Comprehensive IGNOU BMTC-133 Solved Assignment 2024 for B.Sc (G) CBCS Students

IGNOU BMTC-133 Solved Assignment 2024 | B.Sc (G) CBCS

Solved By – Narendra Kr. Sharma – M.Sc (Mathematics Honors) – Delhi University

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IGNOU BMTC-133 Assignment Question Paper 2024

bmtc-133-solved-assignment-2024-10cd7422-4b11-4531-9171-a6445826d855

bmtc-133-solved-assignment-2024-10cd7422-4b11-4531-9171-a6445826d855

BMTC-133 Solved Assignment 2024
Part A (20 Marks)
  1. Which of the following statements are true or false? Give reasons for your answers in the form of a short proof or counter-example, whichever is appropriate:
    i) Every infinite set is an open set.
ii) The negation of p ∧ ∼ q p ∧ ∼ q p^^∼qp \wedge \sim qp∧∼q is p → q p → q p rarr qp \rightarrow qp→q.
iii) -1 is a limit point of the interval 1 − 2 , 1 ] 1 − 2 , 1 ] 1-2,1]1-2,1]1−2,1].
iv) The necessary condition for a function f f fff to be integrable is that it is continuous.
v) The function f : R → R f : R → R f:RrarrRf: \mathbb{R} \rightarrow \mathbb{R}f:R→R defined by f ( x ) = | x − 2 | + | 3 − x | f ( x ) = | x − 2 | + | 3 − x | f(x)=|x-2|+|3-x|f(x)=|x-2|+|3-x|f(x)=|x−2|+|3−x| is differentiable at x = 5 x = 5 x=5x=5x=5.
  1. Give an example for each of the following.
    i) A set in R R R\mathbb{R}R with a unique limit point.
ii) A set in R R R\mathbb{R}R whose all points except the one are its limit points.
iii) A set having no limit point.
iv) A set S S SSS with S ∘ = S ¯ S ∘ = S ¯ S^(@)= bar(S)S^{\circ}=\bar{S}S∘=S¯.
v) A bijection from N odd N odd N_(“odd “)\mathbb{N}_{\text {odd }}Nodd to Z Z Z\mathbb{Z}Z.
Part B (30 Marks)
  1. a) Give an example of a divergent sequence which has two convergent subsequences. Justify your claim.
b) The product of two divergent sequences is divergent. True or false? Justify.
c) Let ( a n ) n ∈ N a n n ∈ N (a_(n))_(n inN)\left(a_n\right)_{n \in \mathbb{N}}(an)n∈N be any sequence. Show that lim n → ∞ a n = L lim n → ∞   a n = L lim_(n rarr oo)a_(n)=L\lim _{n \rightarrow \infty} a_n=Llimn→∞an=L iff for every ε > 0 ε > 0 epsi > 0\varepsilon>0ε>0, there exists some N ∈ N N ∈ N N inNN \in \mathbb{N}N∈N such that n ≥ N n ≥ N n >= Nn \geq Nn≥N implies a n ∈ N ε ( L ) a n ∈ N ε ( L ) a_(n)inN_(epsi)(L)a_n \in N_{\varepsilon}(L)an∈Nε(L).
d) Show that ( 1 n 2 + n + 1 ) n ∈ N 1 n 2 + n + 1 n ∈ N ((1)/(n^(2)+n+1))_(n inN)\left(\frac{1}{n^2+n+1}\right)_{n \in \mathbb{N}}(1n2+n+1)n∈N is a Cauchy sequence.
e) Evaluate
lim n → ∞ [ n 1 + n 2 + n 4 + n 2 + n 9 + n 2 + ⋯ + n 2 n 2 ] lim n → ∞   n 1 + n 2 + n 4 + n 2 + n 9 + n 2 + ⋯ + n 2 n 2 lim_(n rarr oo)[(n)/(1+n^(2))+(n)/(4+n^(2))+(n)/(9+n^(2))+cdots+(n)/(2n^(2))]\lim _{n \rightarrow \infty}\left[\frac{n}{1+n^2}+\frac{n}{4+n^2}+\frac{n}{9+n^2}+\cdots+\frac{n}{2 n^2}\right]limn→∞[n1+n2+n4+n2+n9+n2+⋯+n2n2]
  1. a) Determine the points of discontinuity of the function f f fff and the nature of discontinuity at each of those points:
f ( x ) = { − x 2 , when x ≤ 0 4 − 5 x , when 0 < x ≤ 1 3 x − 4 x 2 , when 1 < x ≤ 2 − 12 x + 2 x , when x > 2 f ( x ) = − x 2 ,      when x ≤ 0 4 − 5 x ,      when 0 < x ≤ 1 3 x − 4 x 2 ,      when 1 < x ≤ 2 − 12 x + 2 x ,      when x > 2 f(x)={[-x^(2)”,”,” when “x <= 0],[4-5x”,”,” when “0 < x <= 1],[3x-4x^(2)”,”,” when “1 < x <= 2],[-12 x+2x”,”,” when “x > 2]:}f(x)= \begin{cases}-x^2, & \text { when } x \leq 0 \\ 4-5 x, & \text { when } 0<x \leq 1 \\ 3 x-4 x^2, & \text { when } 1<x \leq 2 \\ -12 x+2 x, & \text { when } x>2\end{cases}f(x)={−x2, when x≤04−5x, when 0<x≤13x−4x2, when 1<x≤2−12x+2x, when x>2
Also check whether the function f f f\mathrm{f}f is derivable at x = 1 x = 1 x=1\mathrm{x}=1x=1.
b) Find the following limit.
lim x → 0 1 − cos ⁡ x 2 x 2 sin ⁡ x 2 lim x → 0   1 − cos ⁡ x 2 x 2 sin ⁡ x 2 lim_(x rarr0)(1-cos x^(2))/(x^(2)sin x^(2))\lim _{x \rightarrow 0} \frac{1-\cos x^2}{x^2 \sin x^2}limx→01−cos⁡x2x2sin⁡x2
c) Prove that a strictly decreasing function is always one-one.
d) Determine the local minimum and local maximum values of the function f f fff defined by f ( x ) = 3 − 5 x 3 + 5 x 4 − x 5 f ( x ) = 3 − 5 x 3 + 5 x 4 − x 5 f(x)=3-5x^(3)+5x^(4)-x^(5)f(x)=3-5 x^3+5 x^4-x^5f(x)=3−5x3+5x4−x5.
Part C (50 Marks)
5. a) Let f : [ 0 , 1 ] → R f : [ 0 , 1 ] → R f:[0,1]rarrRf:[0,1] \rightarrow \mathbb{R}f:[0,1]→R be a function defined by f ( x ) = x m ( 1 − x ) n f ( x ) = x m ( 1 − x ) n f(x)=x^(m)(1-x)^(n)f(x)=x^m(1-x)^nf(x)=xm(1−x)n, where m , n ∈ N m , n ∈ N m,n inNm, n \in \mathbb{N}m,n∈N. Find the values of m m mmm and n n nnn such that the Rolle’s Theorem holds for the function f f fff.
b) Let f f fff be a differentiable function on [ α , β ] [ α , β ] [alpha,beta][\alpha, \beta][α,β] and x ∈ [ α , β ] x ∈ [ α , β ] x in[alpha,beta]x \in[\alpha, \beta]x∈[α,β]. Show that, if f ′ ( x ) = 0 f ′ ( x ) = 0 f^(‘)(x)=0f^{\prime}(x)=0f′(x)=0 and f ′ ′ ( x ) > 0 f ′ ′ ( x ) > 0 f^(”)(x) > 0f^{\prime \prime}(x)>0f′′(x)>0, then f f fff must have a local maximum at x x xxx.
c) Suppose that f : [ 0 , 2 ] → R f : [ 0 , 2 ] → R f:[0,2]rarrRf:[0,2] \rightarrow \mathbb{R}f:[0,2]→R is continuous on [ 0 , 2 ] [ 0 , 2 ] [0,2][0,2][0,2] and differentiable on ] 0 , 2 [ ] 0 , 2 [ ]0,2[] 0,2[]0,2[, and that f ( 0 ) = 0 , f ( 1 ) = 1 , f ( 2 ) = 1 f ( 0 ) = 0 , f ( 1 ) = 1 , f ( 2 ) = 1 f(0)=0,f(1)=1,f(2)=1f(0)=0, f(1)=1, f(2)=1f(0)=0,f(1)=1,f(2)=1.
(i) Show that there exists c 1 ∈ ( 0 , 1 ) c 1 ∈ ( 0 , 1 ) c_(1)in(0,1)c_1 \in(0,1)c1∈(0,1) such that f ′ ( c 1 ) = 1 f ′ c 1 = 1 f^(‘)(c_(1))=1f^{\prime}\left(c_1\right)=1f′(c1)=1.
(ii) Show that there exists c 2 ∈ ( 1 , 2 ) c 2 ∈ ( 1 , 2 ) c_(2)in(1,2)c_2 \in(1,2)c2∈(1,2) such that f ′ ( c 2 ) = 0 f ′ c 2 = 0 f^(‘)(c_(2))=0f^{\prime}\left(c_2\right)=0f′(c2)=0.
(iii) Show that there exists c ∈ ( 0 , 2 ) c ∈ ( 0 , 2 ) c in(0,2)c \in(0,2)c∈(0,2) such that f ′ ( c ) = 1 3 f ′ ( c ) = 1 3 f^(‘)(c)=(1)/(3)f^{\prime}(c)=\frac{1}{3}f′(c)=13.
  1. a) Test the following series for convergence.
    (i) ∑ n = 1 ∞ n x n − 1 , x > 0 ∑ n = 1 ∞   n x n − 1 , x > 0 quadsum_(n=1)^(oo)nx^(n-1),x > 0\quad \sum_{\mathrm{n}=1}^{\infty} \mathrm{n} \mathrm{x}^{\mathrm{n}-1}, \mathrm{x}>0∑n=1∞nxn−1,x>0.
    (ii) ∑ n = 1 ∞ [ n 4 + 9 − n 4 − 9 ] ∑ n = 1 ∞   n 4 + 9 − n 4 − 9 sum_(n=1)^(oo)[sqrt(n^(4)+9)-sqrt(n^(4)-9)]\sum_{n=1}^{\infty}\left[\sqrt{n^4+9}-\sqrt{n^4-9}\right]∑n=1∞[n4+9−n4−9]
b) Show that ∑ n = 1 ∞ ( − 1 ) n + 1 5 7 n + 2 ∑ n = 1 ∞   ( − 1 ) n + 1 5 7 n + 2 sum_(n=1)^(oo)(-1)^(n+1)(5)/(7n+2)\sum_{n=1}^{\infty}(-1)^{n+1} \frac{5}{7 n+2}∑n=1∞(−1)n+157n+2 is conditionally convergent.
  1. a) Use Cauchy’s Mean Value Theorem to prove that:
cos ⁡ α − cos ⁡ β sin ⁡ α − sin ⁡ β = tan ⁡ θ , 0 < α < θ < β < π 2 cos ⁡ α − cos ⁡ β sin ⁡ α − sin ⁡ β = tan ⁡ θ , 0 < α < θ < β < π 2 (cos alpha-cos beta)/(sin alpha-sin beta)=tan theta,0 < alpha < theta < beta < (pi)/(2)\frac{\cos \alpha-\cos \beta}{\sin \alpha-\sin \beta}=\tan \theta, 0<\alpha<\theta<\beta<\frac{\pi}{2}cos⁡α−cos⁡βsin⁡α−sin⁡β=tan⁡θ,0<α<θ<β<π2
b) Using Weiestrass M-test, show that the following series converges uniformly.
∑ n = 1 ∞ n 3 x n , x ∈ [ − 1 3 , 1 3 ] . ∑ n = 1 ∞   n 3 x n , x ∈ − 1 3 , 1 3 . sum_(n=1)^(oo)n^(3)x^(n),x in[-(1)/(3),(1)/(3)].\sum_{n=1}^{\infty} n^3 x^n, x \in\left[-\frac{1}{3}, \frac{1}{3}\right] .∑n=1∞n3xn,x∈[−13,13].
  1. a) Use the Fundamental Theorem of Integral Calculus to evaluate the integral
∫ 0 1 ( 2 x sin ⁡ 1 x − cos ⁡ 1 x ) d x . ∫ 0 1   2 x sin ⁡ 1 x − cos ⁡ 1 x d x . int_(0)^(1)(2x sin (1)/(x)-cos (1)/(x))dx.\int_0^1\left(2 x \sin \frac{1}{x}-\cos \frac{1}{x}\right) d x .∫01(2xsin⁡1x−cos⁡1x)dx.
b) Show that the function f : R → R f : R → R f:RrarrRf: \mathbb{R} \rightarrow \mathbb{R}f:R→R defined by f ( x ) = 2 x + 7 f ( x ) = 2 x + 7 f(x)=2x+7\mathrm{f}(\mathrm{x})=2 \mathrm{x}+7f(x)=2x+7 has an inverse by applying the inverse function theorem. Find its inverse also.
  1. a) Check whether the series ∑ n = 1 ∞ n 2 x 5 n 4 + x 3 , x ∈ [ 0 , α ] ∑ n = 1 ∞   n 2 x 5 n 4 + x 3 , x ∈ [ 0 , α ] sum_(n=1)^(oo)(n^(2)x^(5))/(n^(4)+x^(3)),x in[0,alpha]\sum_{n=1}^{\infty} \frac{n^2 x^5}{n^4+x^3}, x \in[0, \alpha]∑n=1∞n2x5n4+x3,x∈[0,α] is uniformly convergent or not, where α ∈ R + α ∈ R + alpha inR^(+)\alpha \in \mathbb{R}^{+}α∈R+.
b) Show that the series ∑ n = 1 ∞ sin ⁡ n θ n ∑ n = 1 ∞   sin ⁡ n θ n sum_(n=1)^(oo)(sin n theta)/(n)\sum_{n=1}^{\infty} \frac{\sin n \theta}{n}∑n=1∞sin⁡nθn does not converge uniformly on the interval ] 0 , 2 π [ ] 0 , 2 π [ ]0,2pi[] 0,2 \pi[]0,2π[.
c) If the power series ∑ n = 0 ∞ a n x n ∑ n = 0 ∞   a n x n sum_(n=0)^(oo)a_(n)x^(n)\sum_{n=0}^{\infty} a_n x^n∑n=0∞anxn converges uniformly in ] α , β [ ] α , β ]alpha,beta[:}] \alpha, \beta\left[\right.]α,β[, then so does ∑ n = 0 ∞ a n ( − x ) n ∑ n = 0 ∞   a n ( − x ) n sum_(n=0)^(oo)a_(n)(-x)^(n)\sum_{n=0}^{\infty} a_n(-x)^n∑n=0∞an(−x)n. True or false? Justify.
\(\sec ^2 \theta=1+\tan ^2 \theta\)

BMTC-133 Sample Solution 2024

bmtc-133-solved-assignment-2024-10cd7422-4b11-4531-9171-a6445826d855

bmtc-133-solved-assignment-2024-10cd7422-4b11-4531-9171-a6445826d855

BMTC-133 Solved Assignment 2024
Part A (20 Marks)
  1. Which of the following statements are true or false? Give reasons for your answers in the form of a short proof or counter-example, whichever is appropriate:
    i) Every infinite set is an open set.
Answer:
The statement "Every infinite set is an open set" is not necessarily true. To evaluate this statement, we need to consider the context of topology, as the concept of open sets is defined within topological spaces.

Background

  1. Open Set: In topology, an open set is a set that, for every point in the set, includes some neighborhood of that point within the set. The exact definition of "neighborhood" depends on the topology defined on the space.
  2. Topological Space: A topological space is a set equipped with a topology, a collection of open sets that satisfies certain properties (such as the union of open sets is open, and the finite intersection of open sets is open).

Analysis

The statement implies that in any topological space, every infinite set is open. This is not true in general, as the openness of a set depends on the specific topology defined on the space, not on whether the set is finite or infinite.

Counterexample

Consider the real numbers R R R\mathbb{R}R with the standard topology, where open sets are those that can be expressed as unions of open intervals.
  • Take the set S = Z S = Z S=ZS = \mathbb{Z}S=Z, the set of all integers. This is an infinite set.
  • However, S S SSS is not an open set in R R R\mathbb{R}R with the standard topology. For any integer z ∈ Z z ∈ Z z inZz \in \mathbb{Z}z∈Z, there is no open interval around z z zzz that is entirely contained within Z Z Z\mathbb{Z}Z, as any open interval around z z zzz will contain non-integer real numbers.

Conclusion

The statement "Every infinite set is an open set" is false. Whether a set is open depends on the topology of the space it is in, not on the finiteness or infiniteness of the set. The set of integers Z Z Z\mathbb{Z}Z in the real numbers R R R\mathbb{R}R with the standard topology serves as a counterexample.
ii) The negation of p ∧ ∼ q p ∧ ∼ q p^^∼qp \wedge \sim qp∧∼q is p → q p → q p rarr qp \rightarrow qp→q.
Answer:
To evaluate the statement "The negation of p ∧ ∼ q p ∧ ∼ q p^^∼qp \wedge \sim qp∧∼q is p → q p → q p rarr qp \rightarrow qp→q", we need to understand the logical structure of these expressions.

Background

  1. Logical AND ( ∧ ∧ ^^\wedge∧): The statement p ∧ q p ∧ q p^^qp \wedge qp∧q is true if and only if both p p ppp and q q qqq are true.
  2. Negation ( ∼ ∼ ∼\sim∼): The negation ∼ p ∼ p ∼p\sim p∼p is true if and only if p p ppp is false.
  3. Logical Implication ( → → rarr\rightarrow→): The implication p → q p → q p rarr qp \rightarrow qp→q is true if either p p ppp is false or q q qqq is true (or both).

Analysis

  1. Original Statement: p ∧ ∼ q p ∧ ∼ q p^^∼qp \wedge \sim qp∧∼q is true if and only if p p ppp is true and q q qqq is false.
  2. Negation of the Original Statement: The negation of p ∧ ∼ q p ∧ ∼ q p^^∼qp \wedge \sim qp∧∼q is true if and only if either p p ppp is false or q q qqq is true. This is denoted as ∼ ( p ∧ ∼ q ) ∼ ( p ∧ ∼ q ) ∼(p^^∼q)\sim (p \wedge \sim q)∼(p∧∼q).
  3. Comparing with p → q p → q p rarr qp \rightarrow qp→q: The statement p → q p → q p rarr qp \rightarrow qp→q is true if either p p ppp is false or q q qqq is true. This is exactly the condition for the negation of p ∧ ∼ q p ∧ ∼ q p^^∼qp \wedge \sim qp∧∼q.

Conclusion

The statement "The negation of p ∧ ∼ q p ∧ ∼ q p^^∼qp \wedge \sim qp∧∼q is p → q p → q p rarr qp \rightarrow qp→q" is true. The logical structure of the negation of p ∧ ∼ q p ∧ ∼ q p^^∼qp \wedge \sim qp∧∼q matches exactly with the logical structure of p → q p → q p rarr qp \rightarrow qp→q. Both statements are true in the same scenarios: either when p p ppp is false or when q q qqq is true.
iii) -1 is a limit point of the interval 1 − 2 , 1 ] 1 − 2 , 1 ] 1-2,1]1-2,1]1−2,1].
Answer:
The statement to evaluate is: "-1 is a limit point of the interval ( 1 − 2 , 1 ] ( 1 − 2 , 1 ] (1-2,1](1 – 2, 1](1−2,1]".

Background

  1. Limit Point: A point x x xxx is a limit point of a set S S SSS in a topological space if every neighborhood of x x xxx contains at least one point of S S SSS different from x x xxx itself.
  2. Interval ( 1 − 2 , 1 ] ( 1 − 2 , 1 ] (1-2,1](1 – 2, 1](1−2,1]: The interval ( 1 − 2 , 1 ] ( 1 − 2 , 1 ] (1-2,1](1 – 2, 1](1−2,1] is equivalent to ( − 1 , 1 ] ( − 1 , 1 ] (-1,1](-1, 1](−1,1], which includes all real numbers greater than − 1 − 1 -1-1−1 and less than or equal to 1 1 111.

Analysis

To determine if − 1 − 1 -1-1−1 is a limit point of the interval ( − 1 , 1 ] ( − 1 , 1 ] (-1,1](-1, 1](−1,1], we need to check if every neighborhood of − 1 − 1 -1-1−1 contains at least one point from the interval ( − 1 , 1 ] ( − 1 , 1 ] (-1,1](-1, 1](−1,1] other than − 1 − 1 -1-1−1 itself.
  1. Neighborhood of − 1 − 1 -1-1−1: A neighborhood of − 1 − 1 -1-1−1 is any interval that contains − 1 − 1 -1-1−1. For example, consider the neighborhood ( − 1 − ϵ , − 1 + ϵ ) ( − 1 − ϵ , − 1 + ϵ ) (-1-epsilon,-1+epsilon)(-1 – \epsilon, -1 + \epsilon)(−1−ϵ,−1+ϵ) for any ϵ > 0 ϵ > 0 epsilon > 0\epsilon > 0ϵ>0.
  2. Points in the Interval: The interval ( − 1 , 1 ] ( − 1 , 1 ] (-1,1](-1, 1](−1,1] contains points arbitrarily close to − 1 − 1 -1-1−1 but does not include − 1 − 1 -1-1−1 itself. For any ϵ > 0 ϵ > 0 epsilon > 0\epsilon > 0ϵ>0, there are points in ( − 1 , 1 ] ( − 1 , 1 ] (-1,1](-1, 1](−1,1] that lie within the neighborhood ( − 1 − ϵ , − 1 + ϵ ) ( − 1 − ϵ , − 1 + ϵ ) (-1-epsilon,-1+epsilon)(-1 – \epsilon, -1 + \epsilon)(−1−ϵ,−1+ϵ) of − 1 − 1 -1-1−1.

Conclusion

The statement "-1 is a limit point of the interval ( − 1 , 1 ] ( − 1 , 1 ] (-1,1](-1, 1](−1,1]" is true. Every neighborhood of − 1 − 1 -1-1−1 contains points from the interval ( − 1 , 1 ] ( − 1 , 1 ] (-1,1](-1, 1](−1,1] other than − 1 − 1 -1-1−1 itself, satisfying the definition of a limit point.
iv) The necessary condition for a function f f fff to be integrable is that it is continuous.
Answer:
The statement to evaluate is: "The necessary condition for a function f f fff to be integrable is that it is continuous."

Background

  1. Integrable Function: A function is said to be integrable (in the Riemann sense) if the Riemann integral of the function over a certain interval exists. This means that it is possible to calculate the area under the curve of the function over that interval.
  2. Continuous Function: A function is continuous if, intuitively, you can draw its graph without lifting your pen from the paper. Formally, a function f f fff is continuous at a point c c ccc if the limit of f ( x ) f ( x ) f(x)f(x)f(x) as x x xxx approaches c c ccc is equal to f ( c ) f ( c ) f(c)f(c)f(c).

Analysis

The statement claims that continuity is a necessary condition for integrability. This is not true. While continuous functions on a closed interval are always integrable (a result from the fundamental theorem of calculus), there are many examples of functions that are not continuous everywhere on their domain but are still integrable.

Counterexample

A classic counterexample is the function defined by
f ( x ) = { 1 if x ∈ Q (the set of rational numbers) , 0 if x ∉ Q (the set of irrational numbers) . f ( x ) = 1      if x ∈ Q (the set of rational numbers) , 0      if x ∉ Q (the set of irrational numbers) . f(x)={[1,”if “x inQ” (the set of rational numbers)””,”],[0,”if “x!inQ” (the set of irrational numbers)”.]:}f(x) = \begin{cases} 1 & \text{if } x \in \mathbb{Q} \text{ (the set of rational numbers)}, \\ 0 & \text{if } x \notin \mathbb{Q} \text{ (the set of irrational numbers)}. \end{cases}f(x)={1if x∈Q (the set of rational numbers),0if x∉Q (the set of irrational numbers).
This function, known as the Dirichlet function, is discontinuous at every point in its domain. However, it is Riemann integrable over any closed interval in R R R\mathbb{R}R. The integral of this function over any interval is 0, as the set of points where the function is not zero (the rationals) has measure zero.

Conclusion

The statement "The necessary condition for a function f f fff to be integrable is that it is continuous" is false. While continuous functions are indeed integrable, there are also discontinuous functions that are integrable, as demonstrated by the Dirichlet function.
v) The function f : R → R f : R → R f:RrarrRf: \mathbb{R} \rightarrow \mathbb{R}f:R→R defined by f ( x ) = | x − 2 | + | 3 − x | f ( x ) = | x − 2 | + | 3 − x | f(x)=|x-2|+|3-x|f(x)=|x-2|+|3-x|f(x)=|x−2|+|3−x| is differentiable at x = 5 x = 5 x=5x=5x=5.
Answer:
To determine whether the function f : R → R f : R → R f:RrarrRf: \mathbb{R} \rightarrow \mathbb{R}f:R→R defined by f ( x ) = | x − 2 | + | 3 − x | f ( x ) = | x − 2 | + | 3 − x | f(x)=|x-2|+|3-x|f(x) = |x – 2| + |3 – x|f(x)=|x−2|+|3−x| is differentiable at x = 5 x = 5 x=5x = 5x=5, we need to understand the concept of differentiability and analyze the behavior of the function at that point.

Background

  1. Differentiable Function: A function f f fff is differentiable at a point x = a x = a x=ax = ax=a if the derivative f ′ ( a ) f ′ ( a ) f^(‘)(a)f'(a)f′(a) exists. This means that the limit of the difference quotient f ( x ) − f ( a ) x − a f ( x ) − f ( a ) x − a (f(x)-f(a))/(x-a)\frac{f(x) – f(a)}{x – a}f(x)−f(a)x−a as x x xxx approaches a a aaa exists and is finite.
  2. Absolute Value Function: The absolute value function | x | | x | |x||x||x| is not differentiable at x = 0 x = 0 x=0x = 0x=0 because it has a sharp corner at that point. The derivative from the left does not equal the derivative from the right.

Analysis

The function f ( x ) = | x − 2 | + | 3 − x | f ( x ) = | x − 2 | + | 3 − x | f(x)=|x-2|+|3-x|f(x) = |x – 2| + |3 – x|f(x)=|x−2|+|3−x| involves two absolute value expressions. To analyze differentiability, we can break it down into cases based on the intervals defined by the points where the absolute value expressions change, which are x = 2 x = 2 x=2x = 2x=2 and x = 3 x = 3 x=3x = 3x=3.
  1. For x < 2 x < 2 x < 2x < 2x<2: f ( x ) = − ( x − 2 ) + ( 3 − x ) = − 2 x + 5 f ( x ) = − ( x − 2 ) + ( 3 − x ) = − 2 x + 5 f(x)=-(x-2)+(3-x)=-2x+5f(x) = -(x – 2) + (3 – x) = -2x + 5f(x)=−(x−2)+(3−x)=−2x+5.
  2. For 2 ≤ x < 3 2 ≤ x < 3 2 <= x < 32 \leq x < 32≤x<3: f ( x ) = ( x − 2 ) + ( 3 − x ) = 1 f ( x ) = ( x − 2 ) + ( 3 − x ) = 1 f(x)=(x-2)+(3-x)=1f(x) = (x – 2) + (3 – x) = 1f(x)=(x−2)+(3−x)=1.
  3. For x ≥ 3 x ≥ 3 x >= 3x \geq 3x≥3: f ( x ) = ( x − 2 ) − ( 3 − x ) = 2 x − 5 f ( x ) = ( x − 2 ) − ( 3 − x ) = 2 x − 5 f(x)=(x-2)-(3-x)=2x-5f(x) = (x – 2) – (3 – x) = 2x – 5f(x)=(x−2)−(3−x)=2x−5.
At x = 5 x = 5 x=5x = 5x=5, which falls in the interval x ≥ 3 x ≥ 3 x >= 3x \geq 3x≥3, the function simplifies to f ( x ) = 2 x − 5 f ( x ) = 2 x − 5 f(x)=2x-5f(x) = 2x – 5f(x)=2x−5. This is a linear function, which is differentiable everywhere, including at x = 5 x = 5 x=5x = 5x=5.

Conclusion

The statement "The function f : R → R f : R → R f:RrarrRf: \mathbb{R} \rightarrow \mathbb{R}f:R→R defined by f ( x ) = | x − 2 | + | 3 − x | f ( x ) = | x − 2 | + | 3 − x | f(x)=|x-2|+|3-x|f(x) = |x – 2| + |3 – x|f(x)=|x−2|+|3−x| is differentiable at x = 5 x = 5 x=5x = 5