Consider the function $f ( x ) = x ^ { n } ( 1 - x ) ^ { n } / n !$, where $n \geq 1$ is a fixed integer. Let $f ^ { ( k ) }$ denote the $k$-th derivative of $f$. Which of the following is true for all $k \geq 1$?
(a) $f ^ { ( k ) } ( 0 )$ and $f ^ { ( k ) } ( 1 )$ are integers.
(b) $f ^ { ( k ) } ( 0 )$ is an integer, but not $f ^ { ( k ) } ( 1 )$
(c) $f ^ { ( k ) } ( 1 )$ is an integer, but not $f ^ { ( k ) } ( 0 )$
(d) Neither $f ^ { ( k ) } ( 1 )$ nor $f ^ { ( k ) } ( 0 )$ is an integer.

Let $f(x) = e^{-1/x}$ for $x > 0$ and $f(x) = 0$ for $x \leq 0$. Which of the following is true?
(A) $f$ is not differentiable at $x = 0$
(B) $f$ is differentiable at $x = 0$ but $f'$ is not differentiable at $x = 0$
(C) $f$ is differentiable at $x = 0$ and $f'$ is differentiable at $x = 0$
(D) $f$ is differentiable everywhere and $f'$ is also differentiable everywhere

Let $f(x) = x|x|^n$ for $n \geq 1$ a positive integer. Which of the following is true?
(A) $f$ is differentiable everywhere except at $x = 0$
(B) $f$ is continuous but not differentiable at $x = 0$
(C) $f$ is differentiable everywhere
(D) None of the above

The limit $$\lim _ { x \rightarrow 0 } \frac { \left( e ^ { x } - 1 \right) \tan ^ { 2 } x } { x ^ { 3 } }$$ (A) does not exist
(B) exists and equals 0
(C) exists and equals $2/3$
(D) exists and equals 1

Let $f _ { 1 } ( x ) = e ^ { x } , f _ { 2 } ( x ) = e ^ { f _ { 1 } ( x ) }$ and generally $f _ { n + 1 } ( x ) = e ^ { f _ { n } ( x ) }$ for all $n \geq 1$. For any fixed $n$, the value of $\frac { d } { d x } f _ { n } ( x )$ is equal to
(A) $f _ { n } ( x )$
(B) $f _ { n } ( x ) f _ { n - 1 } ( x )$
(C) $f _ { n } ( x ) f _ { n - 1 } ( x ) \cdots f _ { 1 } ( x )$
(D) $f _ { n + 1 } ( x ) f _ { n } ( x ) \cdots f _ { 1 } ( x ) e ^ { x }$

Let $f(x) = a_0 + a_1 |x| + a_2 |x|^2 + a_3 |x|^3$, where $a_0, a_1, a_2, a_3$ are constants. Then
(A) $f(x)$ is differentiable at $x = 0$ whatever be $a_0, a_1, a_2, a_3$
(B) $f(x)$ is not differentiable at $x = 0$ whatever be $a_0, a_1, a_2, a_3$
(C) $f(x)$ is differentiable at $x = 0$ only if $a_1 = 0$
(D) $f(x)$ is differentiable at $x = 0$ only if $a_1 = 0, a_3 = 0$

$\lim _ { x \rightarrow 0 } \frac { \left( e ^ { x } - 1 \right) \tan ^ { 2 } x } { x ^ { 3 } }$
(a) does not exist
(b) exists and equals 0
(c) exists and equals $\frac { 2 } { 3 }$
(d) exists and equals 1.

Let $f _ { 1 } ( x ) = e ^ { x } , f _ { 2 } ( x ) = e ^ { f _ { 1 } ( x ) }$ and generally $f _ { n + 1 } ( x ) = e ^ { f _ { n } ( x ) }$ for all $n \geq 1$. For any fixed $n$, the value of $\frac { d } { d x } f _ { n } ( x )$ is:
(a) $f _ { n } ( x )$
(b) $f _ { n } ( x ) f _ { n - 1 } ( x )$
(c) $f _ { n } ( x ) f _ { n - 1 } ( x ) \ldots f _ { 1 } ( x )$
(d) $f _ { n + 1 } ( x ) f _ { n } ( x ) \ldots f _ { 1 } ( x ) e ^ { x }$.

Let $f _ { 1 } ( x ) = e ^ { x } , f _ { 2 } ( x ) = e ^ { f _ { 1 } ( x ) }$ and generally $f _ { n + 1 } ( x ) = e ^ { f _ { n } ( x ) }$ for all $n \geq 1$. For any fixed $n$, the value of $\frac { d } { d x } f _ { n } ( x )$ is:
(a) $f _ { n } ( x )$
(b) $f _ { n } ( x ) f _ { n - 1 } ( x )$
(c) $f _ { n } ( x ) f _ { n - 1 } ( x ) \ldots f _ { 1 } ( x )$
(d) $f _ { n + 1 } ( x ) f _ { n } ( x ) \ldots f _ { 1 } ( x ) e ^ { x }$.

The limit $$\lim _ { x \rightarrow 0 } \frac { \left( e ^ { x } - 1 \right) \tan ^ { 2 } x } { x ^ { 3 } }$$ (A) does not exist
(B) exists and equals 0
(C) exists and equals $2 / 3$
(D) exists and equals 1

Let $f _ { 1 } ( x ) = e ^ { x } , f _ { 2 } ( x ) = e ^ { f _ { 1 } ( x ) }$ and generally $f _ { n + 1 } ( x ) = e ^ { f _ { n } ( x ) }$ for all $n \geq 1$. For any fixed $n$, the value of $\frac { d } { d x } f _ { n } ( x )$ is equal to
(A) $f _ { n } ( x )$
(B) $f _ { n } ( x ) f _ { n - 1 } ( x )$
(C) $f _ { n } ( x ) f _ { n - 1 } ( x ) \cdots f _ { 1 } ( x )$
(D) $f _ { n + 1 } ( x ) f _ { n } ( x ) \cdots f _ { 1 } ( x ) e ^ { x }$

Suppose $f : \mathbb { R } \rightarrow \mathbb { R }$ is a function given by
$$f ( x ) = \begin{cases} 1 & \text { if } x = 1 \\ e ^ { \left( x ^ { 10 } - 1 \right) } + ( x - 1 ) ^ { 2 } \sin \left( \frac { 1 } { x - 1 } \right) & \text { if } x \neq 1 \end{cases}$$
(a) Find $f ^ { \prime } ( 1 )$.
(b) Evaluate $\lim _ { u \rightarrow \infty } \left[ 100 u - u \sum _ { k = 1 } ^ { 100 } f \left( 1 + \frac { k } { u } \right) \right]$.

Consider the function $f : \mathbb{R} \rightarrow \mathbb{R}$ defined as $$f(x) = \begin{cases} \frac{x}{e^x - 1} & \text{if } x \neq 0 \\ 1 & \text{if } x = 0 \end{cases}$$ Then which one of the following statements is correct?
(A) $f$ is not continuous at $x = 0$.
(B) $f$ is continuous but not differentiable at $x = 0$.
(C) $f$ is differentiable at $x = 0$ and $f'(0) = -\frac{1}{2}$.
(D) $f$ is differentiable at $x = 0$ and $f'(0) = \frac{1}{2}$.

Prove that the family of curves
$$\frac{x^{2}}{a^{2} + \lambda} + \frac{y^{2}}{b^{2} + \lambda} = 1$$
satisfies
$$\frac{dy}{dx}\left(a^{2} - b^{2}\right) = \left(x + y\frac{dy}{dx}\right)\left(x\frac{dy}{dx} - y\right).$$

Define $f : \mathbb { R } \rightarrow \mathbb { R }$ by $$f ( x ) = \begin{cases} ( 1 - \cos x ) \sin \left( \frac { 1 } { x } \right) , & x \neq 0 \\ 0 , & x = 0 \end{cases}$$ Then,
(A) $f$ is discontinuous.
(B) $f$ is continuous but not differentiable.
(C) $f$ is differentiable and its derivative is discontinuous.
(D) $f$ is differentiable and its derivative is continuous.

Let $f , g$ be continuous functions from $[ 0 , \infty )$ to itself, $$h ( x ) = \int _ { 2 ^ { x } } ^ { 3 ^ { x } } f ( t ) d t , x > 0$$ and $$F ( x ) = \int _ { 0 } ^ { h ( x ) } g ( t ) d t , x > 0$$ If $F ^ { \prime }$ is the derivative of $F$, then for $x > 0$,
(A) $F ^ { \prime } ( x ) = g ( h ( x ) )$.
(B) $F ^ { \prime } ( x ) = g ( h ( x ) ) \left[ f \left( 3 ^ { x } \right) - f \left( 2 ^ { x } \right) \right]$.
(C) $F ^ { \prime } ( x ) = g ( h ( x ) ) \left[ x 3 ^ { x - 1 } f \left( 3 ^ { x } \right) - x 2 ^ { x - 1 } f \left( 2 ^ { x } \right) \right]$.
(D) $F ^ { \prime } ( x ) = g ( h ( x ) ) \left[ 3 ^ { x } f \left( 3 ^ { x } \right) \ln 3 - 2 ^ { x } f \left( 2 ^ { x } \right) \ln 2 \right]$.

15. $\operatorname { Lim } x \rightarrow 1 \sqrt { } ( 1 - \cos 2 ( x - 1 ) ) / ( x - 1 )$ :
(A) exists and it equals $\sqrt { } 2$.
(B) exists and it equals $- \sqrt { } 2$
(C) does not exist because $x - 1 - - > 0$
(D) does not exist because left hand limit is not equal to right hand limit
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1. If in a triangle $P Q R , \sin P , \sin Q , \sin R$ are in $A$. $P$., then :
   (A) the altitudes are in A.P.
   (B) the altitudes are in H.P.
   (C) the medians are in G.P.
   (D) the medians are in A.P.
2. If $a n = \sum r = 0 n 1 / n \mathrm { Cr }$, then $\sum r = 0 n \mathrm { r } / \mathrm { n }$ Cr equals:
   (A) (n - 1) an
   (B) $n$ an
   (C) $1 / 2$ nan
   (D) none of these
3. If the vertices $P , Q , R$ of a triangle $P Q R$ are rational points, which of the following points of the triangle PQR is/(are) always rational point(s).
   (A) centroid \&
   (B) incentre
   (C) circumcentre
   (D) orthocenter
   (A rational point is a point both of whose co-ordinates are rational numbers).
4. The number of values of $c$ such that the straight line $y = 4 x + c$ touches the curve $x 2 / 4 + \mathrm { y } 2 = 1$ is :
   (A) 0
   (B) 1
   (C) 2
   (D) infinite.
5. If $x > 1 , y > 1 , z > 1$ are in G.P., then $1 / ( 1 + \operatorname { In } x ) , 1 / ( 1 + \operatorname { In } y ) , 1 / ( 1 + \operatorname { In } z )$ are in :
   (A) A.P.
   (B) H.P.
   (C) G.P.
   (D) none of these
6. The number of values of $x$ in the interval $[ 0,5 p ]$ satisfying the equation $3 \sin 2 x - 7 x + 2 = 0$ is:
   (A) 0
   (B) 5
   (C) 6
   (D) 10
7. The order of the differential equation whose general solution is given by

$$\begin{aligned} & y = \left( C _ { 1 } + C _ { 2 } \right) \cos \left( x + C _ { 3 } \right) - C _ { 4 } e ^ { x + C s } \\ & \text { where } C _ { 1 } , C _ { 2 } , C _ { 3 } , C _ { 4 } , C _ { 5 } \end{aligned}$$
are arbitrary constants, is:

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(A) 5
(B) 4
(C) 3
(D) 2

27. Which of the following functions is differentiable at $x = 0$ :
(A) $\cos ( | x | ) + | x |$
(B) $\cos ( | x | ) - | x |$
(C) $\sin ( | x | ) + | x |$
(D) $\sin ( | x | ) - | x |$

22. The integer $n$ for which $\lim _ { x \rightarrow 0 } ( \cos x - 1 ) \left( \cos x - e ^ { x } \right) / x ^ { n }$ is a finite non-zero number is
(A) 1

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(B) 2
(C) 3
(D) 4

23. Let $f : R \rightarrow R$ be such that $f ( 1 ) = 3$, and $f ^ { \prime } ( 1 ) = 6$ Then $\lim _ { x \rightarrow 0 } ( f ( 1 + x ) / f ( 1 ) ) ^ { 1 / x }$ equals
(A) 1
(B) $\quad \mathrm { e } ^ { 1 / 2 }$
(C) $\quad \mathrm { e } ^ { 2 }$
(D) $\quad \mathrm { e } ^ { 3 }$

9. If $\lim _ { ( x \rightarrow 0 ) } ( ( ( a - n ) n x - \tan x ) \sin n x ) / x ^ { 2 } = 0$, where $n$ is non zero real number, then $a$ is equal to:
(a) 0
(b) $( n + 1 ) / n$
(c) $n$
(d) $n + 1 / n$

7. If $f : [ - 1,1 ] \rightarrow R$ and $f ^ { \prime } ( 0 ) = \lim _ { n \rightarrow \infty } n f \left( \frac { 1 } { n } \right)$ and $f ( 0 ) = 0$. Find the value of $\lim _ { n \rightarrow \infty } \frac { 2 } { \pi } ( n + 1 ) \cos ^ { - 1 } \left( \frac { 1 } { n } \right) - n$.
Given that $0 < \left| \lim _ { \mathrm { n } \rightarrow \infty } \cos ^ { - 1 } \left( \frac { 1 } { \mathrm { n } } \right) \right| < \frac { \pi } { 2 }$. Sol. $\lim _ { n \rightarrow \infty } \frac { 2 } { \pi } ( n + 1 ) \cos ^ { - 1 } \frac { 1 } { n } - n = \lim _ { n \rightarrow \infty } n \left[ \frac { 2 } { \pi } \left( 1 + \frac { 1 } { n } \right) \cos ^ { - 1 } \frac { 1 } { n } - 1 \right]$ $= \lim _ { n \rightarrow \infty } n f \left( \frac { 1 } { n } \right) = f ^ { \prime } ( 0 )$ where $f ( x ) = \frac { 2 } { \pi } ( 1 + x ) \cos ^ { - 1 } x - 1$. Clearly, $\mathrm { f } ( 0 ) = 0$.
Now, $f ^ { \prime } ( x ) = \frac { 2 } { \pi } \left[ ( 1 + x ) \frac { - 1 } { \sqrt { 1 - x ^ { 2 } } } + \cos ^ { - 1 } x \right]$ $\Rightarrow \mathrm { f } ^ { \prime } ( 0 ) = \frac { 2 } { \pi } \left[ - 1 + \frac { \pi } { 2 } \right] = \frac { 2 } { \pi } \left[ \frac { \pi - 2 } { 2 } \right] = 1 - \frac { 2 } { \pi }$.

7. If $f : [ - 1,1 ] \rightarrow R$ and $f ^ { \prime } ( 0 ) = \lim _ { n \rightarrow \infty } n f \left( \frac { 1 } { n } \right)$ and $f ( 0 ) = 0$. Find the value of $\lim _ { n \rightarrow \infty } \frac { 2 } { \pi } ( n + 1 ) \cos ^ { - 1 } \left( \frac { 1 } { n } \right) - n$.
Given that $0 < \left| \lim _ { \mathrm { n } \rightarrow \infty } \cos ^ { - 1 } \left( \frac { 1 } { \mathrm { n } } \right) \right| < \frac { \pi } { 2 }$. Sol. $\lim _ { n \rightarrow \infty } \frac { 2 } { \pi } ( n + 1 ) \cos ^ { - 1 } \frac { 1 } { n } - n = \lim _ { n \rightarrow \infty } n \left[ \frac { 2 } { \pi } \left( 1 + \frac { 1 } { n } \right) \cos ^ { - 1 } \frac { 1 } { n } - 1 \right]$ $= \lim _ { n \rightarrow \infty } n f \left( \frac { 1 } { n } \right) = f ^ { \prime } ( 0 )$ where $f ( x ) = \frac { 2 } { \pi } ( 1 + x ) \cos ^ { - 1 } x - 1$. Clearly, $\mathrm { f } ( 0 ) = 0$.
Now, $f ^ { \prime } ( x ) = \frac { 2 } { \pi } \left[ ( 1 + x ) \frac { - 1 } { \sqrt { 1 - x ^ { 2 } } } + \cos ^ { - 1 } x \right]$ $\Rightarrow \mathrm { f } ^ { \prime } ( 0 ) = \frac { 2 } { \pi } \left[ - 1 + \frac { \pi } { 2 } \right] = \frac { 2 } { \pi } \left[ \frac { \pi - 2 } { 2 } \right] = 1 - \frac { 2 } { \pi }$.

16. $f ( x ) = \begin{cases} b \sin ^ { - 1 } \left( \frac { x + c } { 2 } \right) , & - \frac { 1 } { 2 } < x < 0 \\ \frac { 1 } { 2 } , & x = 0 \\ \frac { e ^ { \frac { a } { 2 } x } - 1 } { x } , & 0 < x < \frac { 1 } { 2 } \end{cases}$
If $\mathrm { f } ( \mathrm { x } )$ is differentiable at $\mathrm { x } = 0$ and $| \mathrm { c } | < \frac { 1 } { 2 }$ then find the value of ' a ' and prove that $64 \mathrm {~b} ^ { 2 } = \left( 4 - \mathrm { c } ^ { 2 } \right)$. Sol. $\mathrm { f } \left( 0 ^ { + } \right) = \mathrm { f } \left( 0 ^ { - } \right) = \mathrm { f } ( 0 )$ Here $\mathrm { f } \left( 0 ^ { + } \right) = \lim _ { \mathrm { x } \rightarrow \infty } \frac { \mathrm { e } ^ { \frac { \mathrm { ax } } { 2 } } - 1 } { \mathrm { x } } = \lim _ { \mathrm { x } \rightarrow \infty } \frac { \mathrm { e } ^ { \frac { \mathrm { ax } } { 2 } } - 1 } { \frac { \mathrm { ax } } { 2 } } \cdot \frac { \mathrm { a } } { 2 } = \frac { \mathrm { a } } { 2 }$. $\Rightarrow \mathrm { b } \sin ^ { - 1 } \frac { \mathrm { c } } { 2 } = \frac { \mathrm { a } } { 2 } = \frac { 1 } { 2 } \Rightarrow \mathrm { a } = 1$. $\mathrm { L } \mathrm { f } ^ { \prime } \left( 0 _ { - } \right) = \lim _ { \mathrm { h } \rightarrow 0 ^ { - } } \frac { \mathrm { b } \sin ^ { - 1 } \frac { ( \mathrm {~h} + \mathrm { c } ) } { 2 } - \frac { 1 } { 2 } } { \mathrm {~h} } = \frac { \mathrm { b } / 2 } { \sqrt { 1 - \frac { \mathrm { c } ^ { 2 } } { 4 } } }$ $R f ^ { \prime } \left( 0 _ { + } \right) = \lim _ { h \rightarrow 0 ^ { + } } \frac { \frac { e ^ { h / 2 } - 1 } { h } - \frac { 1 } { 2 } } { h } = \frac { 1 } { 8 }$ Now L $f ^ { \prime } \left( 0 _ { - } \right) = \mathrm { R } \mathrm { f } ^ { \prime } \left( 0 _ { + } \right) \Rightarrow \frac { \frac { \mathrm { b } } { 2 } } { \sqrt { 1 - \frac { \mathrm { c } ^ { 2 } } { 4 } } } = \frac { 1 } { 8 }$ $4 b = \sqrt { 1 - \frac { c ^ { 2 } } { 4 } } \Rightarrow 16 b ^ { 2 } = \frac { 4 - c ^ { 2 } } { 4 } \Rightarrow 64 b ^ { 2 } = 4 - c ^ { 2 }$.

16. $\mathrm { f } ( \mathrm { x } ) = \begin{cases} \mathrm { b } \sin ^ { - 1 } \left( \frac { \mathrm { x } + \mathrm { c } } { 2 } \right) , & - \frac { 1 } { 2 } < \mathrm { x } < 0 \\ \frac { 1 } { 2 } , & \mathrm { x } = 0 \\ \frac { \mathrm { e } ^ { \frac { \mathrm { a } } { 2 } \mathrm { x } } - 1 } { \mathrm { x } } , & 0 < \mathrm { x } < \frac { 1 } { 2 } \end{cases}$
If $\mathrm { f } ( \mathrm { x } )$ is differentiable at $\mathrm { x } = 0$ and $| \mathrm { c } | < \frac { 1 } { 2 }$ then find the value of ' a ' and prove that $64 \mathrm {~b} ^ { 2 } = \left( 4 - \mathrm { c } ^ { 2 } \right)$. Sol. $f \left( 0 ^ { + } \right) = f \left( 0 ^ { - } \right) = f ( 0 )$ Here $\mathrm { f } \left( 0 ^ { + } \right) = \lim _ { \mathrm { x } \rightarrow \infty } \frac { \mathrm { e } ^ { \frac { \mathrm { ax } } { 2 } } - 1 } { \mathrm { x } } = \lim _ { \mathrm { x } \rightarrow \infty } \frac { \mathrm { e } ^ { \frac { \mathrm { ax } } { 2 } } - 1 } { \frac { \mathrm { ax } } { 2 } } \cdot \frac { \mathrm { a } } { 2 } = \frac { \mathrm { a } } { 2 }$. $\Rightarrow \mathrm { b } \sin ^ { - 1 } \frac { \mathrm { c } } { 2 } = \frac { \mathrm { a } } { 2 } = \frac { 1 } { 2 } \Rightarrow \mathrm { a } = 1$. $\mathrm { L } \mathrm { f } ^ { \prime } \left( 0 _ { - } \right) = \lim _ { \mathrm { h } \rightarrow 0 ^ { - } } \frac { \mathrm { b } \sin ^ { - 1 } \frac { ( \mathrm {~h} + \mathrm { c } ) } { 2 } - \frac { 1 } { 2 } } { \mathrm {~h} } = \frac { \mathrm { b } / 2 } { \sqrt { 1 - \frac { \mathrm { c } ^ { 2 } } { 4 } } }$ $R f ^ { \prime } \left( 0 _ { + } \right) = \lim _ { h \rightarrow 0 ^ { + } } \frac { \frac { e ^ { h / 2 } - 1 } { h } - \frac { 1 } { 2 } } { h } = \frac { 1 } { 8 }$ Now $\mathrm { L } ^ { \prime } \left( 0 _ { - } \right) = \mathrm { Rf } ^ { \prime } \left( 0 _ { + } \right) \Rightarrow \frac { \frac { \mathrm { b } } { 2 } } { \sqrt { 1 - \frac { \mathrm { c } ^ { 2 } } { 4 } } } = \frac { 1 } { 8 }$ $4 b = \sqrt { 1 - \frac { c ^ { 2 } } { 4 } } \Rightarrow 16 b ^ { 2 } = \frac { 4 - c ^ { 2 } } { 4 } \Rightarrow 64 b ^ { 2 } = 4 - c ^ { 2 }$.