Let $f ( x )$ be a polynomial of degree 5 such that $x = \pm 1$ are its critical points. If $\lim _ { x \rightarrow 0 } \left( 2 + \frac { f ( x ) } { x ^ { 3 } } \right) = 4$, then which one of the following is not true?
(1) $f$ is an odd function
(2) $f ( 1 ) - 4 f ( - 1 ) = 4$
(3) $x = 1$ is a point of local minimum and $x = - 1$ is a point of local maximum
(4) $x = 1$ is a point of local maxima of $f$

The function, $f ( x ) = ( 3 x - 7 ) x ^ { \frac { 2 } { 3 } } , x \in \mathrm { R }$, is increasing for all $x$ lying in
(1) $( - \infty , 0 ) \cup \left( \frac { 14 } { 15 } , \infty \right)$
(2) $( - \infty , 0 ) \cup \left( \frac { 3 } { 7 } , \infty \right)$
(3) $\left( - \infty , \frac { 14 } { 15 } \right)$
(4) $\left( - \infty , - \frac { 14 } { 15 } \right) \cup ( 0 , \infty )$

Let a function $f : [ 0,5 ] \rightarrow R$ be continuous, $f ( 1 ) = 3$ and $F$ be defined as: $F ( x ) = \int _ { 1 } ^ { x } t ^ { 2 } g ( t ) d t$, where $g ( t ) = \int _ { 1 } ^ { t } f ( u ) d u$. Then for the function $F ( x )$, the point $x = 1$ is:
(1) a point of local minima
(2) not a critical point
(3) a point of local maxima
(4) a point of inflection

If $p(x)$ be a polynomial of degree three that has a local maximum value 8 at $x = 1$ and a local minimum value 4 at $x = 2$ then $p(0)$ is equal to
(1) 6
(2) $-12$
(3) 24
(4) 12

If $x = 1$ is a critical point of the function $f(x) = (3x^2 + ax - 2 - a)e^x$, then
(1) $x = 1$ and $x = -\frac{2}{3}$ are local minima of $f$
(2) $x = 1$ and $x = -\frac{2}{3}$ is a local maxima of $f$
(3) $x = 1$ is a local maxima and $x = -\frac{2}{3}$ is a local minima of $f$
(4) $x = 1$ is a local minima and $x = -\frac{2}{3}$ are local maxima of $f$

Let $f : ( - 1 , \infty ) \rightarrow R$ be defined by $f ( 0 ) = 1$ and $f ( x ) = \frac { 1 } { x } \log _ { e } ( 1 + x ) , x \neq 0$. Then the function $f$
(1) Decreases in $( - 1,0 )$ and increases in $( 0 , \infty )$
(2) Increases in $( - 1 , \infty )$
(3) Increases in $( - 1,0 )$ and decreases in $( 0 , \infty )$
(4) Decreases in $( - 1 , \infty )$

The area (in sq. units) of the largest rectangle $ABCD$ whose vertices $A$ and $B$ lie on the $x$-axis and vertices $C$ and $D$ lie on the parabola, $y = x ^ { 2 } - 1$ below the $x$-axis, is:
(1) $\frac { 2 } { 3 \sqrt { 3 } }$
(2) $\frac { 1 } { 3 \sqrt { 3 } }$
(3) $\frac { 4 } { 3 }$
(4) $\frac { 4 } { 3 \sqrt { 3 } }$

If $P$ is a point on the parabola $y = x ^ { 2 } + 4$ which is closest to the straight line $y = 4 x - 1$, then the coordinates of $P$ are:
(1) $( - 2,8 )$
(2) $( 1,5 )$
(3) $( 2,8 )$
(4) $( 3,13 )$

The triangle of maximum area that can be inscribed in a given circle of radius ' $r$ ' is :
(1) An equilateral triangle having each of its side of length $\sqrt { 3 } r$.
(2) An isosceles triangle with base equal to $2 r$.
(3) An equilateral triangle of height $\frac { 2 r } { 3 }$.
(4) A right angle triangle having two of its sides of length $2 r$ and $r$.

Let $f$ be a real valued function, defined on $R - \{ - 1,1 \}$ and given by $f ( x ) = 3 \log _ { \mathrm { e } } \left| \frac { x - 1 } { x + 1 } \right| - \frac { 2 } { x - 1 }$. Then in which of the following intervals, function $f ( x )$ is increasing?
(1) $( - \infty , - 1 ) \cup \left( \left[ \frac { 1 } { 2 } , \infty \right) - \{ 1 \} \right)$
(2) $( - \infty , \infty ) - \{ - 1,1 \}$
(3) $\left( - 1 , \frac { 1 } { 2 } \right]$
(4) $\left( - \infty , \frac { 1 } { 2 } \right] - \{ - 1 \}$

The sum of all the local minimum values of the twice differentiable function $f : R \rightarrow R$ defined by $f ( x ) = x ^ { 3 } - 3 x ^ { 2 } - \frac { 3 f ^ { \prime \prime } ( 2 ) } { 2 } x + f ^ { \prime \prime } ( 1 )$ is:
(1) - 22
(2) 5
(3) - 27
(4) 0

A box open from top is made from a rectangular sheet of dimension $a \times b$ by cutting squares each of side $x$ from each of the four corners and folding up the flaps. If the volume of the box is maximum, then $x$ is equal to: (1) $\frac { a + b + \sqrt { a ^ { 2 } + b ^ { 2 } - a b } } { 6 }$ (2) $\frac { a + b - \sqrt { a ^ { 2 } + b ^ { 2 } - a b } } { 12 }$ (3) $\frac { a + b - \sqrt { a ^ { 2 } + b ^ { 2 } - a b } } { 6 }$ (4) $\frac { a + b - \sqrt { a ^ { 2 } + b ^ { 2 } + a b } } { 6 }$

Let $A = \left[ a _ { i j } \right]$ be a $3 \times 3$ matrix, where $a _ { i j } = \left\{ \begin{array} { c c } 1 , & \text { if } i = j \\ - x , & \text { if } | i - j | = 1 \\ 2 x + 1 , & \text { otherwise } \end{array} \right.$ Let a function $f : R \rightarrow R$ be defined as $f ( x ) = \operatorname { det } ( A )$. Then the sum of maximum and minimum values of $f$ on $R$ is equal to:
(1) $- \frac { 20 } { 27 }$
(2) $\frac { 88 } { 27 }$
(3) $\frac { 20 } { 27 }$
(4) $- \frac { 88 } { 27 }$

If Rolle's theorem holds for the function $f ( x ) = x ^ { 3 } - a x ^ { 2 } + b x - 4 , x \in [ 1,2 ]$ with $f ^ { \prime } \left( \frac { 4 } { 3 } \right) = 0$, then ordered pair $( a , b )$ is equal to :
(1) $( - 5 , - 8 )$
(2) $( - 5,8 )$
(3) $( 5,8 )$
(4) $( 5 , - 8 )$

The function $f ( x ) = \left| x ^ { 2 } - 2 x - 3 \right| \cdot \mathrm { e } ^ { 9 x ^ { 2 } - 12 x + 4 }$ is not differentiable at exactly :
(1) Four points
(2) Two points
(3) three points
(4) one point

Let $M$ and $m$ respectively be the maximum and minimum values of the function $f ( x ) = \tan ^ { - 1 } ( \sin x + \cos x )$ in $\left[ 0 , \frac { \pi } { 2 } \right]$. Then the value of $\tan ( M - m )$ is equal to: (1) $2 - \sqrt { 3 }$ (2) $3 - 2 \sqrt { 2 }$ (3) $3 + 2 \sqrt { 2 }$ (4) $2 + \sqrt { 3 }$

Let $a$ be a real number such that the function $f ( x ) = a x ^ { 2 } + 6 x - 15 , x \in R$ is increasing in $( - \infty , \frac { 3 } { 4 } )$ and decreasing in $\left( \frac { 3 } { 4 } , \infty \right)$. Then the function $g ( x ) = a x ^ { 2 } - 6 x + 15 , x \in R$ has a
(1) local maximum at $x = - \frac { 3 } { 4 }$
(2) local minimum at $x = - \frac { 3 } { 4 }$
(3) local maximum at $x = \frac { 3 } { 4 }$
(4) local minimum at $x = \frac { 3 } { 4 }$

The range of $a \in R$ for which the function $f ( x ) = ( 4 a - 3 ) \left( x + \log _ { e } 5 \right) + 2 ( a - 7 ) \cot \left( \frac { x } { 2 } \right) \sin ^ { 2 } \left( \frac { x } { 2 } \right) , x \neq 2 n \pi , n \in N$, has critical points, is :
(1) $( - 3,1 )$
(2) $\left[ - \frac { 4 } { 3 } , 2 \right]$
(3) $[ 1 , \infty )$
(4) $( - \infty , - 1 ]$

Let $f : [ - 1,1 ] \rightarrow R$ be defined as $f ( x ) = a x ^ { 2 } + b x + c$ for all $x \in [ - 1,1 ]$, where $a , b , c \in R$ such that $f ( - 1 ) = 2 , f ^ { \prime } ( - 1 ) = 1$ and for $x \in ( - 1,1 )$ the maximum value of $f ^ { \prime \prime } ( x )$ is $\frac { 1 } { 2 }$. If $f ( x ) \leq \alpha , x \in [ - 1,1 ]$, then the least value of $\alpha$ is equal to

If $R$ is the least value of $a$ such that the function $f ( x ) = x ^ { 2 } + \mathrm { a } x + 1$ is increasing on $[ 1,2 ]$ and $S$ is the greatest value of $a$ such that the function $f ( x ) = x ^ { 2 } + a x + 1$ is decreasing on $[ 1,2 ]$, then the value of $| R - S |$ is

Let $f : [ - 1,1 ] \rightarrow R$ be defined as $f ( x ) = a x ^ { 2 } + b x + c$ for all $x \in [ - 1,1 ]$, where $a , b , c \in R$ such that $f ( - 1 ) = 2 , f ^ { \prime } ( - 1 ) = 1$ and for $x \in ( - 1,1 )$ the maximum value of $f ^ { \prime \prime } ( x )$ is $\frac { 1 } { 2 }$. If $f ( x ) \leq \alpha , x \in [ - 1,1 ]$, then the least value of $\alpha$ is equal to

Let $f ( x )$ be a polynomial of degree 6 in $x$, in which the coefficient of $x ^ { 6 }$ is unity and it has extrema at $x = - 1$ and $x = 1$. If $\lim _ { x \rightarrow 0 } \frac { f ( x ) } { x ^ { 3 } } = 1$, then $5 \cdot f ( 2 )$ is equal to

Let a function $g : [ 0,4 ] \rightarrow R$ be defined as $g ( x ) = \left\{ \begin{array} { c c } \max \left\{ t ^ { 3 } - 6 t ^ { 2 } + 9 t - 3 \right\} , & 0 \leq x \leq 3 \\ 0 \leq t \leq x & \\ 4 - x, & 3 < x \leq 4 \end{array} \right.$ then the number of points in the interval $( 0,4 )$ where $g ( x )$ is NOT differentiable, is $\underline{\hspace{1cm}}$.

Let $x , y > 0$. If $x ^ { 3 } y ^ { 2 } = 2 ^ { 15 }$, then the least value of $3 x + 2 y$ is
(1) 30
(2) 32
(3) 36
(4) 40

If the minimum value of $f(x) = \frac{5x^2}{2} + \frac{\alpha}{x^5}$, $x > 0$, is 14, then the value of $\alpha$ is equal to
(1) 32
(2) 64
(3) 128
(4) 256