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Section A
Q. 1
(a)Prove that every non-empty set of real numbers that has an upper bound also has an supremum in R. [10]
(b)If x x∈R , set of real numbers, then there exists n n∈N such that x < n x < n x < n . [10]
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Q. 2
(a)Define continuity of a function at a point and also prove that if f and g be functions on A A⊆R then f + g f + g f + g and f f⋅g are continuous at c c∈A . [10]
(b)If f : I f:I→R is differentiable at c c∈I , then f is continuous at c . [10]
(c)f'
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Q. 3
(a)Evaluate ∫ d x x − 2 \int \frac{dx}{\sqrt{x}-2} ∫ x − 2 d x [8]
(b)i) Define Complete metric space. ii) Prove that a sequence of real numbers is convergent iff it is a Cauchy sequence. This theorem is not in metric space, for justification give one example. [12]
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Q. 4
(a)Let (X,d) be a metric space and A a subset of X . Then prove that: i) Interior A of A is an open subset of X . ii) A ‾ \overline{A} A is the largest subset of X contained in A . [10]
(b)State and prove Mean value theorem. [10]
(c)= f ( b ) − f ( a ) b − a f'
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Q. 5
(a)If ∑an converges absolutely then ∑an converges. [10]
(b)Find the area enclosed by the parabola y2+16x−71=0 and the line 4x+y+7=0 . [10]
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SECTION–B
Q. 6
(a)Let Z = ( cos θ + i sin θ ) Z = (\cos\theta + i \sin\theta) Z = ( cos θ + i sin θ ) . Then prove that Z n = cos n θ + i sin n θ Z^n = \cos n\theta + i \sin n\theta Z n = cos n θ + i sin n θ for all n . [10]
(b)Using De Moivre's Theorem evaluate ( 3 − i 3 + i ) 6 \left(\frac{\sqrt{3}-i}{\sqrt{3}+i}\right)^6 ( 3 + i 3 − i ) 6 . [10]
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Q. 7
(a)Expand f ( x ) = f(x)=x2 , 0 < x < 0<x<2π in a Fourier series if period is 2π . [10]
(b)If f(z) is analytic inside a circle C with centre at a , then for all z inside C f ( z ) = f ( a ) + f ′ ( a ) ( z − a ) + f ′ ′ ( a ) 2 ! ( z − a ) 2 + . f(z) = f(a) + f'(a)(z-a) + \frac{f''(a)}{2!}(z-a)^2 + ... f ( z ) = f ( a ) + f ′ ( a ) ( z − a ) + 2 ! f ′′ ( a ) ( z − a ) 2 + ... [10]
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Q. 8
(a)Evaluate the integral by using Cauchy integral Formula ∮Cz(z−1)(z−2)(4−3z)dz where C is a circle ∣ z ∣ = 3 2 |z| = \frac{3}{2} ∣ z ∣ = 2 3 . [10]
(b)Prove that ∫ 0 2 π d θ 1 − 2 p cos θ + p 2 = 2 π 1 − p 2 \int_0^{2\pi} \frac{d\theta}{1-2p\cos\theta+p^2} = \frac{2\pi}{1-p^2} ∫ 0 2 π 1 − 2 p c o s θ + p 2 d θ = 1 − p 2 2 π [10]
The 2011 CSS Pure Mathematics paper set by the FPSC. Question wording only; questions marked “Not yet checked” have not been compared with the official paper yet.
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