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Application of Derivatives question

2021 · 27 Aug · Shift 1 · Q35
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  5. /2021 · 27 Aug · Shift 1 · Q35

Application of Derivatives question

2021 · 27 Aug · Shift 1 · Q35

JEE MainMathematicsApplication of DerivativesNumerical+4 / −1
The number of distinct real roots of the equation 3x4 + 4x3 −-− 12x2 + 4 = 0 is ‾\underline{\hspace{2cm}}​.
Numerical answer
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Correct answer: 4

We need the number of distinct real roots of 3x4+4x3−12x2+4=0.3x^4+4x^3-12x^2+4=0.3x4+4x3−12x2+4=0.

1. Let

f(x)=3x4+4x3−12x2+4.f(x)=3x^4+4x^3-12x^2+4.f(x)=3x4+4x3−12x2+4. We will factor it.

2. Try factorization

Assume 3x4+4x3−12x2+4=(x2+ax+b)(3x2+cx+d).3x^4+4x^3-12x^2+4=(x^2+ax+b)(3x^2+cx+d).3x4+4x3−12x2+4=(x2+ax+b)(3x2+cx+d). Expanding,

(x2+ax+b)(3x2+cx+d)=3x4+(c+3a)x3+(d+ac+3b)x2+(ad+bc)x+bd.(x^2+ax+b)(3x^2+cx+d) =3x^4+(c+3a)x^3+(d+ac+3b)x^2+(ad+bc)x+bd.(x2+ax+b)(3x2+cx+d)=3x4+(c+3a)x3+(d+ac+3b)x2+(ad+bc)x+bd.

Comparing coefficients with 3x4+4x3−12x2+0x+4,3x^4+4x^3-12x^2+0x+4,3x4+4x3−12x2+0x+4, we get:

  1. c+3a=4c+3a=4c+3a=4
  2. d+ac+3b=−12d+ac+3b=-12d+ac+3b=−12
  3. ad+bc=0ad+bc=0ad+bc=0
  4. bd=4bd=4bd=4

Now try b=−2, d=−2b=-2,\, d=-2b=−2,d=−2 so that bd=4bd=4bd=4. Then ad+bc=−2a−2c=0  ⟹  a+c=0  ⟹  c=−a.ad+bc=-2a-2c=0 \implies a+c=0 \implies c=-a.ad+bc=−2a−2c=0⟹a+c=0⟹c=−a. Using c+3a=4c+3a=4c+3a=4, −a+3a=4  ⟹  2a=4  ⟹  a=2, c=−2.-a+3a=4 \implies 2a=4 \implies a=2,\, c=-2.−a+3a=4⟹2a=4⟹a=2,c=−2. Check the x2x^2x2 coefficient: d+ac+3b=−2+(2)(−2)+3(−2)=−2−4−6=−12,d+ac+3b=-2+(2)(-2)+3(-2)=-2-4-6=-12,d+ac+3b=−2+(2)(−2)+3(−2)=−2−4−6=−12, which works.

Hence, 3x4+4x3−12x2+4=(x2+2x−2)(3x2−2x−2).3x^4+4x^3-12x^2+4=(x^2+2x-2)(3x^2-2x-2).3x4+4x3−12x2+4=(x2+2x−2)(3x2−2x−2).

3. Solve each quadratic

(i) x2+2x−2=0x^2+2x-2=0x2+2x−2=0

Using quadratic formula, x=−2±4+82=−2±122=−1±3.x=\frac{-2\pm\sqrt{4+8}}{2}=\frac{-2\pm\sqrt{12}}{2}=-1\pm\sqrt{3}.x=2−2±4+8​​=2−2±12​​=−1±3​. These are two distinct real roots.

(ii) 3x2−2x−2=03x^2-2x-2=03x2−2x−2=0

Using quadratic formula,

=\frac{2\pm\sqrt{4+24}}{6} =\frac{2\pm\sqrt{28}}{6} =\frac{1\pm\sqrt{7}}{3}.$$ These are also two distinct real roots. ## 4. Count distinct real roots The four roots are $$-1+\sqrt{3},\quad -1-\sqrt{3},\quad \frac{1+\sqrt{7}}{3},\quad \frac{1-\sqrt{7}}{3}.$$ They are all real and distinct. Therefore, the number of distinct real roots is $$\boxed{4}.$$
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