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Chain Rule is one of the most important concepts in Differentiations That's why we will discuss it in upcoming concepts till then you can take some basic idea of this concept Thanking You !!
Case 1 :                  Case 2 :                =>    correct option (2)     Option 1) Option 2) Option 3) Option 4)
 has exterme points at  So,  at    we can write   given that                                                Option 1) four irrational numbers. Option 2) four rational numbers. Option 3) two irrational and two rational numbers. Option 4) two irrational and one rational number.
Find shortest distance   the tangent at point P is parallel to So putiing the value of in curve                                              Shortest distance between two parallel lines or perpendicular distance from       Option 1) Option 2) Option 3)   Option 4)
  Indeterminate forms - The form   are known as indeterminate form means they do not exist directly -     L - Hospital Rule - - wherein Given  and    This is in the form of L.D.E I.F. =        =  Now, So,  Option 1)exists and equals 0Option 2)does not existOption 3)exist and equals Option 4)exists and equals 4
@Shekhar suman The equation of a line perpendicular to the line x/a -y/b =1 and passing through the point where it meets x axis is slope of given line is b/a so the slope of perpendicular line is -a/b so required line is  given perpendicular line passes through (a,0) 
# Engineering
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As learned in concept

Eccentricity -

e= \sqrt{1-\frac{b^{2}}{a^{2}}}

- wherein

For the ellipse  

\frac{x^{2}}{a^{2}}+ \frac {y^{2}}{b^{2}}= 1

 

 and

Equation of directrices -

x= \pm \frac{a}{e}

- wherein

For the ellipse  

\frac{x^{2}}{a^{2}}+ \frac {y^{2}}{b^{2}}= 1

 

 

we have e= \frac{1}{2} &  x=-4

 

From this we can calculate the value of a

-\frac{a}{e}=-4 \Rightarrow a=4e

 so a=4

Now,  b^{2}=a^{2}(1-e^{2})=3

Hence, Equation of ellipse is-

\frac{x^{2}}{4}+\frac{y^{2}}{3}=1

Differentiate to get the value of slope- 

\frac{x}{2}+\frac{2y}{3}*y^{'}=0 \Rightarrow y^{'}=\frac{3x}{4y}

So slope at given point     (1'\frac{3}{2}) is -\frac{1}{2}

Now by using the formula of normal of a equation-

y-y_{1}=-\frac{1}{m}(x-x_{1})   normal at point   (1'\frac{3}{2}) is   y-\frac{3}{2}=2*(x-1)

we get the equation of the normal to be    4x-2y=1

 


Option 1)

4x-2y=1

Correct

Option 2)

4x+2y=7

Incorrect

Option 3)

x+2y=4

Incorrect

 

 

Option 4)

2y-x=2

Incorrect

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# Engineering
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It is not a function of only x- variable.    if we want to find differentiation of this function,we need partial differentiation w.r.t (x).      Hence, Final answer will be .
# Engineering
112 Views   |  
As we have learned Lagrange's mean value theorem - If  a  function  f(x)  1.   is continuous in the closed interval [a,b] and  2.   is differentiable in the open interval [a, b] then  -   (A),(C), (D) are discontinous in [1,2] but (B) satisfies all condition of L.M.V.T           Option 1) Option 2) Option 3) Option 4)
# Engineering
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As we have learned Length of Tangent - - wherein Where      Length of tangent at (x,y)   Length =            Option 1) 1 Option 2) 2 Option 3) 3 Option 4) 4
# Engineering
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As we have learned Properties of differentiable functions - The composition of a differentiable function is a differentiable functions. -      is composition of sin x with sin x so it will  be diffrentiable      is compostion of  with  (both diffrentiable ) so it will be also diffrentiable    so sum of two diffrentiable function will also be diffrentiable for all            Option 1) 0 Option...
# Engineering
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As we have learned Irremovable discontinuity - A function f is said to possess irremovable discontinuity if at  x = a the left hand limit is not equal to the right hand limit so limit does not exist   -     Limit doesn't exist  , f(x) has irremovable discontinuty , so ffor no 'k' it will be continous     Option 1) If k= 1 , f(x ) becomes continous at x= 0 Option 2) If k= -1 , f(x ) becomes...
# Engineering
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As we have learned Removal discontinuity - A function  f is said to possess removable discontinuity if at x = a :     - wherein       f(0)=1 Limit exists but  not equal to f(0)  removable discontinuity at x= 0       Option 1) f(x ) is continous at x= 0 Option 2) f(x) has non - exsiting limit at x= 0   Option 3) f(x) has LHL=RHL = f(0) Option 4) f(x) has removable discontinuty at x=0
# Engineering
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As we learnt in  Differentiation - Derivative  of a function  f(x) is defined as  f'(x) means  small increment    in x  corresponding increment in the value of y  be   - wherein               Now, Now, Put x=-1 From (i) and (ii) it is zero Option 1) This option is incorrect Option 2) 0 This option is correct Option 3) 1 This option is incorrect Option 4) This option is incorrect
# Engineering
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As we learnt in Condition for differentiable - A function  f(x) is said to be differentiable at    if      both exist and are equal otherwise non differentiable -      and      So f(x) is continuous at x = 0   So, g(x) is differentiable at        Option 1)  Both statements I and II are false.     This option is incorrect. Option 2)  Both statements I and II are true. This option is...
# Engineering
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As we learnt in Slope of a line - If  is the angle at which a straight line is inclined to a positive direction of x-axis, then the slope is defined by . - wherein    and,   Mid-point formula -   - wherein If the point P(x,y) is the mid point of line joining A(x1,y1) and B(x2,y2) .    As well as,   Slope – point from of a straight line - - wherein slope point through which line...
# Engineering
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As we learnt in General form of a circle -   - wherein centre = radius =     Mid-point formula -   - wherein If the point P(x,y) is the mid point of line joining A(x1,y1) and B(x2,y2) .   Centre is (-1, -2)  Option 1) (3, 4) This option is incorrect. Option 2) (3, – 4) This option is incorrect. Option 3) (– 3, 4) This option is incorrect. Option 4) (– 3, – 4) This option is correct.
# Engineering
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As we learnt in 

Greatest distance of a point A from a circle -

AC+r

- wherein

A is the point and circle has centre C and radius r.

 Q is point on the circle x^{2}+y^{2}-5x-y+5=0

r=\sqrt{\left ( \frac{5}{2} \right )^{2}+\left(\frac{1}{2} \right )^{2}-5}

=\frac{\sqrt{6}}{2}

PQ=Maximum distance =PC + r

=\frac{5}{2}\sqrt{2}+\frac{\sqrt{6}}{2}

PQ^{2}=\frac{(5\sqrt{2}+\sqrt{6})^{2}}{4}=\frac{50+6+20\sqrt{3}}{4}

PQ^{2}=14+5\sqrt{3}


Option 1)

\frac{25 + \sqrt{6} }{2}

This option is incorrect

Option 2)

14 + 5\sqrt{3}

This option is correct

Option 3)

\frac{47 + 10\sqrt{6}}{2}

This option is incorrect

Option 4)

8 + 5\sqrt{3}

This option is incorrect

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# Engineering
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As we learnt in  Equation of directrices - - wherein For the Hyperbola     or for Hyperbola     Also   Also, Option 1) 7 Incorrect Option 2) -7 Correct Option 3) 5 Incorrect Option 4) -5 Incorrect
# Engineering
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As we learnt in  Eccentricity - - wherein For the ellipse       Coordinates of foci - - wherein For the ellipse       Also  Hence ellipse passes through    doesn't lie on this ellipse Option 1) Incorrect option     Option 2) Incorrect option     Option 3) Correct option Option 4) Incorrect option    
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