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All Current Mid Term Papers Spring 2013 (25 May 2013 ~ 06 June 2013) at One Place
From 25 May 22, 2013 to 06 June 2013 Spring 2013
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mth 101 midterm paper 29.05.2013 time 3:30
Only 4 to 5question from past papers and remaning from book. Paper was conceptual so read properly book.
1) Limt sinx/x=?
3)tan 5-3x option sy select karna tha
agar concept theak ho to paper easily solve ho jata hay.
remember me in ur prayers
Mostly question on concept basis not from past paper.
Raheel thanks for sharing
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What are the critical points of f(x)= x^3+2x^2-4x+5?
to find the critical points of a function, we need to find the first derivative and equal it to zero as the following:
f(x) = x^3+2x^2-4x+5
f ' (x) = 3x^2 + 4x - 4 ===>equal it to zero
0 = 3x^2 + 4x - 4
-b +/- [ sqrt( b^2 - 4*a*c ) ] / [ 2a ]
-4 +/- [ sqrt( 4^2 - 4*3*-4 ) ] / [ 2*3 ]
-4 +/- [ sqrt( 16 + 48) ] / [ 6 ]
-4 +/- [ sqrt( 64) ] / [ 6 ]
[ -4+/- 8 ] / 6 =====> [ -4+8 / 6 ] = 2/3 =====> & [ -4-8 / 6 ] = -2 ( critical points )
Find the equation of the line that passes through (6,5) and (-3, -1).?
First you want to find the slope. The formula is:
y2 - y1 (divided by)
x2 - x1 (the numbers are sub 1 and sub 2, not exponents)
in the first ordered pair (x1, y1) is (6,5). the second (x2, y2) is (-3, -1)
now fill in the formula.
y2 is -1. y1 is 5. So the numerator is -1 minus 5. Which is -6.
x2 is -3. x1 is 6. So the denominator is -3 minus 6. Which is -9.
So our slope is -6/-9 which can be simplified to 2/3.
NOW. You know the slope is 2/3. Now you must find a line with this slope that runs through those points (either set of points will work).
y - y1 = 2/3 (x - x1).
Fill in for y1 and x1 with either set of points (but be consistent. if you use from the first set for x, you must use from the first set for y).
We'll use (-3, -1)
y - -1 = 2/3 (x - -3)
When subtracting a negative, you add. So:
y + 1 = 2/3 (x + 3)
Now distribute 2/3.
y + 1 = 2/3x + 2. (2/3 times 3 is 2)
now subract 1 from both sides.
y = 2/3x + 1
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