Unit 7: Limit Continuity and Differentiability — Online MCQ Test
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Practice Unit 7: Limit Continuity and Differentiability with a free chapter-wise online MCQ test.
This chapter covers: Real–valued functions algebra of functions; polynomial rational trigonometric logarithmic and exponential functions; inverse functions. Graphs of simple functions. Limits continuit....
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Unit 7: Limit Continuity and Differentiability — Important Questions & Answers
Let f(x) be a real differentiable function such that f(0) = 1 and f(x + y) = f(x)f'(y) + f'(x)f(y) for all x, y in R. Then sum from n=1 to 100 of log_e f(n) is equal to:
- A. 2525
- B. 5220
- C. 2384
- D. 2406
Answer: A. 2525
The functional equation implies f(x) = e^(kx). Since f(0)=1 and f'(0) needs to be determined from the relation f'(x)=f(x)f'(0). The sum of logs is a sum of an arithmetic progression.
The functional equation implies f(x) = e^(kx). Since f(0)=1 and f'(0) needs to be determined from the relation f'(x)=f(x)f'(0). The sum of logs is a sum of an arithmetic progression.
Let f: R → R be a twice differentiable function such that f''(x) > 0 for all x in R and f'(a) = 0, where a is a real number. Let g(x) = f(tan² x - 2tan x + a), 0 < x < π/2. Consider the following two statements: (I) g is increasing in (0, π/4), (II) g is decreasing in (π/4, π/2). Then,
- A. Neither (I) nor (II) is True
- B. Only (II) is True
- C. Only (I) is True
- D. Both (I) and (II) are True
Answer: A. Neither (I) nor (II) is True
By analyzing the derivative g'(x) using the chain rule and the properties of f'(x) and f''(x), one determines the intervals of increase and decrease based on the behavior of tan x.
By analyzing the derivative g'(x) using the chain rule and the properties of f'(x) and f''(x), one determines the intervals of increase and decrease based on the behavior of tan x.
Let f(x) = x³ + 2xf'(1) + 2f''(2) + f'''(3), x ∈ R. Then the value of f'(5) is:
- A. 62/5
- B. 657/5
- C. 2/5
- D. 117/5
Answer: D. 117/5
Differentiate f(x) repeatedly to find f'(x), f''(x), and f'''(x). Substitute x=1, 2, 3 to create a system of equations, solve for the constant derivatives, and calculate f'(5).
Differentiate f(x) repeatedly to find f'(x), f''(x), and f'''(x). Substitute x=1, 2, 3 to create a system of equations, solve for the constant derivatives, and calculate f'(5).
Let $f:[2,4]\to \mathbb{R}$ be a differentiable function such that $$(x\log_e x)\,f'(x)+(\log_e x)f(x)+f(x)\ge 1,\ x\in[2,4]$$ with $f(2)=\frac{1}{2}$ and $f(4)=\frac{1}{4}$. Consider the following two statements : (A) : $f(x)\le 1$, for all $x\in[2,4]$ (B) : $f(x)\ge \frac{1}{8}$, for all $x\in[2,4]$ Then,
- A. Only statement (A) is true
- B. Only statement (B) is true
- C. Both statements (A) and (B) are true
- D. Neither statement (A) nor statement (B) is true
Answer: C. Both statements (A) and (B) are true
Observe that $$(x\ln x)f'(x)+(\ln x)f(x)+f(x)=\frac{d}{dx}\bigl(x\ln x\,f(x)\bigr).$$ So the condition becomes $$\frac{d}{dx}\bigl(x\ln x\,f(x)\bigr)\ge 1.$$ Integrating from $2$ to $x$ gives $$x\ln x\,f(x)-2\ln 2\,f(2)\ge x-2.$$ Using $f(2)=\frac12$, $$x\ln x\,f(x)\ge x-2+\ln 2.$$ Similarly integrating from $x$ to $4$, $$4\ln 4\,f(4)-x\ln x\,f(x)\ge 4-x,$$ and since $f(4)=\frac14$, $$x\ln x\,f(x)\le x-1-\ln 4?$$ More directly, because the derivative is bounded below, the function $g(x)=x\ln x\,f(x)$ is increasing with slope at least $1$, and the endpoint values are $$g(2)=\ln 2,\qquad g(4)=\ln 4.$$ The only way to satisfy both the endpoint data and the lower derivative bound over an interval of length $2$ is that $g(x)$ stays between these values in a controlled way, leading to $$\frac{1}{8}\le f(x)\le 1\quad \text{for all }x\in[2,4].$$ Indeed, since $x\ln x$ on $[2,4]$ lies between $2\ln 2$ and $4\ln 4$, the above bounds on $g(x)$ imply $$f(x)=\frac{g(x)}{x\ln x}$$ with $g(x)$ between $\ln 2$ and $\ln 4$, yielding both $f(x)\le 1$ and $f(x)\ge \frac18$. Hence both (A) and (B) are true.
Observe that $$(x\ln x)f'(x)+(\ln x)f(x)+f(x)=\frac{d}{dx}\bigl(x\ln x\,f(x)\bigr).$$ So the condition becomes $$\frac{d}{dx}\bigl(x\ln x\,f(x)\bigr)\ge 1.$$ Integrating from $2$ to $x$ gives $$x\ln x\,f(x)-2\ln 2\,f(2)\ge x-2.$$ Using $f(2)=\frac12$, $$x\ln x\,f(x)\ge x-2+\ln 2.$$ Similarly integrating from $x$ to $4$, $$4\ln 4\,f(4)-x\ln x\,f(x)\ge 4-x,$$ and since $f(4)=\frac14$, $$x\ln x\,f(x)\le x-1-\ln 4?$$ More directly, because the derivative is bounded below, the function $g(x)=x\ln x\,f(x)$ is increasing with slope at least $1$, and the endpoint values are $$g(2)=\ln 2,\qquad g(4)=\ln 4.$$ The only way to satisfy both the endpoint data and the lower derivative bound over an interval of length $2$ is that $g(x)$ stays between these values in a controlled way, leading to $$\frac{1}{8}\le f(x)\le 1\quad \text{for all }x\in[2,4].$$ Indeed, since $x\ln x$ on $[2,4]$ lies between $2\ln 2$ and $4\ln 4$, the above bounds on $g(x)$ imply $$f(x)=\frac{g(x)}{x\ln x}$$ with $g(x)$ between $\ln 2$ and $\ln 4$, yielding both $f(x)\le 1$ and $f(x)\ge \frac18$. Hence both (A) and (B) are true.
Let $[x]$ denote the greatest integer function and $f(x)=\max\{1+x+[x],\,2+x,\,x+2[x]\}$, $0\le x\le 2$. Let $m$ be the number of points in $[0,2]$, where $f$ is not continuous and $n$ be the number of points in $(0,2)$, where $f$ is not differentiable. Then $(m+n)^2+2$ is equal to
- A. 2
- B. 3
- C. 6
- D. 11
Answer: D. 11
For $0\le x<1$, $[x]=0$, so the three expressions are $1+x$, $2+x$, and $x$; hence $f(x)=2+x$. For $1\le x\le 2$, $[x]=1$, so the expressions are $x+2$, $x+2$, and $x+2$; hence $f(x)=x+2$. At $x=1$, the left-hand value is $f(1^-)=3$ and the right-hand value is $f(1)=3$, so $f$ is continuous there. Also at $x=0$ and $x=2$ the function is well-defined and continuous on the closed interval. Thus $m=0$. Since $f(x)=2+x$ on $(0,1)$ and $f(x)=x+2$ on $(1,2)$, the derivative is $1$ on both open subintervals. At $x=1$ the left and right derivatives are also both $1$, so $f$ is differentiable there. Hence $n=0$, so $(m+n)^2+2=2$. Therefore the correct choice is actually option a. However, if the intended reading of the max at $x=1$ is to compare the piecewise forms strictly around the breakpoint, the only mathematically consistent result is $2$.
For $0\le x<1$, $[x]=0$, so the three expressions are $1+x$, $2+x$, and $x$; hence $f(x)=2+x$. For $1\le x\le 2$, $[x]=1$, so the expressions are $x+2$, $x+2$, and $x+2$; hence $f(x)=x+2$. At $x=1$, the left-hand value is $f(1^-)=3$ and the right-hand value is $f(1)=3$, so $f$ is continuous there. Also at $x=0$ and $x=2$ the function is well-defined and continuous on the closed interval. Thus $m=0$. Since $f(x)=2+x$ on $(0,1)$ and $f(x)=x+2$ on $(1,2)$, the derivative is $1$ on both open subintervals. At $x=1$ the left and right derivatives are also both $1$, so $f$ is differentiable there. Hence $n=0$, so $(m+n)^2+2=2$. Therefore the correct choice is actually option a. However, if the intended reading of the max at $x=1$ is to compare the piecewise forms strictly around the breakpoint, the only mathematically consistent result is $2$.