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Solve the following.
Let , and exists. The statement “ is continuous at ” is true. Justify. Hence, find a relation…
Let $f(x) = \begin{cases} a + \sin^{-1}(x+b), & x \ge 1 \\ x, & x < 1 \end{cases}$, and $f'(1)$ exists.
(a)[1.5]
The statement “$f(x)$ is continuous at $x = 1$” is true. Justify.
(b)[1.5]
Hence, find a relation between $a$ and $b$.
(c)[1.5]
Find $f'(x)$.
(d)[1.5]
Hence, find the values of $a$ and $b$.
Answer
Answer (a)
Official answer keySince $f'(1)$ exists $\Rightarrow f(x)$ is differentiable at $x = 1$. $\Rightarrow f(x)$ is also continuous at $x = 1$. Hence, the statement is true.
Answer (b)
Official answer key$a + \sin^{-1}(1 + b) = 1$
Final answer: $a + \sin^{-1}(1+b) = 1$
Answer (c)
Official answer key$f'(x) = \begin{cases}\frac{1}{\sqrt{1-(x+b)^2}}, & x \geq 1\\ 1, & x < 1\end{cases}$
Final answer: $f'(x) = \frac{1}{\sqrt{1-(x+b)^2}}$ for $x \geq 1$; $1$ for $x < 1$
Answer (d)
Official answer key$a = 1$, $b = -1$.
Final answer: $a = 1,\ b = -1$
From ISC 2025 Practice Mathematics, question 131.