MTH101 MCQs October 2024 by Miss Mehwish

Past Papers for Mids./Finals are also Available

MTH101

 The solution interval of the inequality 12<x+512 < x + 512<x+5 is:

(−∞, 7) ✔️

(5, 12]

(−∞, 0)

•  Which of the following is the solution of the equation ∣y∣=3|y| = 3∣y∣=3: y=±3y = \pm 3y=±3 ✔️

y=−3y = -3y=−3 · y=3y = 3y=3 · Does not exist

 Which of the following describes the equation y=2y = 2y=2:

A line parallel to y-axis · A point (0, 2) · A point (2, 0) · A line parallel to x-axis ✔️

•  Using the midpoint formula, what are the coordinates of the midpoint between (-5,3) and (7, -

1):

(1, 1) ✔️

(2, -1) · (4, 0) · (0, 0) ·  In the xy-plane, both x and y are positive in: · Third · Fourth

Second ✔️

 If a<ba < ba<b, then on a coordinate line: · aaa is to the left of bbb ✔️ · aaa is to the right of bbb

•  Consider the inequality: ∣z+2∣>6|z + 2| > 6∣z+2∣>6. What is the solution for zzz: z<−8 or z>4z < -8 \text{ or } z > 4z<−8 or z>4 ✔️

z=0z = 0z=0 · z=6z = 6z=6 · None of these

 Is the graph of the equation 4y=3x4y = 3x4y=3x symmetric about the x-axis? No

Yes ✔️

•  Which of the following is the equation of a circle centered at the origin and radius 1? x2+y2=1x^2 + y^2 = 1x2+y2=1 ✔️

x+y=1x + y = 1x+y=1 · x2+y2+1=0x^2 + y^2 + 1 = 0x2+y2+1=0 · None of these

 The domain of the function f(x)=tan⁡(z)f(x) = \tan(z)f(x)=tan(z) is:

None of these

R−{z∈R:z=π2+nπ,n∈Z}\mathbb{R} - \{ z \in \mathbb{R} : z = \frac{\pi}{2} + n\pi, n \in \mathbb{Z} \}R−{z∈R:z=2π+nπ,n∈Z} ✔️

(−∞,2]∪[2,∞)(-∞, 2] \cup [2, ∞)(−∞,2]∪[2,∞)

Q1. Let \( f(x) = x - 3 \) and \( g(x) = x^3 \), then \( f(g(x)) = ?

- \( x^3 - 2 \) · - \( x^3 - 3 \) · - \( 3x^3 - 3 \) · - \( \mathbf{x^3 + 3} \) ✔️ · 2. At each point of domain, the function __ · - \( \mathbf{Is defined} \) ✔️ · - Is infinite

3. The limit of \( f(x) \) as \( x \) approaches 1 from the right is equal to \( L \) is represented by the notation

- \( \lim_{{x \to 1}} f(x) = L \) · - \( \lim_{{x \to 1}} f(x) = L \) · - \( \lim_{{x \to L}} f(x) = 1 \) · - \( \mathbf{\lim_{{x \to 1^+}} f(x) = L} \) ✔️

Q4. Let \( f(x) = 2 \) and \( g(x) = x^2 + 1 \), then \( (f - g)(x) = ?

- \( 2 - x^2 \) · - \( 1 + x^2 \) · - \( \mathbf{3 - x^2} \) ✔️