JEE Advanced 2016 · Paper 2 · Mathematics Q46

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MathematicsMultiple correct+4 / -2Question 46 of 54

Let $f:\left[-\frac12,2\right]\to\mathbb R$ and $g:\left[-\frac12,2\right]\to\mathbb R$ be functions defined by $f(x)=[x^2-3]$ and $g(x)=|x|f(x)+|4x-7|f(x)$ , where $[y]$ denotes the greatest integer less than or equal to $y$ for $y\in\mathbb R$ . Then

f

is discontinuous exactly at three points in

\left[-\frac12,2\right]

f

is discontinuous exactly at four points in

\left[-\frac12,2\right]

g

is NOT differentiable exactly at four points in

\left(-\frac12,2\right)

g

is NOT differentiable exactly at five points in

\left(-\frac12,2\right)

Answer and worked solution

Official answer: B, C

f. On $[-\frac12,2]$ , $x^2-3$ runs over $[-3,1]$ . $f$ jumps where $x^2-3$ crosses an integer: $x=1,\ \sqrt2,\ \sqrt3,\ 2$ . (At $x=0$ , $x^2-3$ touches $-3$ but $f$ stays $-3$ .) That is four points. (B) is true.

g. $g=(|x|+|4x-7|)f$ , and the factor $|x|+|4x-7|$ never vanishes.

Jumps of f: $g$ is discontinuous at $1,\sqrt2,\sqrt3$ , so not differentiable there.
At $x=0$ : $f=-3$ nearby and $|x|$ has a corner, so $g$ is not differentiable.
At $x=\frac74$ : $f=0$ on $(\sqrt3,2)$ , so $g\equiv0$ there and the corner of $|4x-7|$ is harmless.

That is exactly four points in $(-\frac12,2)$ . (C) is true.

(B), (C)

Topics:Continuity Differentiability

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JEE Advanced 2016 · Paper 2, Mathematics question 46