Consider a function $f:\left[0, \frac{\pi}{2}\right] \rightarrow \mathbf{R}$ given by $f(x)=\sin x$ and $g:\left[0, \frac{\pi}{2}\right] \rightarrow \mathbf{R}$ given by $g(x)=\cos x$. Show that $f$ and $g$ are one-one, but $f+g$ is not one-one.

Step 1: Show f(x) is one-one

To show that a function is one-one, we assume that $f(x_1) = f(x_2)$ for two values $x_1$ and $x_2$ in the domain. If this assumption implies that $x_1 = x_2$, then the function is one-one. For $f(x) = \sin x$ on the interval $\left[0, \frac{\pi}{2}\right]$, if $\sin x_1 = \sin x_2$, then $x_1 = x_2$ because the sine function is strictly increasing in this interval.

Step 2: Show g(x) is one-one

Similarly, for $g(x) = \cos x$ on the interval $\left[0, \frac{\pi}{2}\right]$, if $\cos x_1 = \cos x_2$, then $x_1 = x_2$. This is because the cosine function is strictly decreasing in this interval, meaning each output value corresponds to a unique input value.

Step 3: Define (f+g)(x)

The sum of the two functions, $(f+g)(x)$, is defined as the sum of their individual expressions. So, $(f+g)(x) = \sin x + \cos x$. We need to check if this new function is one-one on the given domain.

Step 4: Test if (f+g)(x) is one-one

To show that $(f+g)(x)$ is not one-one, we need to find two distinct values $x_1$ and $x_2$ in the domain $\left[0, \frac{\pi}{2}\right]$ such that $(f+g)(x_1) = (f+g)(x_2)$ but $x_1 \neq x_2$. Let's evaluate the function at the endpoints of the domain. We find that $(f+g)(0) = 1$ and $(f+g)\left(\frac{\pi}{2}\right) = 1$. Since $0 \neq \frac{\pi}{2}$ but their function values are equal, $(f+g)(x)$ is not one-one.

More from this chapter

• Show that an onto function f:1,2,3 →1,2,3 is always one-one.
• Let A=-2,-1,0,1,2,3. Define a relation R on A by xRy if and only if y=max(x,1). If l is the number of elements in R, and m and n are the minimum numbers of orde
• 32. Show that f: R -> R defined as f(x) = x / sqrt(1 + x²) is one-one but not onto. (function, one-one, onto, real numbers, domain, codomain, range)
• Let A be the set of all students of a boys school. Show that the relation R in A given by R=(a, b): a is sister of b is the empty relation and R^()=(a, b): the
• Consider the identity function I_N: N → N defined as I_N(x)=x x N. Show that although I_N is onto but I_N+I_N: N → N defined as (I_N+I_N)(x)=I_N(x)+I_N(x)=x+x=2
• Prove that the relation R in the set A = 1, 2, 3, 4, 5 given by R = (a, b): |a - b| is even is an equivalence relation.
• Let f(x) = log_e x and g(x) = (x^4 - 2x³ + 3x² - 2x + 2)/(2x² - 2x + 1). Then the domain of f ° g is
• Let the domain of the function f(x) = cos^(-1) ( (4x + 5)/(3x - 7)) be [α, β] and the domain of g(x) = log_2(2 - 6log_7(2x + 5)) be [, ]. Then |7(α + β) + 4( +)

All Relations and Functions solutions →