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\(\sqrt{3^{\frac{1}{x}}}\) = \(\sqrt{9}\) 3\(^{\frac{1}{2x}}\) = \(9^{\frac{1}{2}}\) 3\(^{\frac{1}{2x}}\) = 3\(^{2 \times \frac{1}{2}}\) 3\(\frac{1}{2x}\) = 3\(\frac{2}{2}\) = 3 \(3^{\frac{1}{2x}}\) = \(3^{1}\) \(\frac{1}{2x}\) = \(\frac{1}{1}\) x = \(\frac{1}{2}\)

x\(^2\) + 4 = 0 x\(^2\) = -4 You can't apply a square root to a negative number. So the answer is None of the above.

surface area of formula = \pir\(^2\) If the radius is reduced then let its radius be x. Its area is \pix\(^2\) . x : r = 2 : 5, so \(\frac{x}{r}\) = \(\frac{2}{5}\) → 5x = 2r Hence, x = 0.4r. Hence the area of the new circle is \pi (0.4r)\(^2\) = 0.16\pi r\(^2\) . The ratio of the two areas is 0.16\pir\(^2\) : \pir\(^2\) = 0.16 : 1 = 16 : 100 = 4 : 25.

\(3\frac{7}{8} + 1\frac{1}{3} = 4\frac{21 + 8}{24}\) = \(4\frac{29}{24}\) \(\equiv 5\frac{5}{24}\) \(1\frac{2}{3} - \frac{1}{8} = \frac{5}{3} - \frac{1}{8}\) = \(\frac{40 - 3}{24}\) = \(\frac{37}{24}\) \(5\frac{5}{24} - \frac{37}{24} = \frac{125}{24} - \frac{37}{24}\) = \(\frac{11}{3}\) = 3\(\frac{2}{3}\)

\((x + 3y + 5)(2x^{2} + 5y + 2)\) = \(2x^{3} + 5xy + 2x + 6yx^{2} + 15y^{2} + 6y + 10x^{2} + 25y + 10\) = \(2x^{3} + 5xy + 2x + 6yx^{2} + 15y^{2} + 31y + 10x^{2} + 10\)

\(\frac{3}{5}\) = 0.60, \(\frac{9}{16}\) = 0.56, \(\frac{34}{59}\) = 0.58, \(\frac{71}{97}\) = 0.73 Hence, \(\frac{9}{16} < \frac{34}{59} < \frac{3}{5} < \frac{71}{97}\)

Sum of n terms of Geometric progression is \(S_{n} = \frac{a(1 - r^n)}{1 - r}\) In the given series, a (the first term) = 1 and r (the common ratio) = x. \(S_{n} = \frac{1(1 - x^{n})}{1 - x}\) a = 1, and as n tends to infinity \(S_{n} = \frac{1}{1 - x}\)

\(\sqrt{170 - 20 \sqrt{30}} = \sqrt{a} - \sqrt{b}\) Squaring both sides, \(170 - 20\sqrt{30} = a + b - 2\sqrt{ab}\) Equating the rational and irrational parts, we have \(a + b = 170 ... (1)\) \(2 \sqrt{ab} = 20 \sqrt{30}\) \(2 \sqrt{ab} = 2 \sqrt{30 \times 100} = 2 \sqrt{3000} \) \(ab = 3000 ... (2)\) From (2), \(b = \frac{3000}{a}\) \(a + \frac{3000}{a} = 170 \implies a^{2} + 3000 = 170a\) \(a^{2} - 170a + 3000 = 0\) \(a^{2} - 20a - 150a + 3000 = 0\) \(a(a - 20) - 150(a - 20) = 0\) \(\text{a = 20 or a = 150}\) \(\therefore b = \frac{3000}{20} = 150 ; b = \frac{3000}{150} = 20\) \(\sqrt{170 - 20\sqrt{30}} = \sqrt{20} - \sqrt{150}\) or \(\sqrt{150} - \sqrt{20}\) = \(2\sqrt{5} - 5\sqrt{6}\) or \(5\sqrt{6} - 2\sqrt{5}\)

\(\begin{array}{c|c} Nos. & log \\ \hline 222.9 & 2.3482\end{array}\) N : b Look for the antilog of .3482 which gives 222.9

Let the present ages of father be x and son = y five years ago, father = x - 5 son = y - 5 x - 5 = 3(y - 5) x - 5 = 3y - 15 x - 3y = -15 + 5 = -10 ......(i) x + y = 110......(ii) eqn(ii) - eqn(i) 4y = 120 y = 30 sub for y = 30 in eqn(i) x -10 = 3(30) x -10 = 90 x = 80

y = 2x2 + 9x - 35 2x2 + 9x = 35 x2 + \(\frac{9}{2}\) = \(\frac{35}{2}\) x2 + \(\frac{9}{2}\) + \(\frac{81}{16}\) = \(\frac{35}{2}\) = \(\frac{81}{16}\) (x + \(\frac{9}{4}\))2 = \(\frac{361}{16}\) x = \(\frac{-9}{4}\) + \(\frac{\sqrt{361}}{16}\) x = \(\frac{-9}{4}\) + \(\frac{19}{4}\) = 2.5 or -7 -7 < x < \(\frac{5}{2}\)

Let the cost of living = y. The new cost of living = \(y + \frac{15y}{100} = 1.15y\) The food bill now = \((1 - \frac{90}{100})(1.15y)\) = \(1.035y\) The fractional increase in food bill = \((1.035 - 1) \times 100% = 3.5%\) = \(\frac{35}{1000} = \frac{7}{200}\)

Soln. a*b = ab + a + b, a ♦ b = a + b + 1 a*c = ac + a + c (a*b) ♦ (a*c) = (ab + a + b + ac + a + c + 1) = ab + ac + 2a + b + c + 1

\(\frac{a - b}{a + b}\) - \(\frac{a + b}{a - b}\) = \(\frac{(a - b)^2}{(a + b)}\) - \(\frac{(a + b)^2}{(a - b0}\) applying the principle of difference of two sqrt. Numerator = (a - b) + (a + b) (a - b) - (a + b) = (a = b + a = b)(a = b - a = b) 2a(-2b) = -4ab = \(\frac{-4ab}{(a + b)(a - b)}\) = \(\frac{-4ab}{a^2 - b^2}\)

\(3x - (\frac{1}{4})^{-\frac{1}{2}} > \frac{1}{4} - x\) = \(3x - 4^{\frac{1}{2}} > \frac{1}{4} - x\) = \(3x - 2 > \frac{1}{4} - x\) = \(3x + x > \frac{1}{4} + 2 \implies 4x > \frac{9}{4}\) \(x > \frac{9}{16}\)