Correct Answer: D. 1

Explanation

Lim(\(\frac{1 - x}{x^2 - 3x + 2}\))

= \(^{lim}_{x \to 1} \begin{pmatrix} \frac{1 - x}{x^2 - 3x + 2} \end {pmatrix}\)

= \(^{lim}_{x \to 1} \begin{pmatrix} \frac{x - 1}{(x - 2)(x + 1)} \end {pmatrix}\)

= \(\frac{-1}{1 - 2}\)

= \(\frac{-1}{-1}\)

= 1

Explanation

(a) Maximum height reached

Using the equation, \(v^{2} = u^{2} - 2gs\) (Upward movement against gravity)

velocity at maximum height = 0 m/s

\(0 = 80^{2} - 2(10)s\)

\(0 = 6400 - 20s \implies 6400 = 20s \)

\(s = 320 m\)

(b) Time to reach the maximum height

\(v = u + at = u - gt\)

\(0 = 80 - 10t \implies 80 = 10t\)

\(t = 8s\)

Correct Answer: B. \(-1\)

Explanation

\(f(x) = \frac{3x + 1}{x^{2} - 1}\)

\(f(-3) = \frac{3(-3) + 1}{(-3)^{2} - 1} = \frac{-8}{8} = -1\)

Explanation

(a) \(y = 6 - x - x^{2}\)

When x = 0, y = 6.

When y = 0, \(x^{2} + x - 6 = 0\)

\(x^{2} + 3x - 2x - 6 = 0 \implies x(x + 3) - 2(x + 3) = 0\)

\(x = -3 ; x = 2\)

Graph passes through (0, 6), (-3, 0) (2, 0).

Turning point : \(\frac{\mathrm d y}{\mathrm d x} = - 1 - 2x\)

\(-1 - 2x = 0 \implies x = \frac{-1}{2}\)

When \(x = \frac{-1}{2}\), \(y = 6 - (-\frac{1}{2}) - (\frac{-1}{2})^{2} = 6\frac{1}{4}\).

Turning point : \((-\frac{1}{2}, 6\frac{1}{4})\).

\(y = 3x^{2} - 2x + 3\)

When x = 0, y = 3.

\(b^{2} - 4ac = (-2)^{2} - 4(3)(3) = 4 - 36 < 0\)

Graph does not cut x - axis.

\(\frac{\mathrm d y}{\mathrm d x} = 6x - 2\)

At turning point, \(6x - 2 = 0 \implies x = \frac{2}{6} = \frac{1}{3}\)

When \(x = \frac{1}{3}, y = 3(\frac{1}{3})^{2} - 2(\frac{1}{3}) + 3 = 2\frac{2}{3}\)

Turning point : \((\frac{1}{3}, 2\frac{2}{3})\).

(b) For the x- coordinates, we use the equation.

\(3x^{2} - 2x + 3 = 6 - x - x^{2}\)

\(4x^{2} - x - 3 = 0 \implies 4x^{2} - 4x + 3x - 3 = 0\)

\(4x(x - 1) + 3(x - 1) = 0 \implies (4x + 3)(x - 1) = 0\)

\(x = -\frac{3}{4} ; 1\)

(c) Area of shaded portion :

\(\int_{-\frac{3}{4}} ^{1} [(6 - x - x^{2}) - (3x^{2} - 2x + 3)] \mathrm {d} x\)

= \(\int_{-\frac{3}{4}} ^{1} (3 + x - 4x^{2}) \mathrm {d} x\)

= \([3x + \frac{x^{2}}{2} - \frac{4}{3}x^{3}]|_{-\frac{3}{4}} ^{1}\)

= \((3(1) + \frac{1}{2} - \frac{4}{3}) - (3(-\frac{3}{4}) + \frac{(\frac{-3}{4})^{2}}{2} - \frac{4}{3} (-\frac{3}{4})^{3}\)

= \(\frac{13}{6} - (-\frac{45}{32})\)

= \(\frac{343}{96} sq. units\)

Correct Answer: D. -\( \frac{7}{36}\)

Explanation

The general form of a quadratic equation is:

\(x^2\) -(sum of roots)\(x\) +(product of roots) = 0

\(7x^2+12x-4=0\)

Divide through by 7

=\(x^2+\frac{12}{7}x-\frac{4}{7}=0\)

=\(x^2-(-\frac{12}{7})x+(-\frac{4}{7})=0\)

\(\therefore\) sum of roots = \(-\frac{12}{7}\), and products of roots =\(-\frac{4}{7}\)

α + β = \(-\frac{12}{7}, αβ = -\frac{4}{7}\)

\(\frac{αβ}{(α + β)^2} = \frac{\frac{-4}{7}}{(\frac{-12}{7})^2}\)

=\(\frac{\frac{-4}{7}}{\frac{144}{49}}=-\frac{4}{7}\times\frac{49}{144}\)

\(\therefore - \frac{7}{36}\)

Correct Answer: D. \(-1 \leq r \leq 1\)

Correct Answer: B. \(\frac{1}{45}\)

Explanation

P(winning) = \(\frac{2}{10}\)

P(both tickets winning) = \(\frac{2}{10} \times \frac{1}{9} = \frac{1}{45}\)

Explanation

a. a = (10t - 4t\(^2\))m/s\(^2\), t = t seconds, u = 0

from first equation of motion,

v = u + at

v = 0 + (10t - 4t\(^2\))t

v = (10t\(^2\) - 4t\(^3\))m/s\(^2\)

ii. average acceleration of the particle during the 4th second = the average acceleration between the t = 3 and t = 4

Δa4 = 10(t4-t3) - 4(t4-t3)\(^2\)

Δa4 = 10(4-1) - 4(4-3)\(^2\)

Δa4 = 10(1) - 4(12)

Δa4 = (10-4)m/s\(^2\)

Δa4 = 6ms-2

b. R = m(g+a)

When the lift moves with a constant velocity, a = 0

R = mg

R = 120 × 10

R = 1200N

ii. when the lift moves upward with an acceleration of 3m/s\(^2\)

R = m(g+a)

R = 120(10+3)

R = 120 × 13

R = 1560N

Correct Answer: D. {-1, 5, 13, 49}

Explanation

\(P = {x : \text{x is a factor of 6}} \implies P = {1, 2, 3, 6}\)

\(g(x) = x^{2} + 3x - 5\)

\(g(1) = 1^{2} + 3(1) - 5 = 1 + 3 - 5 = -1\)

\(g(2) = 2^{2} + 3(2) - 5 = 4 + 6 - 5 = 5\)

\(g(3) = 3^{2} + 3(3) - 5 = 9 + 9 - 5 = 13\)

\(g(6) = 6^{2} + 3(6) - 5 = 36 + 18 - 5 = 49\)

\(\therefore Range(g(x)) = {-1, 5, 13, 49}\)

Correct Answer: C. 6 ms\(^{-2}\)

Explanation

V = 3t\(^2\) - 6t

\(\frac{dx}{dt}\) = 6r - 6

at t = 4

\(\frac{dx}{dt}\) = 6 x 4 - 6 = 24 - 6

= 18

at t = 3

\(\frac{dy}{dt}\) = 6 x 4 - 6

= 24 - 6

= 18

at t = 3

\(\frac{dx}{dt}\) = 6 x 3 - 6 = 18 - 6

= 12

\(\frac{dy}{dt}\) = 18 - 12

= 6ms\(^{-2}\)