Correct Answer: C. 122^o

Explanation

PRQ = 90 - 58 = 32^o(angle in a semi-circle)

Since PRS = 90^o(radius angular to tangent)

QRS = 90 + 32

= 122^o

Correct Answer: C. {1, 3, 5, 7, 9, 11, 13}

Correct Answer: B. 1/2

Explanation

\(\frac{\log \sqrt{8}}{\log 8}\)

= \(\frac{\log 8^{\frac{1}{2}}}{log 8}\)

= \(\frac{\frac{1}{2} \log 8}{\log 8}\)

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

Correct Answer: D. 0

Explanation

∠PQR + ∠PSR = 180o (opp. angles of cyclic quad. are supplementary)

⇒ 5\(x\) - \(y\) + 10 + (-2\(x\) + 3\(y\) + 145) = 180

⇒ 5\(x\) - \(y\) + 10 - 2\(x\) + 3\(y\) + 145 = 180

⇒ 3\(x\) + 2\(y\) + 155 = 180

⇒ 3\(x\) + 2\(y\) = 180 - 155

⇒ 3\(x\) + 2\(y\) = 25 ----- (i)

∠QPS + ∠QRS = 180o (opp. angles of cyclic quad. are supplementary)

⇒ -4\(x\) - 7\(y\) + 150 + (2\(x\) + 8\(y\) + 105) = 180

⇒ -4\(x\) - 7\(y\) + 75 + 2\(x\) + 8\(y\) + 180 = 180

⇒ -2\(x\) + \(y\) + 255 = 180

⇒ -2\(x\) + y = 180 - 255

⇒ -2\(x\) + \(y\) = -75 ------- (ii)

⇒ \(y\) = -75 + 2\(x\) -------- (iii)

Substitute (-75 + 2\(x\)) for \(y\) in equation (i)

⇒ 3\(x\) + 2(-75 + 2\(x\)) = 25

⇒ 3\(x\) - 150 + 4\(x\) = 25

⇒ 7\(x\) = 25 + 150

⇒ 7\(x\) = 175

⇒ \(x = \frac{175}{7} = 25\)

From equation (iii)

⇒ \(y\) = -75 + 2(25) = -75 + 50

⇒ \(y\) = -25

∴ \(x\) + \(y\) = 25 + (-25) = 0

Explanation

(a) (i) \(y = px^{2} - 5x + q\)

From the table,

y = -12 when x = 0, i.e. \(-12 = p(0^{2}) - 5(0) + q\)

\(q = -12\).

y = 0 when x = 4, i.e. \(0 = p(4^{2}) - 5(4) + q\)

\(0 = 16p - 20 + q \implies 20 = 16p + q .... (1) \)

Substitute q = -12 in (1),

\(20 = 16p - 12\)

\(20 + 12 = 16p \implies p = \frac{32}{16} = 2\)

Equation : \(y = 2x^{2} - 5x - 12 \).

(ii) Given the equation \(y = 2x^{2} - 5x - 12\). The table becomes

(b)

(c) From the graph,

(i) When x = 1.8, y = -14.5 ; (ii) When y = -8, x = -0.65 or 3.15 or 3.2.

Correct Answer: A. (3 √ 15+11) / 2

Correct Answer: B. \(60\pi cm\)

Correct Answer: C. 30cm²

Explanation

Area of circle = πr\(^2\) = \(\frac{22}{7}\) * 7\(^2\) = 154cm\(^2\)

Area o Trapezium = \(\frac{(a + b) h}{2}\) = \(\frac{(10 + 21)8}{2}\) = 124cm\(^2\)

Area of the region not covered = Area of (Circle - Trapezium)

= 154 - 124

= 30cm\(^2\)

Correct Answer: C. 7/4

Correct Answer: D. πr/3