Question

Let $f:R\to R$ be defined as f(x) = mx + p sinx $(m\ne 0,p>0)$. Then which statement(s) is/are true?

• A. f(x) is discontinuous at $x=n\pi$
• B. f(x) is continuous for all real values of m and p
• C. f(x) is bijective if $m\in (-p,p)$
• D. f(x) is bijective if $m\in (-\infty ,-p]\cup [p,\infty )$

Answer 1

Answer: (B), (D)

The correct options are
B f(x) is continuous for all real values of m and p
D f(x) is bijective if $m\in (-\infty ,-p]\cup [p,\infty )$

$f\left(x\right)=mx+psinx(m\ne 0,p>0)$ is a sum of two continuous functions $y_1$ = mx and $y_2=p$ sinx. Hence f(x) is continuous $\forall \{m,p\}\in R.f\left(x\right)$ will be bijective if it is monotonic (either strictly increasing or strictly decreasing)
f'(x) = m + pcosx. Here we have 2 cases
Case – I
$f^{\prime }\left(x\right)<0\forall x\in R$, then even the maximum value of f'(x) should be negative
$\Rightarrow m+p<0$ (given p > 0) $\Rightarrow m<-p$
Check at critical points $\to$ when m = -p,
$f^{\prime }\left(x\right)=-p+pcosx=0at{x}_0=2n\pi$
But we observe that x0 is not a turning point, instead it is a horizontal inflexion point as $f^{\prime }\left(x_0^{+}\right)<0,f^{\prime }\left(x_0^{-}\right)<0$
Hence required range is $m\in (-\infty ,-p]$
Case – II
$f^{\prime }\left(x\right)>0\forall x\in R,$ then even the minimum value of f¢(x) should be positive.
$\Rightarrow$ 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 $m\in (-\infty ,-p]\cup [p,\infty )$