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Application of Derivatives

Name: Zigya - For The Curious Learner
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Short Answer Type

121.

Find points on the curve $straight x squared over 4 plus straight y squared over 25 space equals space 1$ at which the tangents are (i) parallel to the x-axis (ii) parallel to the y-axis.

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Long Answer Type

122.

Find points on the curve $straight x squared over 9 plus straight y squared over 16 space equals space 1$ at which the tangents are (i) parallel to the x-axis (ii) parallel to the y-axis.

82 Views

123.

For the curve y = 4x³ – 2x⁵, find all the points at which the tangent passes through the origin.

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Multiple Choice Questions

124. The points on the curve 9y² = x³, where the normal to the curve make equal intercepts with the axes are

$open parentheses 4 comma space plus-or-minus 8 over 3 close parentheses$
$open parentheses 4 comma space space minus 8 over 3 close parentheses$
$open parentheses 4 comma space plus-or-minus 3 over 8 close parentheses$
$open parentheses 4 comma space plus-or-minus 3 over 8 close parentheses$

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Long Answer Type

125. The curve y = ax³ + bx² + cx + 5 touches the x -axis at P (– 2, 0) and cuts the y-axis at a point Q where its gradient is 3. Find a. b, c.

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126.
Show that the curves 2x = y² and 2xy = k cut at right angles if k² = 8

The equation of two curves are
2x = y²...(1)
and    $2 xy space equals space straight k$ ...(2)
From (1) and (2), $straight y squared. space space straight y space equals space straight k space space space rightwards double arrow space space space space space straight y cubed space equals space straight k space space space space rightwards double arrow space space space straight y space equals space straight k to the power of 1 third end exponent$
$therefore space space space space from space left parenthesis 1 right parenthesis comma space space space space 2 straight x space equals space straight k to the power of 2 over 3 end exponent space space space space rightwards double arrow space space space space straight x space equals space 1 half straight k to the power of 2 over 3 end exponent$
$therefore$ point of intersection of curves (1) and (2) is $open parentheses 1 half straight k to the power of 2 over 3 end exponent comma space space straight k to the power of 1 third end exponent close parentheses$
From (1),    $2 space equals space 2 straight y dy over dx space space space space space rightwards double arrow space space space space space dy over dx space equals 1 over straight y$
$At space open parentheses 1 half straight k to the power of 2 over 3 end exponent comma space space space straight k to the power of 1 third end exponent close parentheses. space dy over dx space equals space 1 over straight k to the power of begin display style 1 third end style end exponent$
$therefore space space space space space space straight m subscript 1 space equals space 1 over straight k to the power of begin display style 1 third end style end exponent space where space straight m subscript 1 space is space slope space of space tangent space to space the space curve space left parenthesis 1 right parenthesis space at space the space point space of space intersection. space$
From (2),    $2 straight y plus 2 straight x dy over dx space equals space 0 space space space space space space rightwards double arrow space space space space dy over dx space equals space minus straight y over straight x$
At $open parentheses 1 half straight k to the power of 2 over 3 end exponent comma space space straight k to the power of 1 third end exponent close parentheses space dy over dx space equals space minus fraction numerator straight k to the power of begin display style 1 third end style end exponent over denominator begin display style 1 half end style straight k to the power of begin display style 2 over 3 end style end exponent end fraction space equals negative 2 over straight k to the power of begin display style 1 third end style end exponent$
$therefore space space space space space straight m subscript 2 space equals space minus 2 over straight k to the power of begin display style 1 third end style end exponent space where space straight m subscript 2 space is space slope space of space tangent space to space the space curve space left parenthesis 2 right parenthesis space at space the space point space of space intersection.$
Curves (1) and (2) cut at right angles if $straight m subscript 1 straight m subscript 2 space equals space minus 1$
i.e., if $1 over straight k to the power of begin display style 1 third end style end exponent. space fraction numerator negative 2 over denominator straight k to the power of begin display style 1 third end style end exponent end fraction space equals negative 1$
i.e. if $straight k to the power of negative 2 over 3 end exponent space equals space 2 space space space space space space space space space straight i. straight e. space space space space if space space space space straight k squared space equals 8$
Hence the result.