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JAMB Mathematics Past Questions & Answers - Page 80

396.

Given m = N $\sqrt{\frac{SL}{T}}$ make T the subject of the formula

$\frac{\text{NSL}}{M}$

$\frac{N^2SL}{M^2}$

$\frac{N^2SL}{M}$

$\frac{NSL}{M^2}$

View answer & explanation

Correct answer is B

M = N $\sqrt{\frac{SL}{T}}$ ,

make T subject of formula square both sides

M $^{2}$ = $\frac{N^2SL}{T}$

TM $^{2}$ = N $^{2}$ SL

T = $\frac{N^2SL}{M^2}$

397.

Evaluate 1 - ( $\frac{1}{5}$ x $\frac{2}{3}$ ) + ( 5 + $\frac{2}{3}$ )

2 $\frac{2}{3}$

$\frac{98}{15}$

View answer & explanation

Correct answer is D

No explanation has been provided for this answer.

398.

If temperature t is directly proportional to heat h, and when t = 20^oC, h = 50 J, find t when h = 60J

24^oC

20^oC

34^oC

30^oC

View answer & explanation

Correct answer is A

t ∝ h, t = 20, h

t = ? h = 60

t = kh where k is constant

20 = 50k

k = $\frac{20}{50}$

k = $\frac{2}{5}$

when h = 60, t = ?

t = $\frac{2}{5}$ × 60

t = 24^oC

399.

If y = x Sin x, find $\frac{dy}{dx}$ when x = $\frac{\pi}{2}$

$\frac{- \pi}{2}$

-1

$\frac{ \pi}{2}$

View answer & explanation

Correct answer is C

y = xsinx

$\frac{dy}{dx}$ = $1 \sin x + x \cos x$

= $\sin x + x \cos x$

At x = $\frac{\pi}{2}$

= sin $\frac{\pi}{2}$ + $\frac{\pi}{2} \cos {\frac{\pi}{2}}$

= 1 + $\frac{\pi}{2}$ × 0

= 1

400.

Evaluate ( $\sin$ 45º + $\sin$ 30º ) in surd form

$\frac{\sqrt{3}}{2\sqrt{2}}$

√3 − $\frac{1}{2}$

$\frac{1}{2}$ √2

1 + $\frac{\sqrt{2}}{2}$

View answer & explanation

Correct answer is D

hypotenuse
sin = $\frac{1}{2}$

$\sin45 = \frac{1}{\sqrt{2}}$

= $\frac{2}{2}$

∴ (sin45 + sin30)

= $\frac{1}{\sqrt{2}} + \frac{1}{2}$

= $\frac{\sqrt{2}}{2}$ + $\frac{1}{2}$

= $\frac{\sqrt{2} + 1}{2}$

= $\frac{1 + \sqrt{2}}{2}$