Find the equation of tangents to the curve y= x3+2x-4, which are

Find the value(s) of x for which y = open square brackets straight x left parenthesis straight x minus 2 right parenthesis close square brackets squared is an increasing function.


Given function is
straight f left parenthesis straight x right parenthesis space equals space open square brackets straight x left parenthesis straight x minus 2 right parenthesis close square brackets squared
rightwards double arrow space straight f apostrophe left parenthesis straight x right parenthesis space equals space straight x squared cross times space 2 left parenthesis straight x minus 2 right parenthesis plus left parenthesis straight x minus 2 right parenthesis squared cross times space 2 straight x
rightwards double arrow straight f apostrophe left parenthesis straight x right parenthesis space equals space 2 straight x left parenthesis straight x minus 2 right parenthesis space open square brackets straight x plus left parenthesis straight x minus 2 right parenthesis close square brackets
rightwards double arrow straight f apostrophe left parenthesis straight x right parenthesis space equals space 2 straight x left parenthesis straight x minus 2 right parenthesis left square bracket 2 straight x minus 2 right square bracket
rightwards double arrow straight f apostrophe left parenthesis straight x right parenthesis space equals space 2 straight x left parenthesis straight x minus 2 right parenthesis thin space open square brackets 2 left parenthesis straight x minus 1 right parenthesis close square brackets
rightwards double arrow straight f apostrophe left parenthesis straight x right parenthesis space equals 4 straight x left parenthesis straight x minus 1 right parenthesis left parenthesis straight x minus 2 right parenthesis

Since space straight f apostrophe left parenthesis straight x right parenthesis space is space an space increasing space function comma space straight f apostrophe left parenthesis straight x right parenthesis greater than 0
rightwards double arrow space space straight f apostrophe left parenthesis straight x right parenthesis space equals space 4 straight x left parenthesis straight x minus 1 right parenthesis thin space left parenthesis straight x minus 2 right parenthesis greater than 0
rightwards double arrow straight x left parenthesis straight x minus 1 right parenthesis space left parenthesis straight x minus 2 right parenthesis thin space greater than 0
rightwards double arrow 0 less than straight x less than 1 space or space straight x greater than 2
rightwards double arrow straight x space element of space left parenthesis 0 comma space 1 right parenthesis space union space left parenthesis 2 comma space infinity right parenthesis

 
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Find the equation of tangents to the curve y= x3+2x-4, which are perpendicular to line x+14y+3 =0.


Taking the given equation,
y = x3+2x-4
Differentiating the above function with respect to x, we have,
dy over dx equals space 3 straight x squared plus 2
rightwards double arrow space straight m subscript 1 equals space 3 straight x squared plus 2

Given that the tangents to the given curve are perpendicular to the line x+ 14y + 3 = 0
Slope of this line, m2=-1/14
Since the given line and the tangents to the curve are perpendicular, we have,

m1 x m2 =-1

rightwards double arrow space left parenthesis 3 straight x squared plus 2 right parenthesis open parentheses fraction numerator negative 1 over denominator 14 end fraction close parentheses space equals negative 1
rightwards double arrow space 3 straight x squared space plus 2 space equals space 14
rightwards double arrow space 3 straight x squared space equals space 12
rightwards double arrow space straight x squared space equals 4
rightwards double arrow space straight x space equals plus-or-minus 2
If space straight x equals 2 comma space straight y equals straight x cubed space plus 2 straight x minus 4
rightwards double arrow space straight y equals left parenthesis negative 2 right parenthesis cubed plus 2 straight x space left parenthesis negative 2 right parenthesis minus 4
rightwards double arrow straight y equals negative 16

Equation space of space the space tangent space having space slope space straight m space at space the space point space left parenthesis straight x subscript 1 comma straight y subscript 1 right parenthesis space is
left parenthesis straight y minus straight y subscript 1 right parenthesis space equals straight m left parenthesis straight x minus straight x subscript 1 right parenthesis

Equation space of space the space tangent space at space straight P space left parenthesis 2 comma 8 right parenthesis space with space slope space 14
left parenthesis straight y minus 8 right parenthesis equals 14 left parenthesis straight x minus 2 right parenthesis
rightwards double arrow space straight y minus 8 space equals space 14 space straight x minus 28
rightwards double arrow 14 straight x minus straight y equals 20

Equation space of space the space tangent space at space straight P left parenthesis negative 2 comma negative 16 right parenthesis space with space slope space 14
left parenthesis straight y minus 8 right parenthesis equals 14 left parenthesis straight x minus 2 right parenthesis
rightwards double arrow space straight y minus 8 space equals space 14 space straight x minus 28
rightwards double arrow 14 space straight x minus straight y space equals 20

Equation space of space the space tangent space at space straight P left parenthesis negative 2 comma negative 16 right parenthesis space with space slope space 1
left parenthesis straight y plus 16 right parenthesis equals 14 left parenthesis straight x plus 2 right parenthesis
rightwards double arrow straight y plus 16 space equals space 14 straight x plus 28
rightwards double arrow 14 space straight x minus straight y space equals space minus 12
thus equation of the tangent is 14 x- y =-12

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Find the absolute maximum and absolute minimum values of the function f given by
straight f left parenthesis straight x right parenthesis space equals sin squared straight x minus cosx comma space straight x space element of space left parenthesis 0 comma space straight pi right parenthesis


straight f left parenthesis straight x right parenthesis space equals space sin squared straight x minus cosx comma
straight f apostrophe left parenthesis straight x right parenthesis equals 2 sinx. cosx space plus space sinx
equals space sin space straight x left parenthesis 2 cosx plus 1 right parenthesis
Equating space straight f apostrophe left parenthesis straight x right parenthesis space to space zero.
straight f apostrophe left parenthesis straight x right parenthesis space equals space 0
sinx left parenthesis 2 cosx plus 1 right parenthesis space equals space 0
sinx space equals space 0
therefore space straight x space equals space 0 comma space straight pi
2 cosx plus 1 space equals 0
rightwards double arrow cosx space equals space minus 1 half
therefore straight x space equals space fraction numerator 5 straight pi over denominator 6 end fraction
straight f left parenthesis 0 right parenthesis space equals space sin squared 0 minus cos 0 equals space minus 1
straight f open parentheses fraction numerator 5 straight pi over denominator 6 end fraction close parentheses equals sin squared open parentheses fraction numerator 5 straight pi over denominator 6 end fraction close parentheses minus cos open parentheses fraction numerator 5 straight pi over denominator 6 end fraction close parentheses
equals sin squared straight pi over 6 plus cos straight pi over 6
equals 1 fourth minus fraction numerator square root of 3 over denominator 2 end fraction
equals open parentheses fraction numerator 1 minus 2 square root of 3 over denominator 4 end fraction close parentheses
straight f left parenthesis straight pi right parenthesis equals sin squared straight pi minus cosπ space equals 1
Of these values, the maximum value is 1, and the minimum value is −1.
Thus, the absolute maximum and absolute minimum values of f(x) are 1 and −1, which it attains at x = 0 and straight x equals straight pi
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Show that semi-vertical angle of a cone of maximum volume and given slant height is cos-1fraction numerator 1 over denominator square root of 3 end fraction.


Volume space of space cone space equals 1 third πr squared straight h
space equals 1 third straight pi left parenthesis space straight l space sin space straight alpha right parenthesis squared space left parenthesis space straight l space space cos space straight alpha right parenthesis
equals 1 third space straight pi space straight l cubed space sin squared space straight alpha space cos space straight alpha
dv over dα equals πl cubed over 3 left square bracket negative sin cubed space straight alpha space plus 2 sin space straight alpha space cosx space. cos space straight alpha right square bracket
equals fraction numerator straight pi space straight l cubed space sin space straight alpha over denominator 3 end fraction left parenthesis negative sin squared space straight alpha space plus space 2 space cos squared space straight alpha right parenthesis
For space maximum space volume


dv over dα space equals 0
fraction numerator πl cubed space sin space straight alpha over denominator 3 end fraction left parenthesis negative s i n space squared space alpha space plus 2 space c o s squared space alpha right parenthesis equals 0
s i n space alpha space not equal to 0
space 2 space c o s squared space alpha space equals space s i n squared space alpha
t a n squared space alpha space equals 2
t a n alpha space equals square root of 2
c o s space alpha space equals space fraction numerator 1 over denominator square root of 3 end fraction
alpha space equals space c o s to the power of negative 1 end exponent fraction numerator 1 over denominator square root of 3 end fraction

again space diff space straight w. straight r. straight t space straight alpha comma space we space get
fraction numerator straight d squared straight v over denominator dα squared end fraction space equals 1 third πl cubed cos squared space straight alpha left parenthesis 2 minus 7 space tan squared space straight alpha right parenthesis
at space cos space straight alpha space equals fraction numerator 1 over denominator square root of 3 end fraction
fraction numerator straight d squared straight v over denominator dα squared end fraction space less than 0
straight V space is space maximum space when space cos space straight alpha space equals fraction numerator 1 over denominator square root of 3 end fraction space or space straight alpha space equals cos to the power of negative 1 end exponent fraction numerator 1 over denominator square root of 3 end fraction
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Prove space that space straight y equals fraction numerator 4 space sin space straight theta over denominator 2 space plus cos space straight theta end fraction space minus straight theta space is space an space increasing space function space of space straight theta space on space open square brackets 0 comma straight pi over 2 close square brackets

considering space the space equation
straight y space equals fraction numerator 4 space sin space straight theta over denominator 2 space plus cos space straight theta end fraction minus straight theta space

differentiation space straight y space straight w. straight r. straight t space straight theta

dy over dθ space equals space space fraction numerator left parenthesis 2 space plus cos space straight theta right parenthesis space plus space 4 space sin squared space straight theta over denominator left parenthesis 2 space plus space cos space straight theta right parenthesis squared end fraction space space minus space 1

equals space space fraction numerator 8 space cos space space plus 4 space cos squared space space plus 4 space sin squared space straight theta over denominator left parenthesis 2 space plus cos space straight theta right parenthesis squared end fraction space space minus space 1

space equals space fraction numerator 8 space cos space space plus 4 over denominator left parenthesis 2 space plus cos space straight theta space right parenthesis squared end fraction minus 1

equals space fraction numerator 4 space cos space space minus space cos squared space straight theta over denominator left parenthesis 2 space plus space cos space straight theta right parenthesis squared end fraction

dy over dθ space space equals space fraction numerator cos space space left parenthesis space 4 minus space cos space straight theta space right parenthesis over denominator left parenthesis 2 plus cos space straight theta right parenthesis squared end fraction

for space increasing space dy over dθ space greater than space 0 space comma space straight theta space element of space open parentheses 0 comma space straight pi over 2 close parentheses

0 space less than for minus of space cos space straight theta space less than for minus of space space 1

left parenthesis 2 plus space cos space straight theta right parenthesis squared space always space greater space than space 0

so space comma space dy over dθ space is space increasing space on open square brackets 0 comma straight pi over 2 close square brackets
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