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Multiple Choice Questions

391.

If f (x) = 3x² + 15x + 5, then find the approximate value of f (3.02) is

47.66
57.66
67.66
67.66

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Short Answer Type

392.

Find the approximate change in the volume V of a cube of side v meters caused by increasing the side by 2%.

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393.

Find the approximate change in the volume V of a cube of side x metres caused by increasing the side by 1%.

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394. Find the approximate change in the surface area of a cube of side x metres caused by decreasing the side by 1%.

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395.

If the radius of a sphere is measured as 9 cm with an error of 0.03 cm, then find the approximate error in calculating its volume.

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396.

If the radius of a sphere is measured as 7 m with an error of 0.02 m, then find the approximate error in calculating its volume.

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397.

If the radius of a sphere is measured as 9 m with an error of 0.03 m. then find the approximate error in calculating its surface area.

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398.

A circular disc of radius 3 cm is being heated. Due to expanison, its radius increases at the rate of 0.05 cm/s. Find the rate at which its area is increasing when radius is 3.2 cm.

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Multiple Choice Questions

399. The approximate change in the volume of a cube of side x metres caused by increasing the side by 3% is

0.06 x³ m³
0.6 x³ m³
0.09 x³ m³
0.09 x³ m³

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Long Answer Type

400.
If the sum of the lengths of hypotenuse and a side of a right angled triangle is given, show that the area is maximum when the angle between them is 60°.

Let ABC be a triangle in which AB = x, AC – y, ∠BAC = 6

straight theta

therefore space space space space space space BC space equals space square root of AC squared minus AB squared end root space equals space square root of straight y squared minus straight x squared end root

From given conditon,

straight x plus straight y space equals space straight k

...(1)
where k is constant.
Let ∆ be area of ∆ABC.

Let ∆ be area of ∆ABC.

therefore space space space space space increment space equals space 1 half cross times AB cross times BC space equals space 1 half open square brackets straight x square root of straight y squared minus straight x squared end root close square brackets space space space space space space space space space space space space space space space equals 1 half open square brackets straight x square root of left parenthesis straight k minus straight x right parenthesis squared minus straight x squared end root close square brackets 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 open square brackets because space of space left parenthesis 1 right parenthesis close square brackets therefore space space space space space increment space equals space 1 half open square brackets straight x square root of straight k squared minus 2 kx end root close square brackets therefore space space space space space space fraction numerator straight d increment over denominator dx end fraction space space equals 1 half open square brackets straight x. space fraction numerator negative 2 straight k over denominator 2 square root of straight k squared minus 2 kx end root end fraction plus square root of straight k squared minus 2 kx end root.1 close square brackets space space space space space space space space space space space space space space space space space space space space space space equals space 1 half open square brackets negative fraction numerator kx over denominator square root of straight k squared minus 2 kx end root end fraction plus fraction numerator square root of straight k squared minus 2 kx end root over denominator 1 end fraction close square brackets space space equals space 1 half open square brackets fraction numerator negative kx plus straight k squared minus 2 kx over denominator square root of straight k squared minus 2 kx end root end fraction close square brackets 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

therefore space space space space space fraction numerator straight d increment over denominator dx end fraction space equals space 1 half open square brackets fraction numerator straight k squared minus 3 kx over denominator square root of straight k squared minus 2 kx end root end fraction close square brackets

Now,

fraction numerator straight d increment over denominator dx end fraction space equals space 0

gives us

1 half open square brackets fraction numerator straight k squared minus 3 kx over denominator square root of straight k squared minus 2 kx end root end fraction close square brackets equals 0 space space space space or space space space space straight k squared minus 3 kx space equals space 0

therefore space space space space space space straight k space minus space 3 space straight x space equals space 0 space space space space space space space space space space space rightwards double arrow space space space straight x space equals space straight k over 3 comma space space space space straight y space equals space fraction numerator 2 straight k over denominator 3 end fraction

open square brackets because space space of space left parenthesis 1 right parenthesis close square brackets

Now space fraction numerator straight d squared increment over denominator dx squared end fraction equals space 1 half open square brackets fraction numerator square root of straight k squared minus 2 straight k space straight x end root. space left parenthesis negative 3 straight k right parenthesis minus left parenthesis straight k squared minus 3 kx right parenthesis. space begin display style fraction numerator negative 2 straight k over denominator 2 square root of straight k squared minus 2 kx end root end fraction end style over denominator open parentheses square root of straight k squared minus 2 kx end root close parentheses squared end fraction close square brackets space space space space space space space space space space space space space space space space space space space space space equals space 1 half open square brackets fraction numerator begin display style fraction numerator negative 3 straight k square root of straight k squared minus 2 kx end root over denominator 1 end fraction end style plus begin display style fraction numerator straight k left parenthesis straight k squared minus 3 kx right parenthesis over denominator square root of straight k squared minus 2 kx end root end fraction end style over denominator straight k squared minus 2 kx end fraction close square brackets therefore space space space space space fraction numerator straight d squared increment over denominator dx squared end fraction space equals 1 half open square brackets fraction numerator negative 3 straight k left parenthesis straight k squared minus 2 kx right parenthesis plus straight k left parenthesis straight k squared minus 3 kx right parenthesis over denominator left parenthesis straight k squared minus 2 kx right parenthesis to the power of begin display style 3 over 2 end style end exponent end fraction close square brackets At space space straight x space equals space straight k over 3 comma space we space have

fraction numerator straight d squared increment over denominator dx squared end fraction space equals space 1 half open square brackets fraction numerator negative 3 straight k open parentheses straight k squared minus begin display style fraction numerator 2 straight k squared over denominator 3 end fraction end style close parentheses plus straight k left parenthesis straight k squared minus straight k squared right parenthesis over denominator open parentheses straight k squared minus begin display style fraction numerator 2 straight k squared over denominator 3 end fraction end style close parentheses to the power of begin display style 3 over 2 end style end exponent end fraction close square brackets space space space space space space space space space space space space space space equals 1 half open square brackets fraction numerator negative 3 straight k open parentheses begin display style straight k squared over 3 end style close parentheses plus 0 over denominator open parentheses begin display style straight k squared over 3 end style close parentheses to the power of begin display style 3 over 2 end style end exponent end fraction close square brackets space equals space 1 half open square brackets fraction numerator negative straight k cubed over denominator open parentheses begin display style straight k squared over 3 end style close parentheses to the power of begin display style 3 over 2 end style end exponent end fraction close square brackets less than 0 therefore space space space space space space increment space is space maximum space when space straight x space equals space straight k over 3 comma space space space straight y space equals space fraction numerator 2 straight k over denominator 3 end fraction therefore space space space space space space space space space space space space cos space straight theta space equals space AB over AC space equals space straight x over straight y equals space straight k over 3 cross times fraction numerator 3 over denominator 2 space straight k end fraction space equals space 1 half therefore space space space space space space space space space space space space space space space space straight theta space equals space 60 degree therefore space space space space space space area space is space maximum space when space straight theta space equals space 60 degree.

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Previous Year Papers

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Test Yourself

Multiple Choice Questions

Short Answer Type

Multiple Choice Questions

Long Answer Type

400. If the sum of the lengths of hypotenuse and a side of a right angled triangle is given, show that the area is maximum when the angle between them is 60°.

400.
If the sum of the lengths of hypotenuse and a side of a right angled triangle is given, show that the area is maximum when the angle between them is 60°.