The area of a circle of radius 4cm is equal to (Take π = 3.142 )
10.3cm2
15.7cm2
50.3cm2
17.4cm2
Correct answer is C
Area of a circle (A) = πr2
Radius = 4cm
π = 3.142
A = 3.142 × (4)2
= 3.142 × 16
= 50.272
A = 50.3cm2 (1dp)
Simplify \( \frac{1}{(x + 1)} + \frac{1}{(x − 1)} \)
2x/(x + 1)(x−3)
2x/(x + 1)(x−1)
2x/(x + 1)2
2x(x+1)2
Correct answer is B
[1 ÷ (x+1)] + [1 ÷ (x − 1)]
= ((x − 1) + [(x + 1)) ÷ (x+1)(x − 1)]
Using the L.C.M.
= (x − 1 + x + 1) ÷ (x + 1)(x − 1)
= (x + 2 − 1 + 1) ÷ (x + 1)(x − 1)
= 2x ÷ (x + 1)(x − 1) =2x ÷ (x + 1)(x − 1)
17.1cm2
27.2cm2
47.1cm2
37.3cm2
Correct answer is C
Find the slant height
\( l^2 = h^2 + r^2(h = 4cm,r = 3cm)\)
\( l^2 = 4^2 + 3^2 = 16 + 9 = 25 \)
\( l^2 = √ 25 \)
Squaring both sides
l = 5cm
The area of curved surface (s) =π(3)(5)
15π = 15 × 3.14
= 47.1cm2
Integral ∫\( (5x^3 + 7x^2 − 2x + 5)\)dx
\( \frac{5x^4}{4} + \frac{7x^3}{3} + 2x + C \)
\( \frac{5x^4}{4} + \frac{7x^3}{3} - x^2 + 5x + C \)
\( \frac{5x^3}{3} + \frac{7x^2}{x} - x + C \)
\( \frac{2x^2}{3} + \frac{x}{5} - C \)
Correct answer is B
\(\int (5x^{3} + 7x^{2} - 2x + 5) \mathrm d x\)
= \(\frac{5x^{4}}{4} + \frac{7x^{3}}{3} - \frac{2x^{2}}{2} + 5x + c\)
= \(\frac{5x^{4}}{4} + \frac{7x^{3}}{3} - x^{2} + 5x + c\)
Evaluate log5(\( y^2x^5 ÷ 125b½) \)
2 log5y + 5log5 y2 − 3
log5 y2 + 5log5 x + 3
25logy 5 + 3
2log5y + 5log5x − ½ log5b −3
Correct answer is D
\(\log_{5}(y^{2} x^{5} \div 125b^{\frac{1}{2}})\)
= \(\log_{5} y^{2} + \log_{5} x^{5} - [\log_{5} 125 + \log_{5} b^{\frac{1}{2}}\)
= \(2\log_{5} y + 5\log_{5} x - \log_{5} 5^{3} - \frac{1}{2} \log_{5} b\)
= \(2\log_{5} y + 5\log_{5} x - 3 - \frac{1}{2}\log_{5} b\)