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# MTH101 Calculus And Analytical Geometry Assignment No 1 Solution & Discussion Due Date:12-11-2012

Assignment No. 1

MTH 101 (Fall 2012)

Maximum Marks: 15

Due Date: 12th Nov. 2012

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Question # 01:                                                                                        05

Find the equation for circle with center (1,3) that passes through (4, -1).

Question # 02:                                                                                        05

Given that . Find . Also find the domains of the composite functions.

Question # 03:                                                                                        05

Find . (Use rationalization technique to simplify)

Hint: (How to rationalize)

Rationalize the numerator of the expression .

Step I: find the conjugate of the numerator which is and vice versa.

Step II: multiply  the numerator and the denominator of the expression with the conjugate

Step III: make sure all the radicals are simplified

Step IV: simplify the fraction if needed.

In the same manner we can rationalize denominator too.

(Note: In order to get full marks, do all necessary steps)

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MTH101 Current Assignment Solution No.1 Fall 2012

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Question # 02: 05
Given that . Find . Also find the domains of the composite functions.
Sol.
fog(x) = f ( g(x) )
= ( g(x) )2 – 4
= x-2- 4
= x-6
gof(x) = g( f(x) )

Calculus And Analytical Geometry assignment solution.

Calculus And Analytical Geometry assignment solution.

Question # 01:
Find the equation for circle with center (1,3) that passes through (4, -1)
Sol.
Radius of circle r is the distance between (4, -1) and (1, 3)

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Question # 02:

Given that . Find . Also find the domains of the composite functions.
Sol.

fog(x) = f ( g(x) )

= ( g(x) )2 – 4
= x-2- 4
= x-6
gof(x) = g( f(x) )

Question # 03:
Find . (Use rationalization technique to simplify)

dear how to write symbols like underroot in Ms word??

Squroot ka domain hai 1 or -6 infinity  abe dono ka unione ley loo.  (- infinity ,+ infinity)U(1, - infinity)

(1, - infinity ) jawab hai , ap key sewal kei mutabiq, ager value change hai to jawab per dosra hoga.

what is the final domain you got of Q.2? and also tell me is the part 2 of a question possess composite function?

plz uplaod the complete solution. I'm facing some prb in Q.2. I've done it bt I'm not sure I did right

Question # 02: 05

Given that . Find . Also find the domains of the composite functions.

Sol.

fog(x) = f ( g(x) )

= ( g(x) )2 – 4

= x-2- 4
= x-6

gof(x) = g( f(x) )

Anything in a square root cannot be negative because it would be undefined. And the denominator in a fraction cannot be zero

So the domain is where x/(x-6) ≥ 0 and where x - 6 ≠ 0

So then x ≥ 0 and x ≠ 6

So the domain would be: 0 ≤ x < 6 and 6 < x < ∞

It can also be written this way: [0,6) and (6,∞)

The domain is the x the range is the y. The domain is an independent variable because it is not influenced by anything. The y is dependent because it is influenced by the independent variable.

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