WorksheetsQuiz: Bisection Method
Total questions: 50
Worksheet time: 17mins
A slope-stability model gives the factor-of-safety F(x) where x is a pore-pressure multiplier. You know F(0)=1.6 and F(1)=-0.2. Which statement about using the bisection method to find the root F(x)=0 is correct?
Root not guaranteed because F is not differentiable.
A root is guaranteed in (0,1) because F changes sign and is continuous.
The method will fail because F has different units at endpoints.
Bisection works only if F(x) is nonzero on (0,1).
For a beam design equation g(c)=0 you determine g(2)=-0.8 and g(5)=0.2 . Using bisection, what is the maximum possible absolute error in the root after 3 iterations? (Use error bound ≤(b–a)/2³ )
You apply bisection on f(x)=x³–x–2 with initial bracket [1,2]. Which of the following is true?
No sign change so bisection cannot start.
There is a sign change; root lies in [1,2] and bisection converges linearly.
Function has multiple roots in the bracket so bisection fails.
Bisection converges quadratically for this cubic.
In practice, why is choosing a narrow initial bracket that still contains the root often preferable?
It reduces the number of iterations required to meet a given interval-length tolerance.
It increases chance of sign-change failure.
It increases round-off error dramatically.
It guarantees quadratic convergence.
For the same interval [a,b], which stopping criterion typically requires fewer iterations to satisfy?
|f(c)|≤ε always requires fewer iterations.
|b–a|/2≤ε always requires fewer iterations.
It depends on the derivative f′ near the root; either may require fewer.
Both are equivalent in iteration count always.
A drainage design leads to equation h+sin(h)–1.5=0 . You bracket a root in [0,2]. If you require an interval-radius tolerance of 10⁻³, how many bisection iterations are guaranteed? (Use n≥log₂((b–a)/tol))
In solving Manning’s flow formula rearranged to root-find a depth y, you observe that the function is nearly flat near the root (|f′| small). Which statement about bisection is correct?
A. Bisection is unaffected by small derivative and converges at the same rate.
B. Bisection may require many iterations because interval halving is independent of derivative, but function-value criterion may be slow.
C. Bisection fails if derivative is zero at root.
D. Bisection becomes quadratic if derivative small.
You use bisection for a pipe friction equation with endpoints giving f(a)=1e-6 and f(b)=-1e-6. What practical issue might you face?
No convergence because f is too small.
Floating-point rounding may make sign detection unreliable; need robust sign test.
Bisection will converge in one step.
This guarantees multiple roots inside.
For the interval [0,1], after 5 bisection iterations the interval is [0.5,0.625]. What is the exact maximum absolute error bound at this stage?
A soil consolidation equation yields a function with a simple root. Which property ensures bisection will converge to that root when bracketed?
The function is differentiable at the root.
The function is continuous on the bracket and sign change occurs at endpoints.
The function is monotonic over the real line.
The function is linear near the root.
Which of the following is a correct expression for the error after nnn bisection iterations starting from [a,b]?
|r - c_n| ≤ (b−a)/2n−1
|r - c_n| ≤ (b−a)/2n
|r - c_n| ≤ (b−a)/2n+1
|r - c_n| ≤ (b - a)/n
In a foundation bearing-capacity formula you bracket the root but find that f(a) and f(b) are both extremely close to zero and opposite sign due to measurement noise. Best practice?
Proceed normally; noise is irrelevant.
Use enhanced precision and possibly widen the bracket or verify continuity; consider method robustness to noise.
Switch to Newton immediately.
Declare root at midpoint.
For a bracket [a,b]=[1,2] and tolerance δ= 10−3 , calculate the minimum iterations nnn needed: n≥ . Compute n.
A function describing the intersection of load and capacity curves has two distinct roots in [0,10]. If you use bisection on [0,10] you may:
Always find both roots in one run.
Find at most one root because bisection requires an initial sign change per root; you must bracket each root separately.
Fail because multiple roots invalidate the Intermediate Value Theorem.
Converge to the average of the two roots.
You attempt bisection and observe the midpoint yields f(mid)=0 exactly (within machine precision). The best next step is:
Stop; root found.
Continue iterations to reduce interval size.
Declare failure; midpoint unlikely to be root.
Switch to secant to refine.
For bisection applied to equation f(x)=Ae−kx−B used in temperature-driven curing models, what convergence rate should you expect?
Quadratic (fast)
Linear (one-half error reduction per iteration)
Cubic
Superlinear but not quadratic
Suppose f(x) has a multiple root of multiplicity 2 inside a bracket but does not change sign across the root. What happens to bisection?
It still finds the root because multiplicity doesn’t matter.
It cannot detect the root via sign change, so plain bisection will fail unless multiple roots cause sign change.
It converges faster.
It produces complex roots.
For a bracketing [3,7], you wish interval half-length ≤ 0.01. How many iterations are required? (Compute: n ≥ ceil(log_2((7-3)/0.01)))
In a cantilever deflection equation, f(x) is continuous but noisy experimental data is used to evaluate f. Which modification improves robustness of bisection?
Use raw data only — noise helps find the root.
Pre-smooth the data or fit an interpolant (polynomial/spline) before applying bisection.
Use larger tolerance only.
Use bisection only on discrete points without interpolation.
Which of the following guarantees that bisection will reduce the interval length by exactly half each iteration?
The function is monotonic.
The function is continuous and sign change occurs at endpoints.
The function is differentiable.
The function is linear.
A groundwater head equation has root near the left endpoint; initial bracket [0,10]. Which strategy lowers iterations?
Keep [0,10] — bisection is independent of root location.
Narrow bracket by searching outward from the endpoint to find a tighter bracket near root before starting bisection.
Use midpoint as initial guess for Newton.
Increase tolerance to avoid many iterations.
Which of these functions can break a simple sign-change bracketing test although a root exists inside?
f(x)=tan(x) on interval crossing a vertical asymptote.
f(x)=x2–4 on [0,3].
f(x)=sin(x) on [0,π].
f(x)=x–1 on [0,2].
In using bisection to solve sin(θ)=0.5=0 for θ in radians near 0.6, you pick bracket [0,π/2]. Which is true?
f(0)=0.5, f(π/2)=0.5, so a root guaranteed.
f(0)=0.5, so no root.
f(π/2) negative so no root.
Sine not continuous so fail.
The theoretical convergence of bisection is linear with factor 1/2. Which practical factor slows convergence further?
A. Round-off and finite precision arithmetic.
B. Increasing bracket width.
C. Using exact arithmetic.
D. Large derivative at root.
In a beam buckling load equation you perform bisection and after many iterations the f(mid) value stops changing in floating-point while midpoint continues to change slightly. This indicates:
Catastrophic cancellation.
Function evaluations are underflowing/insensitive to small changes; you should use interval-length stopping criterion rather than f-value.
Midpoint arithmetic error only; keep iterating.
You found root—stop.
You are solving nonlinear settlement time t from s(t)=5 using bisection on [0,1000]. The required maximum error in t is 0.1 day. How many iterations needed? (Compute n ≥ ceil(log2(1000/0.1))).
For a function f(x)=x5–x–1 on [1,2], which is true about bisection vs Newton for root finding?
Bisection requires fewer function evaluations typically.
Newton may converge faster but needs derivative and a good initial guess; bisection is safer and guaranteed with sign change.
Newton always fails here.
Bisection is quadratic so always faster.
In bisection you find the midpoint c and evaluate f(c)=0.0002 but your tolerance for function value is 1e-6. What is the best immediate action?
Stop because 0.0002 is small enough.
Continue iterations until interval-length criterion is satisfied or f(c) < 1e-6.
Switch to secant method immediately.
Double the bracket.
For bracket [0,5] and tolerance 1e-4, compute minimum iterations: n ≥ ceil(log2(5/1e-4)). Which is correct?
Which of the following scenarios is bisection not appropriate for?
Continuous function with sign change.
Function with discontinuity (vertical asymptote) in bracket.
No sign change but derivative zero at true root (even multiplicity).
Both B and C.
A bisection implementation for a civil problem uses stopping tolerance tol on interval half-width and returns c. The user later needs higher accuracy. Which deterministic next step is best?
Restart bisection with halved tol.
Use c as initial guess for Newton or secant to get fast local convergence (if derivative available/behaves well).
Suppose you have bracketed a root but function evaluation is expensive (finite-element solve). Which bisection variant helps reduce function calls?
Use standard bisection; it minimizes function calls.
Use delayed acceptance: reuse previous function values and avoid redundant evaluation when sign patterns inferred.
Use Newton to reduce function calls without derivatives.
Use Simpson rule instead.
In a pipe-network example, bracket [a,b] is such that f(a) and f(b) are finite but function has a sharp discontinuity inside due to piecewise definition. What happens?
Bisection may converge to a point but root may be spurious — you must ensure continuity first.
Bisection still finds a root.
Bisection avoids discontinuities automatically.
Bisection returns multiple roots.
In structural equilibrium solving, if you encounter oscillatory evaluation due to solver tolerance (f(mid) flips sign randomly near zero), what mitigation is best?
Increase function solver tolerance to make f(mid) stable.
Use interval-length criterion rather than f-value; smooth or refine the underlying solver.
Abandon numerical methods.
Use smaller brackets only.
Compute the guaranteed maximum error after 10 bisection iterations starting from [0,1]. (Error bound = (b−a)/210 ).
10241≈9.7656×10−4
5121≈1.9531×10−3
20481≈4.8828×10−4
2561≈3.9063×10−3
For a function with steep slope crossing the axis, how does bisection compare to Newton in terms of robustness and efficiency?
Bisection is less robust and always slower.
Newton is always better; bisection unnecessary.
Bisection is more robust (guaranteed) but may require more iterations; Newton can be faster but may diverge if initial guess poor.
Both have identical properties.
You are solving for root in [a,b] and find after many iterations that b–ab–a has stabilized due to floating point; midpoint repeats. Best practice:
Return current midpoint as best estimate and document floating-point limit.
Keep iterating forever.
Switch to bisection in double-double arithmetic.
Use symbolic algebra.
In a rainfall-runoff model solved by bisection, how would you bracket a root that is known to be near zero but function values are extremely small?
Use symmetric small bracket like [–ε,ε] with ε tiny.
Use physical insight to choose realistic bracket (e.g., [0,upper bound]) and avoid too-small brackets that amplify noise; consider scaling.
Use random bracket.
Use secant only.
In implementing bisection in code, which safeguard helps when function evaluations occasionally return NaN for some arguments?
Ignore NaN and proceed.
Treat NaN as sign-change trigger.
Detect NaN, shrink bracket, or choose alternate bracket and ensure continuity before proceeding.
Replace NaN with zero.
You need the root to 6 significant digits and your bracket length is 0.5. How many iterations at minimum? (Compute: n ≥ ceil(log2(0.5 / 0.5×10^{-6}))) — note: for 6 sig figs, you need interval radius ≤ 0.5×10^{-6} times magnitude? Choose closest reasonable option.
Which statement about the bisection midpoint computation c=(a+b)/2 in finite precision is true?
It can overflow for large a and b — better to use c=a+(b–a)/2 .
It is always exact in floating-point.
Using c=(a+b)/2 avoids any round-off.
Midpoint formula never overflows.
In a braced frame equilibrium, if the root is extreme (near endpoint), and you initially bracket [0,1000], bisection will:
A common variant of bisection is to choose the next subinterval by keeping the point with the largest |f| (farthest from zero) to possibly improve convergence. Why is this questionable?
It violates sign-change guarantee.
It may bias the search and lose theoretical error-halving guarantee; classical bisection is simpler and safer.
It always accelerates convergence so is never questionable.
It reduces round-off.
If you apply bisection to a transcendental equation from wave propagation, and the root is irrational, what will the algorithm produce?
The exact irrational root.
A rational approximation within specified tolerance.
Failure due to irrationality.
A symbolic expression.
For bracket [2,3], the first midpoint is 2.5. If f(2)=-0.4, f(2.5)=0.05, f(3)=0.3, which subinterval will bisection choose next?
A. [2,2.5] because signs are opposite there.
B. [2.5,3] because signs are opposite there.
C. [2,3] because both ends same sign.
D. Stop — root found exactly.
Which is true regarding complexity: to reduce interval width from 1 to 1e-6 using bisection you need about how many iterations? (Compute log2(1e6) ≈ 19.93)
~20 iterations
~10 iterations
~60 iterations
~100 iterations
In using bisection on f(x)= e−x−x on [0,1], why is bisection preferred initially over Newton?
Because the function is discontinuous.
Because bisection guarantees bracketed convergence without derivative and provides safe initial guess for Newton.
Newton is illegal here.
Bisection converges faster than Newton here.
For a practical civil-engineering routine where evaluations take minutes, which workflow minimizes wall-clock time while ensuring reliability?
Pure bisection to required tolerance.
Do a few bisection iterations to bracket and get a safe starting interval, then switch to Newton or secant to accelerate convergence.
Use only secant from random guess.
Use binary search of parameters without evaluating function.
Suppose you bracket root with [a,b] and compute c. You detect f(c)=NaN due to internal overflow in f. Which is a safe next step?
Abort and widen the bracket searching for safe evaluation points.
Treat NaN as sign of root and stop.
Replace NaN with zero and continue.
Swap a and b and continue.
A final practical question: you are asked to code a robust bisection solver for many civil problems. Which feature should not be included?
Input validation and continuity/sign-change check.
Maximum iteration cap and reporting of convergence metrics.
Automatic switching to Newton with optional derivative estimates when safe.
Blind application to any bracket without sign checking.
