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

2021 · Shift 1 · Q36
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 Advanced
  3. /Mathematics
  4. /Quadratic Equation and Inequalities
  5. /2021 · Shift 1 · Q36

Quadratic Equation and Inequalities question

2021 · Shift 1 · Q36

JEE AdvancedMathematicsQuadratic Equation and InequalitiesNumerical+4 / −1
For x ∈\in∈ R, the number of real roots of the equation 3x2−4∣x2−1∣+x−1=03{x^2} - 4\left| {{x^2} - 1} \right| + x - 1 = 03x2−4​x2−1​+x−1=0 is ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 4

We need to find the number of real solutions of 3x2−4∣x2−1∣+x−1=0.3x^2-4|x^2-1|+x-1=0.3x2−4∣x2−1∣+x−1=0.

1. Split using the absolute value

The expression ∣x2−1∣|x^2-1|∣x2−1∣ depends on whether x2−1≥0orx2−1<0.x^2-1\ge 0 \quad \text{or} \quad x^2-1<0.x2−1≥0orx2−1<0. That is:

  • Case 1: x2≥1⇒∣x∣≥1x^2\ge 1 \Rightarrow |x|\ge 1x2≥1⇒∣x∣≥1
  • Case 2: x2<1⇒∣x∣<1x^2<1 \Rightarrow |x|<1x2<1⇒∣x∣<1

2. Case 1: x2≥1x^2\ge 1x2≥1

Then ∣x2−1∣=x2−1.|x^2-1|=x^2-1.∣x2−1∣=x2−1. So the equation becomes 3x2−4(x2−1)+x−1=0.3x^2-4(x^2-1)+x-1=0.3x2−4(x2−1)+x−1=0. Simplify: 3x2−4x2+4+x−1=03x^2-4x^2+4+x-1=03x2−4x2+4+x−1=0 −x2+x+3=0-x^2+x+3=0−x2+x+3=0 x2−x−3=0.x^2-x-3=0.x2−x−3=0. Now solve: x=1±1+122=1±132.x=\frac{1\pm\sqrt{1+12}}{2}=\frac{1\pm\sqrt{13}}{2}.x=21±1+12​​=21±13​​. Check whether these satisfy x2≥1x^2\ge 1x2≥1:

  • x=1+132≈2.30x=\dfrac{1+\sqrt{13}}{2} \approx 2.30x=21+13​​≈2.30, valid.
  • x=1−132≈−1.30x=\dfrac{1-\sqrt{13}}{2} \approx -1.30x=21−13​​≈−1.30, valid.

So this case gives 2 real roots.

3. Case 2: x2<1x^2<1x2<1

Then ∣x2−1∣=−(x2−1)=1−x2.|x^2-1|=-(x^2-1)=1-x^2.∣x2−1∣=−(x2−1)=1−x2. So the equation becomes 3x2−4(1−x2)+x−1=0.3x^2-4(1-x^2)+x-1=0.3x2−4(1−x2)+x−1=0. Simplify: 3x2−4+4x2+x−1=03x^2-4+4x^2+x-1=03x2−4+4x2+x−1=0 7x2+x−5=0.7x^2+x-5=0.7x2+x−5=0. Now solve: x=−1±1+14014=−1±14114.x=\frac{-1\pm\sqrt{1+140}}{14}=\frac{-1\pm\sqrt{141}}{14}.x=14−1±1+140​​=14−1±141​​. Check whether these satisfy x2<1x^2<1x2<1, i.e. ∣x∣<1|x|<1∣x∣<1:

  • x=−1+14114≈0.777x=\dfrac{-1+\sqrt{141}}{14} \approx 0.777x=14−1+141​​≈0.777, valid.
  • x=−1−14114≈−0.920x=\dfrac{-1-\sqrt{141}}{14} \approx -0.920x=14−1−141​​≈−0.920, valid.

So this case also gives 2 real roots.

4. Total number of real roots

Total real roots =2+2=4.=2+2=4.=2+2=4.

Therefore, the number of real roots is 4.\boxed{4}.4​.

PreviousNext

More from Quadratic Equation and Inequalities

  • Suppose a, b denote the distinct real roots of the quadratic polynomial x2 + 20x − 2020 and suppose c, d denote the distinct complex roots of the quadratic polynomial x2 − 20x + 2020. Then the value of ac(a − c) + ad(a − d) + bc(b −…2020 · MCQ
  • Let α and β be the roots of x2−x−1=0, with α>β. For all positive integers n, define an​=α−βαn−βn​,n≥1b1​=1andbn​=an−1​+an+1​,n≥2…2019 · Multiple correct
  • Let a, b, c three non-zero real numbers such that the equation 3​acosx+2bsinx=c,x∈[−2π​,2π​], has two distinct real roots α and β with $\alpha + \beta = {\pi \over…2018 · Numerical
  • Let p, q be integers and let α, β be the roots of the equation, x2 − x − 1 = 0 where αeβ. For n = 0, 1, 2, ........, let an = p α n + q β n. FACT : If a and b are rational numbers and a + b 5​…2017 · MCQ
  • Let p, q be integers and let α, β be the roots of the equation, x2 − x − 1 = 0 where αeβ. For n = 0, 1, 2, ........, let an = p α n + q β n. FACT : If a and b are rational numbers and a + b 5​…2017 · MCQ
  • Let −6π​<θ<−12π​. Suppose α1​ and β1​ are the roots of the equation x2−2xsecθ+1=0 and α2​ and β2​ are the roots of the equation x2+2xtanθ−1=0.Ifα1​>β1​…2016 · MCQ
  • Let S be the set of all non-zero real numbers α such that the quadratic equation αx2−x+α=0 has two distinct real roots x1​ and x2​ satisfying the inequality ∣x1​−x2​∣<1.…2015 · Multiple correct
  • The quadratic equation p(x)=0 with real coefficients has purely imaginary roots. Then the equation p(p(x))=0 has2014 · MCQ