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Application of Derivatives

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Short Answer Type

91. Find the equation of the tangent and normal to the given curves at the points given:
y = x⁴ – 6x³ + 13x² – 10x + 5 at ( 0 , 5 )

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92. Find the equation of the tangent and normal to the given curves at the points given:
y = x⁴ –6x³+ 13x²– 10x + 5 at (1 ,3)

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93. Find the equation of the tangent and normal to the given curves at the points given:

straight y squared space equals space fraction numerator straight x cubed over denominator 4 minus straight x end fraction space space at space space left parenthesis 2 comma space space minus 2 right parenthesis

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94. Find the equation of the tangent and normal to the given curves at the points given:

straight y space equals fraction numerator 5 straight x squared over denominator 1 plus straight x squared end fraction

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Long Answer Type

95. Find the equation of the tangent and normal to the given curves at the points given:

square root of straight x plus square root of straight y space equals space 5

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96. Find the equation of the tangent and normal to the given curves at the points given:

16 straight x squared plus 9 straight y squared space equals space 144 space space space at space left parenthesis straight x subscript 1 comma space space straight y subscript 1 right parenthesis space space space where space straight x subscript 1 space equals space 2 space space space and space straight y subscript 1 greater than 0

Also find the points of intersection where both tangent and normal cut the x-axis.

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97. Find the equation of the tangent and normal to the given curves at the points given:

square root of straight x plus square root of straight y space equals space straight a space space space at space space open parentheses straight a squared over 4 comma space space straight a squared over 4 close parentheses

Also find the points of intersection where both tangent and normal cut the x-axis.

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98. Find the equation of the tangent to the curve y = (x³ – 1) (x – 2) at points where the curve cuts the x-axis.

The equation of given curve is
$straight y space equals space left parenthesis straight x cubed minus 1 right parenthesis thin space left parenthesis straight x minus 2 right parenthesis space equals space left parenthesis straight x minus 1 right parenthesis thin space left parenthesis straight x minus 2 right parenthesis thin space left parenthesis straight x squared plus straight x plus 1 right parenthesis$
It meets x-axis where y = 0
Putting y = 0, we get,
$0 space equals space left parenthesis straight x minus 1 right parenthesis thin space left parenthesis straight x minus 2 right parenthesis thin space left parenthesis straight x squared plus straight x plus 1 right parenthesis$

$rightwards double arrow space space space space space space straight x space equals space 1 comma space 2 comma space fraction numerator negative 1 plus-or-minus square root of 1 minus 4 end root over denominator 2 end fraction space space space space space space rightwards double arrow space space space straight x space equals 1 comma space space 2 comma space space fraction numerator negative 1 plus-or-minus square root of negative 3 end root over denominator 2 end fraction therefore space space space space space space straight x space equals space 1 comma space space space 2 space space space are space only space real space values therefore space space space space space space space space space straight y space equals space 0 comma space space 0 therefore space space space space space space points space are space left parenthesis 1 comma space 0 right parenthesis comma space left parenthesis 2 comma space 0 right parenthesis. Now space dy over dx space equals space left parenthesis straight x cubed minus 1 right parenthesis. space 1 space plus space left parenthesis straight x plus 2 right parenthesis.3 space straight x squared space equals space straight x cubed minus 1 plus 3 straight x cubed minus 6 straight x squared space space space space space space space space space space space space space space space space space space equals space 4 straight x cubed minus 6 straight x squared minus 1 comma space space which space is space slope space of space tangent.$

At (1, 0), slope of tangent = 4 (1)³ – 6 (1)²–1 = 4 – 6 – 1 = – 3
At (2, 0), slope of tangent = 4 (2)³ – 6 (2)² – 1 = 4 × 8 – 6 × 4 – 1 = 7
∴ equation of tangent at (1, 0) with slope – 3 is
y – 0 = – 3(x – 1) or y = – 3x + 3 or 3x + y – 3 = 0
The equation of tangent at (2, 0) with slope 7 is
y – 0 = 7(x – 2), or y = 7x – 14 or 7x – y - 14 = 0

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Short Answer Type

99. Find the equation of the tangent to the curve

straight y equals space fraction numerator straight x minus 7 over denominator left parenthesis straight x minus 2 right parenthesis thin space left parenthesis straight x minus 3 right parenthesis end fraction

at the point where it cuts the x-axis.

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Long Answer Type

100.

Prove that the line $straight x over straight a plus straight y over straight b space equals space 1$ is a tangent to the curve $straight y space equals be to the power of negative straight x over straight a end exponent$ at the point where the curve cuts y-axis.