IISER Mathematics - Differentiability and Differentiation
Exam Duration: 45 Mins Total Questions : 30
If f and g are differentiable functions in [0,1] satisfying f(0)=2=g(1), g(0)=0 and f(1)=6 then for some c\(\epsilon\)]o,1[
- (a)
2f'(c)=g'(c)
- (b)
2f'(c)=3g'(c)
- (c)
f'(c)=g'(c)
- (d)
f'(c)=2g'(c)
Let h(x)=f(x)-2g(x) as h(0)=h(1)=2
Hence, using Rolle's theorem, we get
h'(c)=0\(\Rightarrow\)f'(c)=2g'(c)
If y=sec(tan-1 x), then \(dy\over dx\) at x=1 is equal to
- (a)
\(1\over2\)
- (b)
1
- (c)
\(\sqrt{2}\)
- (d)
\(1\over\sqrt{2}\)
Given, y=sec(tan-1x)
\(dy\over dx\)=sec(tan-1x)tan(tan-1x).\(1\over1+x^2\)
\({dy\over dx}|_{x=1}=\sqrt{2}\times1\times{1\over2}={1\over\sqrt{2}}\)
Let f(x)=x|x| and g(x)=sin x
Statement I go f isw differentiable at x=0 and its derivative is continuopus at that point.
Statement II go f is twice differentiable atx=0.
- (a)
Statement I is true, Statement II is true; Statement II is the correct explanation for Statement I
- (b)
Statement I is true, Statement II is true; Statement II is not the correct explanation for Statement I
- (c)
Statement I is true, Statement II is false
- (d)
Statement I is false, Statement II is true
Given f(x)=x|x| and get(x) =sinx
g f(x)=sin(x|X|)= \(\begin{cases} -sinx^{ 2 },x<0 \\ sinx^{ 2 },x\ge 0 \end{cases}\)
\(\therefore (gof)x=\begin{cases} -2xcosx^{ 2 },x<0 \\ 2xcosx^{ 2 },x\ge 0 \end{cases}\)
Clearly L(gof)(0) =0=R(gof)(0)
Hence, gof is differentiable at x = 0 and also its derivative is continuous at x = 0
Now \(\therefore (gof)x=\begin{cases} -2xcosx^{ 2 },+4x^{ 2 }sinx^{ 2 },x<0 \\ 2xcosx^{ 2 }-4x^{ 2 }sinx^{ 2 }x\ge 0 \end{cases}\)
\(\because L(gof)^{ '' }(0)=-2\) and R(gof) ''(0)=2
\(\because L(gof)^{ '' }(0)\neq R(gof)^{ '' }(0)\)
Hence gof(0) \(\neq \) R(gof)''(0)
Consider the greatest integer function, defined by f(x)=[x],0\(\le \)x<2.Then,
- (a)
f is derivable at x=1
- (b)
f is not derivable at x=1
- (c)
f is derivable at x=2
- (d)
None of these
The given function is \(f(x)=\begin{cases} 0,0\le x<1 \\ 1,1\le x<2 \end{cases}\)
Now, LHD=\(\underset { x\rightarrow 1^{ - } }{ lim } \frac { f(x)-f(1) }{ x-1 } =\underset { x\rightarrow 1^{ - } }{ lim } \frac { 0-1 }{ x-1 } \) , which does not exist.
\(\because\)LHD does not exist.
\(\Rightarrow\)f is not derivable at x=1
If f(x)=\(\left| \begin{matrix} x+{ a }^{ 2 } & ab & ac \\ ab & x+b^{ 2 } & bc \\ ac & bc & x+c^{ 2 } \end{matrix} \right| \), then f'(x) is
- (a)
3x2+2x(a2+b2+c2)
- (b)
3x+2x2(a2+b2+c2)
- (c)
3x2+2x2(a2-b2-c2)
- (d)
None of the above
Given that f(x) = \(\left| \begin{matrix} x+a^{ 2 } & ab & ac \\ ab & x+b^{ 2 } & bc \\ ac & bc & x+c^{ 2 } \end{matrix} \right| \)
So f(x)= \(\left| \begin{matrix} 1 & 0 & 0 \\ ab & x+b^{ 2 } & bc \\ ac & bc & x+c^{ 2 } \end{matrix} \right| +\left| \begin{matrix} x+a^{ 2 } & ab & ac \\ 0 & 1 & 0 \\ ac & bc & x+c^{ 2 } \end{matrix} \right| +\left| \begin{matrix} x+a^{ 2 } & ab & ac \\ ab & x+b^{ 2 } & bc \\ 0 & 0 & 1 \end{matrix} \right| \)
= [(x+b2)(x+c2) -b2c2]+[(x+a2)(x+c2)-a2b2]+[(x+a2)(x+b2)-a2b2]
= [x2+b2c2+x(b2+c2)-b2c2]+[x2+c2+a2+c2)-a2b2] [x2+a2b2+x(a2+b2)-a2b2]
=x2+x(b2+c2)+x2+x(a2+c2)+x2+x(a2+b2)
=3x2+2x(a2+b2+c2)
If f is a real - valued differentiable function satisfying \(\left| f(x)-f(y) \right| \le (x-y)^{ 2 },x,y\in R\) and f(0) =0, then f(1) equals
- (a)
1
- (b)
2
- (c)
0
- (d)
-1
since, \(=\underset { x\rightarrow 2 }{ lim } \frac { \left| f(x)-f(y) \right| }{ \left| x-y \right| } \le \underset { x\rightarrow y }{ lim } \left| x-y \right| \)
\(\Rightarrow \left| f'(y) \right| \le 0\Rightarrow f'(y)=0\Rightarrow f(y)=\) constant
As \(f(0)=0\Rightarrow f(y)=0\Rightarrow f(1)=0\)
If sec \(\left( \frac { { x }^{ 2 }-2x }{ { x }^{ 2 }+1 } \right) =y,\) then \(\frac { dy }{ dx } \) is equal to
- (a)
\(\frac { { y }^{ 2 } }{ x^{ 2 } } \)
- (b)
\(\frac { 2y\sqrt { { y }^{ 2 }-1 } ({ x }^{ 2 }+x-1) }{ { (x }^{ 2 }+{ 1 })^{ 2 } } \)
- (c)
\(,\frac { { (x }^{ 2 }+x-1) }{ y\sqrt { { y }^{ 2 }-1 } } \)
- (d)
\(\frac { { x }^{ 2 }-{ y }^{ 2 } }{ { x }^{ 2 }+{ y }^{ 2 } } \)
\(y=sec\left( \frac { { x }^{ 2 }-2x }{ { x }^{ 2 }+1 } \right) \)
\(\Rightarrow \frac { dy }{ dx } =sec\left( \frac { { x }^{ 2 }-2x }{ { x }^{ 2 }+1 } \right) tan=\left( \frac { { x }^{ 2 }-2x }{ { x }^{ 2 }+1 } \right) .\)
\(\left\{ \frac { ({ x }^{ 2 }+1)(2x-2)-({ x }^{ 2 }-2x)(2x) }{ { (x }^{ 2 }+{ 1) }^{ 2 } } \right\} \)
\(=sec\left( \frac { { x }^{ 2 }-2x }{ { x }^{ 2 }+1 } \right) tan=\left( \frac { { x }^{ 2 }-2x }{ { x }^{ 2 }+1 } \right) .\frac { { 2x }^{ 2 }+2x-2 }{ (x^{ 2 }+{ 1) }^{ 2 } } \)
\(\Rightarrow \frac { dy }{ dx } =\frac { 2y\sqrt { { y }^{ 2 }-1 } ({ x }^{ 2 }+x-1) }{ { (x }^{ 2 }+1)^{ 2 } } \)
If \(\sqrt { (x+y) } +\sqrt { (y-x) } \)=a, then \(\frac{dy}{dx}\)=
- (a)
\(\frac { \sqrt { (x+y) } -\sqrt { (y-x) } }{ \sqrt { (x+y) } +\sqrt { (y-x) } } \)
- (b)
\(\frac { 2\sqrt { (x-y) } }{ \sqrt { (x+y) } -\sqrt { (y-x) } } \)
- (c)
\(\frac { x+y+\sqrt { xy } }{ \sqrt { x+y } } \)
- (d)
\(\frac { { x }^{ 2 }+{ y }^{ 2 }+2xy }{ { x }^{ 2 }+y^{ 2 } } \)
We have, \(\sqrt { x+y } +\sqrt { y-x } \)
Differentiating w.r.t. x we get
\(\frac { 1 }{ 2\sqrt { x+y } } \left( 1+\frac { dy }{ dx } \right) +\frac { 1 }{ 2\sqrt { y-x } } \left( \frac { dy }{ dx } -1 \right) \)
\(\Rightarrow \frac { 1 }{ \sqrt { x+y } } -\frac { 1 }{ \sqrt { y-x } } =\left( \frac { 1 }{ \sqrt { x+y } } +\frac { 1 }{ \sqrt { y-x } } \right) \frac { dy }{ dx } \)
\(\Rightarrow \frac { -(\sqrt { y-x } -\sqrt { x+y } ) }{ (\sqrt { y-x } +\sqrt { x+y } ) } =\frac { dy }{ dx } \Rightarrow \frac { dy }{ dx } =\frac { \sqrt { x+y } -\sqrt { y-x } }{ \sqrt { y-x } +\sqrt { x+y } } \).
\(\frac { d }{ dx } \left[ { tan }^{ -1 }\left( \frac { a-x }{ 1+ax } \right) \right] \)is equal to
- (a)
\(-\frac { 1 }{ 1+{ x }^{ 2 } } \)
- (b)
\(\frac { 1 }{ 1+a^{ 2 } } -\frac { 1 }{ 1+x^{ 2 } } \)
- (c)
\(\frac { 1 }{ 1+\left( \frac { a-x }{ 1+ax } \right) ^{ 2 } } \)
- (d)
\(\frac { -1 }{ \sqrt { 1-\left( \frac { a-x }{ 1+ax } \right) ^{ 2 } } } \)
\(\frac { d }{ dx } \left[ { tan }^{ -1 }\left( \frac { a-x }{ 1+ax } \right) \right] =\frac { d }{ dx } [tan^{ -1 }a-tan^{ -1 }x]\)
\(=0-\frac { 1 }{ 1+{ x }^{ 2 } } =-\frac { 1 }{ 1+{ x }^{ 2 } } \)
\(\frac { d }{ dx } \left( x\sqrt { { a }^{ 2 }-{ x }^{ 2 } } +{ a }^{ 2 }{ sin }^{ -1 }\left( \frac { x }{ a } \right) \right) \) is equal to
- (a)
\(\sqrt { { a }^{ 2 }-{ x }^{ 2 } } \)
- (b)
\(2\sqrt { { a }^{ 2 }-{ x }^{ 2 } } \)
- (c)
\(\frac { 1 }{ \sqrt { { a }^{ 2 }-{ x }^{ 2 } } } \)
- (d)
none of these
\(\frac { d }{ dx } \left\{ x\sqrt { { a }^{ 2 }-{ x }^{ 2 } } +{ a }^{ 2 }{ sin }^{ -1 }\left( \frac { x }{ a } \right) \right\} \)
\(=\frac { x\times 1\times (-2x) }{ 2\sqrt { { a }^{ 2 }-{ x }^{ 2 } } } +\sqrt { { a }^{ 2 }-{ x }^{ 2 } } +{ a }^{ 2 }\frac { 1 }{ \sqrt { 1-\frac { { x }^{ 2 } }{ { a }^{ 2 } } } } \times \frac { 1 }{ a } \)
\(=\frac { { -x }^{ 2 } }{ \sqrt { { a }^{ 2 }-{ x }^{ 2 } } } +\sqrt { { a }^{ 2 }-{ x }^{ 2 } } +\frac { { a }^{ 2 } }{ \sqrt { { a }^{ 2 }-{ x }^{ 2 } } } \)
\(=\sqrt { { a }^{ 2 }-{ x }^{ 2 } } +\frac { ({ a }^{ 2 }-{ x }^{ 2 }) }{ \sqrt { { a }^{ 2 }-{ x }^{ 2 } } } =2\sqrt { { a }^{ 2 }-{ x }^{ 2 } } \)
If f(x)=(logcotxtanx)(logtanxcotx)-1+\(tan^{ -1 }\left( \frac { 4x }{ 4-{ x }^{ 2 } } \right) \) then f'(2) is equal to
- (a)
\(\frac{1}{2}\)
- (b)
-\(\frac{1}{2}\)
- (c)
1
- (d)
-1
(logcotxtanx)(logtanxcotx)-1+\(tan^{ -1 }\frac { 4x }{ 4-{ x }^{ 2 } } \)
=\(\frac { logtanx }{ logcotx } .\frac { logtanx }{ logcotx } +tan^{ -1 }\left( \frac { 4x }{ 4-{ x }^{ 2 } } \right) \)
=\(\frac { (logtanx)^{ 2 } }{ (-logtanx)^{ 2 } } +tan^{ -1 }\left( \frac { 4x }{ 4-{ x }^{ 2 } } \right) \)
=\(1+tan^{ -1 }\left( \frac { 4x }{ 4-{ x }^{ 2 } } \right) \)
\(\therefore f'(x)=\frac { 1 }{ 1+\left( \frac { 4x }{ 4-{ x }^{ 2 } } \right) ^{ 2 } } .\frac { (4-x^{ 2 })-4x(-2x) }{ (4-x^{ 2 })^{ 2 } } \)
=\(\frac { 16-4x^{ 2 }+8{ x }^{ 2 } }{ (4-x^{ 2 })^{ 2 }+16{ x }^{ 2 } } =\frac { 4(4+{ x }^{ 2 }) }{ (4-{ x }^{ 2 })^{ 2 }+(4x)^{ 2 } } \)
Hence, f'(2)=\(\frac { 4(4+4) }{ 0+(8)2 } =\frac { 32 }{ 64 } =\frac { 1 }{ 2 } \)
If y=\(log\left[ { e }^{ x }\left( \frac { x-1 }{ x+2 } \right) ^{ 1/2 } \right] \), then \(\frac{dy}{dx}\) is equal to
- (a)
7
- (b)
\(\frac{3}{x-2}\)
- (c)
\(\frac{3}{(x-1)}\)
- (d)
None of these
We have, y=\(log\left[ { e }^{ x }\left( \frac { x-1 }{ x+2 } \right) ^{ 1/2 } \right] \)
=logex+\(\frac{1}{2}\)[log(x-1)-log(x+2)]
=x+\(\frac{1}{2}\)[log(x-1)-log(x+2)]
\(\therefore \frac { dy }{ dx } =1+\frac { 1 }{ 2 } \left[ \frac { 1 }{ x-1 } -\frac { 1 }{ x+2 } \right] =1+\frac { 3 }{ 2(x-1)(x+2) } \).
The derivative of y =(1-x) (2-x)....(n-x) at x = 1 is equal to
- (a)
0
- (b)
(-1) (n-1)!
- (c)
n! - 1
- (d)
(-1)n-1(n-1)!
\(\because \) y = ( 1-x) (2-x) ....(n-x)
Taking log on both sides, we get
logy = log (1-x) + log(2-x) +...+ log(n-x)
\(\frac { y }{ 1 } \frac { dy }{ dx } =\frac { 1 }{ (1-x) } (-1)+\frac { 1 }{ (2-x) } (-1)+...+\frac { 1 }{ (n-x) } (-1)\)
\(\frac { dy }{ dx } =-y\left[ \frac { [(2-x)(3-x)....(n-x)]+... }{ y } \right] \)
\(\therefore \left( \frac { dy }{ dx } \right) _{ x=1 }=1.2...(n-1)(-1)=(-1)(n-1)!\)
If x =\(\frac { 1-t^{ 2 } }{ 1+{ t }^{ 2 } } \)and y= \(\frac { 2t }{ 1+{ t }^{ 2 } } \),then \(\frac { dy }{ dx } \) is equal to
- (a)
\(-\frac { y }{ x } \)
- (b)
\(\frac { y }{ x } \)
- (c)
\(-\frac { x }{ y } \)
- (d)
\(\frac { x }{ y } \)
\(\because x=\frac { 1-t^{ 2 } }{ 1+{ t }^{ 2 } } andy=\frac { 2t }{ 1+{ t }^{ 2 } } \)
Put t = tan\(\theta \) in both the equations, we get
\(x=\frac { { 1-tan }^{ 2 }\theta }{ { 1+tan }^{ 2 }\theta } =cos2\theta ...(i)\)
and \(y=\frac { { 2tan }\theta }{ { 1+tan }^{ 2 }\theta } =sin2\theta ..(ii)\)
Differentiating both the equations (i) and (ii), we get
\(\frac { dx }{ d\theta } =-2sin2\theta \quad and\frac { dy }{ d\theta } 2cos2\theta \)
Therefore, \(\frac { dy }{ dx } =\frac { \frac { dy }{ d\theta } }{ \frac { dx }{ d\theta } } =-\frac { cos2\theta }{ sin2\theta } =-\frac { x }{ y } \)
If u=x2+y2 and x=s+3t, y=2x-t, then \(\frac { d^{ 2 }u }{ ds^{ 2 } } \) is equal to
- (a)
12
- (b)
32
- (c)
36
- (d)
10
Given, u=x2+y2, x=s+3t, y=2s-t
⇒ \(\frac { dx }{ ds } =1,\frac { dy }{ ds } =2\)
Now, u=x2+y2 ⇒ \(\frac { du }{ dx } =2x\frac { dx }{ dx } +2y\frac { dy }{ ds } \)=2x+4y
\(\frac { { d }^{ 2 }u }{ ds^{ 2 } } =2\left( \frac { dx }{ ds } \right) +4\left( \frac { dy }{ ds } \right) \quad \Rightarrow \left( \frac { { d }^{ 2 }u }{ ds^{ 2 } } \right) \)=2(1)+4(2)=10
The derivative of \({ sin }^{ -1 }\left( \frac { 2x }{ 1+{ x }^{ 2 } } \right) \)with respect to \(cos^{ -1 }\left( \frac { 1-{ x }^{ 2 } }{ 1+x^{ 2 } } \right) is\)
- (a)
-1
- (b)
1
- (c)
2
- (d)
4
Let \(P={ sin }^{ -1 }\left[ \frac { 2x }{ 1+{ x }^{ 2 } } \right] =2tan^{ -1 }x\quad andq=cos^{ -1 }\left[ \frac { 1-{ x }^{ 2 } }{ 1+{ x }^{ 2 } } \right] =2tan^{ -1 }x\)
\(\Rightarrow \frac { dp }{ dx } =\frac { 2 }{ 1+{ x }^{ 2 } } ,\frac { dq }{ dx\\ } =\frac { 2 }{ 1+{ x }^{ 2 } } \)
\(\therefore \frac { dp }{ dq } =\frac { \frac { dp }{ dx } }{ \frac { dq }{ dx\\ } } =\frac { \frac { 2 }{ (1+{ x }^{ 2 }) } }{ \frac { 2 }{ (1+{ x }^{ 2 }) } } =1\)
The derivative of sin-1\(\left( \frac { 2x }{ 1+x^{ 2 } } \right) \) with respect to tan-1\(\left( \frac { 2x }{ 1-x^{ 2 } } \right) \) is
- (a)
0
- (b)
1
- (c)
\(\frac { 1 }{ 1-{ x }^{ 2 } } \)
- (d)
\(\frac { 1 }{ 1+{ x }^{ 2 } } \)
Let u=sin-1\(\left( \frac { 2x }{ 1+x^{ 2 } } \right) \)=2 tan-1x
and v=tan-1\(\left( \frac { 2x }{ 1-x^{ 2 } } \right) \)=2 tan-1x
⇒ u=v ⇒ \(\frac { du }{ dx } =\frac { dv }{ dx } \) ∴ \(\frac { du }{ dv } =\frac { (du/dx) }{ (dv/dx) } \)=1
If f(x) = xn, then the value of \(f(1)-\frac { f'(1) }{ 1! } +\frac { f''(1) }{ 2! } -\frac { f'''(1) }{ 3! } +...+\frac { { (-1) }^{ n }{ f }^{ n }(1) }{ n! } is\)
- (a)
2n
- (b)
2n-1
- (c)
0
- (d)
1
Since, \(f(x)={ x }^{ n }\Rightarrow f(1)=1\)
\(f'(x)=nx^{ x-1 }\Rightarrow f'(1)=n\)
\(f''(x)=n(n-1)^{ n-2 }\Rightarrow f''(1)=n(n-1)\)
... .... ... .... ...
... .... .... ... ...
Fn(x)= n (n-1) (n-2)...2.1
\(\Rightarrow { f }^{ n }(1)=n(n-1)(n-2)...2.1\)
We have,
\(f(1)-\frac { f'(1) }{ 1! } +\frac { f''(1) }{ 2! } -\frac { f'''(1) }{ 3! } +...+\frac { { (-1) }^{ n }{ f }^{ n }(1) }{ n! } \)
\(=-1\frac { n }{ 1! } +\frac { n(n-1) }{ 2! } -\frac { n(n-1)(n-2) }{ 3! } +.....+\frac { (-1)^{ n }n(n-1)(n-2)....2.1 }{ n! } \)
\(=(1-1)^{ n }=0\)
\(\frac { { d }^{ 2 } }{ { dx }^{ 2 } } (cos^{ -1 }(1-x))\) is equal to
- (a)
\(\frac { x-1 }{ (2x-{ x }^{ 2 })^{ 3/2 } } \)
- (b)
\(\frac { 1-x }{ (2x-{ x }^{ 2 })^{ 3/2 } } \)
- (c)
\(\frac { 1 }{ 2(2x-{ x }^{ 2 })^{ 3/2 } } \)
- (d)
none of these
Here,\(\frac { d }{ dx } \left( { cos }^{ -1 }(1-x) \right) \)
\(=\frac { 1 }{ \sqrt { 1-{ (1-x) }^{ 2 } } } \times (-1)=\frac { 1 }{ \sqrt { 2x-{ x }^{ 2 } } } \)
\(\Rightarrow \frac { { d }^{ 2 } }{ { dx }^{ 2 } } \left( { cos }^{ -1 }(1-x) \right) =\frac { d }{ dx } (2x-{ x }^{ 2 }{ ) }^{ -1/2 }\)
\(=-\frac { 1 }{ 2 } \frac { (2-2x) }{ ({ 2x-x^{ 2 }) }^{ 3/2 } } =\frac { x-1 }{ ({ 2x-x^{ 2 }) }^{ 3/2 } } \)
For what choice of a and b, is the function \(f(x)=\left\{ \begin{matrix} { x }^{ 2 }, \\ ax+b, \end{matrix}\begin{matrix} x\le c \\ x>c \end{matrix} \right\} \) is differential at x=?
- (a)
a=c,b=c
- (b)
a=c,b=-c
- (c)
a=-c2,b=2c
- (d)
a=2c,b=-c2
since f is differentiable at the point x=c,it must be continuous at this point
\(\therefore \begin{matrix} lim \\ x\rightarrow { c }^{ - } \end{matrix}f(x)=f(c)=\begin{matrix} lim \\ h\rightarrow { c }^{ + } \end{matrix}f(x)\)
\(\Rightarrow \begin{matrix} lim \\ x\rightarrow { c }^{ - } \end{matrix}{ x }^{ 2 }=c^{ 2 }=\begin{matrix} lim \\ h\rightarrow 0 \end{matrix}(ax+b)\)
\(\Rightarrow { c }^{ 2 }={ c }^{ 2 }=ac+b\Rightarrow { c }^{ 2 }=ac+b\)
Also f is diiferential at x=c
\(\Rightarrow Lf'(c)=Rf'(C)\)
\(\Rightarrow \begin{matrix} lim \\ h\rightarrow { 0 }^{ - } \end{matrix}\cfrac { f(c+h)-f(c) }{ h } =\begin{matrix} lim \\ h\rightarrow { 0 }^{ + } \end{matrix}\cfrac { f(c+h)-f(c) }{ h } \)
\(\Rightarrow \begin{matrix} lim \\ h\rightarrow { 0 }^{ - } \end{matrix}\cfrac { \left( c+h \right) ^{ 2 }-{ c }^{ 2 } }{ h } =\begin{matrix} lim \\ h\rightarrow 0^{ + } \end{matrix}\cfrac { a(c+h)+b-{ c }^{ 2 } }{ h } \)
\(\Rightarrow \begin{matrix} lim \\ h\rightarrow { 0 }^{ - } \end{matrix}\cfrac { { c }^{ 2 }+{ h }^{ 2 }+2ch-{ c }^{ 2 } }{ h } \) = \(\Rightarrow \begin{matrix} lim \\ h\rightarrow { 0 }^{ + } \end{matrix}\cfrac { ac+ah+b-{ c }^{ 2 } }{ h } \)
\(\Rightarrow \begin{matrix} lim \\ h\rightarrow { 0 }^{ - } \end{matrix}\cfrac { h(h+2c) }{ h } =\begin{matrix} lim \\ h\rightarrow { 0 }^{ + } \end{matrix}\cfrac { ah }{ h } \left( \therefore ac+b={ c }^{ 2 }by(i) \right) \)
\(\Rightarrow \begin{matrix} lim \\ h\rightarrow { 0 }^{ - } \end{matrix}\left( h+2c \right) =\begin{matrix} lim \\ h\rightarrow { 0 }^{ + } \end{matrix}\Rightarrow 2c=a\)
and then from (i) \({ c }^{ 2 }2c\cdot c+b\Rightarrow { c }^{ 2 }-{ 2c }^{ 2 }=b\)
\(\Rightarrow b=-{ c }^{ 2 }\)
Hence, a=2c and b=-c2
If f(x) = x2g(X) and g(x) is twice differentiable, then f"(x) is equal to
- (a)
2g"(x)
- (b)
x2g"(x) + 2xg'(x) + 2g(x)
- (c)
x2g"(x) + 4g'(x) + 2g(x)
- (d)
None of these
\(f'x=\frac { d }{ dx } ({ x }^{ 2 }g(x))={ x }^{ 2 }g'(x)+2xg(x)\)
Now, f"(x)=\(\frac{d}{dx}\)(x2g'(x)+2xg(x))
= x2g"(x) +g'(x) 2x + 2{xg'(x) + g(x) ·1}
= x2g"(x) +4xg'(x) + 2g(x)
A value of c for which the Mean value theorem holds for the function f(x) = logex on the interval [1,3] is
- (a)
2log3e
- (b)
\(\frac { 1 }{ 2 } { log }_{ e }3\)
- (c)
log3e
- (d)
loge3
Using Mean Value theorem
\(f'(c)=\frac { f(3)-f(1) }{ 3-1 } \Rightarrow c=\frac { 2 }{ { log }_{ e }3 } =2{ log }_{ 3 }e\)
The value of c in Mean value theorem for the function f(x) = x(x - 2), x \(\in \) [1, 2] is
- (a)
\(\frac { 3 }{ 2 } \)
- (b)
\(\frac { 2 }{ 3 } \)
- (c)
\(\frac { 1 }{ 2 } \)
- (d)
\(\frac { 5 }{ 2 } \)
Since mean value theorem is satisfied.
\(\therefore f'(c)=\frac { f(2)-f(1) }{ 2-1 } \Rightarrow 2c-2=1\Rightarrow c=\frac { 3 }{ 2 } \)
Fill in the blanks.
(i) The value of c in LMVT for the function f(x)=2x3 - 5x2 - 4x + 3, \(x\in \left[ \frac { 1 }{ 3 } ,3 \right] \)is ____P____
(ii) The value of c in Rolle's theorem for the function f(x) = (x - 2) (x - 3) in [2,3] is ____Q____
(iii) The value of c in Rolle's theorem for the function f(x) = x2- 5x + 9, x \(\in \) [1,4] is ____R____
(iv) The value of c in LMVT for the function f(x) = 6x2 - x3, x \(\in \) [0,6] is _____S____
- (a)
P Q R S 1.5 3/2 5/0 4 - (b)
P Q R S 1.9 5/2 3/2 2 - (c)
P Q R S 1.5 2 1 2 - (d)
P Q R S 1.9 5/2 5/2 4
(i) f(x) = 2x3 - 5x2 - 4x + 3 being a polynomial function is continous on \(\left[ \frac { 1 }{ 3 } ,3 \right] \) and differentiable on \(\left( \frac { 1 }{ 3 } ,3 \right) \)
\(\therefore By\ LMVT,\ we\ have\ c\left( \in \frac { 1 }{ 3 } ,3 \right) 8.t\)
\(\frac { f(3)-f\left( \frac { 1 }{ 3 } \right) }{ 3-\frac { 1 }{ 3 } } =f'(c)\Rightarrow f'(c)=\frac { 0-\frac { 32 }{ 27 } }{ \frac { 8 }{ 3 } } =\frac { -4 }{ 9 } \)
\(\Rightarrow { 6c }^{ 2 }-10c-4=\frac { -4 }{ 9 } \Rightarrow 6{ c }^{ 2 }-10c-\frac { 32 }{ 9 } =0\)
\(\Rightarrow c=\frac { 45\pm 61.26 }{ 54 } \Rightarrow c=\frac { 45+61.26 }{ 54 } \approx 1.9\left( \because c\in \left( \frac { 1 }{ 3 } ,3 \right) \right) \)
(ii) f(x) = (x - 2)(x - 3), x \(\in \) [2,3]
\(\therefore\) f(x) = x2 - 5x + 6 and f'(x) = 2x - 5 ...(i)
Now,f(x) is continuous on [2,3] and differentiable on (2, 3).
Also,f(2) = f(3) = 0
\(\therefore\) By Rolle's theorem we have e E (2,3) such that \(f'(c)=0\Rightarrow 2c-5=0\Rightarrow c=\frac { 5 }{ 2 } \)
(iii) The function f (x) = x2 - 5x + 9 being a polynomial function is continuous on [1,4] and differentiable on (1,4).
Now,f(1) = 12 - 5(1) + 9 = 1- 5 + 9 = 5
and f(4) = 42 - 5(4) + 9 = 16 - 20 + 9 = 5
\(\therefore\) f(1) = f(4)
Thus, the function satisfies all the conditions of the Rolle's theorem.
\(\therefore\) We have c\(\in \) (1,4) such that f'(c) = 0.
Now,f(x) = x2 - 5x + 9 \(\Rightarrow\) f'(x) = 2x - 5
\(\therefore f'(c)=0\Rightarrow 2c-5=0\Rightarrow c=\frac { 5 }{ 2 } \in (1,4)\)
(iv) f(x) being polynomial function is continuous on [0,6] and differentiable on (0, 6).
\(\therefore\) By LMVT, we have e \(\in \) (0,6) such that
\(\frac { f(6)-f(0) }{ 6-0 } =f'(c)\Rightarrow \frac { 0-0 }{ 6-0 } =f'(c)\Rightarrow f'(c)=0\)
Now f(x) = 6x2 -.0 \(\Rightarrow\) f'(x) = 12x - 3x2
\(\Rightarrow\) f'(c) = I2c - 3c2 0 = 12c - 3c2
\(\Rightarrow\) 32 = I2e \(\Rightarrow\) c = 4 which lies in the interval (0, 6).
State T for true and F for false.
(i) The exponential function aX is continuous
everywhere.
(ii) The function f{x) = sinlxl is continuous for all x \(\epsilon \) R.
(iii) The function f{x) = \(\sqrt{x-2}\) is nowhere continuous.
- (a)
(i) (ii) (iii) T T F - (b)
(i) (ii) (iii) T F F - (c)
(i) (ii) (iii) F F T - (d)
(i) (ii) (iii) F T T
If y=\(\sqrt{sinx+y}\), then is equal to
- (a)
\(\frac{cosx}{2y-1}\)
- (b)
\(\frac{cosx}{1-2y}\)
- (c)
\(\frac{sinx}{1-2y}\)
- (d)
\(\frac{sinx}{2y-1}\)
y=\(\sqrt{sinx+y}\) y2=sinx+y
Differentiating w.r.t. to x, we get
\(2y\frac { dy }{ dx } =cosx+\frac { dy }{ dx } \Rightarrow \frac { dy }{ dx } =\frac { cosx }{ 2y-1 } \)
If f(x) = x2sin\(\frac { 1 }{ x } \), wnere x\(\neq \)0, then the value of the function f at x = 0, so that the function is continuous at x = 0, is
- (a)
0
- (b)
-1
- (c)
1
- (d)
None of these
\(f(x)={ x }^{ 2 }sin\frac { 1 }{ x } forx\neq 0\)
\(\lim _{ x\rightarrow 0 }{ f(x)=\lim _{ x\rightarrow 0 }{ { x }^{ 2 }sin\frac { 1 }{ x } } } \)
Since the value of sin \(\frac { 1 }{ x } \) is between -1 to 1
\(\lim _{ x\rightarrow 0 }{ { x }^{ 2 }sin\left( \frac { 1 }{ x } \right) } \)
=(0)2 \(\times\) (value between -1 to 1.) = 0
So f(x) is continuous at x = 0
If \(\lim _{ x\rightarrow 0 }{ f(x)=f(0)\Rightarrow f(0)=0 } \)
If x = t2, y = t3, then \(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \) is
- (a)
\(\frac { 3 }{ 2 } \)
- (b)
\(\frac { 3 }{ 4t } \)
- (c)
\(\frac { 3 }{ 2t } \)
- (d)
\(\frac { 3 }{ 4 } \)
x = t2 .....(i), y = t3 ....(ii)
From (i) and (ii), we get y = x3/2
Differentiating w.r.t. to x, we get
\(\frac { dy }{ dx } \frac { 3 }{ 2 } { x }^{ \frac { 1 }{ 2 } }\)
Again differentiating w.r.t. to x, we get
\(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =\frac { 3 }{ 4\sqrt { x } } =\frac { 3 }{ 4t } \) (from eqn. (i))
For thefunction f(x) = x + \(\frac { 1 }{ x } \), x\(\in \)[1,3], the value of c for mean value theorem is
- (a)
1
- (b)
\(\sqrt { 3 } \)
- (c)
2
- (d)
None of these
\(f'(c)=\frac { f(3)-f(1) }{ 3-1 } \Rightarrow 1-\frac { 1 }{ { c }^{ 2 } } =\left( \frac { \frac { 10 }{ 3 } -2 }{ 2 } \right) \)
\(\Rightarrow 1-\frac { 1 }{ { c }^{ 2 } } =\frac { 4 }{ 3\times 2 } \Rightarrow \frac { 1 }{ { c }^{ 2 } } =1-\frac { 2 }{ 3 } \)
\(\Rightarrow { c }^{ 2 }=3\Rightarrow c=\pm \sqrt { 3 } \)
\(\Rightarrow c=\sqrt { 3 } \left( \because -\sqrt { 3 } \notin [1,3] \right) \)
Statement-I: If y =log10x + logex,then
\(\frac { dy }{ dx } =\frac { { log }_{ 10 }e }{ x } +\frac { 1 }{ x } \)
Statement-II:\(\frac { d }{ dx } ({ log }_{ 10 }x)=\frac { logx }{ log10 } \) and \(\frac { d }{ dx } ({ log }_{ e }x)=\frac { logx }{ loge } \)
- (a)
If both Statement-I and Statement-II are true and Statement-II is the correct explanation of Statement -I.
- (b)
If both Statement-I and Statement-II are true but Statement-II is not the correct explanation of Statement -I.
- (c)
If Statement-I is true but Statement-II is false.
- (d)
If Statement-I is false and Statement-II is true.
We have y = log10x+ loggx
\(\frac { dy }{ dx } =\frac { 1 }{ x } { log }_{ 10 }e+\frac { 1 }{ x } \)