Question

# Let f:R→R be defined as f(x) = mx + p sinx (m≠0,p>0). Then which statement(s) is/are true?f(x) is discontinuous at x=nπf(x) is continuous for all real values of m and pf(x) is bijective if m∈(−p,p)f(x) is bijective if m∈(−∞,−p]∪[p,∞)

Solution

## The correct options are B f(x) is continuous for all real values of m and p D f(x) is bijective if m∈(−∞,−p]∪[p,∞)f(x)=mx+psinx(m≠0,p>0) is a sum of two continuous functions y1 = mx and y2=p sinx. Hence f(x) is continuous  ∀{m,p}∈R.f(x) will be bijective if it is monotonic (either strictly increasing or strictly decreasing) f'(x) = m + pcosx. Here we have 2 cases Case – I f′(x)<0∀x∈R, then even the maximum value of f'(x) should be negative ⇒m+p<0 (given p > 0) ⇒m<−p Check at critical points →  when m = -p, f′(x)=−p+pcosx=0atx0=2nπ But we observe that x0 is not a turning point, instead it is a horizontal inflexion point as f′(x+0)<0,f′(x−0)<0 Hence required range is m∈(−∞,−p] Case – II f′(x)>0∀x∈R, then even the minimum value of f¢(x) should be positive. ⇒ m - p > 0 i.e. m > p. Similar to above deduction, f(x) will also be bijective if m = p as it is an inflexion point. Hence combining we get m∈(−∞,−p]∪[p,∞)

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