# 158 - Mathematics Questions Answers

Maths >> Calculus >> Functions IIT JEE

best calculus books for iitjee.

Joshi sir comment

its I. E. Maron

Maths >> Algebra >> Binomial Theorem Board Exam

what is the range of    log 203?

Joshi sir comment

There should be any variabe in the question

if x=2Cosθ-Cos2θ and y=2Sinθ-Sin2θ prove that dy/dx=tan(3θ/2)

Joshi sir comment

calculate dx/dθ and dy/dθ

then dy/dx = [2cosθ-2cos2θ]/[-2sinθ+2sin2θ]

now solve

use cosC-cosD = 2sin[(C+D)/2]sin[(D-C)/2]

and sinC+sinD = 2sin[(C+D)/2]cos[(C-D)/2]

sinA + sinB + sinC = 4cosA/2.cosB/2.cosC/2

Joshi sir comment

sinA + sinB + sinC = 2sin[(A+B)/2]cos[(A-B)/2] + 2sinC/2cosC/2

= 2sin[(π-C)/2]cos[(A-B)/2] + 2sinC/2cosC/2

= 2cosC/2cos[(A-B)/2] + 2sinC/2cosC/2

= 2cosC/2* [cos[(A-B)/2]+[cos(A+B)/2]                     here sinC/2 = cos(A+B)/2

now use cosC + cosD = 2 cos[(C+D)/2]cos[(C-D)/2]

You should remember that the given formula is based on a triangle.

lim n--> ∞  4^n/n!

Joshi sir comment

limn->∞ 4n/ n!

= limn->∞ [4/1][4/2][4/3][4/4][4/5][4/6][4/7]................................[4/n]

besides first four bracket, all brackets are real numbers less than 1 so their product will be zero for n tends to infinite

Sir,is the function (2x+5) is differentiable everywhere in its domain set. If yes the what is the case with l2x+5l.
Joshi sir comment

since first one is a straight line so at every point only one tangent is possible so it is differentiable

but second one which is modulus has a sharp turn at x = -5/2 so two tangents at x = -5/2 are possible so is not differentiable.

A window frame is shaped like a rectangle with an arch forming the top ( ie; a box with a semicircle on the top)

The vertical straight side is Y, the width at the base and at the widest part of the arch (Diam) is X.

I know that the perimeter is 4.

Find X if the area of the frame is at a maximum?

Joshi sir comment

perimeter = 4 so 2Y + X + πX/2 = 4

and area = XY + [π(X/2)2]/2

put Y from 1st equation to 2nd we get A = X[4-X{1+(π/2)}]/2 + πX2/8

now calculate dA/dX and then compare it to zero

X will be 8/(4+π)

For which integers n ≥ 3 does there exist a regular n-gon in the plane such that
all of its vertices have integer coordinates?

Joshi sir comment

for n = 4, we can form a square with integral coordinates of all the vertices.

Prove that the four projections of vertex A of the triangle ABC onto the exterior
and interior angle bisectors of  B and  C are collinear.
Joshi sir comment

in the given diagram red lines are angle bisectors (interior and exterior)and X, Y, Z, W are the projections of A on these red lines

by angle bisector property all these 4 points will lie on the line BC (either externally or internally) so these points will be in a line

If cos-1x - cos-1(y/2) = α , then 4x2-4xycosα + y2 is equal to ?

Joshi sir comment

according to the given condition

cos-1x  =  cos-1(y/2)  +  a

so x = cos ( cos-1y/2   +  a)

so x = y/2 cosa  -  sin cos-1y/2 + sina

so x = y/2 cosa  - [1-(y2/4)]1/2 + sina

so x - sina - y/2 cosa = -[1-(y2/4)]1/2

now square both side and solve