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Let $A = [a, \infty)$ denotes the domain, then $f: [a, \infty) \to B$, which is defined by $f(x) = 2x^3 - 3x^2 + 6$ will have an inverse for the smallest real value of '$a$' if

This question was previously asked in
WBJEE 2026 Physics and Chemistry Question Paper (24-May-2026)
The correct answer is
$a=2, B=[10, \infty)$

Finding Smallest Domain Value for Inverse Function

For a function $f(x)$ to have an inverse on a domain $[a, \infty)$, it must be strictly monotonic (either strictly increasing or strictly decreasing) throughout that domain.

Analyzing Function Monotonicity

We are given the function $f(x) = 2x^3 - 3x^2 + 6$. To determine its monotonicity, we first find its derivative, $f'(x)$.

$ f'(x) = \frac{d}{dx}(2x^3 - 3x^2 + 6) = 6x^2 - 6x $

Now, we find the critical points by setting the derivative to zero:

$ 6x^2 - 6x = 0 $ $ 6x(x - 1) = 0 $

This gives us critical points at $x = 0$ and $x = 1$. We analyze the sign of $f'(x)$ in the intervals defined by these points:

  • For $x < 0$, $f'(x) > 0$ (e.g., $f'(-1) = 6(-1)(-2) = 12 > 0$). The function $f(x)$ is increasing.
  • For $0 < x < 1$, $f'(x) < 0$ (e.g., $f'(0.5) = 6(0.5)(-0.5) = -1.5 < 0$). The function $f(x)$ is decreasing.
  • For $x > 1$, $f'(x) > 0$ (e.g., $f'(2) = 6(2)(1) = 12 > 0$). The function $f(x)$ is increasing.

Determining the Domain for Invertibility

The function $f(x)$ must be strictly monotonic on the domain $[a, \infty)$ for it to have an inverse.

  • If $a=0$, the domain is $[0, \infty)$. $f(x)$ decreases on $(0, 1)$ and increases on $(1, \infty)$, so it's not monotonic.
  • If $a=-1$, the domain is $[-1, \infty)$. $f(x)$ increases on $[-1, 0)$ and decreases on $(0, 1)$, so it's not monotonic.
  • For the function to be strictly increasing on $[a, \infty)$, we need $f'(x) > 0$ for all $x \in [a, \infty)$. Since $f'(x) = 0$ at $x=1$, we must start the domain after $x=1$ to ensure strict increase. Thus, we require $a > 1$.

The smallest value of '$a$' from the given options that satisfies $a > 1$ is $a=2$.

Calculating Range for $a=2$

If $a=2$, the domain is $[2, \infty)$. On this interval, $f'(x) > 0$, so the function is strictly increasing and has an inverse.

We need to find the range $B = [f(a), \infty)$.

Calculate $f(a)$ for $a=2$:

$ f(2) = 2(2)^3 - 3(2)^2 + 6 $ $ f(2) = 2(8) - 3(4) + 6 $ $ f(2) = 16 - 12 + 6 $ $ f(2) = 10 $

Therefore, the range is $B = [10, \infty)$.

Conclusion

The smallest value of '$a$' among the options for which $f(x)$ is strictly monotonic on $[a, \infty)$ and thus has an inverse is $a=2$. The corresponding range is $B = [10, \infty)$.

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