Question

f(x) = \begin{cases} \sin^{-1}(\frac{2x}{1+x^2}), & x \leq 0 \ [3x] & 0 < x < k \text{ Where } [p] \text{ denotes greatest integer} \ (x-\alpha)^2 + \beta, & x \geq k \end{cases}

less than or equal to p and $k \in N$. If $f(x)$ is non-derivable at exactly 7 points, then the value of $(\alpha + \beta + k) =$

• A. 9
• B. 10
• C. 11
• D. 12

Answer 1

Answer: (A)

9

Answer 2

The function $f(x)$ is defined piecewise. We analyze the differentiability in each interval and at the transition points.

1. For $x < 0$, $f(x) = \sin^{-1}(\frac{2x}{1+x^2})$. This function is non-derivable at $x=-1$.
2. For $0 < x < k$, $f(x) = [3x]$. The greatest integer function $[y]$ is non-derivable when $y$ is an integer. Thus, $[3x]$ is non-derivable when $3x = n$, where $n$ is an integer. For $x \in (0, k)$, we have $0 < n < 3k$. The points of non-derivability are $x = 1/3, 2/3, \dots, (3k-1)/3$. There are $3k-1$ such points.
3. For $x > k$, $f(x) = (x-\alpha)^2 + \beta$, which is a polynomial and is differentiable everywhere.

Now, we check the transition points $x=0$ and $x=k$.

• At $x=0$: The left limit is $\lim_{x \to 0^-} \sin^{-1}(\frac{2x}{1+x^2}) = \sin^{-1}(0) = 0$. The right limit is $\lim_{x \to 0^+} [3x] = 0$. The function is continuous at $x=0$. The left derivative at $x=0$ is $\lim_{x \to 0^-} \frac{d}{dx}(\sin^{-1}(\frac{2x}{1+x^2})) = \lim_{x \to 0^-} \frac{2}{1+x^2} = 2$. The right derivative at $x=0$ is $\lim_{x \to 0^+} \frac{d}{dx}([3x])$. For $x \in (0, 1/3)$, $[3x]=0$, so the derivative is 0. Since $2 \neq 0$, the function is non-derivable at $x=0$.

• At $x=k$: For continuity, $\lim_{x \to k^-} [3x] = f(k)$. Since $k \in \mathbb{N}$, $\lim_{x \to k^-} [3x] = 3k-1$. So, $3k-1 = (k-\alpha)^2 + \beta$. The left derivative at $x=k$ is $\lim_{x \to k^-} \frac{d}{dx}([3x])$. For $x \in (k-1/3, k)$, $[3x] = 3k-1$, so the derivative is 0. The right derivative at $x=k$ is $\lim_{x \to k^+} \frac{d}{dx}((x-\alpha)^2 + \beta) = \lim_{x \to k^+} 2(x-\alpha) = 2(k-\alpha)$. For the function to be non-derivable at $x=k$, we need $0 \neq 2(k-\alpha)$, which means $\alpha \neq k$.

The total number of non-derivable points is $1$ (at $x=-1$) $+ 1$ (at $x=0$) $+ (3k-1)$ (from $[3x]$) $+ 1$ (at $x=k$, if $\alpha \neq k$).

Case 1: $\alpha \neq k$. Total non-derivable points = $1 + 1 + (3k-1) + 1 = 3k+2$. Given $3k+2 = 7$, we get $3k=5$, so $k=5/3$. This contradicts $k \in \mathbb{N}$.

Case 2: $\alpha = k$. Total non-derivable points = $1 + 1 + (3k-1) = 3k+1$. Given $3k+1 = 7$, we get $3k=6$, so $k=2$. This is a valid integer. If $k=2$, then $\alpha=k=2$. Using the continuity condition at $x=k=2$: $3k-1 = (k-\alpha)^2 + \beta$. $3(2)-1 = (2-2)^2 + \beta \implies 5 = 0 + \beta \implies \beta = 5$. So, $\alpha=2, \beta=5, k=2$. The value of $(\alpha + \beta + k) = 2 + 5 + 2 = 9$.

The 7 non-derivable points are: $-1, 0, 1/3, 2/3, 1, 4/3, 5/3$.