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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.

The equation of curve is $16 straight x squared plus 9 straight y squared space equals space 144$ ...(1)
Now, $left parenthesis straight x subscript 1 comma space space straight y subscript 1 right parenthesis$ lies on it
$therefore space space space space 16 straight x subscript 1 squared plus 9 straight y subscript 1 squared space equals space 144$
Put $straight x subscript 1 space equals space 2$
$therefore space space space space 64 plus 9 straight y subscript 1 squared space equals space 144 comma space space space space space space or space space space 9 straight y subscript 1 squared space equals space 80 therefore space space space space space space straight y subscript 1 squared space equals space 80 over 9 space space space space space space space rightwards double arrow space space space space space straight y subscript 1 space equals space fraction numerator square root of 80 over denominator 3 end fraction space space space space space space space space space space space space space space space space space space space space space space left parenthesis because space space space straight y subscript 1 greater than 0 right parenthesis space space therefore space space space space space space point space left parenthesis straight x subscript 1 comma space space straight y subscript 1 right parenthesis space is space open parentheses 2 comma space fraction numerator square root of 80 over denominator 3 end fraction close parentheses space space space space Differentiating space both space sides space straight w. straight r. straight t. straight x. comma space space space space space space space space space space space space space space space space 32 straight x plus 18 straight y dy over dx space equals space 0 space space space space space space space space space space space 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 negative fraction numerator 32 straight x over denominator 18 straight y end fraction equals negative 16 over 9 straight x over straight y space space At space open parentheses 2 comma space fraction numerator square root of 80 over denominator 3 end fraction close parentheses comma space space dy over dx space equals space minus 16 over 9 fraction numerator 2 over denominator begin display style fraction numerator square root of 80 over denominator 3 end fraction end style end fraction space equals space minus 16 over 9 cross times fraction numerator 6 over denominator 4 square root of 5 end fraction equals negative fraction numerator 8 over denominator 3 square root of 5 end fraction therefore space space space space space equation space of space tangent space at space open parentheses 2 comma space fraction numerator square root of 80 over denominator 3 end fraction close parentheses space is space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space straight y minus fraction numerator square root of 80 over denominator 3 end fraction space equals space minus fraction numerator 8 over denominator 3 square root of 5 end fraction left parenthesis straight x minus 2 right parenthesis$
or $3 square root of 5 straight y space minus space square root of 400 space equals negative 8 straight x plus 16 space space space space or space space space space 8 straight x plus 3 square root of 5 straight y minus 36 space equals space 0$
Slope of normal = $fraction numerator 3 square root of 5 over denominator 8 end fraction$
$therefore$ the equation of normal at $open parentheses 2 comma space space fraction numerator square root of 80 over denominator 3 end fraction close parentheses space is$
$straight y minus fraction numerator square root of 80 over denominator 3 end fraction space equals space fraction numerator 3 square root of 5 over denominator 8 end fraction left parenthesis straight x minus 2 right parenthesis space space space space space space or space space 24 straight y minus 8 square root of 80 space equals space 9 square root of 5 left parenthesis straight x minus 2 right parenthesis$

$or space space space space 24 straight y minus 32 square root of 5 space equals space 9 square root of 5 straight x minus 18 square root of 5 space space space space or space space space 9 square root of 5 straight x minus 24 straight y plus 14 square root of 5 space equals space 0$
The tangent $8 straight x plus 3 square root of 5 straight y minus 36 space equals space 0$ cuts x-axis where y = 0, So putting y = 0, we get,
$8 straight x minus 36 space equals space 0 space space space space rightwards double arrow space space space space space straight x space equals space 9 over 2 therefore space space space space space space required space point space is space open parentheses 9 over 2 comma space space 0 close parentheses$
The normal $9 square root of 5 straight x minus 24 straight y plus 14 square root of 5 space equals space 0$ cut x-axis where y = 0. So putting y = 0, we get.
$9 square root of 5 straight x plus 14 square root of 5 space equals space 0 space space space space rightwards double arrow space space space space space straight x space equals space minus 14 over 9$
$therefore space space space space space required space point space is space minus open parentheses 14 over 9 comma space 0 close parentheses.$

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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.

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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.