} } \right)^{n - 1}}
                  (1+x2+x)1+x2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{\left( {\sqrt {1 + {x^2}} + x} \right)} \over {\sqrt {1 + {x^2}} }}
=n(1+x2+x)n1+x2= {{n{{\left( {\sqrt {1 + {x^2}} + x} \right)}^n}} \over {\sqrt {1 + {x^2}} }}

or

1+x2dydx=ny\sqrt {1 + {x^2}} {{dy} \over {dx}} = ny

or

1+x2y1=ny\sqrt {1 + {x^2}} {y_1} = ny
(y1=dydx)\left( {{y_1} = {{dy} \over {dx}}} \right)

Squaring,

(1+x2)y12=n2y2\left( {1 + {x^2}} \right){y_1}^2 = {n^2}{y^2}

Differentiating,

(1+x2)2y1y2+y12.2x\left( {1 + {x^2}} \right)2{y_1}{y_2} + {y_1}^2.2x
=n2.2yy1= {n^2}.2y{y_1}

or

(1+x2)y2+xy1=n2y\left( {1 + {x^2}} \right){y_2} + x{y_1} = {n^2}y
Q2
If f(x)=xn,f\left( x \right) = {x^n}, then the value of f(1)−f′(1)1!+f′′(1)2!−f′′′(1)3!+..........(−1)nfn(1)n!f\left( 1 \right) - {{f'\left( 1 \right)} \over {1!}} + {{f''\left( 1 \right)} \over {2!}} - {{f'''\left( 1 \right)} \over {3!}} + ..........{{{{\left( { - 1} \right)}^n}{f^n}\left( 1 \right)} \over {n!}} is
A 11
B 2n{{2^n}}
C 2n−1{{2^n} - 1}
D 00
Correct Answer
Option D
Solution
f(x)=xn⇒f(1)=1f\left( x \right) = {x^n} \Rightarrow f\left( 1 \right) = 1
f′(x)=nxn−1⇒f′(1)=nf'\left( x \right) = n{x^{n - 1}} \Rightarrow f'\left( 1 \right) = n
f′′(x)=n(n−1)xn−2f''\left( x \right) = n\left( {n - 1} \right){x^{n - 2}}
⇒f′′(1)=n(n−1)\Rightarrow f''\left( 1 \right) = n\left( {n - 1} \right)

∴\therefore

fn(x)=n!{f^n}\left( x \right) = n!
⇒fn(1)=n!\Rightarrow {f^n}\left( 1 \right) = n!
=1−n1!+n(n−1)2!n(n−1)(n−2)3!= 1 - {n \over {1!}} + {{n\left( {n - 1} \right)} \over {2!}}{{n\left( {n - 1} \right)\left( {n - 2} \right)} \over {3!}}
                          +....+(−1)nn!n!\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + .... + {\left( { - 1} \right)^n}{{n!} \over {n!}}
=n C0−n C1+n C2−n C3= {}^n\,{C_0} - {}^n\,{C_1} + {}^n\,{C_2} - {}^n\,{C_3}
                          +......+(−1)n nCn=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + ...... + {\left( { - 1} \right)^n}\,{}^n{C_n} = 0
Q3
Let f(x)f\left( x \right) be a polynomial function of second degree. If f(1)=f(−1)f\left( 1 \right) = f\left( { - 1} \right) and a,b,ca,b,c are in A.P,A.P, then f′(a),f′(b),f′(c)f'\left( a \right),f'\left( b \right),f'\left( c \right) are in
A Arithmetic -Geometric Progression
B A.PA.P
C G.PG.P
D H.PH.P
Correct Answer
Option B
Solution
f(x)=ax2+bx+cf\left( x \right) = a{x^2} + bx + c
f(1)=f(−1)f\left( 1 \right) = f\left( { - 1} \right)
⇒a+b+c=a−b+c\Rightarrow a + b + c = a - b + c

or

b=0b = 0

∴\therefore

f(x)=ax2+cf\left( x \right) = a{x^2} + c

or

f′(x)=2axf'\left( x \right) = 2ax

Now

f′(a);f′(b);f'\left( a \right);f'\left( b \right);

and

f′(c)f'\left( c \right)

are

2a(a);2a(b);2a(c)2a\left( a \right);2a\left( b \right);2a\left( c \right)

i.e.

 2a2, 2ab, 2ac.\,2{a^2},\,2ab,\,2ac.

⇒\Rightarrow If

a,b,ca,b,c

are in

A.P.A.P.

then

f′(a);f′(b)f'\left( a \right);f'\left( b \right)

and

f′(c)f'\left( c \right)

are also in

A.P.A.P.
Q4
If x=ey+ey+ey+.....∞x = {e^{y + {e^y} + {e^{y + .....\infty }}}} , x>0,x > 0, then dydx{{{dy} \over {dx}}} is
A 1+xx{{1 + x} \over x}
B 1x{1 \over x}
C 1−xx{{1 - x} \over x}
D x1+x{x \over {1 + x}}
Correct Answer
Option C
Solution
x=ey+ey+.....∞  ⇒x=ey+x.x = {e^{y + {e^{y + .....\infty }}}}\,\, \Rightarrow x = {e^{y + x}}.

Taking log.

log⁡  x=y+x\log \,\,x = y + x
⇒1x=dydx+1\Rightarrow {1 \over x} = {{dy} \over {dx}} + 1
⇒dydx=1x−1=1−xx\Rightarrow {{dy} \over {dx}} = {1 \over x} - 1 = {{1 - x} \over x}
Q5
If xm.yn=(x+y)m+n,{x^m}.{y^n} = {\left( {x + y} \right)^{m + n}}, then dydx{{{dy} \over {dx}}} is
A yx{y \over x}
B x+yxy{{x + y} \over {xy}}
C xyxy
D xy{x \over y}
Correct Answer
Option A
Solution
xm.yn=(x+y)m+n{x^m}.{y^n} = {\left( {x + y} \right)^{m + n}}
⇒mln⁡x+nln⁡y=(m+n)ln⁡(x+y)\Rightarrow m\ln x + n\ln y = \left( {m + n} \right)\ln \left( {x + y} \right)

Differentiating both sides. ∴\therefore

mx+nydydx=m+nx+y(1+dydx){m \over x} + {n \over y}{{dy} \over {dx}} = {{m + n} \over {x + y}}\left( {1 + {{dy} \over {dx}}} \right)
⇒(mx−m+nx+y)=(m+nx+y−ny)dydx\Rightarrow \left( {{m \over x} - {{m + n} \over {x + y}}} \right) = \left( {{{m + n} \over {x + y}} - {n \over y}} \right){{dy} \over {dx}}
⇒my−nxx(x+y)=(my−nxy(x+y))dydx\Rightarrow {{my - nx} \over {x\left( {x + y} \right)}} = \left( {{{my - nx} \over {y\left( {x + y} \right)}}} \right){{dy} \over {dx}}
⇒dydx=yx\Rightarrow {{dy} \over {dx}} = {y \over x}
Q6
Let yy be an implicit function of xx defined by x2x−2xxcot⁡ y−1=0{x^{2x}} - 2{x^x}\cot \,y - 1 = 0. Then y′(1)y'(1) equals
A 11
B log⁡ 2\log \,2
C −log⁡ 2-\log \,2
D −1-1
Correct Answer
Option D
Solution
x2x−2xx cot⁡ y−1=0{x^{2x}} - 2{x^x}\,\cot \,y - 1 = 0
⇒2 cot⁡ y=xx−x−x\Rightarrow 2\,\cot \,y = {x^x} - {x^{ - x}}
⇒2 cot⁡ y =u−1u\Rightarrow 2\,\cot \,y\, = u - {1 \over u}

where

u=xxu = {x^x}

Differentiating both sides with respect to

x,x,

we get

⇒−2cos⁡ec2ydydx\Rightarrow - 2\cos e{c^2}y{{dy} \over {dx}}
=(1+1u2)dudx= \left( {1 + {1 \over {{u^2}}}} \right){{du} \over {dx}}

where

u=xx⇒log⁡ u=x log⁡ xu = {x^x} \Rightarrow \log \,u = x\,\log \,x
⇒1ududx=1+log⁡ x\Rightarrow {1 \over u}{{du} \over {dx}} = 1 + \log \,x
⇒dudx=xx(1+log⁡ x)\Rightarrow {{du} \over {dx}} = {x^x}\left( {1 + \log \,x} \right)

∴\therefore We get

−2cos⁡ec2ydydx- 2\cos e{c^2}y{{dy} \over {dx}}
=(1+x−2x)xx(1+log⁡ x)= \left( {1 + {x^{ - 2x}}} \right){x^x}\left( {1 + \log \,x} \right)
⇒dydx=(xx+x−x)(1+log⁡x)−2(1+cot⁡2y)      ...(i)\Rightarrow {{dy} \over {dx}} = {{\left( {{x^x} + {x^{ - x}}} \right)\left( {1 + \log x} \right)} \over { - 2\left( {1 + {{\cot }^2}y} \right)}}\,\,\,\,\,\,...\left( i \right)

Now when

x=1,x=1,
x2x−2xx cot⁡ y−1=0,{x^{2x}} - 2{x^x}\,\cot \,y - 1 = 0,

gives

1−2 cot⁡y−1=01 - 2\,\cot y - 1 = 0
⇒  cot⁡y =0\Rightarrow \,\,\cot y\, = 0

∴\therefore From equation

(i),(i),

at

x=1x=1

and

cot⁡ y=0,\cot \,y = 0,

we get

y′(1)=(1+1)(1+0)−2(1+0)=−1y'\left( 1 \right) = {{\left( {1 + 1} \right)\left( {1 + 0} \right)} \over { - 2\left( {1 + 0} \right)}} = - 1
Q7
Let f:(−1,1)→Rf:\left( { - 1,1} \right) \to R be a differentiable function with f(0)=−1f\left( 0 \right) = - 1 and f′(0)=1f'\left( 0 \right) = 1. Let g(x)=[f(2f(x)+2)]2g\left( x \right) = {\left[ {f\left( {2f\left( x \right) + 2} \right)} \right]^2}. Then g′(0)=g'\left( 0 \right) =
A −4-4
B 00
C −2-2
D 44
Correct Answer
Option A
Solution
g′(x)=2(f(2f(x)+2))g'\left( x \right) = 2\left( {f\left( {2f\left( x \right) + 2} \right)} \right)
                           (ddx(f(2f(x)+2)))\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{d \over {dx}}\left( {f\left( {2f\left( x \right) + 2} \right)} \right)} \right)
=2f(2f(x)+2)f′(2f(x))= 2f\left( {2f\left( x \right) + 2} \right)f'\left( {2f\left( x \right)} \right)
                           +2).(2f′(x))\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \left. 2 \right).\left( {2f'\left( x \right)} \right)
⇒g′(0)=2f(2f(0)+2).\Rightarrow g'\left( 0 \right) = 2f\left( {2f\left( 0 \right) + 2} \right).
                           f′(2f(0)+2).2f′(0)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,f'\left( {2f\left( 0 \right) + 2} \right).2f'\left( 0 \right)
=4f(0)(f′(0))2= 4f\left( 0 \right){\left( {f'\left( 0 \right)} \right)^2}
=4(−1)(1)2=−4= 4\left( { - 1} \right){\left( 1 \right)^2} = - 4
Q8
d2xdy2{{{d^2}x} \over {d{y^2}}} equals:
A −(d2ydx2)−1(dydx)−3 - {\left( {{{{d^2}y} \over {d{x^2}}}} \right)^{ - 1}}{\left( {{{dy} \over {dx}}} \right)^{ - 3}}
B (d2ydx2)(dydx)−2{\left( {{{{d^2}y} \over {d{x^2}}}} \right)^{}}{\left( {{{dy} \over {dx}}} \right)^{ - 2}}
C −(d2ydx2)(dydx)−3 - \left( {{{{d^2}y} \over {d{x^2}}}} \right){\left( {{{dy} \over {dx}}} \right)^{ - 3}}
D (d2ydx2)−1{\left( {{{{d^2}y} \over {d{x^2}}}} \right)^{ - 1}}
Correct Answer
Option C
Solution
d2xdy2=ddy(dxdy){{{d^2}x} \over {d{y^2}}} = {d \over {dy}}\left( {{{dx} \over {dy}}} \right)
=ddx(dxdy)dxdy= {d \over {dx}}\left( {{{dx} \over {dy}}} \right){{dx} \over {dy}}
=ddx(1dy/dx)dxdy= {d \over {dx}}\left( {{1 \over {dy/dx}}} \right){{dx} \over {dy}}
=−1(dydx)2.d2ydx2.1dydx= - {1 \over {{{\left( {{{dy} \over {dx}}} \right)}^2}}}.{{{d^2}y} \over {d{x^2}}}.{1 \over {{{dy} \over {dx}}}}
=−1(dydx)3d2ydx2= - {1 \over {{{\left( {{{dy} \over {dx}}} \right)}^3}}}{{{d^2}y} \over {d{x^2}}}
Q9
If y=sec⁡(tan⁡−1x),y = \sec \left( {{{\tan }^{ - 1}}x} \right), then dydx{{{dy} \over {dx}}} at x=1x=1 is equal to :
A 12{1 \over {\sqrt 2 }}
B 12{1 \over 2}
C 11
D 2\sqrt 2
Correct Answer
Option A
Solution

Let

y=sec⁡(tan⁡−1x)y = \sec \left( {{{\tan }^{ - 1}}x} \right)

and

tan⁡−1  x=θ.{\tan ^{ - 1}}\,\,x = \theta .
⇒x=tan⁡θ\Rightarrow x = \tan \theta

Thus, we have

y=sec⁡ θy = \sec \,\theta
⇒y=1+x2\Rightarrow y = \sqrt {1 + {x^2}}
(  \left( {\,\,} \right.

As

      sec⁡2θ=1+tan⁡2θ\,\,\,\,\,\,{\sec ^2}\theta = 1 + {\tan ^2}\theta
  )\left. {\,\,} \right)
⇒dydx=121+x2.2x\Rightarrow {{dy} \over {dx}} = {1 \over {2\sqrt {1 + {x^2}} }}.2x

At

x=1,  dydx=12.x = 1,\,\,{{dy} \over {dx}} = {1 \over {\sqrt 2 }}.
Q10
If gg is the inverse of a function ff and f′(x)=11+x5,f'\left( x \right) = {1 \over {1 + {x^5}}}, then g′(x)g'\left( x \right) is equal to:
A 11+{g(x)}5{1 \over {1 + {{\left\{ {g\left( x \right)} \right\}}^5}}}
B 1+{g(x)}51 + {\left\{ {g\left( x \right)} \right\}^5}
C 1+x51 + {x^5}
D 5x45{x^4}
Correct Answer
Option B
Solution

Since

f(x)f(x)

and

g(x)g(x)

are inverse of each other ∴\therefore

g′(f(x))=1f′(x)g'\left( {f\left( x \right)} \right) = {1 \over {f'\left( x \right)}}
⇒g′(f(x))=1+x5\Rightarrow g'\left( {f\left( x \right)} \right) = 1 + {x^5}
( \left( \, \right.

As

 f′(x)=11+x5\,f'\left( x \right) = {1 \over {1 + {x^5}}}
 )\left. \, \right)

Here

x=g(y)x=g(y)

∴\therefore

g′(y)=1+{g(y)}g'\left( y \right) = 1 + \left\{ {g\left( y \right)} \right\}
⇒g′(x)=1+{g(x)}\Rightarrow g'\left( x \right) = 1 + \left\{ {g\left( x \right)} \right\}
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