Intermediate Value Theorem: Exercises — Hinglish deep dive

4.1.8 · D4 · Maths › Calculus I — Limits & Derivatives › Intermediate Value Theorem

Shuru karne se pehle, ek reminder simple shabdon mein. Intermediate Value Theorem (IVT) kehta hai:

Do words jinpar hum lagaataar rely karenge, ek baar ke liye define kar dete hain:

Ek hi move hai jo tum baar baar use karoge: " $f (x) = N$ hota hai kya?" solve karne ke liye, $g (x) = f (x) - N$ banao aur $g$ ka sign change dhundho. Ek continuous function ka sign change ek zero force karta hai — yahi poora game hai.

Level 1 — Recognition

(Kya tum pehchaan sakte ho ki IVT kab apply hota hai aur ingredients read kar sakte ho?)

Exercise 1.1. $f (x) = x^{2}$ ke liye $[0, 3]$ par, IVT ke teen ingredients list karo (continuity, dono endpoint values, aur ek aisi $N$ ki value jo guaranteed hit hogi). Kya $N = 5$ hit hogi? Kya $N = 10$ hit hogi?

Recall Solution 1.1

Continuity: $x^{2}$ ek polynomial hai, toh everywhere continuous — khaas taur par closed interval $[0, 3]$ par. ✓ Endpoint heights: $f (0) = 0$ aur $f (3) = 9$ . Guaranteed values: har woh $N$ jo $min (0, 9) < N < max (0, 9)$ satisfy kare, yaani har $N$ open interval $(0, 9)$ mein. Toh $N = 5$ is $(0, 9)$ mein hai ⇒ hit (IVT se ek $c \in (0, 3)$ hai jahan $c^{2} = 5$ , yaani $c = 5 \approx 2.236$ ). $N = 10$ is not between $0$ aur $9$ (yeh $max = 9$ se upar hai), toh IVT kuch nahi kehta — aur waise bhi $[0, 3]$ par $x^{2} \leq 9$ , toh $10$ hit nahi hoti. ✓

Exercise 1.2. Inme se kaun sa IVT akele se $(a, b)$ mein $f$ ka root guarantee karta hai? Har ek ke liye YES / NO / IVT-SILENT batao. (a) $f$ continuous, $f (a) = - 3$ , $f (b) = 4$ . (b) $f$ continuous, $f (a) = 2$ , $f (b) = 6$ . (c) $f (a) = - 1$ , $f (b) = 1$ , lekin $f$ midpoint par jump karta hai.

Recall Solution 1.2

(a) YES. Sign change ( $- 3 < 0 < 4$ ) aur continuous ⇒ IVT ek root deta hai. (b) IVT-SILENT. Dono positive hain, toh $0$ is not between $2$ aur $6$ . IVT koi guarantee nahi deta (root ho sakta hai, lekin is theorem se nahi). (c) IVT-SILENT. Continuity fail ho rahi hai, toh hypothesis toota hua hai aur IVT yahan kuch bhi apply nahi karta — koi promise nahi karta dono taraf se. (Jump $0$ ko skip kar sakta hai, toh hum root exist karna bhi conclude nahi kar sakte.)

Level 2 — Application

Exercise 2.1. Show karo ki $f (x) = x^{3} - x - 1$ ka $[1, 2]$ mein ek root hai, phir interval ko shrink karne ke liye bisection ka ek step use karo.

Recall Solution 2.1

Continuous: polynomial. ✓ Endpoints: $f (1) = 1 - 1 - 1 = - 1 < 0$ , $f (2) = 8 - 2 - 1 = 5 > 0$ . Yeh ek sign change hai ⇒ IVT se $c \in (1, 2)$ milta hai jahan $f (c) = 0$ . ✓ Bisection step: midpoint $m = 1.5$ . $f (1.5) = 3.375 - 1.5 - 1 = 0.875 > 0$ . Sign change ab $x = 1$ (negative) aur $x = 1.5$ (positive) ke beech hai, toh root $[1, 1.5]$ mein hai. Dekho Bisection Method.

Exercise 2.2. Kya $cos x = x$ ka $[0, 1]$ mein koi solution hai? (Angles radians mein.)

Recall Solution 2.2

$g (x) = cos x - x$ set karo, continuous (cosine aur ek line dono continuous hain). Yeh move kyun? " $cos x = x$ solve karo" asal mein ek disguised root question hai. $g (0) = cos 0 - 0 = 1 > 0$ . $g (1) = cos 1 - 1 \approx 0.5403 - 1 = - 0.4597 < 0$ . Sign change ⇒ IVT se $c \in (0, 1)$ milta hai jahan $g (c) = 0$ , yaani $cos c = c$ . ✓ (Yeh $c \approx 0.739$ famous "Dottie number" hai.)

Exercise 2.3. Show karo ki $f (x) = 2^{x} - 3 x$ $[0, 1]$ par kahin $N = - 1$ value leta hai.

Recall Solution 2.3

$g (x) = f (x) - N = 2^{x} - 3 x - (- 1) = 2^{x} - 3 x + 1$ banao, continuous. $g (0) = 1 - 0 + 1 = 2 > 0$ . $g (1) = 2 - 3 + 1 = 0$ . Right endpoint exactly $0$ hai! Toh $c = 1$ kaam karta hai, lekin $1$ ek endpoint hai. Strictly andar land karne ke liye, ek mid value check karo: $g (0.5) = 2 - 1.5 + 1 = 2 - 0.5 \approx 0.914 > 0$ , abhi bhi positive — yahan sirf boundary par zero hai. Toh $f (1) = 2 - 3 = - 1 = N$ exactly. Value $N = - 1$ attain hoti hai, endpoint par. ✓ Lesson: IVT ko $N$ ki zaroorat hai jo endpoint values ke strictly beech mein ho taaki interior $c$ force ho. Yahan $N = f (1)$ exactly hai, toh yeh value trivially boundary par milti hai, IVT ki interior guarantee se nahi.

Level 3 — Analysis

Exercise 3.1. Real coefficients wala har odd-degree polynomial kam se kam ek real root rakhta hai. IVT use karke prove karo.

Recall Solution 3.1

Maano $p (x) = a_{n} x^{n} + \dots + a_{0}$ jahan $n$ odd hai aur $a_{n} \neq = 0$ . Without loss of generality (WLOG — yaani is case ko chunne se hum kuch nahi kho rahe, kyunki agar $a_{n} < 0$ hota toh hum $- p$ ke baare mein argue karte, jiske same roots hain aur ab positive leading coefficient hai) $a_{n} > 0$ maano. Continuity: polynomials everywhere continuous hote hain. ✓ End behaviour: bade $∣ x ∣$ ke liye leading term $a_{n} x^{n}$ dominate karta hai. Kyunki $n$ odd hai, $x^{n} \to + \infty$ jab $x \to + \infty$ aur $x^{n} \to - \infty$ jab $x \to - \infty$ . Toh ek bada $b$ exist karta hai jahan $p (b) > 0$ aur ek large-negative $a$ jahan $p (a) < 0$ . IVT: $[a, b]$ par, $p$ continuous hai aur $p (a) < 0 < p (b)$ (ek sign change) ⇒ koi $c \in (a, b)$ hai jahan $p (c) = 0$ . $■$ Odd kyun? Even degree ke liye dono ends ek hi taraf jaate hain, toh koi forced sign change nahi hoti (e.g. $x^{2} + 1$ ka koi real root nahi).

Exercise 3.2. Maano $f$ is $[0, 2]$ par continuous, $f (0) = 1$ aur $f (2) = 1$ ke saath. Kya $f$ zaroor value $1.5$ hit karegi? Kya $0.5$ hit karegi? Precisely explain karo IVT kya keh sakta hai aur kya nahi.

Recall Solution 3.2

Yaad karo " $N$ is between $f (0)$ aur $f (2)$ " ka matlab hai $min (f (0), f (2)) < N < max (f (0), f (2))$ . Yahan dono endpoint heights $1$ hain, toh $min = max = 1$ aur condition $1 < N < 1$ impossible hai — guaranteed values ka open interval empty hai. IVT isliye $1$ ke siwa har value ke baare mein silent hai.

$1.5$ : guaranteed nahi. Counterexample: $f (x) = 1$ constant kabhi $1.5$ nahi reach karta.
$0.5$ : yeh bhi guaranteed nahi. Same constant example.
Lekin IVT inhe forbid bhi nahi karta: ek continuous $f$ jo $1$ se shuru aur $1$ par khatam ho, beech mein aasaani se $0.5$ tak dip kar sakti hai ya $1.5$ tak rise kar sakti hai. IVT ek sufficient condition deta hai, kabhi necessary nahi. Equal endpoints ⇒ koi forced intermediate values nahi, lekin doosri values phir bhi occur ho sakti hain.

Level 4 — Synthesis

Exercise 4.1 (Fixed point). Maano $f : [0, 1] \to [0, 1]$ continuous hai. Prove karo ki ek point $c$ hai jahan $f (c) = c$ . Phir ek line mein explain karo ki yeh kyun fail ho jaata hai agar codomain $[0, 2]$ ho jaaye.

Recall Solution 4.1

$g (x) = f (x) - x$ define karo, continuous (continuous functions ka difference). Kyun? "Fixed point $f (x) = x$ " ko "root $g (x) = 0$ " mein convert karo. Dekho Fixed Point Theorems.

$g (0) = f (0) - 0 = f (0) \geq 0$ (outputs $[0, 1]$ mein land karte hain, toh $f (0) \geq 0$ ).
$g (1) = f (1) - 1 \leq 0$ (kyunki $f (1) \leq 1$ ). Cases: agar $g (0) = 0$ toh $c = 0$ ; agar $g (1) = 0$ toh $c = 1$ . Otherwise $g (0) > 0 > g (1)$ : ek sign change, IVT se $c \in (0, 1)$ milta hai jahan $g (c) = 0$ , yaani $f (c) = c$ . $■$ Codomain $[0, 2]$ : ab $f (1) \leq 2$ , toh $g (1) = f (1) - 1$ positive ho sakta hai (e.g. $f \equiv 2$ deta hai $g (0) = 2 > 0$ , $g (1) = 1 > 0$ , koi sign change nahi) — koi fixed point guaranteed nahi. Argument ko zaroorat thi ki outputs usi interval ke andar rahein jisme inputs hain.

Exercise 4.2 (Antipodal temperatures). Equator par, temperature $T (θ)$ angle ka ek continuous function hai, jahan $T (θ + 36 0^{\circ}) = T (θ)$ . Prove karo ki do antipodal points (angles $θ$ aur $θ + 18 0^{\circ}$ ) same temperature share karte hain.

Recall Solution 4.2

$h (θ) = T (θ) - T (θ + 18 0^{\circ})$ define karo, continuous (continuous ka difference). Kyun? "Equal temperatures" ban jaata hai " $h = 0$ ". $θ = 0$ aur $θ = 18 0^{\circ}$ par evaluate karo: $h (0) = T (0) - T (18 0^{\circ}), h (18 0^{\circ}) = T (18 0^{\circ}) - T (36 0^{\circ}) = T (18 0^{\circ}) - T (0) = - h (0) .$ Toh $h (18 0^{\circ})$ exactly $h (0)$ ka negative hai.

Agar $h (0) = 0$ : done, $θ = 0$ kaam karta hai.
Warna $h (0)$ aur $h (18 0^{\circ})$ ke opposite signs hain (ek sign change) ⇒ $[0^{\circ}, 18 0^{\circ}]$ par IVT se $θ^{*}$ milta hai jahan $h (θ^{*}) = 0$ , yaani $T (θ^{*}) = T (θ^{*} + 18 0^{\circ})$ . $■$

Kya dekhna hai: red curve

h

hai. Left par violet dot (

h (0) = + 1.40

) se right par orange dot (

h (18 0^{\circ}) = - 1.40

) tak follow karo — zero ke across mirror image. Kyunki

h

continuous hai aur navy zero-line se upar shuru hoti hai aur neeche khatam hoti hai, toh woh zaroor us line ko navy crossing point par pierce karegi. Woh crossing exactly woh angle hai jahan antipodal temperatures match karte hain.

Level 5 — Mastery

Exercise 5.1 (Monotonicity se sharpen karna). Show karo ki $f (x) = x^{3} - x - 1$ ka exactly ek real root hai. (IVT existence deta hai; uniqueness tum supply karo.)

Recall Solution 5.1

Existence: 2.1 se, $(1, 2)$ mein ek root exist karta hai. Uniqueness via the derivative. Derivative hai $f^{'} (x) = 3 x^{2} - 1$ . $r = \frac{1}{3} \approx 0.577$ set karo; tab $f^{'} (x) > 0$ exactly jab $x^{2} > \frac{1}{3}$ , yaani $(- \infty, - r)$ par aur $(r, \infty)$ par, aur $f^{'} (x) < 0$ middle band $(- r, r)$ par. Toh $f$ strictly increasing hai $(- \infty, - r)$ par, strictly decreasing hai $(- r, r)$ par, aur strictly increasing hai $(r, \infty)$ par.

Ek strictly monotonic piece kisi bhi horizontal level ko at most once cross kar sakta hai, toh $f$ ke teen pieces mein se har ek par at most ek root hai — total at most teen. Ab hum do ko eliminate karte hain.
Local max at $x = - r$ : $f (- r) = (- \frac{1}{3})^{3} + \frac{1}{3} - 1 = \frac{2}{3 3} - 1 \approx 0.385 - 1 = - 0.615 < 0$ . Kyunki $- r$ left-and-middle hump ka sabse uuncha point hai, saara $(- \infty, r]$ is local max value $< 0$ par ya neeche hai. Wahan strictly negative rehte hue, $f$ ka koi root nahi $(- \infty, r]$ par.
Toh sirf last strictly-increasing piece $(r, \infty)$ bachta hai, jahan $f$ $f (r) < 0$ se upar $0$ se ek baar guzarta hai. Strict monotonicity doosra crossing forbid karti hai. Isliye exactly ek real root. $■$ Moral: IVT (existence) + $f^{'}$ se monotonicity (koi doosra crossing nahi) = "exactly one". IVT akele kabhi "ek" nahi keh sakta. Dekho Rolle's Theorem.

Exercise 5.2 (Ek continuous function apna average cross karta hai). Maano $f$ is $[0, 2]$ par continuous, $f (0) = 4$ , $f (1) = 0$ , $f (2) = 4$ ke saath. Show karo ki $f$ value $2$ kam se kam do baar attain karta hai.

Recall Solution 5.2

Value $N = 2$ dekho.

$[0, 1]$ par: $f (0) = 4$ , $f (1) = 0$ , aur $2$ satisfy karta hai $min (0, 4) < 2 < max (0, 4)$ , yaani $0 < 2 < 4$ . Continuous ⇒ IVT se $c_{1} \in (0, 1)$ milta hai jahan $f (c_{1}) = 2$ .
$[1, 2]$ par: $f (1) = 0$ , $f (2) = 4$ , phir se $0 < 2 < 4$ . IVT se $c_{2} \in (1, 2)$ milta hai jahan $f (c_{2}) = 2$ . Kyunki $c_{1} < 1 < c_{2}$ , woh distinct hain. Toh $f = 2$ kam se kam do baar. $■$

Kya dekhna hai: violet curve teen navy dots

(0, 4), (1, 0), (2, 4)

se guzarti hai. Dashed orange line target height

y = 2

hai. Left half par curve

4

0

tak drop karta hai aur dashed line ko ek baar zaroor cross karta hai (magenta dot

c_{1}

); right half par woh

0

4

tak wapas climb karta hai aur phir cross karta hai (magenta dot

c_{2}

). Do crossings, har half par IVT se ek guaranteed.

Exercise 5.3 (IVT kahan toot-ta hai?). Explain karo ki " $g (x) = 1/ x$ $x = - 1$ par $- 1$ se $x = 1$ par $+ 1$ tak jaata hai phir bhi kabhi $0$ nahi hota" IVT ko contradict kyun nahi karta. Phir statement repair karo: kin subinterval(s) par IVT $1/ x$ par sahi tarike se apply hota hai?

Recall Solution 5.3

Koi contradiction nahi: IVT pure closed interval par continuity demand karta hai. $g (x) = 1/ x$ defined nahi hai, isliye continuous bhi nahi, $x = 0 \in [- 1, 1]$ par. Hypothesis fail hoti hai, toh IVT koi promise nahi karta — aur wakai $g$ $- \infty$ se $+ \infty$ tak jump karta hai, $0$ ko skip karte hue. Repair (blow-up ke dono sides par): kisi bhi closed interval par jo $0$ avoid kare, $g$ continuous hai, toh IVT wahan apply hota hai. Discontinuity ke dono sides par do natural families hain.

Positive side, e.g. $[1, 2]$ : $g$ yahan continuous hai (andar koi $0$ nahi), $g (1) = 1$ aur $g (2) = 0.5$ . IVT guarantee karta hai ki open interval $(0.5, 1)$ mein har value hit hoti hai — for instance $g (4/3) = 0.75$ usi range mein land karta hai. ✓
Negative side, e.g. $[- 1, - 0.5]$ : yahan bhi $g$ continuous hai, $g (- 1) = - 1$ aur $g (- 0.5) = - 2$ . IVT guarantee karta hai ki open interval $(- 2, - 1)$ mein har value hit hoti hai — for instance $g (- 4/3) = - 0.75$ usi range mein land karta hai. ✓ Koi bhi interval $0$ cross nahi karta, toh $g$ ko kabhi $0$ attain karne ki zaroorat nahi; koi contradiction nahi. Generally, IVT $1/ x$ par apply hota hai kisi bhi closed interval par jo puri tarah $(0, \infty)$ ke andar ho ya puri tarah $(- \infty, 0)$ ke andar — bas kabhi $0$ stride karne wale interval par nahi. IVT kabhi galat nahi tha — use simply $[- 1, 1]$ par invite hi nahi kiya gaya kyunki uski continuity ticket wahan invalid thi.

Active Recall

Recall One-line answers

IVT existence deta hai ya uniqueness? ::: Sirf existence — uniqueness ke liye monotonicity/Rolle ke saath pair karo. $f (x) = N$ solve karne ke liye tum kaun sa function banate ho? ::: $g (x) = f (x) - N$ , phir ek sign change dhundho. $[a, b]$ par $g$ ka "sign change" define karo. ::: $g (a)$ aur $g (b)$ ke opposite signs hain, yaani $g (a) \cdot g (b) < 0$ . " $N$ is between $p$ and $q$ " define karo. ::: $min (p, q) < N < max (p, q)$ (empty jab $p = q$ ). Odd-degree polynomials ka real root kyun hona zaroori hai? ::: Opposite end behaviours ek sign change deti hain; IVT ek crossing force karta hai. Equal endpoint values — IVT kya guarantee karta hai? ::: Uss value se aage kuch nahi; woh silent hai (lekin doosri values forbid nahi karta). $1/ x$ on $[- 1, 1]$ counterexample kyun nahi hai? ::: Yeh $0$ par discontinuous hai, toh hypothesis fail hoti hai aur IVT kabhi apply hi nahi hota.

Connections

Intermediate Value Theorem — woh parent theorem jise yeh exercises drill karte hain.
Continuity — har problem mein pehle check hone wala entry ticket.
Bisection Method — Exercise 2.1 ka shrink step.
Bolzano's Theorem — Move 1 ke peeche $N = 0$ engine.
Fixed Point Theorems — Exercise 4.1.
Rolle's Theorem / Mean Value Theorem — 5.1 mein uniqueness partners.