HSC Commerce (Marathi Medium) 12th Standard Board Exam [इयत्ता १२ वी] - Maharashtra State Board Question Bank Solutions for Mathematics and Statistics

Find `("d"y)/("d"x)`, if y = [log(log(logx))]² 

Find `("d"y)/("d"x)`, if xy = log(xy)

Find `("d"y)/("d"x)`, if x = `sqrt(1 + "u"^2)`, y = log(1 +u²)

If x = t.logt, y = t^t, then show that `("d"y)/("d"x)` = t^t 

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

Find `("d"y)/("d"x)`, if y = x^x + (7x – 1)^x 

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

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

If x^a .y^b = `(x + y)^((a + b))`, then show that `("d"y)/("d"x) = y/x`

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 = x^x

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 = 20^x

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)` = x^x(1 + log x) + 20^x.log(20)

Divide the number 20 into two parts such that their product is maximum

A rod of 108 m long is bent to form a rectangle. Find it’s dimensions when it’s area is maximum.

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`

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`

Choose the correct alternative:

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

`int ("d"x)/(x - x^2)` = ______

Choose the correct alternative:

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

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

`int 1/x  "d"x` = ______ + c

`int 1/(x^2 - "a"^2)  "d"x` = ______ + c