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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.

The equation of curve is $square root of straight x plus square root of straight y space equals space square root of straight a$
Differentiating both sides w.r.t. x, we get,.
$fraction numerator 1 over denominator 2 square root of straight x end fraction plus fraction numerator 1 over denominator 2 square root of straight y end fraction dy over dx space equals space 0 space space space space space space space space space space space space space space rightwards double arrow space space space space space dy over dx space equals space minus fraction numerator square root of straight y over denominator square root of straight x end fraction$
$At space open parentheses straight a squared over 4 comma straight a squared over 4 space close parentheses space space dy over dx space equals space fraction numerator negative square root of begin display style straight a squared over 4 end style end root over denominator square root of begin display style straight a squared over 4 end style end root end fraction space equals space minus 1 therefore space space space space space slope space of space tangent space space equals space minus 1 therefore space space space space space space the space equation space of space tangent space at space open parentheses straight a squared over 4 comma space straight a squared over 4 close parentheses space is space space space space space space space space space space space space space space space space space space straight y minus straight a squared over 4 space equals space minus open parentheses straight x minus straight a squared over 4 close parentheses space space space or space space space space straight y minus straight a squared over 4 space equals space minus straight x plus straight a squared over 4 space space space space or space space straight x plus straight y minus straight a squared over 2 space equals space 0 or space space space 2 straight x plus 2 straight y minus straight a squared space equals space 0 slope space of space normal space space equals space minus fraction numerator 1 over denominator negative 1 end fraction space equals 1 therefore space space space space space equation space of space normal space at space open parentheses straight a squared over 4 comma space space space straight a squared over 4 close parentheses space space is space space space straight y minus straight a squared over 4 space equals space 1 space open parentheses straight x minus straight a squared over 4 close parentheses space space space space or space space space space straight y minus straight a squared over 4 space equals space straight x minus straight a squared over 4 space space space or space space space straight x minus straight y space equals space 0$

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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.

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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.