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HSC Commerce (English Medium) 12th Standard Board Exam - Maharashtra State Board Question Bank Solutions for Mathematics and Statistics

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Mathematics and Statistics
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Find `("d"y)/("d"x)`, if y = `root(3)(((3x - 1))/((2x + 3)(5 - x)^2)`

[3] Differentiation
Chapter: [3] Differentiation
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If xa .yb = `(x + y)^((a + b))`, then show that `("d"y)/("d"x) = y/x`

[3] Differentiation
Chapter: [3] Differentiation
Concept: undefined >> undefined

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Find `("d"y)/("d"x)`, if y = x(x) + 20(x) 

Solution: Let y = x(x) + 20(x) 

Let u = `x^square` and v = `square^x`

∴ y = u + v

Diff. w.r.to x, we get

`("d"y)/("d"x) = square/("d"x) + "dv"/square`   .....(i)

Now, u = xx

Taking log on both sides, we get

log u = x × log x

Diff. w.r.to x,

`1/"u"*"du"/("d"x) = x xx 1/square + log x xx square`

∴ `"du"/("d"x)` = u(1 + log x)

∴ `"du"/("d"x) = x^x (1 +  square)`    .....(ii)

Now, v = 20x

Diff.w.r.to x, we get

`"dv"/("d"x") = 20^square*log(20)`     .....(iii)

Substituting equations (ii) and (iii) in equation (i), we get

`("d"y)/("d"x)` = xx(1 + log x) + 20x.log(20)

[3] Differentiation
Chapter: [3] Differentiation
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Divide the number 20 into two parts such that their product is maximum

[4] Applications of Derivatives
Chapter: [4] Applications of Derivatives
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A rod of 108 m long is bent to form a rectangle. Find it’s dimensions when it’s area is maximum.

[4] Applications of Derivatives
Chapter: [4] Applications of Derivatives
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A metal wire of 36 cm long is bent to form a rectangle. By completing the following activity, find it’s dimensions when it’s area is maximum.

Solution: Let the dimensions of the rectangle be x cm and y cm.

∴ 2x + 2y = 36

Let f(x) be the area of rectangle in terms of x, then

f(x) = `square`

∴ f'(x) = `square`

∴ f''(x) = `square`

For extreme value, f'(x) = 0, we get

x = `square`

∴ f''`(square)` = – 2 < 0

∴ Area is maximum when x = `square`, y = `square`

∴ Dimensions of rectangle are `square`

[4] Applications of Derivatives
Chapter: [4] Applications of Derivatives
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By completing the following activity, examine the function f(x) = x3 – 9x2 + 24x for maxima and minima

Solution: f(x) = x3 – 9x2 + 24x

∴ f'(x) = `square`

∴ f''(x) = `square`

For extreme values, f'(x) = 0, we get

x = `square` or `square`

∴ f''`(square)` = – 6 < 0

∴ f(x) is maximum at x = 2.

∴ Maximum value = `square`

∴ f''`(square)` = 6 > 0

∴ f(x) is maximum at x = 4.

∴ Minimum value = `square`

[4] Applications of Derivatives
Chapter: [4] Applications of Derivatives
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Choose the correct alternative:

`intx^(2)3^(x^3) "d"x` =

[5] Integration
Chapter: [5] Integration
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`int ("d"x)/(x - x^2)` = ______

[5] Integration
Chapter: [5] Integration
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Choose the correct alternative:

`int ("d"x)/((x - 8)(x + 7))` =

[5] Integration
Chapter: [5] Integration
Concept: undefined >> undefined

`int(x + 1/x)^3 dx` = ______.

[5] Integration
Chapter: [5] Integration
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`int 1/x  "d"x` = ______ + c

[5] Integration
Chapter: [5] Integration
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`int 1/(x^2 - "a"^2)  "d"x` = ______ + c

[5] Integration
Chapter: [5] Integration
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`int"e"^(4x - 3) "d"x` = ______ + c

[5] Integration
Chapter: [5] Integration
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`int (x^2 + x - 6)/((x - 2)(x - 1))  "d"x` = x + ______ + c

[5] Integration
Chapter: [5] Integration
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State whether the following statement is True or False:

If `int((x - 1)"d"x)/((x + 1)(x - 2))` = A log|x + 1|  + B log|x – 2|, then A + B = 1

[5] Integration
Chapter: [5] Integration
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Evaluate `int 1/(x(x - 1))  "d"x`

[5] Integration
Chapter: [5] Integration
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Evaluate `int 1/(x log x)  "d"x`

[5] Integration
Chapter: [5] Integration
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Evaluate `int 1/(4x^2 - 1)  "d"x`

[5] Integration
Chapter: [5] Integration
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Evaluate `int (2x + 1)/((x + 1)(x - 2))  "d"x`

[5] Integration
Chapter: [5] Integration
Concept: undefined >> undefined
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