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Mathematics: Post your doubts here!

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hi,
how would you find the range of a function in this question
q3ii) http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_s10_qp_13.pdf

Plugin the values 0 , pi/2 , pi , 3pi/2 , 2pi into the equation and youll get the range. Set your calculator to radian mode.
 
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We have two equations y= kx^2 +1 and y=kx

Since it says the tangent touches the curve , kx^2+1=kx

Solving this using b^2 - 4ac gives two values of k , k=0 and k=4

Since in the question it says k is a non zero , we ignore k=0 and take k as 4

Now plugin this value into the original equations => y=4x^2 + 1 and y=4x

Solve them simultaneously => 4x^2 + 1 =4x

Therefore x= 1.5 , y=2
 
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We have two equations y= kx^2 +1 and y=kx

Since it says the tangent touches the curve , kx^2+1=kx

Solving this using b^2 - 4ac gives two values of k , k=0 and k=4

Since in the question it says k is a non zero , we ignore k=0 and take k as 4

Now plugin this value into the original equations => y=4x^2 + 1 and y=4x

Solve them simultaneously => 4x^2 + 1 =4x

Therefore x= 1.5 , y=2
yeah okay, got it. :D
thanks alot.
 
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ERMMM Anyone?
look. when you write in the value of f in f you'll get a fraction in the numerator as well as the denominator. find the l.c.m. of both the numerator and the denominator. the (2x-1) will get cancelled out (from both the numerator and denominator) and then when you simplify you'll be left with 7x/7. cancel the 7s and thus you're left with only an x. :D
 
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(tanx^n+2 tanx^n)dx substitute tanx = u

(u^n+2 + u^n)dx now du = sec^2x dx

so when substituting du will give


(u^n+2 + u^n)/sec^2x du
now use identity sec^2x = 1+ tan^2x .. means (1 + U^2)

substitute this value of sec^2x in the eq. above



(u^n+2 + u^n)/ (1 + U^2)du now take U^n in numenator

U^n (U^2 + 1 )/ (1 + U^2)du now cxancel (U^2 +1)
u wil get a simple eq. (U^n)du
inegrate it
u^n+1 / n+1 now convert back into original by substiuting tanx

tanx^n+1/ n+1 put the limits
tan(45)^n+1/n+1 tan45 = 1 so... 1^n+1 = 1 :)


it wil be as follows 1/ n+1
THNX a LOT!!!!
cant believe the thing bugging me was this simple!
 
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what is the difference in f '(x) and
inver0%7Bimage10%7D.gif
??? 9709/13/m/J/11 question 10
 
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some one plzz helpp me out with dis .. :S
oct/nov 2010 p12 question num 11 .. how to do it ..
 
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1.do the LCM in the numerator as well as in the denominator .
in the numerator You'll get something like this ---> ((x+3)+(6x-3)) divided by 2x-1 (which is the LCM of the numerator)
in the denominator you'll get something like this ---> ((2x+6)-(2x+1)) divided by 2x-1 ( which is the LCM of the denominator )
2. both the LCMs gets cancelled therefore, numerator remains as ---> (x+3)+(6x-3) and the denominator as ---> (2x+6)-(2x+1)
3. Solve it ..... you'll get the numerator as 7x and denominator as 7 , which is equal to x.



Hope this helps.
 
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