{"id":22091,"date":"2023-10-28T15:24:23","date_gmt":"2023-10-28T09:54:23","guid":{"rendered":"https:\/\/apboardsolutions.in\/?p=22091"},"modified":"2023-10-28T17:26:45","modified_gmt":"2023-10-28T11:56:45","slug":"inter-2nd-year-maths-2a-quadratic-expressions-important-questions","status":"publish","type":"post","link":"https:\/\/apboardsolutions.in\/inter-2nd-year-maths-2a-quadratic-expressions-important-questions\/","title":{"rendered":"Inter 2nd Year Maths 2A Quadratic Expressions Important Questions"},"content":{"rendered":"

Students get through Maths 2A Important Questions<\/a> Inter 2nd Year Maths 2A Quadratic Expressions Important Questions which are most likely to be asked in the exam.<\/p>\n

Intermediate 2nd Year Maths 2A Quadratic Expressions Important Questions<\/h2>\n

Question 1.
\nForm quadratic equation whose root 7 \u00b1 2\\(\\sqrt{5}\\) (Mar. ’11, ’05)
\nSolution:
\n\u03b1 + \u03b2 = 7 + 2\\(\\sqrt{5}\\) + 7 – 2\\(\\sqrt{5}\\) = 14
\n\u03b1\u03b2 = (7 + 2\\(\\sqrt{5}\\)) (7 – 2\\(\\sqrt{5}\\)) = 49 – 20 = 29
\nThe required equation is
\nx2<\/sup> – (\u03b1 + \u03b2)x + \u03b1\u03b2 = 0
\nx2<\/sup> – 14x + 29 = 0<\/p>\n

Question 2.
\nForm quadratic equation whose root -3 \u00b1 5i. (Mar. ’07)
\nSolution:
\n\u03b1 + \u03b2 = -3 + 5i – 3 – 51 = -6
\n\u03b1\u03b2 = (-3 + 5i)(-3 – 5i)
\n= 9 + 25 = 34
\nThe required equation is
\nx2<\/sup> – (\u03b1 + \u03b2)x + \u03b1\u03b2 = 0
\nx2<\/sup> + 6x + 34 = 0<\/p>\n

\"Inter<\/p>\n

Question 3.
\nFor what values of x, 15 + 4x – 3x2<\/sup> expressions are negative? (AP Mar. ’15)
\nSolution:
\nThe roots of 15 + 4x – 3x2<\/sup> = 0 are
\n\\(\\frac{-4 \\pm \\sqrt{16+180}}{-6}\\) i.e., \\(\\frac{-5}{3}\\), 3
\n\u2234 The expression 15 + 4x – 3x2<\/sup> is negative if
\nx < \\(\\frac{-5}{3}\\) or x > 31 \u2235 a = -3 < 0<\/p>\n

Question 4.
\nIf \u03b1, \u03b2 are the roots of the equation ax2<\/sup> + bx + c = 0, find the value \\(\\frac{1}{\\alpha^{2}}\\) + \\(\\frac{1}{\\beta^{2}}\\) expressions in terms of a, b, c. (AP & TS Mar. \u201816, 08)
\nSolution:
\n\"Inter<\/p>\n

Question 5.
\nForm quadratic equation whose root
\n\\(\\frac{p-q}{p+q}\\), \\(\\frac{-p+q}{p-q}\\), (p \u2260 \u00b1q) (Mar. ’06)
\nSolution:
\n\u03b1 + \u03b2 = \\(\\frac{p-q}{p+q}\\) – \\(\\frac{p+q}{p-q}\\)
\n\"Inter<\/p>\n

Question 6.
\nFind the values of m for which the following equations have equal roots?
\ni) x2<\/sup> – 15 – m(2x – 8) = 0. (AP Mar. ’17) (TS Mar. ’15 13)
\nSolution:
\nGiven equation is x2<\/sup> – 15 – m(2x – 8) = 0
\nx2<\/sup> – 2mx + 8m – 15 = 0
\na = 1, b = -2m, c = 8m – 15
\nb2<\/sup> – 4ac = (-2m)2<\/sup> – 4(1) (8m – 15)
\n= 4m2<\/sup> – 32m + 60
\n= 4(m2<\/sup> – 8m + 15)
\n= 4(m – 3)(m – 5)
\nHint: If the equation ax2<\/sup> + bx + c = 0 has equal roots then its discriminant is zero.
\n\u2235 The roots are equal b2<\/sup> – 4ac = 0
\n\u21d2 4(m – 3) (m – 5) = 0
\n\u21d2 m – 3 = 0 or m – 5 = 0
\n\u2234 m = 3 or 5<\/p>\n

\"Inter<\/p>\n

Question 7.
\n(m + 1)x2<\/sup> + 2(m + 3)x + (m + 8) = 0.
\nSolution:
\nGiven equation is
\n(m + 1)x2<\/sup> + 2(m + 3)x + (m + 8) = 0
\na = m + 1, b = 2(m + 3), c = m + 8
\nb2<\/sup> – 4ac = (2(m + 3)]2<\/sup> – 4(m + 1) (m + 8)]
\n= 4(m2<\/sup> + 6m + 9) – 4(m2<\/sup> + 8m + m + 8)
\n= 4m2<\/sup> + 24m + 36 – 4m2<\/sup> – 36 m – 32
\n= -12m + 4
\n= -4(3m – 1)
\n\u2235 The roots are equal \u21d2 b2<\/sup> – 4ac = 0
\n\u21d2 -4(3m – 1) = 0
\n\u21d2 3m – 1 = 0
\n\u21d2 3m = 1
\n\u2234 m = \\(\\frac{1}{3}\\)<\/p>\n

Question 8.
\nIf x is real, prove that \\(\\frac{x}{x^{2}-5 x+9}\\) lies between 1 and \\(\\frac{-1}{11}\\). (Mar. \u201814, 13, \u201808, \u201802; May 11, \u201807)
\nSolution:
\nLet y = \\(\\frac{x}{x^{2}-5 x+9}\\) \u21d2 yx2<\/sup> – 5yx + 9y = x
\n\u21d2 yx2<\/sup> + (-5y – 1)x + 9y = 0
\nx \u2208 R \u21d2 (-5y – 1)2<\/sup> – 4y(9y) \u2265 0
\n\u21d2 25y2<\/sup> + 1 + 10y – 36y2<\/sup> \u2265 0
\n\u21d2 -11y2<\/sup> + 10y + 1 \u2265 0 —— (1)
\n\u21d2 -11y2<\/sup> + 10y + 1 = 0 \u21d2 -11y2<\/sup> + 11y – y + 1 = 0
\n\u21d2 11y(-y + 1) + 1(-y + 1) = 0
\n\u21d2 (-y + 1)(11y + 1) = 0 \u21d2 y = 1, \\(\\frac{-1}{11}\\)
\n-11y2<\/sup> + 10y + 1 \u2265 0
\n\u2234 y2<\/sup> coeff is be, but the exp is \u2265 0 from (1)
\n\u21d2 \\(\\frac{-1}{11}\\) \u2264 y \u2264 1 \u21d2 y lies between 1 and \\(\\frac{-1}{11}\\)<\/p>\n

Question 9.
\nTheorem : The roots of ax2<\/sup> + bx + c = 0 are
\n\"Inter
\n(Mar. ’02)
\nProof:
\nGiven quadratic equation is ax2<\/sup> + bx + c = 0
\n\u21d2 4a(ax2<\/sup> + bx + c) = 0
\n\u21d2 4a2<\/sup>x2<\/sup> + 4abx + 4ac = 0
\n\u21d2 (2ax)2<\/sup> + 2(2ax) (b) + b2<\/sup> – b2<\/sup> + 4ac = 0
\n\u21d2 (2ax + b)2<\/sup> = b2<\/sup> – 4ac
\n\"Inter<\/p>\n

Question 10.
\nFind the maximum value of the function \\(\\frac{x^{2}+14 x+9}{x^{2}+2 x+3}\\) over R.
\nSolution:
\nLet y = \\(\\frac{x^{2}+14 x+9}{x^{2}+2 x+3}\\)
\n\u21d2 yx2<\/sup> + 2yx + 3y = x2<\/sup> + 14x + 9
\n\u21d2 (y – 1)x2<\/sup> + 2(y – 7)x + 3y – 9 = 0
\nSince x \u2208 R, discriminant \u2265 0
\n\u21d2 [2(y – 7)]2<\/sup> – 4(y – 1) (3y – 9) \u2265 0
\n\u21d2 4[(y2<\/sup> – 14y + 49) [(3y2<\/sup> – 12y + 9)] \u2265 0
\n\u21d2 -2y2<\/sup> – 2y + 40 \u2265 0
\n\u21d2 y2<\/sup> + y – 20 \u2264 0
\n\u21d2 (y + 5) (y -4) \u2264 0
\n\u21d2 -5 \u2264 y \u2264 4
\n\u21d2 y \u2208 [-5, 4]
\n\u21d2 Maximum value of y = 4
\n\u2234 Maximum value of the function
\n\\(\\frac{x^{2}+14 x+9}{x^{2}+2 x+3}\\) over R is 4.<\/p>\n

\"Inter<\/p>\n

Question 11.
\nIf x2<\/sup> – 6x + 5 = 0 and x2<\/sup> – 12x + p = 0 have a common root, then find p. (TS Mar. ’17)
\nSolution:
\nGiven x2<\/sup> – 6x + 5 = 0, x2<\/sup> – 12x + p = 0 have a common root.
\nIf \u03b1 is the common root then
\n\u03b12<\/sup> -6\u03b1 + 5 = 0, \u03b12<\/sup> – 12\u03b1 + p = 0
\n\u03b12<\/sup> – 6\u03b1 + 5 = 0 \u21d2 (\u03b1 – 1) (\u03b1 – 5) = 0
\n\u21d2 \u03b1 = 1 or 5
\nIf \u03b1 = 1 then \u03b12<\/sup> – 12\u03b1 + p = 0
\n\u21d2 1 – 12 + p = 0 \u21d2 p = 11
\nIf \u03b1 = 5 then \u03b12<\/sup> – 12\u03b1 + p = 0
\n\u21d2 25 – 60 + p = 0 \u21d2 p = 35
\n\u2234 p = 11 or 35<\/p>\n

Question 12.
\nIf x1<\/sub>, x2<\/sub> are the roots of the quadratic equation ax2<\/sub> + bx + c = 0 and c \u2260 0, find the value of (ax1<\/sub> + b)-2<\/sup> + (ax2<\/sub> + b)-2<\/sub> in terms of a, b, c. (TS Mar. ’17)
\nSolution:
\nx1<\/sub>, x2<\/sub> are the roots of the equation
\nax2<\/sup> + bx + c = 0
\n\"Inter
\n\"Inter<\/p>\n

Question 13.
\nProve that \\(\\frac{1}{3 x+1}\\) + \\(\\frac{1}{x+1}\\) – \\(\\frac{1}{(3 x+1)(x+1)}\\) does not lie between 1 and 4, if x is real. (AP & TS Mar. ’16, AP Mar. 15, ’11) (AP Mar. \u201817)
\nSolution:
\n\"Inter
\n\u21d2 3yx2<\/sup> + 4yx + y = 4x + 1
\n\u21d2 3yx2<\/sup> + (4y – 4) x + (y – 1) = 0
\nx \u2208 R \u21d2 (4y – 4)2<\/sup> – 4(3y)(y – 1) \u2265 0
\n\u21d2 16y2<\/sup> + 16 – 32y – 12y2<\/sup> + 12y \u2265 0
\n\u21d2 4y2<\/sup> – 20y + 16 \u2265 0
\n4y2<\/sup> – 20y + 16 = 0
\n\u21d2 y2<\/sup> – 5y + 4 = 0
\n\u21d2 (y – 1)(y – 4) = 0
\n\u21d2 y = 1, 4
\n\u21d2 4y2<\/sup> – 20y + 16 \u2265 0.
\n\u21d2 y \u2264 1 or y \u2265 4
\n\u21d2 y does not lie between 1 and 4
\nSince y2<\/sup> coeff is the and exp \u2265 0.<\/p>\n

Question 14.
\nSolve the following equations: (T.S Mar. \u201815)
\n2x4<\/sup> + x3<\/sup> – 11x2<\/sup> + x + 2 = 0
\nSolution:
\nDividing by x2<\/sup>
\n\"Inter
\nSubstituting in (1)
\n2(a2<\/sup> – 2) + a – 11 = 0
\n\u21d2 2a2<\/sup> – 4 + a – 11 = 0
\n\u21d2 2a2<\/sup> + a – 15 = 0
\n\u21d2 (a + 3) (2a – 5) = 0
\n\u21d2 a + 3 = 0 or 2a – 5 = 0
\na = -3 or 2a = 5 a = \\(\\frac{5}{2}\\)<\/p>\n

Case(i) : a = -3
\nx + \\(\\frac{1}{x}\\) = -3
\nx2<\/sup> + 1 = -3
\nx2<\/sup> + 3x + 1 = 0
\nx = \\(\\frac{-3 \\pm \\sqrt{9-4}}{2}\\) = \\(\\frac{-3 \\pm \\sqrt{5}}{2}\\)<\/p>\n

Case (ii) : a = \\(\\frac{5}{2}\\)
\nx + \\(\\frac{1}{x}\\) = \\(\\frac{5}{2}\\)
\n\u21d2 \\(\\frac{x^{2}+1}{x}\\) = \\(\\frac{5}{2}\\)
\n\u21d2 2x2<\/sup> + 2 = 5x
\n\u21d2 2x2<\/sup> – 5x + 2 = 0
\n\u21d2 (2x – 1) (x – 2) = 0
\n\u21d2 2x – 1 = 0 or x – 2 = 0
\n\u21d2 x = \\(\\frac{1}{2}\\), 2
\n\u2234 The roots are \\(\\frac{1}{2}\\), 2, \\(\\frac{-3 \\pm \\sqrt{5}}{2}\\)<\/p>\n

\"Inter<\/p>\n

Question 15.
\nFor what values of x, the following expressions are positive? (May ’11)
\ni) x2<\/sup> – 5x + 6
\nSolution:
\nx2<\/sup> – 5x + 6 = (x – 2) (x – 3)
\nRoots of x2<\/sup> – 5x + 6 = 0 are 2, 3 which are real.
\nThe expression x2<\/sup> – 5x + 6 is positive if x < 2 or x > 3, \u2234 a = 1 > 0.<\/p>\n

ii) 3x2<\/sup> + 4x + 4
\nSolution:
\nHere a = 3, b = 4, c = 4,
\n\u0394 = b2<\/sup> – 4ac
\n= 16 – 48 = -32 < 0
\n\u2234 3x2<\/sup> + 4x + 4 is positive \u2200 x \u2208 R,
\n\u2235 a = 3 >0 and \u0394 < 0
\nHint: ax2<\/sup> + bx + c and ‘a\u2019 have same sign \u2200 x \u2208 R, if \u0394 < 0<\/p>\n

iii) 4x – 5x2<\/sup> + 2
\nSolution:
\nRoots of 4x – 5x2<\/sup> + 2 = 0 are
\n\"Inter<\/p>\n

iv) x2<\/sup> – 5x + 14
\nSolution:
\nHere a = 1, b = -5, c = 14,
\n\u0394 = b2<\/sup> – 4ac
\n= 25 – 56 = -31 < 0
\n\u2234 \u0394 < 0 \u2235 a = 1 > 0 and \u0394 < 0
\n\u21d2 x2<\/sup> – 5x + 14 is positive \u2200 x \u2208 R.<\/p>\n

\"Inter<\/p>\n

Question 16.
\nDetermine the range of the \\(\\frac{x^{2}+x+1}{x^{2}-x+1}\\) expressions. (Mar ’04)
\nSolution:
\nLet y = \\(\\frac{x^{2}+x+1}{x^{2}-x+1}\\)
\n\u21d2 x2<\/sup>y – xy + y = x2<\/sup> + x + 1
\n\u21d2 x2<\/sup>y – xy + y – x2<\/sup> – x – 1 = 0
\n\u21d2 x2<\/sup>(y – 1) – x(y + 1) + (y – 1) = 0
\nx is real \u21d2 b2<\/sup> – 4ac \u2265 0
\n\u21d2 (y + 1)2<\/sup> – 4(y – 1)2<\/sup> \u2265 0
\n\u21d2 (y + 1)2<\/sup> – (2y – 2)2<\/sup> \u2265 0
\n\u21d2 (y + 1 + 2y – 2)(y + 1 – 2y + 2) \u2265 0
\n\u21d2 (3y – 1)(-y + 3) \u2265 0
\n\u21d2 -(3y – 1) (y – 3) \u2265 0
\na = coeff. of y2<\/sup> = -3 < 0., But
\nThe expression \u2265 0
\n\u21d2 y lies between \\(\\frac{1}{3}\\) and 3
\n\u2234 The range of \\(\\frac{x^{2}+x+1}{x^{2}-x+1}\\) is \\(\\left[\\frac{1}{3}, 3\\right]\\)<\/p>\n

Question 17.
\nTheorem : Let \u03b1, \u03b2 be the real roots of ax2<\/sup> + bx + c = 0 and \u03b1 < \u03b2. Then
\ni) x \u2208 R, \u03b1 < x < \u03b2 \u21d2 ax2<\/sup> + bx + c and \u2019a\u2019 have opposite signs.
\nii) x \u2208 R, x < \u03b1 or x > \u03b2 \u21d2 ax2<\/sup> + bx + c and \u2018a\u2019 have the same sign. (Apr. ’96, ’93)
\nProof:
\n\u03b1, \u03b2 are the roots of ax2<\/sup> + bx + c = 0
\n\u21d2 ax2<\/sup> + bx + c = a(x – \u03b1) (x – \u03b2)
\n\u21d2 \\(\\frac{a x^{2}+b x+c}{a}\\) = (x – \u03b1) (x – \u03b2)<\/p>\n

i) Suppose x \u2208 R, \u03b1 < x < \u03b2
\n\u03b1 < x < \u03b2 \u21d2 x – \u03b1 > 0, x – \u03b2 < 0
\n\u21d2 (x – \u03b1) (x – \u03b2) < 0
\n\u21d2 ax2<\/sup> + bx + c, a have opposite signs.<\/p>\n

ii) Suppose x \u2208 R, x < \u03b1
\nx < \u03b1 < 13
\n\u21d2 x – \u03b1 < 0, x – \u03b2 < 0 \u21d2 (x – \u03b1) (x – \u03b2) > 0
\n\u21d2 \\(\\frac{a x^{2}+b x+c}{a}\\) > 0
\n\u21d2 ax2<\/sup> + bx + c and a have the same sign.
\nSuppose x \u2208 R, x > \u03b2
\nx > \u03b2 > \u03b1
\n\u21d2 x – \u03b1 > 0, x – \u03b2 > 0
\n\u21d2 (x – \u03b1)(x – \u03b2) > 0
\n\u21d2 \\(\\frac{a x^{2}+b x+c}{a}\\) > 0
\n\u21d2 ax2<\/sup> + bx + c and a have the same sign,
\n\u2234 x \u2208 R, x < \u03b1 or x > \u03b2
\n\u21d2 ax2<\/sup> + bx + c and a have the same sign.<\/p>\n

Question 18.
\nTheorem: Let f(x) = ax2<\/sup> + bx + c be a quadratic function. (Apr. \u201801)
\ni) If a > 0 then f(x) has minimum value at x = \\(\\frac{-b}{2 a}\\) and the minimum value = \\(\\frac{4 a c-b^{2}}{4 a}\\)
\nii) If a < 0 then f(x) has maximum value at x = \\(\\frac{-b}{2 a}\\) and the maximum value \\(\\frac{4 a c-b^{2}}{4 a}\\)
\nProof:
\nax2<\/sup> + bx + c =
\n\"Inter
\n\u2264 \\(\\frac{4 a c-b^{2}}{4 a}\\),
\nwhen a < 0
\n\u2234 If a < 0, then \\(\\frac{4 a c-b^{2}}{4 a}\\) is the maximum f\u00f3r f when x = \\(\\frac{-b}{2 a}\\)
\nSecond Proof : f(x) = ax2<\/sup> + bx + c
\n\u21d2 f'(x) = 2ax + b
\n\u21d2 f”(x) = 2a
\nIf f'(x) = 0, then 2ax + b = 0 and hence x = \\(-\\frac{b}{2 a}\\)
\nIf a > 0, then f”(x) > 0 and hence \u2018f’ has minimum value at x = \\(-\\frac{b}{2 a}\\).
\nMinimum of
\n\"Inter
\nIf a < 0,
\nthen f”(x) < 0 and hence \u2018f’ has maximum value at x = \\(-\\frac{b}{2 a}\\) maximum of
\n\"Inter<\/p>\n

Question 19.
\nTheorem : The roots of ax2<\/sup> + bx + c = 0 are \\(\\frac{-b \\pm \\sqrt{b^{2}-4 a c}}{2 a}\\).
\nProof:
\nGiven quadratic equation is ax2<\/sup> + bx + c = 0
\n\u21d2 4a (ax2<\/sup> + bx + c) = 0
\n\u21d2 4a2<\/sup>x2<\/sup> + 4abx + 4ac = 0
\n\u21d2 (2ax)2<\/sup> + 2(2ax) (b) + b2<\/sup> – b2<\/sup> + 4ac = 0
\n\u21d2 (2ax + b)2<\/sup> = b2<\/sup> – 4ac
\n\u21d2 2ax + b = \u00b1\\(\\sqrt{b^{2}-4 a c}\\)
\n\u21d2 2ax = -b \u00b1 \\(\\sqrt{b^{2}-4 a c}\\)
\n\u21d2 x = \\(\\frac{-b \\pm \\sqrt{b^{2}-4 a c}}{2 a}\\)
\nThe roots of ax2<\/sup> + bx + c = 0 are
\n\\(\\frac{-b \\pm \\sqrt{b^{2}-4 a c}}{2 a}\\)<\/p>\n

\"Inter<\/p>\n

Question 20.
\nTheorem : Let a, b, c \u2208 R and a \u2260 0. Then the roots of ax2<\/sup> + bx + c = 0 are non-real complex numbers if and only if ax2<\/sup> + bx + c and a have the same sign for all x \u2208 R.
\nProof :
\nThe condition for the equation
\nax2<\/sup> + bx + c = o to have non-real complex roots is b2<\/sup> – 4ac < 0, i.e., 4ac – b2<\/sup> > 0.
\n\"Inter
\nHence 4ac – b2<\/sup> > 0, so that b2<\/sup> – 4ac < 0. Thus b2<\/sup> – 4ac < 0 if and only if ax2<\/sup> + bx + c and a have the same sign for all real x.<\/p>\n

Question 21.
\nTheorem: If the roots of ax2<\/sup> + bx + c = 0 are real and equal to \u03b1 = \\(\\frac{-b}{2 a}\\); then for \u03b1 \u2260 x \u2208 R, ax2<\/sup> + bx + c and \u2018a\u2019 have the same sign.
\nProof:
\nThe roots are equal \u21d2 b2<\/sup> – 4ac = 0
\n\u21d2 4ac – b2<\/sup> = 0
\n\"Inter
\n\u2234 For \u03b1 \u2260 x \u2208 R, ax2<\/sup> + bx + c and a have the same sign.<\/p>\n

Question 22.
\nTheorem : Let \u03b1, \u03b2 be the real roots of ax2<\/sup> + bx + c = 0 and \u03b1 < \u03b2. Then
\ni) x \u2208 R, \u03b1 < x < 13 \u21d2 ax2<\/sup> + bx + c and \u2019a’ have opposite signs.
\nii) x \u2208 R, x < \u03b1 or x > \u03b2 \u21d2 ax2<\/sup> + bx + c and \u2018a\u2019 have the same sign.
\nProof.
\n\u03b1, \u03b2 are the roots of ax2<\/sup> + bx + c = 0
\n\u21d2 ax2<\/sup> + bx + c = a(x – \u03b1)(x – \u03b2)
\n\u21d2 \\(\\frac{a x^{2}+b x+c}{a}\\) = (x – \u03b1)(x – \u03b2)<\/p>\n

i) Suppose x \u2208 R, \u03b1 < x < \u03b2
\n\u03b1 0, x – \u03b2 < 0
\n\u21d2 (x – \u03b1)(x – \u03b2) < 0 \u21d2 \\(\\frac{a x^{2}+b x+c}{a}\\) < 0
\n\u21d2 ax2<\/sup> + bx + c, a have opposite signs<\/p>\n

ii) Suppose x \u2208 R, x < \u03b1
\nx < \u03b1 < \u03b2 \u21d2 x – \u03b1 < 0, x – \u03b2 < 0 \u21d2 (x – \u03b1) (x – \u03b2) > 0 \u21d2 \\(\\frac{a x^{2}+b x+c}{a}\\) > 0
\n\u21d2 ax2<\/sup> + bx + c and a have the same sign.
\nSuppose x \u2208 R, x > 3
\nx > \u03b2 > \u03b1 \u21d2 x – \u03b1 > 0, x – \u03b2 > 0
\n\u21d2 (x – \u03b1) (x – \u03b2) > 0
\n\u21d2 \\(\\frac{a x^{2}+b x+c}{a}\\) > 0
\n\u21d2 ax2<\/sup> + bx + c and a have the same sign,
\nx \u2208 R, x < \u03b1 or x > \u03b2
\n\u21d2 ax2<\/sup> + bx + c and a have the same sign<\/p>\n

Question 23.
\nTheorem : Let f(x) = ax2<\/sup> + bx+ c be a quadratic function.
\ni) If a > 0 then f(x) has minimum value at x = \\(\\frac{-\\mathbf{b}}{2 a}\\) and the minimum value = \\(\\frac{4 a c-b^{2}}{4 a}\\)
\nii) If a < 0 then f(x) has maximum value at x = \\(\\frac{-\\mathbf{b}}{2 a}\\) and the maximum value = \\(\\frac{4 a c-b^{2}}{4 a}\\) (Apr. ’01)
\nProof.
\nax2<\/sup> + bx+ c =
\n\"Inter
\nSecond Proof: f(x) = ax2<\/sup> + bx + c
\n\u21d2 f'(x) = 2ax + b \u21d2 f”(x) = 2a
\nIf f'(x) = 0, then 2ax + b = 0 and hence x = \\(-\\frac{b}{2 a}\\)
\nIf a > 0, then f”(x) > 0 and hence \u2018f\u2019 has minimum value at x = \\(-\\frac{b}{2 a}\\)
\nMinimum of
\n\"Inter
\nIf a < 0 then f”(x) < 0 and hence ‘f’ has maximum value at x = –\\(\\frac{b}{2 a}\\)
\nmaximum of
\n\"Inter<\/p>\n

Question 24.
\nTheorem : A necessary and sufficient condition for the quadratic equations
\na1<\/sub>x2<\/sup> + b1<\/sub>x + c1<\/sub> = 0 and a2<\/sub>x2<\/sup> + b2<\/sub>x + c2<\/sub> = 0 to have a common root is (c1<\/sub>a2<\/sub> – c2<\/sub>a1<\/sub>)2<\/sup> = (a1<\/sub>b2<\/sub> – a2<\/sub>b1<\/sub>) (b1<\/sub>c2<\/sub> – b2<\/sub>c1<\/sub>).
\nProof:
\nNecessity
\nLet \u03b1 be a common root of the given equations.
\nThen a1<\/sub>\u03b12<\/sup> + b1<\/sub>\u03b1 + c1<\/sub> = 0 ——(1)
\na2<\/sub>\u03b12<\/sup> + b2<\/sub>\u03b1 + c2<\/sub> = 0 —— (2)
\nOn multiplying euqation (1) by a2<\/sub>, equation (2) by a1<\/sub> and then subtracting the latter from the former, we get
\n\"Inter
\nOn multiplying euqation (1) by b2<\/sub>, equation (2) by b1<\/sub> and then subtracting the latter from the former, we get
\n\u03b12<\/sup> (a1<\/sub>b2<\/sub> – a2<\/sub>b1<\/sub>) = b1<\/sub>c2<\/sub> – b2<\/sub>c1<\/sub> —— (4)
\nOn squaring both sides of equation (3) and using (4) we obtain
\n\"Inter
\n\"Inter
\nTherefore the given equations have the same roots.<\/p>\n

Case (ii): a1<\/sub>b2<\/sub> – a2<\/sub>b1<\/sub> \u2260 0
\n\"Inter
\nSimilarly we can prove that a2<\/sub>\u03b12<\/sup> + b2<\/sub>\u03b1 + c2<\/sub> = 0
\nThus \u03b1 is a common root of the given equations.<\/p>\n

\"Inter<\/p>\n

Question 25.
\nFind the roots of the equation 3x2<\/sup> + 2x – 5 = 0.
\nSolution:
\nThe roots of the quadratic equation
\nax2<\/sup> + bx + c = 0 are \\(\\frac{-b \\pm \\sqrt{b^{2}-4 a c}}{2 a}\\)
\nHere a = 3, b = 2 and c = -5.
\nTherefore the roots of the given equation are
\n\"Inter
\nHence 1 and –[\/latex]\\frac{5}{3}[\/latex] are the roots of the given equation.
\nAnother method
\nWe can also obtain these roots in th\u00e8 following way.
\n3x2<\/sup> + 2x – 5 = 3x2<\/sup> + 5x – 3x – 5
\n= x(3x + 5) -1 (3x + 5)
\n= (x – 1) (3x + 5)
\n= 3(x – 1) \\(\\left(x+\\frac{5}{3}\\right)\\)
\nSince 1 and –[\/latex]\\frac{5}{3}[\/latex] are the zeros of 3x2<\/sup> + 2x – 5, they are the roots of 3x2<\/sup> + 2x – 5 = 0.<\/p>\n

Question 26.
\nFind the roots of the equation 4x2<\/sup> – 4x + 17 = 3x2<\/sup> – 10x – 17.
\nSolution:
\nGiven equation can be rewritten as x2<\/sup> + 6x + 34 = 0. The roots of the quadratic equation
\nax2<\/sup> + bx + c = 0 are \\(\\frac{-b \\pm \\sqrt{b^{2}-4 a c}}{2 a}\\)
\nHere a = 1, b = 6 and c = 34
\nTherefore the roots of the given equation are
\n\"Inter
\nHence the roots of the given equation are -3 + 5i and -3 – 5i<\/p>\n

Question 27.
\nFind the roots of the equation
\n\\(\\sqrt{3}\\)x2<\/sup> + 10x – 8\\(\\sqrt{3}\\) = 0.
\nSolution:
\nHere a = \\(\\sqrt{3}\\), b = 10, c = -8\\(\\sqrt{3}\\)
\n\"Inter<\/p>\n

Question 28.
\nFind the nature of the roots of 4x2<\/sup> – 20x + 25 = 0
\nSolution:
\nHere a =4, b = -20, c = 25
\nHence \u0394 = b2<\/sup> – 4ac
\n= (-20)2<\/sup> – 4(4) (25) = 0
\nSince \u0394 is zero and a, b, c are real, the roots of the given equation are real and equal.<\/p>\n

Question 29.
\nFind the nature of the roots of 3x2<\/sup> + 7x + 2 = 0
\nSolution:
\nHere a = 3, b = 7, c = 2
\nHence \u0394 = b2<\/sup> – 4ac
\n= 49 – 4(3) (2)
\n= 49 – 24
\n= 25 = (5)2<\/sup> > 0
\nSince a, b, c are rational numbers and \u0394 = 52<\/sup> is a perfect square and positive, the roots of the given equation are rational and unequal.<\/p>\n

Question 30.
\nFor what values of m, the equation x2<\/sup> – 2(1 + 3m)x + 7(3 + 2m) = 0 will have equal roots?
\nSolution:
\nThe given equation will have equal roots iff its discriminant is 0.
\nHere \u0394 = [(-2(1 + 3m)]2<\/sup> – 4(1) [7 + (3 + 2m)]
\n= 4(1 + 9m2<\/sup> + 6m) – 28(3 + 28m)
\n= 9m2<\/sup> – 8m – 20
\n= (m – 2)(9m + 10)
\nHence \u0394 = 0 \u21d4 m = 2, \\(\\frac{-10}{9} .\\)<\/p>\n

\"Inter<\/p>\n

Question 31.
\nIf \u03b1 and \u03b2 are the roots of ax2<\/sup> + bx + c = 0, find the value of \u03b12<\/sup> + \u03b22<\/sup> and \u03b13<\/sup> + \u03b23<\/sup> in terms of a, b, c.
\nSolution:
\n\u03b1, \u03b2 are the roots of ax2<\/sup> + bx + c = 0
\n\"Inter<\/p>\n

Question 32.
\nShow that the roots of the equation x2<\/sup> – 2px + p2<\/sup> – q2<\/sup> + 2qr – r2<\/sup> = 0 are rational, given that p, q, r are rational.
\nSolution:
\nHere a = 1, b = -2p, c = p2<\/sup> – q2<\/sup> + 2qr – r2<\/sup>
\n\u2206 = b2<\/sup> – 4ac
\n= (-2p)2<\/sup> – 4(1) (p2<\/sup> – q2<\/sup> + 2qr – r2<\/sup>)
\n= 4p2<\/sup> – 4p2<\/sup> + 4q2<\/sup> + 8qr + 4r2<\/sup>
\n= 4(q – r)2<\/sup>.
\n\u2235 p, q, r are rational, the coefficient of the given equation are rational and is a square of a rational number 2(q – r).
\n\u2234 The roots of the given equation are rational.<\/p>\n

Question 33.
\nForm a quadratic equation whose roots are 2\\(\\sqrt{3}\\) – 5 and -2\\(\\sqrt{3}\\) – 5.
\nSolution:
\nLet \u03b1 = 2\\(\\sqrt{3}\\) – 5 and \u03b2 = -2\\(\\sqrt{3}\\) – 5
\nThen \u03b1 + \u03b2 = (2\\(\\sqrt{3}\\) – 5) + (-2\\(\\sqrt{3}\\) – 5) = -10
\nand \u03b1\u03b2 = (2\\(\\sqrt{3}\\) – 5) + (-2\\(\\sqrt{5}\\) – 5)
\n= (-5)2<\/sup> – (-2\\(\\sqrt{3}\\))2<\/sup>
\n= 25 – 4 \u00d7 3
\n= 13
\nThe required quadratic equation is
\nx2<\/sup> – (\u03b1 + \u03b2)x + \u03b1\u03b2 = 0
\n\u21d2 x2<\/sup> – (-10)x + 13 = 0
\n\u21d2 x2<\/sup> + 10x + 13 = 0<\/p>\n

Question 34.
\nFind the quadratic equation, the sum of whose roots is 1 and the sum of squares of the roots is 13.
\nSolution:
\nLet \u03b1, \u03b2 be the roots of a required quadratic equation.
\nThen \u03b1 + \u03b2 = 1 and \u03b12<\/sup> + \u03b22<\/sup> = 13
\nNow \u03b1\u03b2 = \\(\\frac{1}{2}\\)[(\u03b1 + \u03b2)2<\/sup> – (\u03b12<\/sup> + \u03b22<\/sup>)]
\n= \\(\\frac{1}{2}\\)[(1)2<\/sup> – (13)]
\nTherefore x2<\/sup> – (\u03b1 + \u03b2)x + \u03b1\u03b2 = 0
\n\u21d2 x2<\/sup> – (1)x + (-6) = 0
\n\u21d2 x2<\/sup> – x – 6 = 0<\/p>\n

Question 35.
\nLet \u03b1 and \u03b2 be the roots of the quadratic equation ax2<\/sup> + bx + c = 0, c \u2260 0, then form the quadratic equation whose roots are \\(\\frac{1-\\alpha}{\\alpha}\\) and \\(\\frac{1-\\beta}{\\beta}\\).
\nSolution:
\nFrom the given equation
\n\"Inter
\n\"Inter<\/p>\n

Question 36.
\nSolve x2\/3<\/sup> + x1\/3<\/sup> – 2 = 0
\nSolution:
\n(x1\/3<\/sup>)2<\/sup> + (x1\/3<\/sup>) – 2 = 0
\nLet x1\/3<\/sup> = a, a2<\/sup> + a – 2 = 0
\n\u21d2 (a +2)(a – 1) = 0 \u21d2 a = 1, -2
\nNow, x1\/3<\/sup> = 1 \u21d2 x = (1)3<\/sup> = 1
\nx1\/3<\/sup> = -2 \u21d2 x = (-2)3<\/sup> = -8
\n\u2234 roots are 1, -8.<\/p>\n

Question 37.
\nSolve 71 + x<\/sup> + 71 – x<\/sup> = 50 for real x.
\nSolution:
\nThe given equation can be written as,
\n7.7x<\/sup> + \\(\\frac{7}{7^{x}}\\) – 50 = 0
\nLet 7x<\/sup> = a
\n7a + \\(\\frac{7}{a}\\) – 50 = 0
\n\u21d2 7a2<\/sup> + 7 – 50a = 0
\n\u21d2 7a2<\/sup> – 49a – a + 7 = 0
\n\u21d2 7a(a – 7) – 1(a – 7) = 0
\n\u21d2 (a – 7)(7a – 1) = 0
\n\u2234 a = 7, \\(\\frac{1}{7}\\)
\nNow, if a = 7 then 7x<\/sup> = 71<\/sup> \u21d2 x = 1
\na = \\(\\frac{1}{7}\\) then 7x<\/sup> = \\(\\frac{1}{7}\\) = 7-1<\/sup>
\nx = -1
\n\u2234 x = -1, 1<\/p>\n

Question 38.
\nSolve
\n\"Inter
\nSolution:
\nOn taking \\(\\sqrt{\\frac{x}{1-x}}\\) = a
\na + \\(\\frac{1}{a}\\) = \\(\\frac{13}{6}\\)
\n\u21d2 \\(\\frac{a^{2}+1}{a}\\) = \\(\\frac{13}{6}\\)
\n\u21d2 \\(\\frac{a^{2}+1}{a}\\) = \\(\\frac{13}{6}\\)
\n\u21d2 6a2<\/sup> + 6 = 13a
\n\u21d2 6a2<\/sup> – 13a + 6 = 0
\n\u21d2 6a2<\/sup> – 9a – 4a + 6 = 0
\n\u21d2 3a(2a – 3) – 2(2a – 3) = 0
\n\u21d2 (2a – 3)(3a – 2) = 0
\na = \\(\\frac{3}{2}\\), a = \\(\\frac{2}{3}\\)
\nIf a = \\(\\frac{3}{2}\\) then \\(\\sqrt{\\frac{x}{1-x}}\\) = \\(\\frac{3}{2}\\)
\n\u21d2 \\(\\frac{x}{1-x}\\) = \\(\\frac{9}{4}\\)
\n\u21d2 4x = 9 – 9x
\n\u21d2 13x = 9
\n\u21d2 x = \\(\\frac{4}{13}\\)
\n\u2234 x = \\(\\frac{9}{13}\\), \\(\\frac{4}{13}\\)<\/p>\n

\"Inter<\/p>\n

Question 39.
\nFind all number which exceed their square root by 12.
\nSolution:
\nLet the required number be \u2018x\u2019
\ngiven, x = \\(\\sqrt{x}\\) + 12
\n\u21d2 x – 12= \\(\\sqrt{x}\\)
\nSquaring on both sides
\n(x – 12)2<\/sup> = (\\(\\sqrt{x}\\))2<\/sup>
\n\u21d2 x2<\/sup> – 24x + 144 = x
\n\u21d2 x2<\/sup> – 25x + 144 = 0
\n\u21d2 (x – 9)(x – 16) = 0
\n\u21d2 x = 9, 16
\nBut x = 9 does not satisfy the given condition
\nx = 16
\n\u2234 The required numb\u00e9r = 16.<\/p>\n

Question 40.
\nIf x2<\/sup> + 4ax + 3 = 0 and 2x2<\/sup> + 3ax – 9 = 0 have a common root, then find the values
\nof a and the common root.
\nSolution:
\nThe quadratic equations
\n\"Inter
\nSubstitute in the above equation
\n[(3) (2) – (-9) (1)]2<\/sup> = [(1) (3a) – (2) (4a)] [(4a) (-9) – (3a) (3)]
\n(15)2<\/sup> = (-5a) (-45a)
\n\u21d2 225 = 225a2<\/sup> \u21d2 a2<\/sup> = 1 \u21d2 a = \u00b11<\/p>\n

Case (i) : If a = 1, the given equations become x2<\/sup> + 4x + 3 = 0 and 2x2<\/sup> + 3x – 9 = 0
\n\u21d2 (x + 1)(x + 3) = 0 and (2x + 3)(x + 3) = 0
\n\u21d2 x = 3, -1 and -3, \\(\\frac{3}{2}\\)
\nIn this case, the common root of the given equations is -3<\/p>\n

Case (ii) : If a = -1, the given equations become x2<\/sup> – 4x + 3 = 0 and 2x2<\/sup> – 3x – 9 = 0
\n\u21d2 (x – 1) (x – 3) = 0 and (2x + 3) (x – 3) = 0
\n\u21d2 x = 1, 3 and x = 3, –\\(\\frac{3}{2}\\)
\nIn this case, the common root of the given equation is 3.<\/p>\n

Question 41.
\nProve that there is unique pair of consecutive positive odd integers such that the sum of their squares is. 290 and find it.
\nSolution:
\nThe difference of two consecutive odd integers is 2.
\nSo, we have to prove that there is a unique positive odd integer \u2018x’ such that
\nx2<\/sup> + (x + 2)2<\/sup> = 290 —– (1)
\nx2<\/sup> + (x + 2)2<\/sup> = 290
\n\u21d2 x2<\/sup> + x2<\/sup> + 4x + 4 = 290
\n\u21d2 2x2<\/sup> + 4x – 286 = 0
\n\u21d2 x2<\/sup> + 2x – 143 = 0
\n\u21d2 (x + 13) (x – 11) = 0
\n\u21d2 x = -13, 11
\nHence 11 is the only positive odd integer satisfying equation (1).
\n\u2234 11, 13 is the unique pair of integers which satisfies the given condition.<\/p>\n

Question 42.
\nThe cost of a piece of cable wire is Rs. 35\/-, If the length of the piece of wire is 4 meters more and each meter costs, Rs. 1\/\u2014 less, the cost would remain unchanged. What is the length of the wire?
\nSolution:
\nLet the length of the piece of the wire be \u2018l\u2019 meters and the cost of each meter be Rs. x.
\ngiven lx. = 35 —— (1)
\nand (l + 4) (x – 1) = 35
\n\u21d2 lx – l + 4x – 4 = 35
\n\u21d2 35 – l + 4x – 4 = 35
\n\u21d2 4x = l + 4
\n\u21d2 x = \\(\\frac{l+4}{4}\\)
\nSubstitute x in (1), we get l[latex]\\frac{l+4}{4}[\/latex] = 35
\n\u21d2 l2<\/sup> + 4l = 140
\n\u21d2 l2<\/sup> + 4l – 140 = 0
\n\u21d2 l2<\/sup> + 14l – 10l – 140 = 0
\n\u21d2 l(l + 14) – 10(l + 14) = 0
\n\u21d2 (l – 10)(l + 14) = 0
\n\u21d2 l = -14, 10
\nSince length cannot be negative, l = 10
\n\u2234 The length of the piece of wire is 10 meters.<\/p>\n

Question 43.
\nOne fourth of a herd of goats was seen in the forest. Twice the square root of the number in the herd had gone up the hill and the remaining 15 goats were on the bank of the river. Find the total number of goats.
\nSolution:
\nLet the number of goats in the herd be \u2018x\u2019.
\nThe number of goats seen in the forest = \\(\\frac{x}{4}\\)
\nThe number of goats gone upto the hill = \\(2 \\sqrt{x}\\)
\nThe number of goats on the bank of a river = 15
\n\u2234 \\(\\frac{x}{4}\\) + 2\\(\\sqrt{x}\\) + 15 = x
\n\u21d2 x + 8\\(\\sqrt{x}\\) + 60 = 4x
\n\u21d2 3x – 8\\(\\sqrt{x}\\) – 60 = 0
\nPut \\(\\sqrt{x}\\) = y
\n\u21d2 3y2<\/sup> – 8y – 60 = 0
\n\u21d2 3y2<\/sup> – 18y + 10y – 60 = 0
\n\u21d2 3y(y – 6) + 10(y – 6) = 0
\n\u21d2 (3y + 10)(y – 6) = 0
\n\u21d2 y = 6, –\\(\\frac{10}{3}\\)
\nSince y cannot be negative, y = 6
\n\u21d2 \\(\\sqrt{x}\\) = 36
\n\u2234 x = 36
\n\u2234 Total number of goats = 36<\/p>\n

Question 44.
\nIn a cricket match Anil took one wicket less than twice the number of wickets taken by Ravi. If the product of the
\nnumber of wickets taken by them is 15, find the number of wickets taken by each of them.
\nSolution:
\nLet the number of wickets taken by Ravi be \u2018x’ and the number of wickets taken by Anil is 2x – 1.
\nGiven x(2x – 1) = 15
\n\u21d2 2x2<\/sup> – x – 15 = 0
\n\u21d2 2x2<\/sup> – 6x + 5x – 15 = 0
\n\u21d2 2x(x – 3) + 5(x – 15) = 0
\n\u21d2 (x – 3)(2x + 5) = 0
\n\u21d2 x = 3, –\\(\\frac{5}{2}\\)
\nSince the number of wickets be integer,
\nx = 3 and 2x – 1 = 2(3) – 1 = 5
\n\u2234 The number of wickets taken by Anil and Ravi are 5 and 3 respectively.<\/p>\n

Question 45.
\nSome points on a plane are marked and they are connected pairwise by line segments. If the total number of line
\nsegments formed is 10, find the number of marked points on the plane.
\nSolution:
\nLet the number of points marked on the plane be \u2019x’.
\nThe total number of line segments actually formed is
\n\"Inter
\nGiven \\(\\frac{x(x-1)}{2}\\) = 10
\n\u21d2 x2<\/sup> – x = 20
\n\u21d2 x2<\/sup> – x – 20 = 0
\n\u21d2 (x – 5)(x + 4) = 0
\nSince x cannot be negatives x = 5
\n\u2234 The number of points marked on the plane is 5.<\/p>\n

Question 46.
\nSuppose that the quadratic equations ax2<\/sup> + bx + c = 0 and bx2<\/sup> + cx + a = 0 have a common root. Then show that a3<\/sup> + b3<\/sup> + c3<\/sup> = 3abc.
\nSolution:
\nLet \u03b1 be the common root of the equations
\n\"Inter<\/p>\n

Question 47.
\nFor what values of x1 the expression x2<\/sup> – 5x – 14 is positive?
\nSolution:
\nSince x2<\/sup> – 5x – 14 = (x + 2) (x – 7), the roots of the equation -2 and 7
\nHere the coefficient of x2<\/sup> is positive.
\nHence x2<\/sup> – 5x – 14 is positive when x < -2 or x > 7.<\/p>\n

Question 48.
\nFor what values ofx. the expression -6x2<\/sup> + 2x3<\/sup> is negative?
\nSolution:
\n-6x2<\/sup> + 2x – 3 = 0 can be written as 6x2<\/sup> – 2x + 3 = 0
\n\"Inter
\n\u2234 The roots of -6x2<\/sup> + 2x – 3 = 0 are non-real complex numbers.
\nHence coefficient of x2<\/sup> is -6, which is negative.
\n\u2234 -6x2<\/sup> + 2x – 3 < 0 for all x \u2208 R.<\/p>\n

Question 49.
\nFind the value of x at which the following expressions have maximum or minimum.<\/p>\n

i) x2<\/sup> + 5x + 6
\nSolution:
\nHere a = 1 > 0, the expression has minimum value at x = \\(\\frac{-b}{2 a}\\)
\n= \\(\\frac{-5}{2(1)}\\)
\n= \\(\\frac{-5}{2}\\)<\/p>\n

ii) 2x – x2<\/sup> + 7
\nSolution:
\nHere a = -1 < 0, the expression has maximum value at x
\n= \\(\\frac{-b}{2 a}\\)
\n= \\(\\frac{-2}{2 x-1}\\) = 1<\/p>\n

Question 50.
\nFind the maximum or minimum value of the quadratic expression
\n(i) 2x – 7 – 5x2<\/sup>
\n(ii) 3x2<\/sup> + 2x + 17 (Mar. ’14)
\nSolution:
\ni) Compare the given equation with ax2<\/sup> + bx + c, we get
\na = -5, b = 2, c = -7
\nHere a = -5 < 0, the given expression has maximum value at x = \\(\\frac{-b}{2 a}\\) = \\(\\frac{-2}{2(-5)}\\)
\n= \\(\\frac{1}{5}\\)
\nand maximum value = \\(\\frac{4 a c-b^{2}}{4 a}\\)
\n= \\(\\frac{4(-5) \\cdot(-7)-(?)^{2}}{4(-5)}\\)
\n= \\(\\frac{140-4}{-20}\\) = \\(\\frac{-34}{5}\\)<\/p>\n

(ii) Compare 3x2<\/sup> + 2x + 11 with
\nax2<\/sup> + bx + c, we get a = 3, b = 2, c = 11
\n\u2235 a = 3 > 0, the given expression has minimum value at
\n\"Inter<\/p>\n

Question 51.
\nFind the changes in the sign of 4x – 5x2<\/sup> + 2 for x \u2208 R and find the extreme value.
\nSolution:
\nThe roots of 4x – 5x2<\/sup> + 2 = 0
\n\u21d2 5x2<\/sup> – 4x – 2 = 0
\nroots are = \\(\\frac{2 \\pm \\sqrt{14}}{5}\\)
\n\u2234 \\(\\frac{2-\\sqrt{14}}{5}\\) < x < \\(\\frac{2+\\sqrt{14}}{5}\\) the sign of 4x – 5x2<\/sup> + 2 is positive.
\nx < \\(\\frac{2-\\sqrt{14}}{5}\\) or x > \\(\\frac{2+\\sqrt{14}}{5}\\), the sign of 4x – 5x2<\/sup> + 2 is negative.
\nSince a = -5 < 0, then maximum value = \\(\\frac{4 a c-b^{2}}{4 a}\\)
\n= \\(\\frac{4(-5)(2)-(4)^{2}}{4(-5)}\\)
\n= \\(\\frac{-56}{-20}\\) = \\(\\frac{14}{5}\\)
\nHence extreme value = \\(\\frac{14}{5}\\)<\/p>\n

\"Inter<\/p>\n

Question 52.
\nFind the solution set of x2<\/sup> + x – 12 \u2264 0 by both algebraic and graphical methods.
\nSolution:
\nAlgebraic Method : We have x2<\/sup> + x – 12 = (x + 4) (x – 3).
\nHence -4 and 3 are the roots of the equation x2<\/sup> + x – 12 = 0.
\nSince the coefficient of x2<\/sup> in the quadratic expression x2<\/sup> + x – 12 = 0 is positive, x2<\/sup> + x – 12 is negative if -4 < x < 3 and positive if either x < -4 or x > 3.
\nHence x2<\/sup> + x – 12 \u2264 0 \u21d4 -4 \u2264 x \u2264 3.
\nTherefore the solution set is
\n{x \u2208 R : -4 \u2264 x \u2264 3}.
\nGraphical Method: Let y = f(x) = x2<\/sup> + x – 12.
\nThe values of y at some selected values of s are given in the following table:
\n\"Inter
\nThe graph of the function y = f(x) is drawn using the above tabulated values. This is shown in Fig.
\n\"Inter
\nTherefore the graph of y = f(x) we observe that
\ny = x2<\/sup> – x – 12 < 0 if f -4 \u2264 x \u2264 3.
\nHence the solution set is {x \u2208 R : -4 \u2264 x \u2264 3}.<\/p>\n

Question 53.
\nFind the set of values of x for which the inequalities x2<\/sup> – 3x – 10 < 0, 10x – x2<\/sup> – 16
\n> 0 hold simultaneously.
\nSolution:
\n\u2235 x2<\/sup> – 3x – 10 < 0
\n\u21d2 x2<\/sup> – 5x + 2x – 10 < 0
\n\u21d2 x(x – 5) + 2(x – 5) < 0
\n\u21d2 (x – 5)(x + 2) < 0
\n\u2234 x2<\/sup> coeff. is the and expression is <0
\n\u21d2 x \u2208 (-2, 5) ——- (1)
\nNow 10x – x2<\/sup> – 16 > 0
\n\u21d2 x2<\/sup> – 10x + 16 < 0
\n\u21d2 (x – 2) (x – 8) < 0
\n\u2234 x2<\/sup> coeff is the ana expression is < 0
\n\u21d2 x \u2208 (2, 8) ——- (2)
\nHence x2<\/sup> – 3x – 10 < 0 and 10x – x2<\/sup> – 16 > 0
\n\u21d4 x \u2208 (-2, 5) \u2229 (2, 8)
\n\u2234 The solution set is {x\/x \u2208 R : 2 < x < 5}<\/p>\n

Question 54.
\nSolve the inequation \\(\\sqrt{x+2}\\) > \\(\\sqrt{8-x^{2}}\\).
\nSolution:
\nWhen a, b \u2208 R and a \u2265 0, b \u2265 0 then \\(\\sqrt{a}\\) > \\(\\sqrt{b}\\) \u21d4 a > b \u2265 0
\n\u2234 \\(\\sqrt{x+2}\\) = \\(\\sqrt{8-x^{2}}\\)
\n\u21d4 x + 2 > 8 – x2<\/sup> \u2265 0 and x > -2, |x| < 2\\(\\sqrt{2}\\)
\nWe have (x + 2) > 8 – x2<\/sup>
\n\u21d4 x2<\/sup> + x – 6 > 0
\n\u21d4 (x + 3)(x – 2) > 0
\n\u21d4 x \u2208 (-\u221e, -3)(2, \u221e) .
\nWe have 8 – x2<\/sup> \u2265 0
\n\u21d4 x2<\/sup> \u2264 8 \u21d4 |x| < 2 \\(\\sqrt{2}\\) \u21d4 x \u2208 [-2 \\(\\sqrt{2}\\), 2 \\(\\sqrt{2}\\)] Also x + 2 > 0 \u21d4 x > -2
\nHence x + 2 > 8 – x2<\/sup> \u2265 0
\n\u21d4 x \u2208 ((-\u221e, -3) \u222a (2, \u221e)) \u2229 [-2 \\(\\sqrt{2}\\), 2 \\(\\sqrt{2}\\)] and x > -2
\n\u21d4 x \u2208 (2, 2 \\(\\sqrt{2}\\))
\n\u2234 The solution set is [x \u2208 R : -2 < x \u2264 2 \\(\\sqrt{2}\\)]<\/p>\n

Question 55.
\nSolve the equation
\n\\(\\sqrt{(x-3)(2-x)}\\) < \\(\\sqrt{4 x^{2}+12 x+11}\\).
\nSolution:
\nThe given inequation is equivalent to the following two inequalities.
\n(x – 3)(2 – x) \u2265 0 and
\n(x – 3)(2 – x) < 4x2<\/sup> + 12x + 11
\n(x – 3)(2 – x) \u2265 0 \u21d4 (x – 2)(x – 3) \u2264 0
\n\u21d4 2 \u2264 x \u2264 3
\n-x2<\/sup> + 5x – 6 < 4x2<\/sup> + 12x + 11
\n\u21d4 5x2<\/sup> + 7x + 17 > 0
\nThe discriminant = b2<\/sup> – 4ac = 49 – 340 < 0 \u21d2 x \u2208 R : a = 5 > 0 and expressive is the
\nHence the solution set of the given inequation is {x \u2208 R : 2 \u2264 x \u2264 3}<\/p>\n

Question 56.
\nSolve the inequation
\n\"Inter
\nSolution:
\n\"Inter
\n\u21d4 either 6 + x – x2<\/sup> = 0, 2x + 5 \u2260 0 and x + 4 \u2260 0 or 6 + x – x2<\/sup> > 0
\nand \\(\\frac{1}{2 x+5}\\) \u2265 \\(\\frac{1}{x+4}\\)
\nWe have 6 + x – x2<\/sup> = -(x2<\/sup> – x – 6)
\n= -(x + 2) (x – 3)
\nHence 6 + x + x2<\/sup> = 0 \u21d4 x = -2 or x = 3
\n2x + 5 = 0 \u21d4 x = –\\(\\frac{5}{2}\\)
\nx + 4 = 0 \u21d4 x = -4
\n\u2234 6 + x – 2x2<\/sup> > 0 \u21d4 -2 < x < 3 x \u2208 (-2, 3), 2x + 5 > -4 + 5 = 1 > 0 and x + 4 > -2 + 4 = 2 > 0
\nx + 4 > -2 + 4 = 2 > 0
\nx \u2208 (-2, 3), \\(\\frac{1}{2 x+5}\\) \u2265 \\(\\frac{1}{x+4}\\)
\n\u21d4 2x + 5 \u2264 x + 4,
\n2x + 5 \u2264 x + 4 \u21d4 x \u2264 1
\nHence 6 + x – x2<\/sup> > 0 and \\(\\frac{1}{2 x+5}\\) \u2265 \\(\\frac{1}{x+4}\\)
\n\u21d4 -2 < x \u2264 -1
\n\u2234 The solution set is {-2, 3} \u222a (-2, -1)
\n= [-2, -1] \u222a {3}<\/p>\n

Question 57.
\nFind the maximum value of the function
\n\\(\\frac{x^{2}+14 x+9}{x^{2}+2 x+3}\\) over R. (May ’05)
\nSolution:
\nLet y = \\(\\frac{x^{2}+14 x+9}{x^{2}+2 x+3}\\)
\n\u21d2 yx2<\/sup> + 2yx + 3y = x2<\/sup> + 14x + 9
\n\u21d2 (y – 1)x2<\/sup> + 2(y – 7)x + 3y – 9 = 0
\nSince x \u2208 R, discriminant \u2265 0
\n\u21d2 [2(y – 7)2<\/sup>] – 4(y – 1) (3y – 9) \u2265 0
\n\u21d2 4[(y2<\/sup> – 14y + 49) [(3y2<\/sup> – 12y + 9)] \u2265 0
\n\u21d2 -2y2<\/sup> – 2y + 40 \u2265 0
\n\u21d2 y2<\/sup> + y – 20 \u2264 0
\n\u21d2 (y + 5) (y – 4) \u2264 0
\n\u21d2 -5 \u2264 y \u2264 4
\n\u21d2 y \u2208 [-5, 4]
\n\u2234 Maximum value of y = 4
\n\u2234 Maximum value of the function
\n\\(\\frac{x^{2}+14 x+9}{x^{2}+2 x+3}\\) over R is 4.<\/p>\n

Question 58.
\nShow that none of the values of the function over \\(\\frac{x^{2}+34 x-71}{x^{2}+2 x-7}\\) over R lies between 5 and 9. (Mar. 2005)
\nSolution:
\nLet y = \\(\\frac{x^{2}+34 x-7}{x^{2}+2 x-7}\\)
\n\u21d2 x2<\/sup> + 2x – 7 = x2<\/sup> + 34x – 71
\n\u21d2 (y – 1)x2<\/sup> + 2(y – 17)x + (-7y + 71) = 0
\nSince x \u2208 R, discriminant \u2265 0
\n\u21d2 [2(y – 17)]2<\/sup> – 4(y – 1)(-7y + 71)] \u2265 0
\n\u21d2 4[(y2<\/sup> – 34y + 289) – (-7y2<\/sup> + 78y – 71)] \u2265 0
\n\u21d2 8y2<\/sup> – 112y + 360 \u2265 0
\n\u21d2 y2<\/sup> – 14y + 45 \u2265 0
\n\u21d2 (y – 5) (y – 9) \u2265 0
\n\u21d2 y \u2264 5 or y \u2265 9
\n\u21d2 y does not lies between 5 and 9, since expression is \u2265 0 and y2<\/sup> coeff is the
\n\u2234 None of the values of the function
\n\\(\\frac{x^{2}+34 x-7}{x^{2}+2 x-7}\\) over R lies between 5 and 9.<\/p>\n

Question 59.
\nSolve the inequation
\n\\(\\sqrt{x^{2}-3 x-10}\\) > (8 – x)
\nSolution:
\n(i) \\(\\sqrt{x^{2}-3 x-10}\\) > (8 – x)
\n\u21d2 x2<\/sup> – 3x – 10 \u2265 0 and (i) (8 – x < 0)
\nor (ii) x2<\/sup> – 3x – 10 > (8 – x)2<\/sup> and (8 – x \u2265 0)
\nNow x2<\/sup> – 3x – 10 = (x – 5) (x + 2)
\nHence x2<\/sup> – 3x – 10 \u2265 0
\n\u21d2 x \u2208 (-\u221e, -2] \u222a [5, \u221e)
\n8 – x < 0 \u21d2 x \u2208 (8, \u221e)
\n\u2234 x2<\/sup> – 3x – 10 \u2265 0 and 8 – x < 0
\n\u21d4 x \u2208 (-\u221e, -2] \u222a [5, \u221e) and x \u2208 (8, \u221e)
\n\u21d4 x \u2208 (8, \u221e)<\/p>\n

(ii) x2<\/sup> – 3x – 10 > (8 – x)2<\/sup>
\n\u21d4 x2<\/sup> – 3x – 10 > 64 + x2<\/sup> – 16x
\n\u21d4 13x > 74
\n\"Inter<\/p>\n","protected":false},"excerpt":{"rendered":"

Students get through Maths 2A Important Questions Inter 2nd Year Maths 2A Quadratic Expressions Important Questions which are most likely to be asked in the exam. Intermediate 2nd Year Maths 2A Quadratic Expressions Important Questions Question 1. Form quadratic equation whose root 7 \u00b1 2 (Mar. ’11, ’05) Solution: \u03b1 + \u03b2 = 7 + … Read more<\/a><\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[15],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/22091"}],"collection":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/comments?post=22091"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/22091\/revisions"}],"predecessor-version":[{"id":22092,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/22091\/revisions\/22092"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/media?parent=22091"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/categories?post=22091"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/tags?post=22091"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}