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Application of Derivatives question

2025 · 23 Jan · Shift 1 · Q47
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Application of Derivatives question

2025 · 23 Jan · Shift 1 · Q47

JEE MainMathematicsApplication of DerivativesNumerical+4 / −1
If the set of all values of aaa, for which the equation 5x3−15x−a=05 x^3-15 x-a=05x3−15x−a=0 has three distinct real roots, is the interval (α,β)(\alpha, \beta)(α,β), then β−2α\beta-2 \alphaβ−2α is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
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Correct answer: 30

  1. We need the cubic 5x3−15x−a=05x^3-15x-a=05x3−15x−a=0 to have three distinct real roots.

  2. Rewrite it as 5(x3−3x)−a=0 ⇒ x3−3x=a5.5(x^3-3x)-a=0 \,\Rightarrow\, x^3-3x=\frac{a}{5}.5(x3−3x)−a=0⇒x3−3x=5a​.

    Let f(x)=x3−3x.f(x)=x^3-3x.f(x)=x3−3x. Then the equation becomes f(x)=a5.f(x)=\frac{a}{5}.f(x)=5a​.

  3. A cubic of this form has three distinct real roots when the horizontal line y=a5y=\frac{a}{5}y=5a​ cuts the graph of y=f(x)y=f(x)y=f(x) at three distinct points. This happens when a5\frac{a}{5}5a​ lies strictly between the local maximum and local minimum values of fff.

  4. Find critical points: f′(x)=3x2−3=3(x2−1).f'(x)=3x^2-3=3(x^2-1).f′(x)=3x2−3=3(x2−1). So, f′(x)=0⇒x=±1.f'(x)=0 \Rightarrow x=\pm 1.f′(x)=0⇒x=±1.

  5. Compute the function values there: f(−1)=(−1)3−3(−1)=−1+3=2,f(-1)=(-1)^3-3(-1)=-1+3=2,f(−1)=(−1)3−3(−1)=−1+3=2, f(1)=1−3=−2.f(1)=1-3=-2.f(1)=1−3=−2.

    Thus,

    • local maximum = 222 at x=−1x=-1x=−1
    • local minimum = −2-2−2 at x=1x=1x=1

  6. Therefore, for three distinct real roots, −2<a5<2.-2<\frac{a}{5}<2.−2<5a​<2.

    Multiplying by 555: −10<a<10.-10<a<10.−10<a<10.

    Hence, (α,β)=(−10,10).(\alpha,\beta)=(-10,10).(α,β)=(−10,10). So, α=−10,β=10.\alpha=-10,\quad \beta=10.α=−10,β=10.

  7. Now compute: β−2α=10−2(−10)=10+20=30.\beta-2\alpha=10-2(-10)=10+20=30.β−2α=10−2(−10)=10+20=30.

Therefore, the required integer is 30.30.30.

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