Limits, Continuity and Differentiability Questions and Answers

Q: Let \(f(x) = x^{2} - 10x + 16, x \in \mathbb{R}\). If \(f'(c)\) is equal to slope of the straight line joining the points (2,0) and (8,0), then the value of \(c\) is

c is 5.The line joining (2, 0) and (8, 0) is horizontal, with slope 0. So we need \(f'(c) = 0\).\[ f'(c) = 2c - 10 = 0 \Rightarrow c = 5 \]This is the mean value theorem: 2 and 8 are the roots of f, and 5, the midpoint, is where the parabola's tangent is horizontal. A sanity check: the vertex of the parabola \(x^{2} - 10x + 16\) is at \(x = -\frac{-10}{2} = 5\), the only place its tangent is horizontal.

Q: Let \([x]\) denote the greatest integer less than or equal to \(x\) . Then \(\displaystyle\lim_{x \to 0^{-}} \frac{\sin[x]}{[x]}\) is equal to

The limit is sin1.As \(x \to 0^{-}\), \(x\) is a small negative number, so \([x] = -1\) throughout the approach. The expression is constant there:\[ \begin{aligned} \frac{\sin[x]}{[x]} &= \frac{\sin(-1)}{-1} \\ &= \frac{-\sin 1}{-1} \\ &= \sin 1 \end{aligned} \]The trap is to use \(\frac{\sin\theta}{\theta} \to 1\), which needs \(\theta \to 0\); here \([x]\) stays at -1 and never approaches 0. Forgetting that the two minus signs cancel gives -sin1.

151.

Let \(f(x) = x^{2} - 10x + 16, x \in \mathbb{R}\). If \(f'(c)\) is equal to slope of the straight line joining the points (2,0) and (8,0), then the value of \(c\) is

Exam Year(s)

KEAM - 2026,

Answer: D

c is 5.

The line joining (2, 0) and (8, 0) is horizontal, with slope 0. So we need \(f'(c) = 0\).

\[ f'(c) = 2c - 10 = 0 \Rightarrow c = 5 \]

This is the mean value theorem: 2 and 8 are the roots of f, and 5, the midpoint, is where the parabola's tangent is horizontal. A sanity check: the vertex of the parabola \(x^{2} - 10x + 16\) is at \(x = -\frac{-10}{2} = 5\), the only place its tangent is horizontal.

Discuss in Forum

152.

Let \([x]\) denote the greatest integer less than or equal to \(x\) . Then \(\displaystyle\lim_{x \to 0^{-}} \frac{\sin[x]}{[x]}\) is equal to

A.1
B.-sin1
C.-2
D.0
E.sin1

Exam Year(s)

KEAM - 2026,

Answer: E

The limit is sin1.

As \(x \to 0^{-}\), \(x\) is a small negative number, so \([x] = -1\) throughout the approach. The expression is constant there:

\[ \begin{aligned} \frac{\sin[x]}{[x]} &= \frac{\sin(-1)}{-1} \\ &= \frac{-\sin 1}{-1} \\ &= \sin 1 \end{aligned} \]

The trap is to use \(\frac{\sin\theta}{\theta} \to 1\), which needs \(\theta \to 0\); here \([x]\) stays at -1 and never approaches 0. Forgetting that the two minus signs cancel gives -sin1.

Limits, Continuity and Differentiability Questions and Answers

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