Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Quadratic Equation and Inequalities question

2006 · Shift 0 · Q71
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Mathematics
  4. /Quadratic Equation and Inequalities
  5. /2006 · Shift 0 · Q71

Quadratic Equation and Inequalities question

2006 · Shift 0 · Q71

JEE MainMathematicsQuadratic Equation and InequalitiesMCQ+4 / −1
If xxx is real, the maximum value of 3x2+9x+173x2+9x+7{{3{x^2} + 9x + 17} \over {3{x^2} + 9x + 7}}3x2+9x+73x2+9x+17​ is
  1. A
    14{1 \over 4}41​
  2. B
    414141
  3. C
    111
  4. D
    177{17 \over 7}717​
View written solutionFree

Correct answer: B

  1. Let f(x)=3x2+9x+173x2+9x+7.f(x)=\frac{3x^2+9x+17}{3x^2+9x+7}.f(x)=3x2+9x+73x2+9x+17​. We want the maximum value of f(x)f(x)f(x) for real xxx.

  2. Substitute t=3x2+9x.t=3x^2+9x.t=3x2+9x. Then f(x)=t+17t+7=1+10t+7.f(x)=\frac{t+17}{t+7}=1+\frac{10}{t+7}.f(x)=t+7t+17​=1+t+710​.

So we now need the range of t=3x2+9xt=3x^2+9xt=3x2+9x for real xxx.

  1. Find the minimum of ttt: t=3x2+9x=3(x2+3x)=3[(x+32)2−94].t=3x^2+9x=3\left(x^2+3x\right)=3\left[\left(x+\frac32\right)^2-\frac94\right].t=3x2+9x=3(x2+3x)=3[(x+23​)2−49​]. Hence, t=3(x+32)2−274.t=3\left(x+\frac32\right)^2-\frac{27}{4}.t=3(x+23​)2−427​. Therefore, t≥−274.t\ge -\frac{27}{4}.t≥−427​.

  2. Now compute t+7t+7t+7: t+7≥−274+7=−274+284=14.t+7\ge -\frac{27}{4}+7=-\frac{27}{4}+\frac{28}{4}=\frac14.t+7≥−427​+7=−427​+428​=41​. So t+7≥14>0.t+7\ge \frac14>0.t+7≥41​>0.

Thus, f(x)=1+10t+7f(x)=1+\frac{10}{t+7}f(x)=1+t+710​ is maximized when t+7t+7t+7 is minimized, i.e. when t+7=14.t+7=\frac14.t+7=41​.

  1. Therefore, fmax⁡=1+101/4=1+40=41.f_{\max}=1+\frac{10}{1/4}=1+40=41.fmax​=1+1/410​=1+40=41.

  2. This occurs when 3(x+32)2−274=−2743\left(x+\frac32\right)^2-\frac{27}{4}=-\frac{27}{4}3(x+23​)2−427​=−427​ which gives x=−32.x=-\frac32.x=−23​.

Hence the maximum value is 41.\boxed{41}.41​.

  1. Checking options:
  • A: 14\frac1441​ — incorrect
  • B: 414141 — correct
  • C: 111 — incorrect
  • D: 177\frac{17}{7}717​ — incorrect
PreviousNext

More from Quadratic Equation and Inequalities

  • The value of a for which the sum of the squares of the roots of the equation x2−(a−2)x−a−1=0 assume the least value is2005 · MCQ
  • If the roots of the equation x2−bx+c=0 be two consecutive integers, then b2−4c equals2005 · MCQ
  • If the roots of the equation x2−bx+c=0 be two consecutive integers, then b2−4c equals2005 · MCQ
  • The value of a for which the sum of the squares of the roots of the equation x2−(a−2)x−a−1=0 assume the least value is :2005 · MCQ
  • In a triangle PQR,∠R=2π​.Iftan(2P​) and tan(2Q​) are the roots of ax2+bx+c=0,ae0 then2005 · MCQ
  • If both the roots of the quadratic equation x2−2kx+k2+k−5=0 are less than 5, then k lies in the interval2005 · MCQ
  • Let two numbers have arithmetic mean 9 and geometric mean 4. Then these numbers are the roots of the quadratic equation2004 · MCQ
  • If (1−p) is a root of quadratic equation x2+px+(1−p)=0 then its root are2004 · MCQ