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Application of Derivatives

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Long Answer Type

351.

Find the largest possible area of a right-angled triangle whose hypotenuse is 5 cm long.

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352. A point on the hypotenuse of a triangle is at distance a and b from the sides of the triangle. Show that the maximum length of the hypotenuse is

open parentheses straight a to the power of 2 over 3 end exponent plus straight b to the power of 2 over 3 end exponent close parentheses to the power of 3 over 2 end exponent.

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353. Find the maximum-area of an isosceles triangle inscribed in the ellipse

straight x squared over straight a squared plus straight y squared over straight b squared space equals space 1

with its vertex at one end of the major axis.

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Short Answer Type

354.

Let AP and BQ be two vertical poles at points A and B, respectively. If AP = 16 m. BQ = 22 m and AB = 20 m, then find the distance of a point R on AB from the point A such that RP² + RQ² is minimum.

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Long Answer Type

355.

If length of three sides of a trapezium other than base are equal to 10 cm. then find the area of the trapezium when it is maximum.

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Short Answer Type

356.

Use differentials to approximate $square root of 25.3 end root.$

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357.

Use differentials to approximate $square root of 0.037 end root$ .

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358.
Use differentials to approximate $square root of 36.6 end root$

Take $straight y space equals space square root of straight x comma space space space space straight x space equals space 36 comma space space space dx space equals space dδ space equals space 0.6 comma space so space that space straight x space plus space δx space equals space 36.6$
Now, $straight y plus δy space equals space square root of straight x plus δx end root$
$therefore space space space space space space space space space space space space space δy space equals space square root of straight x plus δx end root space minus space straight y space equals space square root of 36.6 end root space minus space square root of 36 rightwards double arrow space space space space space space space space space space space space space δy space equals space square root of 36.6 end root space minus space 6 rightwards double arrow space space space space space space space space square root of 36.6 end root space equals space δy minus 6 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis Now space δy space is space approximately space equal space to space dy$
and $dy space equals space dy over dx dx space equals space fraction numerator 1 over denominator 2 square root of straight x end fraction dx$
$equals space fraction numerator 1 over denominator 2 square root of 36 end fraction left parenthesis 0.6 right parenthesis space equals space fraction numerator 0.3 over denominator 6 end fraction space equals space 0.05$
$therefore space space space from space left parenthesis 1 right parenthesis comma space space space square root of 36.6 end root space equals space 0.05 plus 6 space equals space 6.05.$