Consider the function f:(0,2) arrow R defined by f(x)=\fracx2+\frac2x and the function g(x) defined by g(x)= beginarrayll \min \lfloor f(t)\, & 0\lt t \leq x and 0 \lt x \leq 1 \frac32+x, & 1 \lt x \lt 2 \endarray .. Then,

Limits Continuity and Differentiability PYQ — JEE Main 2024 · 27 Jan · Shift 2 · Q33 with Solution

View written solutionFree

Correct answer: D

Given functions

We have $f(x)=\frac{x}{2}+\frac{2}{x}, \qquad x\in(0,2).$

And

\begin{cases} \min \lfloor f(t)\rfloor, & 0<t\le x,\; 0<x\le 1,\\[4pt] \dfrac{3}{2}+x, & 1<x<2. \end{cases}$$ Here, for $0<x\le 1$, the expression means the minimum value of $\lfloor f(t)\rfloor$ for $t\in(0,x]$. --- 2. **Study $f(x)$ on $(0,1]$** To understand $g(x)$ for $x\le 1$, first analyze $f$. $$f(x)=\frac{x}{2}+\frac{2}{x}$$ Differentiate: $$f'(x)=\frac{1}{2}-\frac{2}{x^2}$$ For $0<x\le 1$, we have $x^2\le 1$, so $$\frac{2}{x^2}\ge 2$$ therefore $$f'(x)=\frac{1}{2}-\frac{2}{x^2}<0.$$ So $f$ is strictly decreasing on $(0,1]$. Now evaluate at $x=1$: $$f(1)=\frac{1}{2}+2=\frac{5}{2}=2.5.$$ Also, as $x\to 0^+$, $$f(x)\to +\infty.$$ Hence on $(0,1]$, $f(t)$ decreases from $+\infty$ to $\frac{5}{2}$. Therefore, $$f(t)\ge \frac{5}{2} \quad \text{for } 0<t\le 1.$$ So $$\lfloor f(t)\rfloor \ge 2.$$ And since at $t=1$, $$\lfloor f(1)\rfloor=\left\lfloor \frac{5}{2}\right\rfloor=2,$$ the minimum value of $\lfloor f(t)\rfloor$ on $(0,x]$ will be $2$ whenever the interval includes some point where $f(t)<3$. Let us check that for every $x\in(0,1]$, the minimum on $(0,x]$ is actually attained at $t=x$ because $f$ is decreasing, hence $\lfloor f(t)\rfloor$ is also minimized at the largest $t$, i.e. $t=x$. Thus for $0<x\le 1$, $$g(x)=\lfloor f(x)\rfloor.$$ Now determine where this equals $2$ or larger. Solve $$f(x)<3$$ $$\frac{x}{2}+\frac{2}{x}<3.$$ Multiply by $2x>0$: $$x^2+4<6x$$ $$x^2-6x+4<0.$$ Roots are $$x=3\pm \sqrt{5}.$$ So $$3-\sqrt{5}<x<3+\sqrt{5}.$$ On $(0,1]$, this means $$x>3-\sqrt{5}\approx 0.764.$$ Hence: - if $0<x<3-\sqrt{5}$, then $f(x)>3$, so $g(x)\ge 3$, - if $x\ge 3-\sqrt{5}$, then $f(x)<3$ (or at equality gives floor $3$ at the threshold), and eventually near $1$, $\lfloor f(x)\rfloor=2$. But for continuity at $x=1$, we only need behavior near $1$. Since for $x$ close to $1$ from the left, $$\frac{5}{2}\le f(x)<3,$$ so $$\lfloor f(x)\rfloor=2.$$ Therefore, $$g(x)=2 \quad \text{for all } x \text{ sufficiently close to }1^-.$$ In particular, $$g(1)=2.$$ --- 3. **Study $g(x)$ for $x>1$** For $1<x<2$, $$g(x)=\frac{3}{2}+x.$$ So $$\lim_{x\to 1^+} g(x)=\frac{3}{2}+1=\frac{5}{2}.$$ --- 4. **Check continuity at $x=1$** Left-hand limit: $$\lim_{x\to 1^-} g(x)=2.$$ Function value: $$g(1)=2.$$ Right-hand limit: $$\lim_{x\to 1^+} g(x)=\frac{5}{2}.$$ Since $$2\ne \frac{5}{2},$$ $g$ is **not continuous** at $x=1$. Therefore it also cannot be differentiable at $x=1$. --- 5. **Check options** - **A:** $g$ is continuous but not differentiable at $x=1$ **False**, because $g$ is not continuous at $x=1$. - **B:** $g$ is continuous and differentiable for all $x\in(0,2)$ **False**. - **C:** $g$ is not continuous for all $x\in(0,2)$ Interpreting this as "$g$ is not continuous on $(0,2)$" or "not continuous at every point" is ambiguous. It is certainly false if taken literally as "discontinuous at all points," because for $1<x<2$, $g$ is linear and continuous. In standard MCQ language, this option is not the intended precise statement. - **D:** $g$ is neither continuous nor differentiable at $x=1$ **True**. So the correct option is $$\boxed{D}.$$ --- 6. **Comparison with stored answer** Stored correct answer is **A**, but our derivation shows the correct answer should be **D**. The likely issue is that for $x\le 1$, the left-side definition gives a step-function behavior with value $2$ near $x=1$, not a value matching the right-side expression $\frac{3}{2}+x$ at $x=1$. Hence continuity fails at $x=1$.

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View written solutionFree

Correct answer: D

Given functions

We have $f(x)=\frac{x}{2}+\frac{2}{x}, \qquad x\in(0,2).$

And

\begin{cases} \min \lfloor f(t)\rfloor, & 0<t\le x,\; 0<x\le 1,\\[4pt] \dfrac{3}{2}+x, & 1<x<2. \end{cases}$$ Here, for $0<x\le 1$, the expression means the minimum value of $\lfloor f(t)\rfloor$ for $t\in(0,x]$. --- 2. **Study $f(x)$ on $(0,1]$** To understand $g(x)$ for $x\le 1$, first analyze $f$. $$f(x)=\frac{x}{2}+\frac{2}{x}$$ Differentiate: $$f'(x)=\frac{1}{2}-\frac{2}{x^2}$$ For $0<x\le 1$, we have $x^2\le 1$, so $$\frac{2}{x^2}\ge 2$$ therefore $$f'(x)=\frac{1}{2}-\frac{2}{x^2}<0.$$ So $f$ is strictly decreasing on $(0,1]$. Now evaluate at $x=1$: $$f(1)=\frac{1}{2}+2=\frac{5}{2}=2.5.$$ Also, as $x\to 0^+$, $$f(x)\to +\infty.$$ Hence on $(0,1]$, $f(t)$ decreases from $+\infty$ to $\frac{5}{2}$. Therefore, $$f(t)\ge \frac{5}{2} \quad \text{for } 0<t\le 1.$$ So $$\lfloor f(t)\rfloor \ge 2.$$ And since at $t=1$, $$\lfloor f(1)\rfloor=\left\lfloor \frac{5}{2}\right\rfloor=2,$$ the minimum value of $\lfloor f(t)\rfloor$ on $(0,x]$ will be $2$ whenever the interval includes some point where $f(t)<3$. Let us check that for every $x\in(0,1]$, the minimum on $(0,x]$ is actually attained at $t=x$ because $f$ is decreasing, hence $\lfloor f(t)\rfloor$ is also minimized at the largest $t$, i.e. $t=x$. Thus for $0<x\le 1$, $$g(x)=\lfloor f(x)\rfloor.$$ Now determine where this equals $2$ or larger. Solve $$f(x)<3$$ $$\frac{x}{2}+\frac{2}{x}<3.$$ Multiply by $2x>0$: $$x^2+4<6x$$ $$x^2-6x+4<0.$$ Roots are $$x=3\pm \sqrt{5}.$$ So $$3-\sqrt{5}<x<3+\sqrt{5}.$$ On $(0,1]$, this means $$x>3-\sqrt{5}\approx 0.764.$$ Hence: - if $0<x<3-\sqrt{5}$, then $f(x)>3$, so $g(x)\ge 3$, - if $x\ge 3-\sqrt{5}$, then $f(x)<3$ (or at equality gives floor $3$ at the threshold), and eventually near $1$, $\lfloor f(x)\rfloor=2$. But for continuity at $x=1$, we only need behavior near $1$. Since for $x$ close to $1$ from the left, $$\frac{5}{2}\le f(x)<3,$$ so $$\lfloor f(x)\rfloor=2.$$ Therefore, $$g(x)=2 \quad \text{for all } x \text{ sufficiently close to }1^-.$$ In particular, $$g(1)=2.$$ --- 3. **Study $g(x)$ for $x>1$** For $1<x<2$, $$g(x)=\frac{3}{2}+x.$$ So $$\lim_{x\to 1^+} g(x)=\frac{3}{2}+1=\frac{5}{2}.$$ --- 4. **Check continuity at $x=1$** Left-hand limit: $$\lim_{x\to 1^-} g(x)=2.$$ Function value: $$g(1)=2.$$ Right-hand limit: $$\lim_{x\to 1^+} g(x)=\frac{5}{2}.$$ Since $$2\ne \frac{5}{2},$$ $g$ is **not continuous** at $x=1$. Therefore it also cannot be differentiable at $x=1$. --- 5. **Check options** - **A:** $g$ is continuous but not differentiable at $x=1$ **False**, because $g$ is not continuous at $x=1$. - **B:** $g$ is continuous and differentiable for all $x\in(0,2)$ **False**. - **C:** $g$ is not continuous for all $x\in(0,2)$ Interpreting this as "$g$ is not continuous on $(0,2)$" or "not continuous at every point" is ambiguous. It is certainly false if taken literally as "discontinuous at all points," because for $1<x<2$, $g$ is linear and continuous. In standard MCQ language, this option is not the intended precise statement. - **D:** $g$ is neither continuous nor differentiable at $x=1$ **True**. So the correct option is $$\boxed{D}.$$ --- 6. **Comparison with stored answer** Stored correct answer is **A**, but our derivation shows the correct answer should be **D**. The likely issue is that for $x\le 1$, the left-side definition gives a step-function behavior with value $2$ near $x=1$, not a value matching the right-side expression $\frac{3}{2}+x$ at $x=1$. Hence continuity fails at $x=1$.

Limits Continuity and Differentiability question

Limits Continuity and Differentiability question

More from Limits Continuity and Differentiability

Limits Continuity and Differentiability question

Limits Continuity and Differentiability question

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