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JAMB Mathematics Past Questions & Answers - Page 297

1,481.

If y = 243(4x + 5)^-2, find $\frac{dy}{dx}$ when x = 1

$\frac{-8}{3}$

$\frac{3}{8}$

$\frac{9}{8}$

- $\frac{8}{9}$

View answer & explanation

Correct answer is A

y = 243(4x + 5)^-2, find $\frac{dy}{dx}$

= -1944(4x + 5)_-3

= 1944(9)_-3

$\frac{dy}{dx}$ when x = 1

= - $\frac{1944}{9^3}$

= - $\frac{1944}{729}$

= $\frac{-8}{3}$

1,482.

From the top of a vertical mast 150m high., two huts on the same ground level are observed. One due east and the other due west of the mast. Their angles of depression are 60° and 45° respectively. Find the distance between the huts

150(1 + $\sqrt{3}$ )m

50( $\sqrt{3}$ - $\sqrt{3}$ )m

150 $\sqrt{3}$ m

$\frac{50}{\sqrt{3}}$ m

View answer & explanation

Correct answer is B

$\frac{150}{Z}$ = tan 60o,

Z = $\frac{150}{tan 60^o}$

= $\frac{150}{3}$

= 50 $\sqrt{3}$ cm

$\frac{150}{X x Z}$ = tan45o = 1

X + Z = 150

X = 150 - Z

= 150 - 50 $\sqrt{3}$

= 50( $\sqrt{3}$ - $\sqrt{3}$ )m

1,483.

solve the equation cos x + sin x $\frac{1}{cos x - sinx}$ for values of such that 0 $\leq$ x < 2 $\pi$

$\frac{\pi}{2}$ , $\frac{3\pi}{2}$

$\frac{\pi}{3}$ , $\frac{2\pi}{3}$

0, $\frac{\pi}{3}$

0, $\pi$

View answer & explanation

Correct answer is D

cos x + sin x $\frac{1}{cos x - sinx}$

= (cosx + sinx)(cosx - sinx) = 1

= cos²x + sin²x = 1

cos²x - (1 - cos²x) = 1

= 2cos²x = 2

cos²x = 1

= cosx = $\pm$ 1 = x

= cos^-1x ( $\pm$ , 1)

= 0, $\pi$ $\frac{3}{2}\pi$ , 2 $\pi$

(possible solution)

1,484.

The midpoint of the segment of the line y = 4x + 3 which lies between the x-ax 1 is and the y-ax 1 is

( $\frac{3}{2}$ , $\frac{3}{2}$ )

( $\frac{2}{3}$ , $\frac{3}{2}$ )

( $\frac{3}{8}$ , $\frac{3}{2}$ )

(- $\frac{3}{8}$ , $\frac{3}{2}$ )

View answer & explanation

Correct answer is D

y = 4x + 3

when x = 0, y = 3 $\to$ (0, 3)

when y = 0, x = - $\frac{3}{4}$ $\to$ ( $\frac{3}{4}$ , 0)

mid-point $\frac{0 + (-{\frac{3}{4}})}{2}$ , $\frac{3 + 0}{4}$

- $\frac{3}{8}$ , $\frac{3}{2}$