Introduction to Relations

Correct answer: A. (R-{-4})

Step 1: The denominator of the function is (x+4). Step 2: The denominator cannot be zero, so (x+4\ne0). Step 3: Removing (x=-4), all other real numbers remain in the domain.

Correct answer: A. ([2,\infty))

Step 1: Write (x^2-6x+11=(x-3)^2+2). Step 2: Since ((x-3)^2\ge0), (f(x)\ge2). Step 3: The minimum value is (2), so the range is ([2,\infty)).

Correct answer: A. (12)

Step 1: ((g\circ f)(1)=g(f(1))). Step 2: (f(1)=2\cdot1+1=3). Step 3: (g(3)=3^2+3=12), so the value is (12).

Correct answer: A. (5x-19)

Step 1: ((f\circ g)(x)=f(g(x))). Step 2: Put (g(x)=x-4) into (f). Step 3: (f(x-4)=5(x-4)+1=5x-19).

Correct answer: A. They are generally not equal

Step 1: ((f\circ g)(x)=f(x+2)=(x+2)^2-1). Step 2: ((g\circ f)(x)=g(x^2-1)=x^2+1). Step 3: The expressions are generally different, so order matters in composition.

Correct answer: A. It is one-one and onto

Step 1: (9x-4) gives different outputs for different (x), so it is one-one. Step 2: For any (y\in R), choose (x=\frac{y+4}{9}). Step 3: Hence every real (y) is an image, so the function is onto.

Correct answer: A. It is neither one-one nor onto

Step 1: (f(0)=2) and (f(-2)=2), so the function is not one-one. Step 2: (x^2+2x+2=(x+1)^2+1), so the range is ([1,\infty)). Step 3: With codomain (R), (0) is not an image, so the function is not onto either.

Correct answer: A. One-one and onto

Step 1: For different (x), (x+1) is different, so the function is one-one. Step 2: For every (y\ge0), (x=y-1), which lies in the domain ([-1,\infty)). Step 3: Hence the function is onto as well.

Correct answer: A. It is onto but not one-one

Step 1: For every (y\ge2), (x=\sqrt{y-2}) or (x=-\sqrt{y-2}) can be chosen, so it is onto. Step 2: (f(1)=f(-1)=3), so it is not one-one. Step 3: Reading the codomain correctly is very important while checking onto.

Correct answer: A. ([4,\infty))

Step 1: The expression inside the square root, (3x-12), must be non-negative. Step 2: From (3x-12\ge0), we get (x\ge4). Step 3: At (x=4), the value is (0), so (4) is included.

Correct answer: A. ((-\infty,5))

Step 1: For the square root, (5-x\ge0) is needed. Step 2: But this square root is in the denominator, so it cannot be zero. Step 3: Thus (5-x>0), meaning (x<5), so the domain is ((-\infty,5)).

Correct answer: A. (R-{-2,2})

Step 1: The denominator is (x^2-4=(x-2)(x+2)). Step 2: The denominator cannot be zero, so (x\ne2) and (x\ne-2). Step 3: Even if (x+2) cancels, (x=-2) is not valid in the original function.

Correct answer: A. ({2,8})

Step 1: To find the preimage, solve (|x-5|=3). Step 2: This gives (x-5=3) or (x-5=-3). Step 3: Hence (x=8) or (x=2), so the preimage is ({2,8}).

Correct answer: A. One-one but not onto

Step 1: The images of (1,2,3) are distinct, so the function is one-one. Step 2: (z) is in the codomain but is not an image. Step 3: Therefore the function is not onto.

Correct answer: A. (30)

Step 1: Total functions are (2^5=32). Step 2: There are two non-onto functions, where all elements go only to (a) or only to (b). Step 3: Therefore onto functions are (32-2=30).

Correct answer: A. (120)

Step 1: In a one-one function, the three inputs must have distinct images. Step 2: The first input has (6) choices, the second has (5), and the third has (4). Step 3: Total one-one functions are (6\cdot5\cdot4=120).

Correct answer: A. No, because the domain has fewer elements than the codomain

Step 1: In an onto function, every element of the codomain must be an image. Step 2: Here the domain has (2) elements and the codomain has (4) elements. Step 3: Two inputs cannot cover four different codomain elements, so no onto function exists.

Correct answer: A. (2)

Step 1: To find (f^{-1}(4)), solve (x^3-4=4). Step 2: (x^3=8), so (x=2). Step 3: For a cubic function, the real cube root can be taken directly.

Correct answer: A. (\frac{2x-1}{5})

Step 1: Write (y=\frac{5x+1}{2}). Step 2: Then (2y=5x+1), so (x=\frac{2y-1}{5}). Step 3: Replacing (y) by (x), (f^{-1}(x)=\frac{2x-1}{5}).

Correct answer: A. Because it is not one-one

Step 1: For an inverse function to exist, the original function must be one-one. Step 2: (f(0)=0) and (f(6)=0), while (0\ne6). Step 3: Since one image has two preimages, the inverse is not a well-defined function.

Correct answer: A. ({-10,2})

Step 1: To find the preimage, solve (|x+4|=6). Step 2: This gives (x+4=6) or (x+4=-6). Step 3: Hence (x=2) or (x=-10), so the preimage is ({-10,2}).

Correct answer: A. (4)

Step 1: Write (x^2-8x+20=(x-4)^2+4). Step 2: Since ((x-4)^2\ge0), (f(x)\ge4). Step 3: The minimum value is (4), attained at (x=4).

Correct answer: A. ([4,\infty))

Step 1: Write the function as ((x-4)^2+4). Step 2: A square is always (0) or more. Step 3: Therefore the least value is (4), and the range is ([4,\infty)).

Correct answer: A. No, it has no real solution

Step 1: Assume (\frac{3x}{x+2}=3). Step 2: This gives (3x=3x+6), which is impossible. Step 3: Therefore (3) is not in the range of this function.

Correct answer: A. (R-{3})

Step 1: Write (y=\frac{3x}{x+2}). Step 2: From (y(x+2)=3x), we get (x=\frac{-2y}{y-3}), if (y\ne3). Step 3: Hence every real (y) is possible except (3).