158 - Mathematics Questions Answers

ƒ(x)=|x-2| and g=ƒoƒ so g'(x)=.......? ; 2<x<4

Asked By: BONEY HAVELIWALA
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Joshi sir comment

for 2<x<4,   f(x) = x-2

so g(x) = f(x-2) = x-2-2 = x-4

so g'(x) = 1 

ƒ(x)=3|2+x| so ƒ'(-3)=? ,if its -3 , how?

Asked By: BONEY HAVELIWALA
is this question helpfull: 5 2 read solutions ( 1 ) | submit your answer
Joshi sir comment

ƒ(x)=3|2+x|

for x>-2,  f(x) = 3(2+x)  so  f'(x) = 3 

for x<-2, f(x) = -3(2+x) so f'(x) = -3

Maths >> Calculus >> Integrations IIT JEE

∫√tanx

Asked By: BONEY HAVELIWALA
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Joshi sir comment

let  √tanx = t

so sec2xdx/2√tanx = dt

so (1+t4)dx/2t = dt

so dx = 2tdt/(1+t4)

now solve

What is the value of 0⁄0? if its otherthan 1, then how this can be true:      lim      (ex-1)/x = 1

                                                                                                                                x −−>0

Asked By: BONEY HAVELIWALA
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Joshi sir comment

exact 0 and lim tends to 0 are different

Maths >> Trigonometry >> Complex Numbers Engineering Exam

what is the angular difference between the +j and -j ?

OR 

+4j and -4j  ?

 

Asked By: ADARSH KUMAR
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Joshi sir comment

180 degree

sin‾¹(sin2) + sin‾¹(sin4) + sin‾¹(sin6) = ?

Asked By: BONEY HAVELIWALA
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Joshi sir comment

sin‾¹(sin2) + sin‾¹(sin4) + sin‾¹(sin6)

= (π-2) + (π-4) + (6-2π) 

= 0

Maths >> Calculus >> Functions AIEEE

if ƒ:R−{0} -> R,  2ƒ(x) − 3ƒ(1⁄x) = x²  then ƒ(3)= ?

Asked By: BONEY HAVELIWALA
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Joshi sir comment

2ƒ(x) − 3ƒ(1⁄x) = x²   (1)

so 2ƒ(1/x) − 3ƒ(x) = 1/x²  (2)

multiply (1) to 2 and (2) to 3, then add the two equations, you will get f(x) then calculate f(3)  

Maths >> Calculus >> Functions Others

Sir/Madam,

                    Suddenly a question struck on my mind:   0×∞= 1 or 0?? as 1⁄0=∞.

     

Asked By: BONEY HAVELIWALA
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Joshi sir comment

it will be 0, assume it by considering that ∞ is a big number 

for ex.   0*(1111111111111111111111111) = 0

Solve

tanx-3cotx=2tan3x(0<x<360)

cosx-sinx=cosy-siny

cos2xcotx+1=cos2x+cotx

Asked By: YUMESH
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Joshi sir comment

tanx - 3/tanx = 2[3tanx-tan3x]/[1-3tan2x]

so [tan2x - 3]/tanx = 2tanx[3-tan2x]/[1-3tan2x]

so [tan2x - 3][1-3tan2x] = 2tan2x[3-tan2x]

so [tan2x - 3][1-3tan2x] + 2tan2x[tan2x-3] = 0

so [tan2x - 3][1-tan2x] = 0

so tanx = 1, -1, √3, -√3

now check the values satisfying last eq.

then put these values in 2nd eq. to get answer.

Find the value :-

(tan69 +tan66) / (1 - tan69.tan66)

Asked By: ADARSH
is this question helpfull: 4 1 read solutions ( 2 ) | submit your answer
Joshi sir comment

use formula of tan(A+B)

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