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

2019 · 12 Jan · Shift 2 · Q32
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. /2019 · 12 Jan · Shift 2 · Q32

Quadratic Equation and Inequalities question

2019 · 12 Jan · Shift 2 · Q32

JEE MainMathematicsQuadratic Equation and InequalitiesMCQ+4 / −1
The number of integral values of m for which the quadratic expression, (1 + 2m)x2 – 2(1 + 3m)x + 4(1 + m), x ∈\in∈ R, is always positive, is :
  1. A
    7
  2. B
    8
  3. C
    3
  4. D
    6
View written solutionFree

Correct answer: A

  1. Let f(x)=(1+2m)x2−2(1+3m)x+4(1+m).f(x)=(1+2m)x^2-2(1+3m)x+4(1+m).f(x)=(1+2m)x2−2(1+3m)x+4(1+m). We want f(x)>0f(x)>0f(x)>0 for all real xxx.

  2. For a quadratic ax2+bx+cax^2+bx+cax2+bx+c to be always positive for all x∈Rx\in\mathbb Rx∈R, we need:

    1. a>0a>0a>0
    2. Discriminant Δ<0\Delta<0Δ<0

So here, a=1+2m,a=1+2m,a=1+2m, b=−2(1+3m),b=-2(1+3m),b=−2(1+3m), c=4(1+m).c=4(1+m).c=4(1+m).

  1. First condition: 1+2m>0  ⟹  m>−12.1+2m>0 \implies m>-\frac12.1+2m>0⟹m>−21​. Since mmm is integral, m≥0.m\ge 0.m≥0.

  2. Now compute the discriminant: Δ=b2−4ac.\Delta=b^2-4ac.Δ=b2−4ac. So, Δ=[−2(1+3m)]2−4(1+2m)⋅4(1+m).\Delta=[-2(1+3m)]^2-4(1+2m)\cdot 4(1+m).Δ=[−2(1+3m)]2−4(1+2m)⋅4(1+m).

Simplify: Δ=4(1+3m)2−16(1+2m)(1+m).\Delta=4(1+3m)^2-16(1+2m)(1+m).Δ=4(1+3m)2−16(1+2m)(1+m). Factor out 444: Δ=4((1+3m)2−4(1+2m)(1+m)).\Delta=4\Big((1+3m)^2-4(1+2m)(1+m)\Big).Δ=4((1+3m)2−4(1+2m)(1+m)).

Now, (1+3m)2=1+6m+9m2,(1+3m)^2=1+6m+9m^2,(1+3m)2=1+6m+9m2, (1+2m)(1+m)=1+3m+2m2.(1+2m)(1+m)=1+3m+2m^2.(1+2m)(1+m)=1+3m+2m2. Thus, 4(1+2m)(1+m)=4+12m+8m2.4(1+2m)(1+m)=4+12m+8m^2.4(1+2m)(1+m)=4+12m+8m2.

Therefore, Δ=4[(1+6m+9m2)−(4+12m+8m2)].\Delta=4\big[(1+6m+9m^2)-(4+12m+8m^2)\big].Δ=4[(1+6m+9m2)−(4+12m+8m2)]. Δ=4(m2−6m−3).\Delta=4(m^2-6m-3).Δ=4(m2−6m−3).

  1. For always positive, Δ<0  ⟹  m2−6m−3<0.\Delta<0 \implies m^2-6m-3<0.Δ<0⟹m2−6m−3<0. Solve: m=6±36+122=6±482=3±23.m=\frac{6\pm\sqrt{36+12}}{2}=\frac{6\pm\sqrt{48}}{2}=3\pm 2\sqrt3.m=26±36+12​​=26±48​​=3±23​. Hence, 3−23<m<3+23.3-2\sqrt3<m<3+2\sqrt3.3−23​<m<3+23​.

Now, 3−23≈−0.464,3-2\sqrt3\approx -0.464,3−23​≈−0.464, 3+23≈6.464.3+2\sqrt3\approx 6.464.3+23​≈6.464. So integer values of mmm satisfying this are m=0,1,2,3,4,5,6.m=0,1,2,3,4,5,6.m=0,1,2,3,4,5,6. These are 777 values.

  1. Also check with m≥0m\ge 0m≥0 from the leading coefficient condition — all these values satisfy it.

Therefore, the number of integral values of mmm is 7.\boxed{7}.7​. So the correct option is A.

PreviousNext

More from Quadratic Equation and Inequalities

  • If λ∈ R is such that the sum of the cubes of the roots of the equation, x2 + (2 −λ) x + (10 −λ) = 0 is minimum, then the magnitude of the difference of the roots of this equation is :2018 · MCQ
  • If tanA and tanB are the roots of the quadratic equation, 3x2 − 10x − 25 = 0, then the value of 3 sin2(A + B) − 10 sin(A + B).cos(A + B) − 25 cos2(A + B) is :2018 · MCQ
  • If f(x) is a quadratic expression such that f (1) + f (2) = 0, and − 1 is a root of f (x) = 0, then the other root of f(x) = 0 is :2018 · MCQ
  • Let p, q and r be real numbers (p e q, r e 0), such that the roots of the equation x+p1​+x+q1​=r1​ are equal in magnitude but opposite in sign, then the sum of squares of these roots is equal to :2018 · MCQ
  • If an angle A of a Δ ABC satiesfies 5 cosA + 3 = 0, then the roots of the quadratic equation, 9x2 + 27x + 20 = 0 are :2018 · MCQ
  • Let S = { x∈ R : x≥ 0 and 2∣x​−3∣+x​(x​−6)+6=0}. Then S2018 · MCQ
  • Let p(x) be a quadratic polynomial such that p(0)=1. If p(x) leaves remainder 4 when divided by x − 1 and it leaves remainder 6 when divided by x + 1; then :2017 · MCQ
  • The sum of all the real values of x satisfying the equation 2(x − 1)(x2 + 5x − 50) = 1 is :2017 · MCQ