Subject: Maths, asked on 19/3/18

Q. Find the dimension of a rectangle of perimeter 36 cm, which will sweep out a volume as large as possible , when revolved about one of its sides. Also, find the maximum value.

Subject: Maths, asked on 19/3/18

Prove this as an increasing function

Subject: Maths, asked on 18/3/18

Show that f(x) = sin x (1 + cos x) is maximum when $a=\frac{\pi }{3}$ in the interval [0, π].

Subject: Maths, asked on 18/3/18

Please explain me Q12 (c) of Exercise 6.2 .

Subject: Maths, asked on 18/3/18

Please explain (c ) again.Why we took pie instead of 0?

Subject: Maths, asked on 18/3/18

In this answer the point (4,-8/3) does not satisfy the equation 6y = x^2  So how is this one of the solutions???

Subject: Maths, asked on 18/3/18

The points on the curve 9y2 = x3, where the normal to the curve makes equal intercepts with the axes are (A) $\left(4,±\frac{8}{3}\right)$              (B)              (C) $\left(4,±\frac{3}{8}\right)$               (D)

Subject: Maths, asked on 18/3/18

If f'(x) has no roots, does that mean f(x) will always have at least one root?

Subject: Maths, asked on 17/3/18

In the solution of the following question , I want to know how the marked (in blue bracket) step came ? Q. Find the intervals in which  is strictly increasing or strictly decreasing?

Subject: Maths, asked on 17/3/18

How did we get to know it's "strictly" decreasing or "strictly" increasing ?!.. Why not just decreasing/increasing ?!..

Subject: Maths, asked on 17/3/18

How area is calculated in this question? Please explain .Dont send links .

Subject: Maths, asked on 17/3/18

Please explain how it is done. Don't send links . Solutions: Ellipse, the area of triangle ABC (A) is given by.

Subject: Maths, asked on 17/3/18

Please explain how perpendicular distance came . Dont provide link please. Solution: The equation of the normal at a given point (x,y) is given by, Now, the perpendicular distance of he normal from the origin is. Hence, the perpendicular distance of the normal from the origin is constant.

Subject: Maths, asked on 16/3/18

Q) Why didn't we change the units or keep it same? Let y m be the height of the wall at which the ladder touches. Also, let the foot of the ladder be x m away from the wall. Then, by Pythagoras theorem, we have: x2 + y2 = 25 [Length of the ladder = 5 m] $y=\sqrt{25-{x}^{2}}$ Then, the rate of change of height (y) with respect to time (t) is given by, $\frac{dy}{dx}=\frac{-x}{\sqrt{25-{x}^{2}}}.\frac{dx}{dt}$ It is given that dx/dt = 2 cm/s $\frac{dy}{dt}=\frac{-2x}{\sqrt{25-{x}^{2}}}$ Now, when x = 4 m, we have: $\frac{dy}{dt}=\frac{-2×4}{\sqrt{25-16}}=-\frac{8}{3}$ Hence, the height of the ladder on the wall is decreasing at the rate of 8/3 cm/s.

Subject: Maths, asked on 16/3/18

Please dont provide link.

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