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Application of Derivatives question

2024 · 29 Jan · Shift 1 · Q46
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  5. /2024 · 29 Jan · Shift 1 · Q46

Application of Derivatives question

2024 · 29 Jan · Shift 1 · Q46

JEE MainMathematicsApplication of DerivativesMCQ+4 / −1
Consider the function f:[12,1]→Rf:\left[\frac{1}{2}, 1\right] \rightarrow \mathbb{R}f:[21​,1]→R defined by f(x)=42x3−32x−1f(x)=4 \sqrt{2} x^3-3 \sqrt{2} x-1f(x)=42​x3−32​x−1. Consider the statements (I) The curve y=f(x)y=f(x)y=f(x) intersects the xxx-axis exactly at one point. (II) The curve y=f(x)y=f(x)y=f(x) intersects the xxx-axis at x=cos⁡π12x=\cos \frac{\pi}{12}x=cos12π​. Then
  1. A
    Both (I) and (II) are correct.
  2. B
    Only (I) is correct.
  3. C
    Both (I) and (II) are incorrect.
  4. D
    Only (II) is correct.
View written solutionFree

Correct answer: A

  1. Given function

    f(x)=42 x3−32 x−1,x∈[12,1].f(x)=4\sqrt{2}\,x^3-3\sqrt{2}\,x-1, \qquad x\in \left[\frac12,1\right].f(x)=42​x3−32​x−1,x∈[21​,1].

    We need to check:

    • (I) whether the curve intersects the xxx-axis exactly once in [12,1]\left[\frac12,1\right][21​,1]
    • (II) whether one such intersection is at x=cos⁡π12x=\cos\frac{\pi}{12}x=cos12π​

  1. Check the number of roots in the interval

    First compute the derivative:

    f′(x)=122x2−32=32(4x2−1)=32(2x−1)(2x+1).f'(x)=12\sqrt{2}x^2-3\sqrt{2}=3\sqrt{2}(4x^2-1)=3\sqrt{2}(2x-1)(2x+1).f′(x)=122​x2−32​=32​(4x2−1)=32​(2x−1)(2x+1).

    On the interval [12,1]\left[\frac12,1\right][21​,1]:

    • 2x+1>02x+1>02x+1>0
    • 2x−1≥02x-1\ge 02x−1≥0

    Hence,

    f′(x)≥0for all x∈[12,1],f'(x)\ge 0 \quad \text{for all } x\in \left[\frac12,1\right],f′(x)≥0for all x∈[21​,1],

    and in fact f′(x)>0f'(x)>0f′(x)>0 for x>12x>\frac12x>21​. So fff is increasing on [12,1]\left[\frac12,1\right][21​,1].

    Now check endpoint values:

    =\frac{\sqrt{2}}{2}-\frac{3\sqrt{2}}{2}-1 =-\sqrt{2}-1<0.$$ $$f(1)=4\sqrt{2}-3\sqrt{2}-1=\sqrt{2}-1>0.$$ Since $f$ is continuous and increasing, and changes sign from negative to positive, it crosses the $x$-axis **exactly once**. Therefore, **statement (I) is correct**.

  1. Check whether x=cos⁡π12x=\cos\frac{\pi}{12}x=cos12π​ is a root

    Let

    x=cos⁡π12=cos⁡15∘.x=\cos\frac{\pi}{12}=\cos 15^\circ.x=cos12π​=cos15∘.

    Use the triple-angle identity:

    4cos⁡3θ−3cos⁡θ=cos⁡3θ.4\cos^3\theta-3\cos\theta=\cos 3\theta.4cos3θ−3cosθ=cos3θ.

    Then

    =\sqrt{2}(4x^3-3x).$$ Putting $x=\cos\frac{\pi}{12}$, $$4x^3-3x=\cos\left(3\cdot \frac{\pi}{12}\right)=\cos\frac{\pi}{4}=\frac{1}{\sqrt{2}}.$$ Therefore, $$f\left(\cos\frac{\pi}{12}\right)=\sqrt{2}\cdot \frac{1}{\sqrt{2}}-1=1-1=0.$$ So the curve intersects the $x$-axis at $$x=\cos\frac{\pi}{12}.$$ Hence, **statement (II) is correct**.

  1. Conclusion

    • (I) is correct.
    • (II) is correct.

    So the correct option is:

    A\boxed{\text{A}}A​

  1. Comparison with stored answer

    Stored correct answer: A

    Our derived answer: A

    Hence, the answer agrees with the stored correct answer.

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