Questions · Page 2 of 2

(Each question 1 marks)

Question 511 Mark
Let f(x) = $\sqrt x$ and g(x) = x be two functions defined over the set of non-negative real numbers. Find (f + g) (x), (f – g) (x), (fg) (x) and $\begin{equation} \left(\frac{f}{g}\right)(x) \end{equation}$.
Answer
We have,
$​​​​​​​(f + g) (x)=f(x)+g(x) = $$\sqrt x$ + x,
$(f – g) (x) =f(x)-g(x)=  \sqrt x– x,$
$(fg) x = f(x).g(x)=$$ \sqrt{x}(x)=x^{\frac{3}{2}} $
$\left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)}=\frac{\sqrt{x}}{x}=x^{-\frac{1}{2}}, x \neq 0 $
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Question 521 Mark
Let $f(x) = x^2$ and $g(x) = 2x + 1$ be two real functions. Find (f + g) (x), (f –g) (x), (fg) (x), $\begin{equation} \left(\frac{f}{g}\right)(x) \end{equation}$
Answer
We have,
$(f + g) (x) =f(x)+g(x)= x^2 + 2x + 1,$
$(f – g) (x)=f(x)-g(x) = x^2 – 2x – 1,$
$(fg) (x)=f(x)g(x) = x^2 (2x + 1) = 2x^3 + x^2,$
$\begin{equation} \left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)}=\frac{x^{2}}{2 x+1}, x \neq-\frac{1}{2} \end{equation}$
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Question 531 Mark
Tell whether the given relat is a function or not? Justify : R = {(1, 2),(2, 3),(3, 4), (4, 5), (5, 6), (6, 7)}
Answer
Since every element has one and only one image, therefore this relation is a function.
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Question 541 Mark
Tell whether the given relation is a function or not? Justify : R = {(2, 2),(2, 4),(3, 3), (4, 4)}
Answer
Since the same first element 2 corresponds to two different images 2 and 4, therefore, this relation is not a function.
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Question 551 Mark
Tell whether the given state is a function or not? Justify your answer.
R = {(2,1),(3,1), (4,2)}
Answer
R = {(2,1),(3,1), (4,2)}
Here we have, Since 2, 3, 4 are the elements of domain of R having their unique images, this relation R is a function.
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Question 561 Mark
If (x + 1, y – 2) = (3, 1), find the values of x and y
Answer
Since the ordered pairs are equal, the corresponding elements are equal.Then,we have,
x + 1 = 3 and y – 2 = 1
Solving we get x = 2 and y = 3
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(Each question 1 marks) - Page 2 - MATHS STD 11 Science Questions - Vidyadip