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Question

Let f:RR be defined as f(x) = mx + p sinx (m0,p>0). Then which statement(s) is/are true?
  1. f(x) is discontinuous at x=nπ
  2. f(x) is continuous for all real values of m and p
  3. f(x) is bijective if m(p,p)
  4. 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(m0,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)<0xR, 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(x0)<0
Hence required range is m(,p]
Case – II
f(x)>0xR, 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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