Eamcet Mathematics - Differential Coefficient Chapter Sample Question Paper With Answer Key
Exam Duration: 60 Mins Total Questions : 50
If y=tan-1\(\sqrt { \left( \frac { 1+sin\quad x }{ 1-sin\quad x } \right) } ,\frac { \pi }{ 2 } <x<\pi ,then\frac { dy }{ dx } \)equals
- (a)
-1/2
- (b)
-1
- (c)
1/2
- (d)
1
\(y={ tan }^{ -1 }\sqrt { \left( \frac { 1-cos\left( \frac { \pi }{ 2 } +x \right) }{ 1+cos\left( \frac { \pi }{ 2 } +x \right) } \right) } ={ tan }^{ -1 }\left| tan\left( \frac { \pi }{ 4 } +\frac { x }{ 2 } \right) \right| \quad \quad ...(i)\\ Now,\quad \quad \quad \quad \quad \frac { \pi }{ 2 } <x<\pi \\ \therefore \quad \quad \quad \quad \quad \quad \quad \frac { \pi }{ 4 } <\frac { x }{ 2 } <\frac { \pi }{ 2 } \\ or\quad \quad \quad \quad \quad \frac { \pi }{ 2 } <\frac { \pi }{ 4 } +\frac { x }{ 2 } <\frac { 3\pi }{ 4 } \\ \therefore \quad \left| tan\left( \frac { \pi }{ 4 } +\frac { x }{ 2 } \right) \right| =-tan\left( \frac { \pi }{ 4 } +\frac { x }{ 2 } \right) (\because \quad in\quad II\quad quadrant)\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =tan\left\{ \pi -\left( \frac { \pi }{ 4 } +\frac { x }{ 2 } \right) \right\} \\ From\quad Eq.(i),\\ \quad \quad \quad \quad \quad \quad y={ tan }^{ -1 }tan\left\{ \pi -\left( \frac { \pi }{ 4 } +\frac { x }{ 2 } \right) \right\} \\ \quad \quad \quad \quad \quad \quad \quad =\quad \pi -\left( \frac { \pi }{ 4 } +\frac { x }{ 2 } \right) \\ \quad \quad \quad \quad \quad \quad \quad =\frac { 3\pi }{ 4 } -\frac { x }{ 2 } \\ \left( \because \quad principal\quad value\quad of\quad { tan }^{ -1 }tan\quad x\quad in-\frac { \pi }{ 2 } to\frac { \pi }{ 2 } \right) \\ \therefore \frac { dy }{ dx } =-\frac { 1 }{ 2 } \)
Let f(x) =\(\left| \begin{matrix} sin\quad 3x & 1 & 2\left( cos\left( \frac { 3x }{ 2 } \right) +sin\left( \frac { 3x }{ 2 } \right) \right) ^{ 2 } \\ cos\quad 3x & -1 & 2\left( cos^{ 2 }\left( \frac { 3x }{ 2 } \right) -sin^{ 2 }\left( \frac { 3x }{ 2 } \right) \right) \\ tan\quad 3x & 4 & 1+2\quad tan\quad 3x \end{matrix} \right| \)
then the value of f' (x) at x = (2n + 1) \(\pi \), n \(\in \) I (the set of integers) is equal to
- (a)
(-1)n
- (b)
3
- (c)
(-1)n+1
- (d)
9
\(\because \left[ cos\left( \frac { 3x }{ 2 } \right) +sin\left( \frac { 3x }{ 2 } \right) \right] ^{ 2 }=1+sin3x\)
and \({ cos }^{ 2 }\left( \frac { 3x }{ 2 } \right) -{ sin }^{ 2 }\left( \frac { 3x }{ 2 } \right) =cos3x\)
Then \(f(x)=\left| \begin{matrix} sin3x & 1 & 2+2sin3x \\ cos3x & -1 & 2cos3x \\ tan3x & 4 & 1+2tan3x \end{matrix} \right| \)
Applying, C3➝C3-2C1
Then \(f(x)=\left| \begin{matrix} sin3x & 1 & 2 \\ cos3x & -1 & 0 \\ tan3x & 4 & 1 \end{matrix} \right| \)
\(\Rightarrow \quad \quad f^{ ' }(x)=\left| \begin{matrix} 3cos3x & 1 & 2 \\ -3sin3x & -1 & 0 \\ 3{ sec }^{ 2 }3x & 4 & 1 \end{matrix} \right| \)
\(\therefore \quad f^{ ' }\left[ \left( 2n+1 \right) \pi \right] =\left| \begin{matrix} 3(-1)^{ 3 } & 1 & 2 \\ -3(0) & -1 & 0 \\ 3{ (-1) }^{ 2 } & 4 & 1 \end{matrix} \right| \)
\(\quad =\left| \begin{matrix} -3 & 1 & 2 \\ 0 & \begin{matrix} \vdots \\ \begin{matrix} \cdots & -1 & \cdots \end{matrix} \\ \vdots \end{matrix} & 0 \\ 3 & 4 & 1 \end{matrix} \right| \)
\(=(-1)(-3-6)=9\)
The set of values of a for which the function f(x) = (4a - 3) (x + In 5) + 2 (a -7) cot \(\left( \frac { x }{ 2 } \right) { sin }^{ 2 }\left( \frac { x }{ 2 } \right) \) does not possess critical points is
- (a)
\((1,\infty )\)
- (b)
\([1,\infty )\)
- (c)
\((-\infty ,2)\)
- (d)
\((-8,-4/3)\cup (2,\infty )\)
\(\because { f }^{ ' }(x)=(4a-3)(1)+(a-7)cos\quad x\\ \quad \quad \quad \quad \quad \quad =\left( \frac { \left( 4a-3 \right) }{ (a-7) } +cos\quad x \right) (a-7)\\ \quad \quad \quad \quad \quad \quad { f }^{ ' }(x)\neq 0\\ (for\quad non\quad existence\quad of\quad critical\quad points)\\ \quad \quad \quad \frac { 4a-3 }{ a-7 } >1\quad or\quad \frac { 4a-3 }{ a-7 } <-1\quad (\because \quad -1\le cos\quad x\le 1)\\ \Rightarrow \quad \quad \frac { 3a+4 }{ a-7 } >0\quad or\quad \frac { 5a-10 }{ a-7 } <0\\ \therefore \quad a\in (-\infty ,-\frac { 4 }{ 3 } )\cup (7,\infty )or\quad a\in (2,7)\\ Hence,\quad a\in \left( -\infty ,-4/3 \right) \cup (2,\infty )(\because \quad at\quad a=7,{ f }^{ ' }(x)\neq 0)\)
If f(x) = x + tan x and f is inverse of g' then g' (x) is equal to
- (a)
\(\frac { 1 }{ 1+(g(x)-x)^{ 2 } } \)
- (b)
\(\frac { 1 }{ 1-(g(x)-x)^{ 2 } } \)
- (c)
\(\frac { 1 }{ 2+(g(x)-x)^{ 2 } } \)
- (d)
\(\frac { 1 }{ 2-(g(x)-x)^{ 2 } } \)
Let y = f(x) \(\Rightarrow\) x = f-1 (y)
then f(x) = x + tan x
\(\Rightarrow\) y = f-1(y) + tan (f-1(y))
\(\Rightarrow\) y =g (y)+ tan (g(y)) or x =g(x)+ tan (g(x)) ... (i)
Differentiating both sides, then we get
1=g'(x)+ sec2 g(x).g'(x)
∴ g'(x) = \(\frac{1}{1+sec^2(g(x))}\)=\(\frac{1}{1+1+tan^2(g(x))}\)
=\(\frac{1}{2+(x-g(x))^2}\) [from Eq.(i)]
=\(\frac{1}{2+(g(x)-x)^2}\)
If \(\sqrt { (1-{ x }^{ 2n }) } +\sqrt { (1-y^{ 2n }) } =a({ x }^{ n }-{ y }^{ n }),then\quad \sqrt { \left( \frac { 1-{ x }^{ 2n } }{ 1-{ y }^{ 2n } } \right) } \frac { dy }{ dx }\) is equal to
- (a)
\(\frac { { x }^{ n-1 } }{ { y }^{ n-1 } } \)
- (b)
\(\frac { { y }^{ n-1 } }{ { x }^{ n-1 } } \)
- (c)
\(\frac { x }{ y } \)
- (d)
1
Put x n sin\(\theta \) and y n = sin\(\Phi \)
then, (cos\(\theta \)+cos\(\Phi \))= a (sin\(\theta \)-sin\(\Phi \))
\(\Rightarrow 2cos\left( \frac { \theta +\Phi }{ 2 } \right) cos\left( \frac { \theta +\Phi }{ 2 } \right) \\ =2acos\left( \frac { \theta +\Phi }{ 2 } \right) sin\left( \frac { \theta +\Phi }{ 2 } \right) \\ \Rightarrow cot\left( \frac { \theta +\Phi }{ 2 } \right) =a\\ \Rightarrow \left( \frac { \theta +\Phi }{ 2 } \right) ={ cot }^{ -1 }a\Rightarrow \theta -\Phi =2{ cot }^{ -1 }a\\ or\quad { sin }^{ -1 }\quad { x }^{ n }-{ sin }^{ -1 }{ y }^{ n }=2{ cot }^{ -1 }a\)
Differentiating both sides, we have
\(\frac { { nx }^{ n-1 } }{ \sqrt { (1-{ x }^{ 2n }) } } -\frac { { ny }^{ n-1 } }{ \sqrt { (1-{ y }^{ 2n }) } } \frac { dy }{ dx } =0\\ \therefore \sqrt { \left( \frac { 1-{ x }^{ 2n } }{ 1-{ y }^{ 2n } } \right) } \frac { dy }{ dx } =\frac { { x }^{ n-1 } }{ { y }^{ n-1 } } \)
If y2=ax2+bx+c, then y3.\(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \) is
- (a)
a constant
- (b)
a function of x only
- (c)
a function of y only
- (d)
a function of x and y
∵ y2 = ax2 + bx + c
∴ 2y \(\frac { { d }y }{ { dx } } \)= 2ax + b
⇒ 2y\(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \)+\(\frac { { d }y }{ { dx } } .2\frac { { d }y }{ { dx } } =2a\)
or y3\(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \)=ay2-\(\left( y\frac { dy }{ dx } \right) ^{ 2 }\)
= a (ax2 + bx + c) -\(\left( ax+\frac { b }{ 2 } \right) ^{ 2 }\)
a2+x2+abx+ac-a2x2-\(\frac { { b }^{ 2 } }{ 4 } \)-abx
=ac-\(\frac { { b }^{ 2 } }{ 4 } \)=constant
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)
\(\frac { 1 }{ 1+{ x }^{ 2 } } \)
\(Let\quad u={ sin }^{ -1 }\left( \frac { 2x }{ 1+{ x }^{ 2 } } \right) ={ 2\quad tan }^{ -1 }x\\ \therefore \quad \frac { du }{ dx } =\frac { 2 }{ 1+{ x }^{ 2 } } \\ and\quad v\quad ={ tan }^{ -1 }\left( \frac { 2x }{ 1-{ x }^{ 2 } } \right) ={ 2\quad tan }^{ -1 }x\\ \therefore \quad \frac { dv }{ dx } =\frac { 2 }{ 1+{ x }^{ 2 } } \\ \therefore \quad \frac { du }{ dv } =\frac { \frac { du }{ dx } }{ \frac { dv }{ dx } } =1\)
If f(x) = |x - 2| and g(x) = fof(x), then for x > 20, g' (x) is equal to
- (a)
2
- (b)
1
- (c)
3
- (d)
none of these
We have
\(\quad f(x)=|x-2|=\begin{cases} x-2,x\ge 2 \\ -(x-2),x<2 \end{cases}\\ \because \quad x>20,\quad \therefore \quad f(x)=x-2\\ \therefore \quad g(x)=fof(x)=f\{ f(x)\} =f(x-2)\\ \quad \quad \quad =x-2-2=x-4\\ \therefore \quad { g }^{ ' }(x)=1\)
If \(\Phi \)(x) be a polynomial function of the second degree. If \(\Phi \)(1)=\(\Phi \)(-1) and a1, a2, a3 are in AP, then \(\Phi \)(a1), \(\Phi \)(a2), \(\Phi \)(a3) are in
- (a)
AP
- (b)
GP
- (c)
HP
- (d)
none of these
\(\Phi \)(x) = ax2 + bx + c
∵\(\Phi \)(1) = \(\Phi \)(-1) ⇒ a+ b+ c=a-b+ c⇒ b=0
∵ \(\Phi \)(x) = ax2+c
⇒ \(\Phi \)' (x) = 2ax
∴ \(\Phi \)' (a1) = 2aa1, \(\Phi \)' (a2) = 2aa2, \(\Phi \)' (a3) = 2aa3
∵a1, a2, a3 are in AP
∴ \(\Phi \)' (a1) \(\Phi \)' (a2)\(\Phi \)' (a3) are also in AP
If P(x) is a polynomial such that P(x2 + 1) = {P(x)}2 + 1 and P(0) = 0, then P' (0) is equal to
- (a)
-1
- (b)
0
- (c)
1
- (d)
none of these
Given,P(x2 + 1) = [P(x)]2 + 1
∴P(x)=x
(∵P(x) is an identity function)
P'(x)=1
∴ P' (0) = 1
If y = In \(\left( \frac { x }{ a+bx } \right) ^{ x }\), then x3 \(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \) is equal to
- (a)
\(\left( \frac { dy }{ dx } +x \right) ^{ 2 }\)
- (b)
\(\left( \frac { dy }{ dx } -y \right) ^{ 2 }\)
- (c)
\(\left( x\frac { dy }{ dx } +y \right) ^{ 2 }\)
- (d)
\(\left( x\frac { dy }{ dx } -y \right) ^{ 2 }\)
∵y = x In \(\left( \frac { x }{ a+bx } \right) \)= x (In x -In (a + bx))
or \(\left( \frac { y }{ x } \right) \)=In x-In(a+bx)
Differentiating both sides w.r.t. x, then
\(\left( \frac { x\frac { dy }{ dx } -y.1 }{ { x }^{ 2 } } \right) \quad =\frac { 1 }{ x } -\frac { b }{ a+bx } =\frac { a }{ x(a+bx) } \)
or \(\left( x\frac { dy }{ dx } -y \right) =\left( \frac { ax }{ a+bx } \right) \)
Again taking logarithm on both sides, then
In \(\left( x\frac { dy }{ dx } -y \right) \)= In (ax) - In (a + bx)
Differentiating both sides w.r.t. x; then
\(\frac { x\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } +\frac { dy }{ dx } -\frac { dy }{ dx } }{ \left( x\frac { dy }{ dx } -y \right) } =\frac { 1 }{ x } -\frac { b }{ a+bx } =\frac { a }{ x(a+bx) } \)
\(\quad =\frac { \left( x\frac { dy }{ dx } -y \right) }{ { x }^{ 2 } } \quad \quad [from]Eq.(i)]\)
or \({ { x } }^{ 3 }\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =\left( x\frac { dy }{ dx } -y \right) ^{ 2 }\)
If y2 = P(x) is a polynomial of degree 3, then \(2\frac { d }{ dx } \left[ { y }^{ 3 }\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \right] \)equals
- (a)
P'" (x) + P' x
- (b)
P" (x)· P''' (x)
- (c)
P (x) . P'" (x)
- (d)
none of these
y2=P(x)⇒2yy'=P'(x)
⇒ (2y) y" + y' (2y') = P" (x)
⇒ 2yy" = P" (x) - 2(y')2
⇒ 2y3y" = y2 P" (x) - 2(yy')2
= y2P''(x)-2\(\frac { \left\{ { P }^{ ' }(x) \right\} ^{ 2 } }{ 4 } \) [from Eq. (i)]
⇒2y3 y'' = P(x) P" (x)-\(\frac { 1 }{ 2 } \) {P' (x)}2
∴ \(\frac { d }{ dx } \)(2y3 y") =P(x)P'" (x)+P" (x) P' (x) - P' (x) P" (x)]=P(x)· P'" (x)
⇒ \(2\frac { d }{ dx } \left( { y }^{ 3 }\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \right) \)P(x) P'" (x)
If 5 f(x) + 3f\(\left( \frac { 1 }{ x } \right) \)x + 2 and y = x f(x), then\(\left( \frac { dy }{ dx } \right) _{ x=1 }\) is equal to
- (a)
14
- (b)
7/8
- (c)
1
- (d)
none of these
5f(x) + 3f\(\left( \frac { 1 }{ x } \right) \)= x + 2 ...(i)
Replacing x by \(\frac { 1 }{ x } \)
∴ 5f\(\left( \frac { 1 }{ x } \right) \)+ 3f(x)=\(\left( \frac { 1 }{ x } \right) \)+2 ...(ii)
From Eq. (i),
25f(x) + 15f \(\left( \frac { 1 }{ x } \right) \) = 5x + 10 ...(iii)
and from Eq. (ii),
9f(x) + 15f\(\left( \frac { 1 }{ x } \right) \)=\(\frac { 3 }{ x } \)+6 ...(iv)
Subtracting Eq. (iv) from (iii), we get
∴16f(x)= 5x -\(\frac { 3 }{ x } \)+4
∴ xf(x) = \(\frac { { 5x }^{ 2 }-3+4x }{ 16 } =y\)
∴ \(\frac { dy }{ dx } =\frac { 10x+4 }{ 16 } \)
\(\frac { dy }{ dx } |_{ x=1 }=\frac { 10+4 }{ 16 } \)
=\(\frac { 7 }{ 8 } \)
The differential coefficient of f(loge x) W.r.t. x, where f(x) = loge x is
- (a)
x/loge x
- (b)
loge x/x
- (c)
(x loge x)-1
- (d)
none of these
Let u = f(loge x)
∴ \(\frac { du }{ dx } \)= f' (loge x).\(\frac { 1 }{ x } \) ...(i)
Given, f(x) = loge x
∴f'(x)=\(\frac { 1 }{ x } \)
∴f' (loge x) =\(\frac { 1 }{ { log }_{ e }x } \) ...(ii)
From Eqs. (i) and (ii),
\(\frac { du }{ dx } =\frac { 1 }{ x{ log }_{ e }x } \)
=(x loge x)-1
If y = \(\sqrt { \left( \frac { 1+cos2\theta }{ 1-cos2\theta } \right) } ,\frac { dy }{ dx } at\theta =3\pi /4\quad is\)
- (a)
-2
- (b)
2
- (c)
\(\pm 2\)
- (d)
none of these
\(y=\sqrt { \left( \frac { 1+cos2\theta }{ 1-cos2\theta } \right) } \\ \quad \quad \quad =|cot\theta |=-cot\theta =-cot\theta (\theta =3\pi /4)\\ \therefore \quad \frac { dy }{ d\theta } ={ cosec }^{ 2 }\theta \\ \quad \quad \frac { dy }{ d\theta } |_{ 0=3\pi /4 }=\left( \sqrt { 2 } \right) ^{ 2 }=2\)
If f(x)=(logcotx tan x) (logtanxcot x)-1 + tan \(\left( \frac { x }{ \sqrt { \left( 4-{ x }^{ 2 } \right) } } \right) \) then f' (0) is equal to
- (a)
-2
- (b)
2
- (c)
\(\frac { 1 }{ 2 } \)
- (d)
0
Since, f(x) = (log cot x tan x) (logtan x cot x)-1
\(+{ tan }^{ -1 }\left( \frac { x }{ \left( \sqrt { 4-{ x }^{ 2 } } \right) } \right) \)
= (-1) (-1)-1 + tan-1\(\left( \frac { x }{ \left( \sqrt { 4-{ x }^{ 2 } } \right) } \right) \)
=1+tan-1\(\left( \frac { x }{ \left( \sqrt { 4-{ x }^{ 2 } } \right) } \right) \)
∴f' (x) = 0+\(\frac { 1 }{ 1+\frac { { x }^{ 2 } }{ 4-{ x }^{ 2 } } } \)
\(\times \frac { \sqrt { \left( 4-{ x }^{ 2 } \right) } .1-x\times \frac { 1 }{ 2\sqrt { \left( 4-{ x }^{ 2 } \right) } } \times -2x }{ \left( \sqrt { 4-{ x }^{ 2 } } \right) ^{ 2 } } \)
=\(\frac { \left( 4-{ x }^{ 2 } \right) }{ 4 } \times \frac { 4-{ x }^{ 2 }+{ x }^{ 2 } }{ \sqrt { 4-{ x }^{ 2 } } } \)
=\(\sqrt { 4-{ x }^{ 2 } } \)
∴ f'(0)=2
If X2 + y2 = t-\(\frac { 1 }{ t } \)and x4 + y4 = t2+\(\frac { 1 }{ { t }^{ 2 } } \),then x3y \(\frac { dy }{ dx } \)equals
- (a)
-1
- (b)
0
- (c)
1
- (d)
none of these
x2+y2=t-\(\frac { 1 }{ t } \),x4+y4=t2+\(\frac { 1 }{ { t }^{ 2 } } \)
=\(\left( t-\frac { 1 }{ 2 } \right) ^{ 2 }+2\)
=x4+y4+2x2y2+2
∴x2y2=-1
\(\Rightarrow { x }^{ 2 }.2y\frac { dy }{ dx } +{ y }^{ 2 }.2x=0\\ \Rightarrow { x }^{ 3 }y\frac { dy }{ dx } =-{ x }^{ 2 }{ y }^{ 2 }=1\)
If y1/n = {x + \(\sqrt { \left( 1+{ x }^{ 2 } \right) } \)}, then (1 + x2) y2 +xy1 is equal to
- (a)
n2y
- (b)
ny2
- (c)
n2y2
- (d)
none of these
Since, yi/n = [x +\(\sqrt { \left( 1+{ x }^{ 2 } \right) } \)]
or y=(x+\(\sqrt { \left( 1+{ x }^{ 2 } \right) } \))n
∴y1=n(x+\(\sqrt { \left( 1+{ x }^{ 2 } \right) } \)n-1\(\left( 1+\frac { x }{ \sqrt { 1+{ x }^{ 2 } } } \right) \)
\(\sqrt { \left( 1+{ x }^{ 2 } \right) } \)y1=n(x+)\(\sqrt { \left( 1+{ x }^{ 2 } \right) } \)n
\(\sqrt { \left( 1+{ x }^{ 2 } \right) } \).y1=ny
Squaring both sides
∴ (1+ x2)\({ y }_{ 1 }^{ 2 }\)= n2y2
Differentiating both sides W.r.t. x
∴(1 + x2) 2y1 y2 + \({ y }_{ 1 }^{ 2 }\)· 2x = 2n2yy1
∴(1 + x2)y2 + xy1 = n2y
If f(x) = cot-1\(\left( \frac { { x }^{ x }-{ x }^{ -x } }{ 2 } \right) \)then r (1) is
- (a)
-1
- (b)
1
- (c)
log 2
- (d)
-log 2
f(x)=cot-1\(\left( \frac { { x }^{ x }-{ x }^{ -x } }{ 2 } \right) \)=cot-1\(\left( \frac { \left( x \right) ^{ 2x }-1 }{ { 2x }^{ x } } \right) \)
=tan-1\(\left( \frac { { 2x }^{ x } }{ -1+\left( { x }^{ x } \right) ^{ 2 } } \right) =-2{ tan }^{ -1 }{ x }^{ x }\)
∴f'(x)=-\(\frac { 2 }{ 1+\left( { x }^{ x } \right) ^{ 2 } } \times { x }^{ x }\left( 1+{ log }_{ e }x \right) \)
∴f' (1)=\(\frac { -2 }{ 2 } \times 1=-1\)
Let f(x) = loge\(\left\{ \frac { u\left( x \right) }{ v\left( x \right) } \right\} \),u'(2) = 4, v' (2) = 2,u(2) = 2, v (2) = 1, then f' (2) is equal to
- (a)
0
- (b)
1
- (c)
-1
- (d)
none of these
f(x) = loge\(\left\{ \frac { u\left( x \right) }{ v\left( x \right) } \right\} \)
=loge u(x) +Iog, v(x)
∴f'(x)=\(\frac { { u }^{ ' }(x) }{ u(x) } -\frac { { v }^{ ' }(x) }{ v(x) } \)
f'(2)=\(\frac { { u }^{ ' }(2) }{ u(2) } -\frac { { v }^{ ' }(2) }{ v(2) } \)
=\(\frac { 4 }{ 2 } -\frac { 2 }{ 1 } \)
=2-2=0
If y = logex (x - 2)2 for x ≠ 0, 2, then y' (3) is equal to
- (a)
1/3
- (b)
2/3
- (c)
4/3
- (d)
none of these
\(y=\frac { 2 }{ x } { log }_{ e }\left( x-2 \right) \\ \therefore \frac { dy }{ dx } =\frac { \frac { 2x }{ \left( x-2 \right) } -{ log }_{ e }\left( x-2 \right) }{ { x }^{ 2 } } \\ \therefore \frac { dy }{ dx } |_{ x=3 }=\frac { 6 }{ 9 } =\frac { 2 }{ 3 } \\ or\quad { y }^{ ' }(3)=\frac { 2 }{ 3 } \)
If y=tan-1 \(\left( \frac { 1 }{ \left( 1+x+{ x }^{ 2 } \right) } \right) +{ tan }^{ -1 }\left( \frac { 1 }{ { x }^{ 2 }+3x+3 } \right) +{ tan }^{ -1 }\left( \frac { 1 }{ { x }^{ 2 }+5x+7 } \right) +...+\) up to n terms. then y' (0) is equal to
- (a)
\(-\frac { 1 }{ 1+{ n }^{ 2 } } \)
- (b)
\(-\frac { { n }^{ 2 } }{ 1+{ n }^{ 2 } } \)
- (c)
\(\frac { n }{ 1+{ n }^{ 2 } } \)
- (d)
none of these
\(y={ tan }^{ -1 }\left\{ \frac { 1 }{ 1+x(1+x) } \right\} +{ tan }^{ -1 }\left\{ \frac { 1 }{ 1+(x+1)(x+2) } \right\} \\ +{ tan }^{ -1 }\left\{ \frac { 1 }{ 1+(x+2)(x+3) } \right\} +.....+{ tan }^{ -1 }\left\{ \frac { 1 }{ 1+(x+n-1)(x+n) } \right\} \\ =\sum _{ r=1 }^{ n }{ { tan }^{ -1 } } \left\{ \frac { 1 }{ 1+(x+r-1)(x+r) } \right\} \\ =\sum _{ r=1 }^{ n }{ { tan }^{ -1 } } \left\{ \frac { (x+r)-(x+r-1) }{ 1+(x+r-1)(x+r) } \right\} \\ =\sum _{ r=1 }^{ n }{ \{ { tan }^{ -1 }(x+r)-{ tan }^{ -1 }(x+r-1)\} } \\ ={ tan }^{ -1 }(x+n)-{ tan }^{ -1 }x\\ { y }^{ ' }=\frac { 1 }{ 1+(x+n)^{ 2 } } -\frac { 1 }{ 1+{ x }^{ 2 } } \\ \Rightarrow { y }^{ ' }(0)=\frac { 1 }{ 1+{ n }^{ 2 } } -1=\frac { -{ n }^{ 2 } }{ 1+{ n }^{ 2 } } \)
Let f be a function such that f(x + y) = f(x) + f(y)for all x and y and f(x) = (2x2 + 3x)g(x)for all x where g(x) is continuous and g(0) = 3. Then f' (x) is equal to
- (a)
9
- (b)
3
- (c)
6
- (d)
none of these
\({ f }^{ ' }(x)=\underset { h\rightarrow 0 }{ lim } \frac { f(x+h)-f(x) }{ h } \\ \quad \quad \quad =\underset { h\rightarrow 0 }{ lim } \frac { f(x)+f(h)-f(x) }{ h } \\ \quad \quad \quad =\underset { h\rightarrow 0 }{ lim } \frac { f(h) }{ h } \\ \quad \quad \quad =\underset { h\rightarrow 0 }{ lim } \frac { ({ 2h }^{ 2 }+3h)g(h) }{ h } =\underset { h\rightarrow 0 }{ lim } (2h+3)g(h)\)
= (0 + 3)g(0)
= 3g(0)
=3·3
=9
Let xcos y + y cos x = 5. Then
- (a)
at x = 0, y = 0, y' = 0
- (b)
at x = 0, y = 1, y' = 0
- (c)
at x = y = 1, y' = - 1
- (d)
at x = 1, Y = 0, y' = 1
xcos y + ycos x = 5
⇒ecos y loge x + ecos x loge y = 5
∴ecos y logex\(\left\{ \frac { cos\quad y }{ x } -{ log }_{ e }x\quad sin\quad y\frac { dy }{ dx } \right\} \)
+ecos y logey \(\left\{ \frac { cos\quad x }{ y } \frac { dy }{ dx } -sin\quad x{ log }_{ e }y \right\} =0\)
Put x=y=1,
then (cos 1 - 0)+\(\left( cos\quad 1\frac { dy }{ dx } -0 \right) =0\)
∴\(\frac { dy }{ dx } =-1\)
or y'=-1
If f(x) = (1 + x)n, then the value of f(0) + f' (0) +\(\frac { { f }^{ '' }(0) }{ 2! } +....+\frac { { f }^{ n }(0) }{ n! } \) is
- (a)
n
- (b)
2n
- (c)
2n-1
- (d)
none of these
f' (x) = n (1 + x)n-1,f" (x) = n (n -1)(1 + x)n- 2
∴ fn(x)=n!,fn(0)=n!
⇒ 1+n+\(\frac { n(n-2) }{ 2! } +...+\frac { n! }{ n! } \)
\(^{ n }{ { C }_{ 0 } }+^{ n }{ { C }_{ 1 } }+^{ n }{ { C }_{ 2 } }+...+^{ n }{ { C }_{ n } }={ 2 }^{ n }\)
If f(x) = sin \(\left\{ \frac { \pi }{ 3 } [x]-{ x }^{ 2 } \right\} \) for 2 < x < 3 and [x] denotes the greatest integer less than or equal to x, then \(\left( { f }^{ ' }\sqrt { \pi /3 } \right) \) is equal to
- (a)
\(\sqrt { \pi /3 } \)
- (b)
\(-\sqrt { \pi /3 } \)
- (c)
\(-\sqrt { \pi } \)
- (d)
none of these
∵ 2<x<3
∴[x] = 2
then f(x)= sin \(\left\{ \frac { 2\pi }{ 3 } -{ x }^{ 2 } \right\} \)
∴ f'(x)=cos\(\left\{ \frac { 2\pi }{ 3 } -{ x }^{ 2 } \right\} \)(-2x)
∴\({ f }^{ ' }\left( \sqrt { \frac { \pi }{ 3 } } \right) =cos\left( \frac { 2\pi }{ 3 } -\frac { \pi }{ 3 } \right) \left( -2\sqrt { \frac { \pi }{ 3 } } \right) \)
=\(\frac { 1 }{ 2 } \left( -2\sqrt { \frac { \pi }{ 3 } } \right) \)
=\(-\sqrt { \frac { \pi }{ 3 } } \)
If y=\(\int _{ 0 }^{ x }{ f(t) } \)sin{k(x-t)}dt, then\(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } +{ k }^{ 2 }y\) equals
- (a)
0
- (b)
y
- (c)
k.f(x)
- (d)
k2f(x)
∵ y=\(\int _{ 0 }^{ x }{ f(t) } \)sin{k(x-t)}dt
= Im \(\int _{ 0 }^{ x }{ f(t) } \)eik(x-t) dt
= Im \(\left\{ { e }^{ ikx }\int _{ 0 }^{ x }{ f(t) } { e }^{ -ikt }dt \right\} \) ...(i)
∴\(\frac { dy }{ dx } =Im\left\{ { e }^{ ikx }.f(x){ e }^{ -ikx }+\int _{ 0 }^{ x }{ f(t){ e }^{ -ukt }dt\times { ike }^{ ikx } } \right\} \)
=0+Im\(\left\{ ik.{ e }^{ ikx }\int _{ 0 }^{ x }{ f(t) } { e }^{ -ikt }dt \right\} \)
Now,\(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =Im\left\{ ik.\left( { e }^{ ikx }.f(x){ e }^{ -ikx }+\int _{ 0 }^{ x }{ f(t).{ e }^{ -ikt }dt\times { ike }^{ ikx } } \right) \right\} \)
= k f(x)- ,k2 y [from Eq. (i)]
∴ \(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =+{ k }^{ 2 }y=k.f(x)\)
If y= tan-1\(\left( \frac { { log }_{ e }\left( e/{ x }^{ 2 } \right) }{ { log }_{ e }\left( e{ x }^{ 2 } \right) } \right) +{ tan }^{ -1 }\left( \frac { 3+2{ log }_{ e }x }{ 1-6{ log }_{ e }x } \right) \)then \(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \)is
- (a)
2
- (b)
1
- (c)
0
- (d)
-1
y=tan-1\(\left( \frac { { log }_{ e }\left( e/{ x }^{ 2 } \right) }{ { log }_{ e }\left( e{ x }^{ 2 } \right) } \right) +{ tan }^{ -1 }\left( \frac { 3+2{ log }_{ e }x }{ 1-6{ log }_{ e }x } \right) \)
=tan-1\(\left( \frac { 1-2{ log }_{ e }x }{ 1+2{ log }_{ e }x } \right) +{ tan }^{ -1 }\left( \frac { 3+2{ log }_{ e }x }{ 1-3{ log }_{ e }x } \right) \quad \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \)
= tan-1 (1) - tan-1 (2 loge x) + tan-1 (3)+ tan-1 (2loge, x)
= tan -1(1) + tan -1 (3)\(\frac { dy }{ dx } \)=0
If \(D*f(x)=\underset{h\rightarrow0}{lim}{{f^2(x+h)}-f^2(x)\over h}\)
Where \(f^2(x)=\{f(x)\}^2\)
D* (tan x) is equal to
- (a)
sec2 x
- (b)
2 sec2 x
- (c)
tan x sec2 x
- (d)
2 tan x sec2x
\(D*(tan\ x)=\underset{h\rightarrow0}{lim}{tan^2(x+h)-tan^2x\over h}\)
\(=\underset{h\rightarrow0}{lim}{tan(x+h)-tan\ x\over h}\underset{h\rightarrow0}{lim}(tan(x+h)+tan x)\)
= sec2 x. 2 tan x
= 2tan x sec2 x
If \(D*f(x)=\underset{h\rightarrow0}{lim}{{f^2(x+h)}-f^2(x)\over h}\)
Where \(f^2(x)=\{f(x)\}^2\)
The value of D*f(x) at the point on the curve y = f(x) such that tangent to it are parallel to x-axis, then
- (a)
f(x)
- (b)
zero
- (c)
2f(x)
- (d)
xf(x)
Given, f' (x) = 0
Now \(D*f(x)=\underset{h\rightarrow0}{lim}{f^2(x+h)-f^2(x)\over h}\)
\(=\underset{h\rightarrow0}{lim}{f(x+h)-f(x)\over h}\underset{h\rightarrow0}{lim}(f(x+h)+f(x))\)
=f'(x).f(x)
=0
If \(D*f(x)=\underset{h\rightarrow0}{lim}{{f^2(x+h)}-f^2(x)\over h}\)
Where \(f^2(x)=\{f(x)\}^2\)
The value of D* c, where c is constant is
- (a)
non-zero constant
- (b)
2 constant
- (c)
does not exist
- (d)
zero
\(D*c=\underset{h\rightarrow0}{lim}{c^2-c^2\over h}\)
=0
If Y = f(x) be a differentiable function of x such that whose second, third, ... , nth derivatives exist.
ie, nth derivative of y is denoted by \(y_n{d^ny\over dx^n},D^n_y, f^n(x)\)
⇒ \({d^ny\over dx^n}=\underset{h\rightarrow0}{lim}{f^{n-1}(x+h)-f^{n-1}(x)\over h}\)
If y = e3x + 7, then the value of yn(O) is
- (a)
1
- (b)
3n
- (c)
3n.e7
- (d)
3n.e7.7!
Y = e3x + 7
Y1 = 3e3x + 7, Y2 = 32.e3x + 7...
yn(x)=3n,e3X+7
then yn (0) = 3n, e7
If Y = f(x) be a differentiable function of x such that whose second, third, ... , nth derivatives exist.
ie, nth derivative of y is denoted by \(y_n{d^ny\over dx^n},D^n_y, f^n(x)\)
⇒ \({d^ny\over dx^n}=\underset{h\rightarrow0}{lim}{f^{n-1}(x+h)-f^{n-1}(x)\over h}\)
If \(y={In\ x\over 2-3x}\)then the value of yn(1) is
- (a)
0
- (b)
(-1)n- 3n
- (c)
(-1)n- 3n.n!
- (d)
none of these
y=(2-3x)-1
y1 = (-1)(2-3x)-2 (-3)
y2 = (-1)(-2)(2 - 3x)-3 (-3)2
y3 = (-1)(-2)(-3)(2 - 3x)4,(-3)3
...............................................
yn = (-1)n, n! (2 - 3x)-n-1(-3)n
yn(1) = (-1)n+1, n! (-1)-n-1(-3)n
=(-1)n+13n.n!
If Y = f(x) be a differentiable function of x such that whose second, third, ... , nth derivatives exist.
ie, nth derivative of y is denoted by \(y_n{d^ny\over dx^n},D^n_y, f^n(x)\)
⇒ \({d^ny\over dx^n}=\underset{h\rightarrow0}{lim}{f^{n-1}(x+h)-f^{n-1}(x)\over h}\)
if \(y={In\ x\over x}\)then the value of y" (e) is
- (a)
1
- (b)
-1/e
- (c)
-1/e2
- (d)
-1/e3
\(y={In\ x\over x}\)
\(y'={(1-In\ x)\over x^2}\)
\(y''={x^2\left(-{1\over x}\right)-(1-In\ x)2x\over x^2}\)
\(y''={-e-0\over e^4}={-{1\over e^3}}\)
If Y = f(x) be a differentiable function of x such that whose second, third, ... , nth derivatives exist.
ie, nth derivative of y is denoted by \(y_n{d^ny\over dx^n},D^n_y, f^n(x)\)
⇒ \({d^ny\over dx^n}=\underset{h\rightarrow0}{lim}{f^{n-1}(x+h)-f^{n-1}(x)\over h}\)
If x = sin t, y = sin kt, then the value of(1- X2)Y2 - XY1is
- (a)
k2y
- (b)
-k2y
- (c)
ky2
- (d)
-ky2
\({dy\over dx}={{dy\over dt}\over {dx\over dt}}={k\ cos\ kt\over cos\ t}\)
\(y_1={k\ cos\ kt\over cos\ t}\)
On squaring both sides, we get
(1 - sin2 t)y12 = k2(1 - sin2 kt)
(1 - x2)yf = k2(1 - y2)
Differentiating both sides w.r.t, x, then
(1 - X2) 2y1y2 + y12 (-2x) = 0 - 2k2.yy1
Dividing by 2y, then
(1 - x2)Y2 - .xy1 = -k2y
If Y = f(x) be a differentiable function of x such that whose second, third, ... , nth derivatives exist.
ie, nth derivative of y is denoted by \(y_n{d^ny\over dx^n},D^n_y, f^n(x)\)
⇒ \({d^ny\over dx^n}=\underset{h\rightarrow0}{lim}{f^{n-1}(x+h)-f^{n-1}(x)\over h}\)
If n = 4p+3, p ∈ I and y = tan-1 x, then yn (0) is
- (a)
0
- (b)
n!
- (c)
-(n-1)!
- (d)
(n-1)!
Y = tan-1 x
\(⇒y_1={1\over (1+x^2)}={1\over 2i}\left({1\over x-i}-{1\over x+i}\right)\)
\(={1\over 2iu}\{(x-i)^{-1}-(x+i)^{-1}\}\)
Yn(x) = (n -I)th derivative of y1
\(={1\over 2i}\left\{{(-1)^{n-1}(n-1)!\over 2i}-{(-1)^{n-1}!\over (x+i)^n}\right\}\)
\(y_n(0)={(-1)^{n-1}.(n-1)!\over 2i}\left\{{1\over (-i)^n}{1\over (i)^n}\right\}\)
but n = 4p + 3
then \(y_n(0)={(-1)^{4p+2}(4p+2)!\over 2i}\left\{ {1\over (-i)^{4p+3}}-{1\over {i}^{4p+3}}\right\}\)
\(={1\over 2i}.(4p+2)\left\{{1\over i}+{1\over i}\right\}\)
= - (4p + 2)!
= -. (n -1)!
If x = a cos \(\theta \), y = b sin \(\theta \), then\(\frac { { d }^{ 3 }y }{ { dx }^{ 2 } } \)is equal to
- (a)
\(\left( \frac { -3b }{ { a }^{ 3 } } \right) { cosec }^{ 4 }\theta { cot }^{ 4 }\theta \)
- (b)
\(\left( \frac { 3b }{ { a }^{ 3 } } \right) { cosec }^{ 4 }\theta { cot }^{ 4 }\theta \)
- (c)
\(\left( \frac { -3b }{ { a }^{ 3 } } \right) { cosec }^{ 4 }\theta { cot }\theta \)
- (d)
none of the above
\(\because x=a\quad cos\theta \Rightarrow \frac { dx }{ d\theta } --a\quad sin\theta \\ and\quad y=b\quad sin\theta \Rightarrow \frac { dy }{ d\theta } =b\quad cos\theta \\ \therefore \frac { dy }{ dx } =-\frac { b }{ a } cot\theta \\ \Rightarrow \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =\frac { b }{ a } { cosec }^{ 2 }\theta \frac { d\theta }{ dx } =-\frac { b }{ { a }^{ 2 } } { cosec }^{ 3 }\theta \\ \therefore \frac { { d }^{ 3 }y }{ { dx }^{ 3 } } =-\frac { 3b }{ { a }^{ 2 } } { cosec }^{ 2 }\theta (-{ cosec }\theta cot\theta ).\frac { d\theta }{ dx } \\ =\frac { 3b }{ { a }^{ 2 } } { cosec }^{ 3 }\theta cot\theta \left( -\frac { 1 }{ a\quad sin\theta } \right) =-\frac { 3b }{ { a }^{ 3 } } { cosec }^{ 4 }\theta cot\theta \)
If F(x) =\(\frac { 1 }{ { x }^{ 2 } } \int _{ 4 }^{ x }{ \{ { 4t }^{ 2 }-{ 2F }^{ ' }(t)\} } dt,\)then F'(4) equals
- (a)
32/9
- (b)
64/3
- (c)
64/9
- (d)
none of these
\(\because F(x)=\frac { 1 }{ { x }^{ 2 } } \int _{ 4 }^{ x }{ \{ { 4t }^{ 2 }-2F^{ ' }(t)\} dt } \\ or\quad { }^{ }F(x)\int _{ 4 }^{ x }{ \{ { 4t }^{ 2 }-2F^{ ' }(t)\} dt } \)
Differentiating both sides W.r.t. x, then
x2 F' (x)+ F(x)·2x = 4x2 - 2F' (x)
Put x=4
16F' (4) + 8F(4) = 64 - 2 F' (4)
∴18F'(4)+ 0=64 [':F(4)=0, from Eq. (0]
∴F'(4)=\(\frac { 32 }{ 9 } \)
If y=\(\sqrt { x+\sqrt { y+\sqrt { x+\sqrt { y+...\infty , } } } } then\quad \frac { dy }{ dx } \) is equal to
- (a)
\(\frac { 1 }{ 2y-1 } \)
- (b)
\(\frac { { y }^{ 2 }-x }{ { 2y }^{ 3 }-2xy-1 } \)
- (c)
(2y - 1)
- (d)
none of these
\(\therefore \quad y=\sqrt { x+\sqrt { y+y } } \\ \Rightarrow ({ y }^{ 2 }-x)=\sqrt { 2y } \\ or\quad ({ y }^{ 2 }-x)^{ 2 }=2y\)
Differentiating both sides W.r.t. x, then
\(2({ y }^{ 2 }-x)\left( 2y\frac { dy }{ dx } -1 \right) =2\frac { dy }{ dx } \\ \therefore \frac { dy }{ dx } =\frac { ({ y }^{ 2 }-x) }{ { 2y }^{ 3 }-2xy-1 } \)
The derivative of cos-1\(\left( \frac { { x }^{ -1 }-x }{ { x }^{ -1 }+x } \right) \) at x = - 1 is
- (a)
-2
- (b)
-1
- (c)
0
- (d)
1
Let y = cos-1 \(\left( \frac { { x }^{ -1 }-x }{ { x }^{ -1 }+x } \right) \)
\(\frac { dy }{ dx } =\frac { -1 }{ \sqrt { 1-\left( \frac { { x }^{ -1 }-x }{ { x }^{ -1 }+x } \right) ^{ 2 } } } \\ \times \frac { ({ x }^{ -1 }+x)(-{ x }^{ -2 }-1)-({ x }^{ -1 }-x)(-{ x }^{ -2 }+1) }{ ({ x }^{ -1 }+x)^{ 2 } } \\ =\frac { -1({ x }^{ -1 }+x)(-{ x }^{ -2 }-1)-({ x }^{ -1 }-x)(-{ x }^{ -2 }+1) }{ 2|{ x }^{ -1 }+x| } \\ \therefore \frac { dy }{ dx } |_{ x=-1 }=\frac { -\{ (-1-1)(-1-1)-0\} }{ 2|-1-1| } \\ =-1\)
If \(\int _{ \pi /2 }^{ x }{ \sqrt { (3-2{ sin }^{ 2 }t) } +\int _{ 0 }^{ y }{ cos\quad t\quad dt=0, } } then\left( \frac { dy }{ dx } \right) _{ \pi ,\pi }\)is
- (a)
-3
- (b)
0
- (c)
\(\sqrt { 3 } \)
- (d)
none of these
Differentiating both sides W.r.t. x, then
\(\sqrt { (3-2{ sin }^{ 2 }x) } .1+cos\quad y\frac { dy }{ dx } =0\\ \therefore \frac { dy }{ dx } =-\frac { \sqrt { (3-2sin^{ 2 }x) } }{ cos\quad y } \\ \therefore \frac { dy }{ dx } |_{ (\pi ,\pi ) }=-\frac { \sqrt { 3 } }{ (-1) } =\sqrt { 3 } \)
Let f(x) = xn, n being a non-negative integer, the value of n for which the equality f'(a + b) = f'(a) + f'(b) is valid for all a, b > 0 is
- (a)
0
- (b)
1
- (c)
2
- (d)
none of these
f' (x) = nxn-1
∴ n (a + b )n-1 = nan-1 + nbn-1
⇒n (a + b)n-1 - an-1 - bn-1 = 0
∴n= 0
⇒(a + b )n-1 = an-1 + bn-1 it is true for n = 2
If f(x) = x3 + x2 f' (1) + x f" (2) + f'" (3) for all x \(\in \) R,then
- (a)
f(0) + f(2) = f(1)
- (b)
f(0) + f(3) = 0
- (c)
f(1) + f(3) = f(2)
- (d)
none of these
Given
f(x) = x3 + x2 f' (1) + x f" (2) + f'" (3)
⇒f' (x) = 3x2 + 2x f' (1) + f" (2)
⇒f' (1) = 3 + 2f' (1) + f" (2)
⇒f' (1) + f" (2) = - 3 ...(i)
⇒and f" (x) = 6x + 2f' (1)
∴f" (2) = 12 + 2f' (1)
∴- 2f' (1) + f" (2) = 12 ...(ii)
Solving Eqs. (i) and (ii) we get
f' (1) = - 5 and f" (2) = 2
and f" (x) = 6
∴f'" (3) = 6 ...(iii)
Substituting the values of f' (1), f" (2) and f'" (3) from Eq. (i), (ii)and (iii) in f(x)
∴f(x) = x3 5x2 + 2x + 6
⇒f(0) = 6, f(1) = 4, f(2) = - 2, f(3) = - 6
Hence, f(1) + f(2) = f(1)
f(0) + f(3) = 0
and f(1) + f(3) = f(2)
Let f(x)=(ax+b)cosx+(cx+d)sinx and f' (x) = X cos X be an identity in x, then
- (a)
a=0
- (b)
b=1
- (c)
c=1
- (d)
d=0
f' (x) = a cos x - (ax + iJ) sin x + c sin x + (ex + d) cos x
x cas x = (a + d) cos x + (c - b) sin x
- a x sin x + c x cas x
Comparing the coefficients, we get
a+d=0,c-b=0, -a=0, c=l
∴ a = 0, d = 0, b = 1, c = 1
If f(x) = sin-1(sin x), then
- (a)
\(f'\left(3\pi\over 4\right)=1\)
- (b)
\(f'\left(5\pi\over 4\right)=-1\)
- (c)
\(f'\left(\pi\over2 \right)\)does not exist
- (d)
f' (π) does not exist.
y = sin-1(sin x)
\(=x, -{\pi\over 2}\le x\le {\pi\over 2}\)
\(=\pi-x, {\pi\over 2}< x\le{3\pi\over 2}\)
\({dy\over dx}=1,-{\pi\over 2 }< x< {\pi\over 2}\)
\(=-1,{\pi\over 2}< x< {3\pi\over 2}\)
If F(x) = f(x)g(x) and f'(x) g' (x) = c, then
- (a)
\(F'=c\left[{f\over f'}+{g\over g'}\right]\)
- (b)
\({F''\over F}={f''\over f}+{g''\over g}+{2\over fg}\)
- (c)
\({F'''\over F}={f'''\over f}+{g'''\over g}\)
- (d)
\({F'''\over F''}={f'''\over f''}+{g'''\over ''}\)
Given, F(x) = f(x)· g(x)
Differentiating both sides W.r.t. x, we get
F' (x) = f' (x)· g(x) + g' (x) f(x)
\(⇒F'(X)=f'(X)g'(x)\left[{f(X)\over f'(x)}+{g(x)\over g'(x)}\right]\)
\(⇒F'=c'\left[{f\over f'}+{g\over g'}\right]\)
⇒ (a) is correct
Again differentiating both sides W.r.t. x, we get
F" (x) = f" (x) g(x)
+ g" (x) f(x) + 2f' (x)· g' (x)
⇒ F"(x) = f" (x)· g(x) + g" (x)· f(x) + 2c ...(ii)
Dividing both sides by F(x) = f(x} g(x)
Then \({F''(x)\over F(x)}={F''(x)\over f(x)}+{g''(x)\over g(x)}+{2c\over f(x)g(x)}\)
or \({F''\over F}={f''\over f}+{g''\over g}+{2c\over fg}\)
(b) is correct
Again given, I' (x) g' (x) = c
Differentiating both sides W.r.t. x, we get
f' (x) g" (x) + g' (x) I" (x) = 0
From Eq. eii)
F"(x) = f" (x)· gex) + g" (x)· f(x) + 2c
Differentiating both sides W.r.t. x, we get
F'''(x)=f''(x)·g'(x)+ 1''' (x)· g(x)
+ g" (x)-f' (x) + fex)· g'" (x) + 0
= 1''' (x) g(x) + g'" (x)· f(x) + 0
Now, dividing both sides by F(x)=f(x)g(x)
Then, \({F'''(x)\over F(x)}={f'''(x)\over f(x)}+{g'''\over g(x)}\)
or \({F'''\over F}={f'''\over f}+{g'''\over g}\)
(c) is correct
If 1 is a twice repeated root of the equation a x3 +bx 2 +bx +d = 0 then
- (a)
a = b = d
- (b)
a + b = 0
- (c)
b + d = 0
- (d)
a = d
Let f(x) = ax3 + bx2 + bx + d
Then f(1) = 0 and f' (1) = 0
⇒ a+2b+d=O
and 3a + 2b + b = 0
From Eqs. (i) and (ii)
a=d=-b
Differential coefficient of sin-1 x W.r.t. sin-1(3x - 4 x3)i
- (a)
\({1\over 3}\ if\ -{\pi\over 8}<x<{\pi\over 8}\)
- (b)
\(3\ if\ {-\pi\over 8}<x<{\pi\over 8}\)
- (c)
\({1\over 3}\ if\ -{\pi\over 9}<x<{\pi\over 9}\)
- (d)
\(3\ if\ -{\pi\over 9}<x<{\pi\over 9}\)
u=sin-1x
\(v=sin^{-1}(3x-4x^3)=\begin{cases} -\pi-3sin^{-1}x,-1\le x\le-{1\over 2}\\ 3sin^{-1},-{1\over 2}\le x\le{1\over 2 }\\\pi-3sin^{_1}x,{1\over 2}\le x\le 1 \end{cases}\)
\({du\over dv}=\begin{cases} -{1\over 3},-1\le x\le-{1\over 2}\\{1\over3},-{1\over 2}\le x\le{1\over 2 }\\-{1\over 3},{1\over 2}\le x\le 1 \end{cases}\)
If y=\(\frac { \sqrt { (1+{ t }^{ 2 }) } -\sqrt { (1-{ t }^{ 2 }) } }{ \sqrt { (1+{ t }^{ 2 }) } +\sqrt { (1-{ t }^{ 2 }) } } and\quad x=\sqrt { (1-{ t }^{ 4 }) } ,\quad then\quad \frac { dy }{ dx } \) is equal to
- (a)
\(\frac { -1 }{ { t }^{ 2 }\{ 1+\sqrt { 1-{ t }^{ 4 } } \} } \)
- (b)
\(\frac { \{ \sqrt { (1-{ t }^{ 4 }) } -1\} }{ { t }^{ 6 } } \)
- (c)
\(\frac { 1 }{ { t }^{ 2 }\{ 1+\sqrt { 1-{ t }^{ 4 } } \} } \)
- (d)
\(\frac { 1-\sqrt { (1-{ t }^{ 4 }) } }{ { t }^{ 6 } } \)
\(y=\frac { \left( \sqrt { 1+{ t }^{ 2 } } \right) -\left( \sqrt { 1-{ t }^{ 2 } } \right) ^{ 2 } }{ \left( \sqrt { 1+{ t }^{ 2 } } \right) ^{ 2 }-\left( \sqrt { 1-{ t }^{ 2 } } \right) ^{ 2 } } \\ \quad =\frac { 2-2\sqrt { 1-{ t }^{ 4 } } }{ { 2t }^{ 2 } } =\frac { 1-\sqrt { 1-{ t }^{ 4 } } }{ { t }^{ 2 } } \\ \therefore \frac { dy }{ dt } =\frac { { t }^{ 2 }\left\{ 0-\frac { 1 }{ 2\sqrt { 1-{ t }^{ 4 } } } \times -4{ t }^{ 3 } \right\} -\{ 1-\sqrt { 1-{ t }^{ 4 } } \} 2t }{ { t }^{ 4 } } \\ =\frac { \frac { { 2t }^{ 5 } }{ \sqrt { 1-{ t }^{ 4 } } } -2t\{ 1-\sqrt { 1-{ t }^{ 4 } } \} }{ { t }^{ 4 } } \\ =\frac { 2\{ 1-\sqrt { 1-{ t }^{ 4 } } \} }{ { t }^{ 3 }\sqrt { 1-{ t }^{ 4 } } } \\ \frac { dx }{ dt } =\frac { { -2t }^{ 3 } }{ \sqrt { 1-{ t }^{ 4 } } } \\ \therefore \frac { dy }{ dx } =\frac { \sqrt { 1-{ t }^{ 4 } } -1 }{ { t }^{ 6 } } =\frac { -1 }{ { t }^{ 2 }\{ 1+\sqrt { 1-{ t }^{ 4 } } \} } \)
If f(x) =\(\left| \begin{matrix} { x }^{ n } & sin\quad x & cos\quad x \\ n! & sin(n\pi /2) & cos(n\pi /2) \\ a & { a }^{ 2 } & { a }^{ 3 } \end{matrix} \right| \) then the value of \(\frac { { d }^{ n } }{ { dx }^{ n } } \) (f(x)) at x = 0 for n = 2m + 1 is
- (a)
-1
- (b)
0
- (c)
1
- (d)
independent of a
\(\frac { { d }^{ n } }{ { dx }^{ n } } [f(x)]=\left| \begin{matrix} \frac { { d }^{ n } }{ { dx }^{ n } } { x }^{ n } & \frac { { d }^{ n } }{ { dx }^{ n } } sin\quad x & -\frac { { d }^{ n } }{ { dx }^{ n } } cos\quad x \\ n! & sin\left( n\pi /2 \right) & cos\left( n\pi /2 \right) \\ a & { a }^{ 2 } & { a }^{ 3 } \end{matrix} \right| \\ =\left| \begin{matrix} n! & sin\left( \frac { n\pi }{ 2 } +x \right) & cos\left( \frac { n\pi }{ 2 } +x \right) \\ n! & sin\quad n\pi /2 & cos\quad n\pi /2 \\ a & { a }^{ 2 } & { a }^{ 3 } \end{matrix} \right| \\ \therefore \quad At\quad x=0\quad [{ R }_{ 1 }={ R }_{ 2 }]\\ =0\)