# 158 - Mathematics Questions Answers

best calculus books for iitjee.

**Asked By: KRISHN RAJ**

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**Joshi sir comment**

its I. E. Maron

what is the range of log _{20}^{3}?

**Asked By: PRINCE PAL**

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**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)

**Asked By: SARABJEET**

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**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

**Asked By: AVI GARG**

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**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!

**Asked By: HIMANSHU MITTAL**

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**Joshi sir comment**

limn->∞ 4^{n}/ n!

= lim_{n->}∞ [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

**Asked By: GAURAV MAHATE**

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**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?

**Asked By: ROBERT DEPANGHER**

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**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 + πX^{2}/8

now calculate dA/dX and then compare it to zero

X will be 8/(4+π)

**Asked By: BUDDHI PRAKASH**

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**Joshi sir comment**

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

**Asked By: BUDDHI PRAKASH**

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**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 ^{-1}x - cos^{-1}(y/2) = α , then 4x^{2}-4xycosα + y^{2} is equal to ?**

**Asked By: AMIT DAS**

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**Joshi sir comment**

according to the given condition

cos^{-1}x = cos^{-1}(y/2) + a

so x = cos ( cos^{-1}y/2 + a)

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

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

so x - sina - y/2 cosa = -[1-(y^{2}/4)]^{1/2}

now square both side and solve