Maths-
General
Easy

Question

The value of 'c' in Lagrange's mean value theorem for f space left parenthesis x right parenthesis equals x space left parenthesis x minus 2 right parenthesis squared in [0, 2] is

  1. 0    
  2. 1    
  3. 2/3    
  4. 3/2    

Hint:

Lagrange mean value theorem states that for any two points on the curve there exists a point on the curve such that the tangent drawn at this point is parallel to the secant through the two points on the curve.

The correct answer is: 2/3


     Given : f space left parenthesis x right parenthesis equals x space left parenthesis x minus 2 right parenthesis squared is a polynomial
    We know that polynomials are continuous and differentiable in their range
    rightwards double arrow space f space left parenthesis x right parenthesis equals x space left parenthesis x minus 2 right parenthesis squared space i s space c o n t i n u o u s space a n d space d i f f e r e n t i a b l e space i n space t h e space r a n g e space left square bracket space 0 comma space 2 space right square bracket
    Simplify f(x) for ease of calculation
    rightwards double arrow space f space left parenthesis x right parenthesis equals x space left parenthesis x minus 2 right parenthesis squared space

rightwards double arrow space f space left parenthesis x right parenthesis equals x space left parenthesis x squared plus 4 space minus 4 x right parenthesis

rightwards double arrow space f space left parenthesis x right parenthesis equals x cubed plus 4 x space minus 4 x squared space....... space left parenthesis 1 right parenthesis

N o w comma space s u b s t i t u t e space t h e space v a l u e s space o f space x space a s space 0 space a n d space 2

rightwards double arrow space f left parenthesis 0 right parenthesis space equals space 0
rightwards double arrow space f left parenthesis 2 right parenthesis space equals space space 0 space

B y space L a g r a n g e s space M e a n space V a l u e space T h e o r e m

f apostrophe left parenthesis c right parenthesis space equals space fraction numerator f left parenthesis b right parenthesis space minus space f left parenthesis a right parenthesis over denominator b minus a end fraction

f apostrophe left parenthesis c right parenthesis space equals space fraction numerator f left parenthesis 2 right parenthesis space minus space f left parenthesis 0 right parenthesis over denominator 2 minus 0 end fraction space equals space 0 space space........ space left parenthesis 2 right parenthesis

N o w space d i f f e r e n t i a t i n g space left parenthesis 1 right parenthesis space

f apostrophe space left parenthesis x right parenthesis equals 3 x squared plus 4 space minus 8 x

f apostrophe left parenthesis c right parenthesis space equals space 3 c squared plus 4 space minus 8 c

F r o m space left parenthesis 2 right parenthesis

3 c squared plus 4 space minus 8 c space equals space 0

N o w space b y space u sin g space q u a d r a t i c space f o r m u l a comma space w e space f i n d space t h e space r o o t s space o f space c

c space equals space fraction numerator negative left parenthesis negative 8 right parenthesis space plus-or-minus square root of left parenthesis negative 8 right parenthesis squared space minus space 4 space cross times 3 cross times 4 end root over denominator 2 cross times 3 end fraction space equals space fraction numerator 8 space plus-or-minus space 4 over denominator 6 end fraction space equals space 2 space o r space 2 over 3

T h u s space c space equals space 2 over 3

    Related Questions to study

    General
    Maths-

    The equation r equals a c o s space theta plus b s i n space theta represents

    The equation r equals a c o s space theta plus b s i n space theta represents

    Maths-General
    General
    Maths-

    The polar equation of the circle whose end points of the diameter are open parentheses square root of 2 comma fraction numerator pi over denominator 4 end fraction close parentheses and open parentheses square root of 2 comma fraction numerator 3 pi over denominator 4 end fraction close parentheses is

    The polar equation of the circle whose end points of the diameter are open parentheses square root of 2 comma fraction numerator pi over denominator 4 end fraction close parentheses and open parentheses square root of 2 comma fraction numerator 3 pi over denominator 4 end fraction close parentheses is

    Maths-General
    General
    Maths-

    The radius of the circle r equals 8 s i n space theta plus 6 c o s space theta is

    The radius of the circle r equals 8 s i n space theta plus 6 c o s space theta is

    Maths-General
    parallel
    General
    Maths-

    The adjoining figure shows the graph of y equals a x to the power of 2 end exponent plus b x plus c Then –

    The adjoining figure shows the graph of y equals a x to the power of 2 end exponent plus b x plus c Then –

    Maths-General
    General
    Maths-

    Graph of y = ax2 + bx + c = 0 is given adjacently. What conclusions can be drawn from this graph –

    As we can see from the graph we have a parabola curve and since it is opening in an upward direction. So we can say that a > 0 and
    Hence, the option (ais correct.
    Here, we can see that the vertex of the parabola is located in the fourth quadrant , therefore it will be = fraction numerator b squared over denominator 2 a end fraction space greater than space 0
    On further solving this, we get
    b space less than space 0 
    Therefore, the option (b) is also correct.
    Since, at x=0 , the y intercept will be positive and from this, we can conclude that c < 0 and
    Hence, the option (cwill also be correct
    On checking all the options, and we can see all options are correct and
    Therefore, we conclude that all the options available are correct.

    Graph of y = ax2 + bx + c = 0 is given adjacently. What conclusions can be drawn from this graph –

    Maths-General
    As we can see from the graph we have a parabola curve and since it is opening in an upward direction. So we can say that a > 0 and
    Hence, the option (ais correct.
    Here, we can see that the vertex of the parabola is located in the fourth quadrant , therefore it will be = fraction numerator b squared over denominator 2 a end fraction space greater than space 0
    On further solving this, we get
    b space less than space 0 
    Therefore, the option (b) is also correct.
    Since, at x=0 , the y intercept will be positive and from this, we can conclude that c < 0 and
    Hence, the option (cwill also be correct
    On checking all the options, and we can see all options are correct and
    Therefore, we conclude that all the options available are correct.
    General
    Maths-

    For the quadratic polynomial f (x) = 4x2 – 8kx + k, the statements which hold good are

    For the quadratic polynomial f (x) = 4x2 – 8kx + k, the statements which hold good are

    Maths-General
    parallel
    General
    Maths-

    The graph of the quadratic polynomial y = ax2 + bx + c is as shown in the figure. Then :


    Clearly, y = a x squared space plus space b x space plus space c
 represent a parabola opening  downwards. Therefore, a < 0
     y = a x squared space plus space b x space plus space c
 cuts negative y- axis , Putting x = 0 in the given equation
    rightwards double arrow-y = c
    rightwards double arrowy = -c
    rightwards double arrowc < 0
    Thus, from the above graph c < 0.

    The graph of the quadratic polynomial y = ax2 + bx + c is as shown in the figure. Then :

    Maths-General

    Clearly, y = a x squared space plus space b x space plus space c
 represent a parabola opening  downwards. Therefore, a < 0
     y = a x squared space plus space b x space plus space c
 cuts negative y- axis , Putting x = 0 in the given equation
    rightwards double arrow-y = c
    rightwards double arrowy = -c
    rightwards double arrowc < 0
    Thus, from the above graph c < 0.
    General
    Maths-

    The greatest possible number of points of intersections of 8 straight line and 4 circles is :

     Complete step-by-step answer:
    The number point of intersection between two circles can be counted by finding the number of ways in which one circle and one line can be selected out of the lot multiplied by 2 as one circle and one line can intersect at most two points.
    For selecting r objects from n objects can be done by using the formula as follows
    C presuperscript n subscript r space equals space fraction numerator n factorial over denominator r factorial space left parenthesis n minus r right parenthesis factorial end fraction
     As mentioned in the question, we have to find the total number of intersection points.

    For calculating the points of intersection between two lines, we can use the formula which is mentioned in the hint as follows = C presuperscript 8 subscript 2 cross times 1 = 28

     For calculating the points of intersection between two circles, we can use the formula which is mentioned in the hint as follows = C presuperscript 4 subscript 2 cross times 2 = 12
    For calculating the points of intersection between one line and one circle, we can use the formula which is mentioned in the hint as follows  C presuperscript 4 subscript 1 cross times C presuperscript 8 subscript 1 cross times 2 space= 64
    Hence, the total number of points of intersection is = 28 + 12  + 64 = 104

    The greatest possible number of points of intersections of 8 straight line and 4 circles is :

    Maths-General
     Complete step-by-step answer:
    The number point of intersection between two circles can be counted by finding the number of ways in which one circle and one line can be selected out of the lot multiplied by 2 as one circle and one line can intersect at most two points.
    For selecting r objects from n objects can be done by using the formula as follows
    C presuperscript n subscript r space equals space fraction numerator n factorial over denominator r factorial space left parenthesis n minus r right parenthesis factorial end fraction
     As mentioned in the question, we have to find the total number of intersection points.

    For calculating the points of intersection between two lines, we can use the formula which is mentioned in the hint as follows = C presuperscript 8 subscript 2 cross times 1 = 28

     For calculating the points of intersection between two circles, we can use the formula which is mentioned in the hint as follows = C presuperscript 4 subscript 2 cross times 2 = 12
    For calculating the points of intersection between one line and one circle, we can use the formula which is mentioned in the hint as follows  C presuperscript 4 subscript 1 cross times C presuperscript 8 subscript 1 cross times 2 space= 64
    Hence, the total number of points of intersection is = 28 + 12  + 64 = 104
    General
    Maths-

    How many different nine digit numbers can be formed from the number 223355888 by rearranging its digits so that the odd digits occupy even position ?

    Complete step-by-step answer:
    Here we need to find the total number of nine digit numbers that can be formed using the given digits i.e. 2, 2, 3, 3, 5, 5, 8, 8, 8.
    XXXXX
    Here, symbol ( is for the even places and (Xis for the odd places of the digit number.
    The digits which are even are 2, 2, 8, 8 and 8.
    Number of even digits  5
    The digits which are odd are 3, 3, 5 and 5.
    Number of odd digits 4
    We have to arrange the odd digits in even places.
     
    Number of ways to arrange the odd digits in 4 even places = fraction numerator 4 factorial over denominator 2 factorial space cross times 2 factorial end fraction
    On finding the value of the factorials, we get
    Number of ways to arrange the odd digits in 4 even places = 6

    Now, we have to arrange the even digits in odd places.
    Number of ways to arrange the even digits in 5 odd places = fraction numerator 5 factorial over denominator 2 factorial cross times 3 factorial end fraction
    On finding the value of the factorials, we get
    Number of ways to arrange the even digits in 5 odd places = fraction numerator 5 space cross times 4 cross times 3 cross times 2 cross times 1 over denominator 2 cross times 1 space cross times 3 cross times 2 cross times 1 end fraction

    On further simplification, we get
    Number of ways to arrange the even digits in 5 odd places 
    =10
    Total number of 9 digits number 6×10 60
    Hence, the required number of 9 digit numbers 60
     
     
     
     

    How many different nine digit numbers can be formed from the number 223355888 by rearranging its digits so that the odd digits occupy even position ?

    Maths-General
    Complete step-by-step answer:
    Here we need to find the total number of nine digit numbers that can be formed using the given digits i.e. 2, 2, 3, 3, 5, 5, 8, 8, 8.
    XXXXX
    Here, symbol ( is for the even places and (Xis for the odd places of the digit number.
    The digits which are even are 2, 2, 8, 8 and 8.
    Number of even digits  5
    The digits which are odd are 3, 3, 5 and 5.
    Number of odd digits 4
    We have to arrange the odd digits in even places.
     
    Number of ways to arrange the odd digits in 4 even places = fraction numerator 4 factorial over denominator 2 factorial space cross times 2 factorial end fraction
    On finding the value of the factorials, we get
    Number of ways to arrange the odd digits in 4 even places = 6

    Now, we have to arrange the even digits in odd places.
    Number of ways to arrange the even digits in 5 odd places = fraction numerator 5 factorial over denominator 2 factorial cross times 3 factorial end fraction
    On finding the value of the factorials, we get
    Number of ways to arrange the even digits in 5 odd places = fraction numerator 5 space cross times 4 cross times 3 cross times 2 cross times 1 over denominator 2 cross times 1 space cross times 3 cross times 2 cross times 1 end fraction

    On further simplification, we get
    Number of ways to arrange the even digits in 5 odd places 
    =10
    Total number of 9 digits number 6×10 60
    Hence, the required number of 9 digit numbers 60
     
     
     
     
    parallel
    General
    Maths-

    A person predicts the outcome of 20 cricket matches of his home team. Each match can result either in a win, loss or tie for the home team. Total number of ways in which he can make the predictions so that exactly 10 predictions are correct, is equal to :

    Matches whose prediction are correct can be selected in C presuperscript 20 subscript 10 ways.
    Since each match can result either in a win, loss or tie for the home team.
    Now, each wrong prediction can be made in two ways (i.e. the correct result is win and the person predicts either lose or tie)
    and there are 10 matches for which he predicted wrong.
    Total number of ways in which he can make the predictions so that exactly 10 predictions are correct =
    N u m b e r space o f space c o r r e c t space p r e d i c t i o n s space cross times space N u m b e r space o f space i n c o r r e c t space p r e d i c t i o n s
    C presuperscript 20 subscript 10 cross times space 2 to the power of 10
    Thus, total number of ways in which he can make the predictions so that exactly 10 predictions are correct, is equal to C presuperscript 20 subscript 10 cross times space 2 to the power of 10 

    A person predicts the outcome of 20 cricket matches of his home team. Each match can result either in a win, loss or tie for the home team. Total number of ways in which he can make the predictions so that exactly 10 predictions are correct, is equal to :

    Maths-General
    Matches whose prediction are correct can be selected in C presuperscript 20 subscript 10 ways.
    Since each match can result either in a win, loss or tie for the home team.
    Now, each wrong prediction can be made in two ways (i.e. the correct result is win and the person predicts either lose or tie)
    and there are 10 matches for which he predicted wrong.
    Total number of ways in which he can make the predictions so that exactly 10 predictions are correct =
    N u m b e r space o f space c o r r e c t space p r e d i c t i o n s space cross times space N u m b e r space o f space i n c o r r e c t space p r e d i c t i o n s
    C presuperscript 20 subscript 10 cross times space 2 to the power of 10
    Thus, total number of ways in which he can make the predictions so that exactly 10 predictions are correct, is equal to C presuperscript 20 subscript 10 cross times space 2 to the power of 10 
    General
    Maths-

    The foot of the perpendicular from the point left parenthesis 3 , 3 pi divided by 4 right parenthesis on the line r left parenthesis c o s space theta minus s i n space theta right parenthesis equals 6 square root of 2 is

    The foot of the perpendicular from the point left parenthesis 3 , 3 pi divided by 4 right parenthesis on the line r left parenthesis c o s space theta minus s i n space theta right parenthesis equals 6 square root of 2 is

    Maths-General
    General
    Maths-

    The point of intersection of the lines 2 c o s space theta plus s i n space theta equals 1 over r comma c o s space theta plus s i n space theta equals 1 over r is

    The point of intersection of the lines 2 c o s space theta plus s i n space theta equals 1 over r comma c o s space theta plus s i n space theta equals 1 over r is

    Maths-General
    parallel
    General
    Maths-

    The line passing through open parentheses negative 1 comma fraction numerator pi over denominator 2 end fraction close parentheses and perpendicular to square root of 3 s i n space theta plus 2 c o s space theta equals 4 over r is

    The line passing through open parentheses negative 1 comma fraction numerator pi over denominator 2 end fraction close parentheses and perpendicular to square root of 3 s i n space theta plus 2 c o s space theta equals 4 over r is

    Maths-General
    General
    Maths-

    The equation of the line passing through left parenthesis negative 1 comma pi divided by 6 right parenthesis comma left parenthesis 1 comma pi divided by 2 right parenthesis is

    The equation of the line passing through left parenthesis negative 1 comma pi divided by 6 right parenthesis comma left parenthesis 1 comma pi divided by 2 right parenthesis is

    Maths-General
    General
    Maths-

    The length of the perpendicular from (-1, π/6) to the line r left parenthesis 3 s i n space theta plus square root of 3 c o s space theta right parenthesis equals 3 is

    The length of the perpendicular from (-1, π/6) to the line r left parenthesis 3 s i n space theta plus square root of 3 c o s space theta right parenthesis equals 3 is

    Maths-General
    parallel

    card img

    With Turito Academy.

    card img

    With Turito Foundation.

    card img

    Get an Expert Advice From Turito.

    Turito Academy

    card img

    With Turito Academy.

    Test Prep

    card img

    With Turito Foundation.