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If S_{1}\; and\; S_{2} are respectively the sets of local minimum and local maximum points of the function,f(x)=9x^{4}+12x^{3}-36x^{2}+25,x\: \epsilon \: \mathbb{R}, then :
 

  • Option 1)

    S_{1}=\left \{ -2 \right \};S_{2}=\left \{ 0,1 \right \}

  • Option 2)

    S_{1}=\left \{ -2 ,1\right \};S_{2}=\left \{ 0 \right \}

  • Option 3)

    S_{1}=\left \{ -2 ,0\right \};S_{2}=\left \{ 1 \right \}

     

  • Option 4)

    S_{1}=\left \{ -1\right \};S_{2}=\left \{ 0,2 \right \}

 
                                  Option 1) Option 2) Option 3)   Option 4)

The tangents to the curve y=(x-2)^{2}-1 at its points 

of intersection with the line x-y=3 , intersect at the point :

  • Option 1)

    (\frac{5}{2},1)

  • Option 2)

    (-\frac{5}{2},-1)

  • Option 3)

    (\frac{5}{2},-1)

  • Option 4)

    (-\frac{5}{2},1)

 
Clearly,   and   Tangent to Parabola at B : To find slope, differentiate the given curve  Equation of tangent at B :                                       => ................(1) Equation of tangent at A : .........................(2) Clearly,(1) and (2) intersect at  .   Option 1) Option 2) Option 3) Option 4)

Let f(x)=5-|x-2|   and  g(x)=|x+1| , 

x\epsilon R . If f(x)  attains maximum value at \alpha and g(x) attains minimum

value at \beta , then 

\lim_{x\rightarrow -\alpha \beta }\frac{(x-1)(x^{2}-5x+6)}{x^{2}-6x+8}  is equal to :

  • Option 1)

    1/2

  • Option 2)

    -3/2

  • Option 3)

    -1/2

  • Option 4)

    3/2

 
Maxima  of  occured at   i.e.  Minima  of     occured at   i.e.     Option 1) Option 2) Option 3) Option 4)

\begin{matrix} lim\\ x \to 0 \end{matrix}\frac{x+2\sin x}{\sqrt{x^{2}+2\sin x+1}-\sqrt{\sin ^{2}x-x+1}} is:

 

  • Option 1)

    6

  • Option 2)

    2

  • Option 3)

    3

  • Option 4)

    1

 
Option 1) 6 Option 2) 2 Option 3) 3 Option 4) 1

The derivative of \tan ^{-1}\left ( \frac{\sin x-\cos x}{\sin x+\cos x} \right ),

with respect to \frac{x}{2}, where \left ( c\epsilon \left ( 0,\frac{\pi }{2} \right ) \right ) is :

 

  • Option 1)

    1

     

     

  • Option 2)

    \frac{2}{3}

  • Option 3)

    \frac{1}{2}

  • Option 4)

    2

 
Derivation of             Now,  Option 1) 1     Option 2) Option 3) Option 4) 2

If m is the minimum value of k for which the function f(x)=x\sqrt{kx-x^{2}} is increasing in the interval \left [ 0,3 \right ] and M is the maximum value of f in \left [ 0,3 \right ] when k=m, then the ordered pair \left ( m,M \right ) is equal to : 


 

  • Option 1)

    \left ( 4,3\sqrt{3} \right ) 

  • Option 2)

       \left ( 3,3\sqrt{3} \right )     

  • Option 3)

      \left ( 5,3\sqrt{6} \right )           

  • Option 4)

    \left ( 4,3\sqrt{2} \right )

 
For                                                                so  &                                                   minimum value of  minimum value of k is                          Option 1)   Option 2)          Option 3)               Option 4)

Let f:R\rightarrow R be a continuously differentiable function such that f(2)=6 and f'(2)=\frac{1}{48}. If \int_{6}^{f\left ( x \right )}4t^{3}dt=(x-2)g(x), then \lim_{x\rightarrow 2}g(x) is equal to : 

 

 

 

 

  • Option 1)

    36

  • Option 2)

    24

  • Option 3)

    12

  • Option 4)

    18

                            and                                                                                              Option 1)Option 2)Option 3)Option 4)

2m ladder leans against a vertical wall. If the top of the ladder begins to slide down the wall at the rate 25cm/sec, then the rate \left ( in\: \: cm/sec\right ) at which the bottom of the ladder slides away from the wall on the horizontal ground when the top of the ladder is 1m above the ground is : 


 

  • Option 1)

    25 

  • Option 2)

     \frac{25}{3} 

  • Option 3)

     \frac{25}{\sqrt{3}} 

  • Option 4)

     25\sqrt{3}

 
                                    Option 1)   Option 2)    Option 3)    Option 4)  

If e^{y}+xy=e, the ordered pair \left ( \frac{\mathrm{d} y}{\mathrm{d} x},\frac{\mathrm{d}^{2}y }{\mathrm{d} x^{2}} \right ) at x=0 is equal to : 

  • Option 1)

    \left ( -\frac{1}{e},-\frac{1}{e^{2}} \right ) 

  • Option 2)

      \left ( \frac{1}{e},-\frac{1}{e^{2}} \right ) 

  • Option 3)

        \left (- \frac{1}{e},\frac{1}{e^{2}} \right )  

  • Option 4)

    \left ( \frac{1}{e},\frac{1}{e^{2}} \right )

Again differentiate w.r.t.   x Option 1) Option 2)   Option 3)      Option 4)

Let a_1,a_2,a_3,......... be an A.P. with a_6=2 . Then the 

common difference of this A.P., which maximises the product 

a_1a_4a_5 , is :

  • Option 1)

    \frac{3}{2}

  • Option 2)

  • Option 3)

    \frac{6}{5}

  • Option 4)

    \frac{2}{3}

 
Assuming the first term of A.P. is a and difference is d. Then, Let  =>  So,  will be maximum at  So, option (2) is correct. Option 1) Option 2) Option 3) Option 4)

If \lim_{x\rightarrow 1}\frac{x^{2}-ax+b}{x-1}=5 , then a+b is equal to : 

  • Option 1)

    -4

  • Option 2)

    5

  • Option 3)

    -7

  • Option 4)

    1

As  denominator will become 0 => for finite limit numerator must also approach to zero as . So, L' Hospital Law will be applicable. 1-a+b=0........................(1) Now,    and    So, option(3) is correct.Option 1)-4Option 2)5Option 3)-7Option 4)1

A spherical iron ball of rdius 10cm is coated with a layer of ice of uniform 

thickness that melts at a rate of  50 \: cm^{3}/min. When the thickness of 

the ice is 5 cm , then the rate at which the thickness ( in cm / min ) of the ice

decreases, is :

  • Option 1)

    \frac{1}{18\pi}

  • Option 2)

    \frac{1}{36\pi}

  • Option 3)

    \frac{5}{6\pi}

  • Option 4)

    \frac{1}{9\pi}

 
Volume of ice ,                                                                         at x = 5,                           So, option (1) is correct.                         Option 1) Option 2) Option 3) Option 4)

If the tangent to the curve  y=\frac{x}{x^{2}-3},x\epsilon R,(x\neq \pm \sqrt3),

at a point (\alpha ,\beta )\neq( 0,0) on it is parallel to the line 2x+6y-11=0, then : 

  • Option 1)

    |6\alpha +2\beta |=19

  • Option 2)

    |6\alpha +2\beta |=9

  • Option 3)

    |2\alpha +6\beta |=19

  • Option 4)

    |2\alpha +6\beta |=11

 
the curve  ,point  parallel  line 2x+6y-11=0 Slope of given line =>  For tangent at   =>  =>  =>  =>  =>   as its given  So, for ,  and  Now,  for  ,  So, option (1)    is correct answer.     Option 1) Option 2) Option 3) Option 4)

Let f(x)=\log_{e}(sinx),(0<x<\pi) and 

g(x)=\sin^{-1}(e^{-x}),(x\geq 0). 

If \alpha is a positive real number such that a=(fog)'(\alpha )

and b=(fog)(\alpha ), then: 

  • Option 1)

    a\alpha ^{2}+b\alpha +a=0

     

  • Option 2)

    a\alpha ^{2}-b\alpha -a=1

  • Option 3)

    a\alpha ^{2}-b\alpha -a=0

  • Option 4)

    a\alpha ^{2}+b\alpha -a=-2\alpha ^{2}

 
  and                                                                                  Now, and   at    is   Now, lets check every option (1)  put the value in LHS (2)  put the value in LHS (3)   put the value in LHS (4)   put the value in LHS   So, option (2) is correct as  is true. Option 1)   Option 2) Option 3) Option 4)

The tangent and normal to the ellipse 3x^{2}+5y^{2}=32 at the point P(2,2)

meet the x-axis at Q and R, respectively, The the area(in sq. units) of the 

triangle PQR is :

  • Option 1)

    \frac{34}{15}

  • Option 2)

    \frac{14}{3}

  • Option 3)

    \frac{16}{3}

  • Option 4)

    \frac{68}{15}

 
 => Equation of tangent at point (2,2)  => Equation of normal at point (2,2)  Area of  =  So, option (4) is correct. Option 1) Option 2) Option 3) Option 4)

If f(x)=\left\{\begin{matrix} \frac{sin(p+1)x+sinx}{x}, x<0\\ q,\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: x=0\\ \frac{\sqrt{x+x^{2}}-\sqrt{x}}{x^{\frac{3}{2}}}, \: \: \: \: \: \: \: x>0 \end{matrix}\right.

is continuous at x = 0 , then the oredered pair ( p , q) is equal to : 

  • Option 1)

    (-\frac{3}{2},-\frac{1}{2})

  • Option 2)

    (-\frac{1}{2},\frac{3}{2})

  • Option 3)

    (-\frac{3}{2},\frac{1}{2})

  • Option 4)

    (\frac{5}{2},\frac{1}{2})

f(x) is continuous at x = 0 then LHL, for continuity LHL = RHL correct option is (3)      Option 1)Option 2)Option 3)Option 4)

Let f(x)=e^{x}-x and  g(x)=x^{2}-x\vee x\epsilon R.

Then the set of all x\epsilon R , where the function h(x)=(fog)(x) 

is increasing , is :

  • Option 1)

    [-1,\frac{-1}{2}]\cup [\frac{1}{2},\infty )

  • Option 2)

    [0,\frac{1}{2}]\cup [1,\infty )

  • Option 3)

    [0,\infty )

  • Option 4)

    [\frac{-1}{2},0]\cup [1,\infty )

 
Case 1 :                  Case 2 :                =>    correct option (2)     Option 1) Option 2) Option 3) Option 4)

If \lim_{x\rightarrow 1}\frac{x^{4}-1}{x-1}=\lim_{x\rightarrow k}\frac{x^{3}-k^{3}}{x^{2}-k^{2}} , then k is:

  • Option 1)

    \frac{8}{3}

  • Option 2)

    \frac{3}{8}

  • Option 3)

    \frac{3}{2}

  • Option 4)

    \frac{4}{3}

 
So, option (1) is correct.   Option 1) Option 2) Option 3) Option 4)

Let f:R\rightarrow R be differentiable at c\epsilon R and f(c) = 0.

If g(x)=\left | f(x) \right |, then at x=c , g is :

  • Option 1)

    not differentiable if f{}'(c)=0

  • Option 2)

    differentiable if f{}'(c)\neq0

  • Option 3)

    differentiable if f{}'(c)=0

  • Option 4)

    not differentiable 

 
Given g(x) = | f(x) | and also given f(c)=0                                             For  g(x)  to be differentiable at C     correct option is (3) Option 1) not differentiable if  Option 2) differentiable if  Option 3) differentiable if  Option 4) not differentiable 

If  f:R\rightarrow R  is a differentiable function and F(2)=6   , then \lim_{x\rightarrow 2}\int_{6}^{f(x)}\frac{2t\:dt}{(x-2)}   is :

 

 

 

  • Option 1)

    24f^{'}(2)

  • Option 2)

    2f^{'}(2)

  • Option 3)

    0

  • Option 4)

    12f^{'}(2)

 
Option 1) Option 2) Option 3) Option 4)
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