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.