Multiple Choice Questions Rio Mike. Q_1 . The value of 6 + 2 × 4 −15 ÷ 3 is; A) 3 B) 9 C) 4 D) 27 Q_2 . 80 as a product of its prime factor is. A) 2^3 × 5 B) 2 × 5^(3 ) C) 2^2 × 5^2 D) 2^4 × 5 Q_3 . The deteminant of (((6 4)),((3 2)) ) is A) 18 B) 0 C) 24 D) −24 Q_4 . the value of ((1/(16)))^(−(1/2)) is A) −(1/4) B) 4 C) −4 D) (1/2) Q_5 . if −8,m,n,19 are in AP then (m,n) is A) (1,10) B) (2,10) C) (3,13) D) (4,16)o Q_6 . if cos θ = ((12)/(13)) then cot^2 θ is; A) ((169)/(25)) B) ((25)/(169)) C)((169)/(144)) D) ((144)/(169)) Q_7 . the solution set of 3 ≤ 2x−1≤5 is; A) ]x≥−2,x≤(8/3)[ B) [x≥2,x≤−(8/3)[ C) ]x>−2,x≤(3/8)] D) [x≥−2,x≤(8/3)] Q_8 . Which is a factor of f(x)= x^2 −3x + 2 A) (x−1) B) (x−2) C) (x+1) D)(x+3) Q_(9 ) . The sum of the sum of roots and product of roots of the quadratic equation 3x^2 + 6x + 9=0 A) 1 B) −1 C) 32 D) 5 Q_(10) . 2log2−log2 = A) log 8 B) log 6 C) log 3 D) log 2. Q_(11) . Σ_(r=1) ^∞ 3^(2−r) = A) (9/4) B) (9/2) C) ((13)/3) D) (1/2) Q_(12) . The constand term in the binomial expansion of (x^2 + (1/x^2 ))^8 is A) 3^(rd) B) 4^(th) C) 5^(th) D) 6^(th) Q_(13) . cos(180−x) = A) −sinx B) sin x C) cos x D) −cos x Q_(14) . The value of p for which (2,1) , (6,3), and (4,p) are collinear is ; A) 2 B) 1 C)−1 D) −2 Q_(15) . ∫_1 ^2 x^3 dx = A) −sin 7x B) sin 7x C) −7sin7x D) 7sin7x Q_(16) . Given that g : → px −5, the value of p for which g^(−1) (3)=4 is A) (1/2) B) −(1/2) C) −2 D) 2 Q_(17) . The value of θ in the range 0°≤θ≤90° for which sinθ = cos θ is A) 90° B) 60° C) 30° D) 45° Questions


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Question Number 35811 by Rio Mike last updated on 23/May/18

$M u l t i p l e C h o i c e Q u e s t i o n s$ $R i o M i k e .$ $Q_{1} . T h e v a l u e o f 6 + 2 \times 4 - 15 \div 3$ $i s;$ $A) 3 B) 9 C) 4 D) 27$ $Q_{2} . 80 a s a p r o d u c t o f i t s p r i m e$ $f a c t o r i s .$ $A) 2^{3} \times 5 B) 2 \times 5^{3} C) 2^{2} \times 5^{2}$ $D) 2^{4} \times 5$ $Q_{3} . T h e d e t e m i n a n t o f (\begin{matrix} 6 4 \\ 3 2 \end{matrix}) i s$ $A) 18 B) 0 C) 24 D) - 24$ $Q_{4} . t h e v a l u e o f {(\frac{1}{16})}^{- \frac{1}{2}} i s$ $A) - \frac{1}{4} B) 4 C) - 4 D) \frac{1}{2}$ $Q_{5} . i f - 8, m, n, 19 a r e i n A P t h e n$ $(m, n) i s$ $A) (1, 10) B) (2, 10) C) (3, 13) D) (4, 16) o$ $Q_{6} . i f c o s θ = \frac{12}{13} t h e n {c o t}^{2} θ i s;$ $A) \frac{169}{25} B) \frac{25}{169} C) \frac{169}{144} D) \frac{144}{169}$ $Q_{7} . t h e s o l u t i o n s e t o f 3 ⩽ 2 x - 1 ⩽ 5$ $i s;$ $A)] x ⩾ - 2, x ⩽ \frac{8}{3} [B) [x ⩾ 2, x ⩽ - \frac{8}{3} [$ $C)] x > - 2, x ⩽ \frac{3}{8}] D) [x ⩾ - 2, x ⩽ \frac{8}{3}]$ $Q_{8} . W h i c h i s a f a c t o r o f f (x) =$ $x^{2} - 3 x + 2$ $A) (x - 1) B) (x - 2) C) (x + 1) D) (x + 3)$ $Q_{9} . T h e s u m o f t h e s u m o f$ $r o o t s a n d p r o d u c t o f r o o t s o f t h e$ $q u a d r a t i c e q u a t i o n 3 x^{2} + 6 x + 9 = 0$ $A) 1 B) - 1 C) 32 D) 5$ $Q_{10} . 2 l o g 2 - l o g 2 =$ $A) l o g 8 B) l o g 6 C) l o g 3$ $D) l o g 2 .$ $Q_{11} . \underset{r = 1}{\sum^{\infty}} 3^{2 - r} =$ $A) \frac{9}{4} B) \frac{9}{2} C) \frac{13}{3} D) \frac{1}{2}$ $Q_{12} . T h e c o n s t a n d t e r m i n t h e$ $b i n o m i a l e x p a n s i o n o f {(x^{2} + \frac{1}{x^{2}})}^{8} i s$ $A) 3^{r d} B) 4^{t h} C) 5^{t h} D) 6^{t h}$ $Q_{13} . c o s (180 - x) =$ $A) - s i n x B) s i n x C) c o s x$ $D) - c o s x$ $Q_{14} . T h e v a l u e o f p f o r w h i c h$ $(2, 1), (6, 3), a n d (4, p) a r e c o l l i n e a r$ $i s;$ $A) 2 B) 1 C) - 1 D) - 2$ $Q_{15} . \int_{1}^{2} x^{3} d x =$ $A) - s i n 7 x B) s i n 7 x C) - 7 s i n 7 x$ $D) 7 s i n 7 x$ $Q_{16} . G i v e n t h a t g : \to p x - 5, t h e$ $v a l u e o f p f o r w h i c h g^{- 1} (3) = 4 i s$ $A) \frac{1}{2} B) - \frac{1}{2} C) - 2 D) 2$ $Q_{17} . T h e v a l u e o f θ i n t h e r a n g e$ $0 ° ⩽ θ ⩽ 90 ° f o r w h i c h s i n θ = c o s θ i s$ $A) 90 ° B) 60 ° C) 30 ° D) 45 °$ $Q u e s t i o n s$
Commented by Rasheed.Sindhi last updated on 24/May/18

$How did you upload so large image ?!$
Commented by Joel579 last updated on 24/May/18

$Actually he typed it, not uploaded it$
Commented by Rasheed.Sindhi last updated on 24/May/18

$Sir I misunderstood . But his other$ $uploaded lengthy post is an image .$
Commented by Rio Mike last updated on 24/May/18

$yeah your right . i typed this but$ $the other was uploaded a s a n i m a g e$
Commented by Rasheed.Sindhi last updated on 25/May/18

$Questioners are requested to send at most$ $(3 short questions) / post$ $or$ $(1 long question) / post$ ${It}^{'} s better practice, I think .$
Commented by rahul 19 last updated on 26/May/18

$I a g r e e .$
	Answered by Rasheed.Sindhi last updated on 25/May/18

	$Q_{1}$ $6 + 2 \times 4 - 15 \div 3$ $= 6 + 8 - 5 = 9$ $B) 9$ $Q_{2}$ $D) 2^{4} . 5$ $Q_{3} T h e d e t e m i n a n t o f (\begin{matrix} 6 4 \\ 3 2 \end{matrix})$ $6 \times 2 - 3 \times 4 = 0$ $B) 0$ $Q_{4} {(\frac{1}{16})}^{- \frac{1}{2}}$ $= \frac{1}{{(4^{2})}^{- \frac{1}{2}}} = \frac{1}{4^{- 1}} = 4$ $B) 4$ $Q_{5} - 8, m, n, 19 are in AP$ $First term = a = - 8, Common difference = d$ $n t h term = a_{n} = a + (n - 1) d$ $a_{4} = - 8 + (4 - 1) d = 19$ $3 d = 19 + 8 = 27$ $d = 9$ $m = - 8 + 9 = 1, n = 1 + 9 = 10$ $(m, n) = (1, 10)$ $A) (1, 10)$
	Answered by Rasheed.Sindhi last updated on 25/May/18

	$\cos θ = \frac{12}{13}$ $\cos^{2} θ + \sin^{2} θ = 1$ $\frac{\cos^{2} θ}{\sin^{2} θ} + 1 = \frac{1}{\sin^{2} θ}$ $\cot^{2} θ + 1 = \frac{1}{1 - \cos^{2} θ} = \frac{1}{1 - {(\frac{12}{13})}^{2}}$ $\cot^{2} θ = \frac{1}{\frac{13^{2} - 12^{2}}{13^{2}}} - 1$ $= \frac{13^{2}}{13^{2} - 12^{2}} - 1 = \frac{144}{25}$ $D) \frac{144}{25}$ $Q_{7} 3 ⩽ 2 x - 1 ⩽ 5$ $4 ⩽ 2 x ⩽ 6$ $2 ⩽ x ⩽ 3$ $Neither option is correct!$ $Q_{8} Factor of x^{2} - 3 x + 2$ $x^{2} - 2 x - x + 2$ $x (x - 2) - 1 (x - 2)$ $(x - 1) (x - 2)$ $Two options are correct ?!$
	Answered by Rasheed.Sindhi last updated on 26/May/18

	${3 x}^{2} + 6 x + 9$ $α + β = - \frac{6}{3} = - 2, α β = \frac{9}{3} = 3$ $(α + β) + α β = - 2 + 3 = 1$ $A) 1$ $Q_{10} . 2 l o g 2 - l o g 2 =$ $l o g 2^{2} - l o g 2 =$ $l o g (4 \div 2) = l o g 2$ $D) l o g 2$ $Q_{11} . \underset{r = 1}{\sum^{\infty}} 3^{2 - r} =$ $a = 3^{2 - 1} = 3$ $r = \frac{3^{2 - 2}}{3^{2 - 1}} = \frac{1}{3}$ $\underset{r = 1}{\sum^{\infty}} 3^{2 - r} = \frac{3}{1 - \frac{1}{3}} = 3 \times \frac{3}{2} = \frac{9}{2}$ $B) \frac{9}{2}$
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