21. If given two points A(3, 12) and B(5, 22) on a x-y plane. Find the equation of the straight line with intercept at 2.
- A. y = 5x + 2
- B. y = 5x + 3
- C. y = 12x + 2
- D. y = 22x + 3
Correct Option: Answer is A
The equation of a straight line is given as \(y = mx + b\)
where m = the slope of the line
b = intercept
Given points A(3, 12) and B(5, 22), the slope = \(\frac{22 – 12}{5 – 3}\)
= \(\frac{10}{2}\) = 5
Hence, the equation of the line is \(y = 5x + 2\).
22. If P(2, m) is the midpoint of the line joining Q(m, n) and R(n, -4), find the values of m and n.
- A. m = 0, n = 4
- B. m = 4, n = 0
- C. m = 2, n = 2
- D. m = -2, n = 4
Correct Option: Answer is A
Q(m, n) and R(n, -4)
Midpoint : P(2, m)
\(\implies (\frac{m + n}{2}, \frac{n – 4}{2}) = (2, m)\)
\(m + n = 2 \times 2 \implies m + n = 4 … (i)\)
\(n – 4 = 2 \times m \implies n – 4 = 2m … (ii)\)
Solving (i) and (ii) simultaneously,
m = 0 and n = 4.
23. If \(\begin{vmatrix} 2 ~~~ -4 \\ x ~~~ 9 \end{vmatrix} = 58\), find the value of x.
- A. 10
- B. 30
- C. 14
- D. 28
Correct Option: Answer is A
\(\begin{vmatrix} 2 & -4 \\ x & 9 \end{vmatrix} = 58\)
\(\implies (2 \times 9) – (-4 \times x) = 58\)
\(18 + 4x = 58 \implies 4x = 58 – 18 = 40\)
\(x = 10\)
24. If \(y = 6x^3 + 2x^{-2} – x^{-3}\), find \(\frac{\mathrm d y}{\mathrm d x}\).
- A.
\(\frac{\mathrm d y}{\mathrm d x} = 15x^2 – 4x^{-2} – 3x^{-2}\) - B.
\(\frac{\mathrm d y}{\mathrm d x} = 6x + 4x^{-1} – 3x^{-4}\) - C.
\(\frac{\mathrm d y}{\mathrm d x} = 18x^2 – 4x^{-3} + 3x^{-4}\) - D.
\(\frac{\mathrm d y}{\mathrm d x} = 12x^2 + 4x^{-1} – 3x^{-2}\)
Correct Option: Answer is C
\(y = 6x^3 + 2x^{-2} – x^{-3}\)
\(\frac{\mathrm d y}{\mathrm d x} = 18x^2 – 4x^{-3} + 3x^{-4}\)
25. \(\frac{d}{dx} [\log (4x^3 – 2x)]\) is equal to
- A.
\(\frac{12x – 2}{4x^2}\) - B.
\(\frac{43x^2 – 2x}{7x}\) - C.
\(\frac{4x^2 – 2}{7x + 6}\) - D.
\(\frac{12x^2 – 2}{4x^3 – 2x}\)
Correct Option: Answer is D
\(\frac{d}{dx} [\log (4x^3 – 2x)]\) … (1)
Let u = 4x\(^3\) – 2x.
\(\frac{\mathrm d}{\mathrm d x} (\log (4x^3 – 2x)) = (\frac{\mathrm d}{\mathrm d u})(\frac{\mathrm d u}{\mathrm d x})\)
\(\frac{\mathrm d}{\mathrm d u} (\log u)\) = \(\frac{1}{u}\)
\(\frac{\mathrm d u}{\mathrm d x} = 12x^2 – 2\)
\(\therefore \frac{d}{dx} [\log (4x^3 – 2x)] = \frac{12x^2 – 2}{u}\)
= \(\frac{12x^2 – 2}{4x^3 – 2x}\)
26. If \(f(x) = 3x^3 + 4x^2 + x – 8\), what is the value of f(-2)?
- A. -24
- B. 30
- C. -18
- D. -50
Correct Option: Answer is C
\(f(x) = 3x^3 + 4x^2 + x – 8\)
\(f(-2) = 3(-2)^3 + 4(-2)^2 + (-2) – 8\)
= \(-24 + 16 – 2 – 8\)
= -18
Solve for x in \(\frac{4x – 6}{3} \leq \frac{3 + 2x}{2}\)
- A. \(x \leq 1\frac{1}{2}\)
- B. \(x \leq \frac{21}{2}\)
- C. \(x \geq \frac{21}{2}\)
- D. \(x \geq 1\frac{1}{2}\)
Correct Option: Answer is B
\(\frac{4x – 6}{3} \leq \frac{3 + 2x}{2}\)
2(4x – 6) \(\leq\) 3(3 + 2x)
8x – 12 \(\leq\) 9 + 6x
8x – 6x \(\leq\) 9 + 12
2x \(\leq\) 21
\(x \leq \frac{21}{2}\)
- A. -1 \(\leq\) x \(\leq\) 2
- B. -1 \(\leq\) x < 2
- C. -1 < x < 2
- D. -1 < x \(\leq\) 2
Correct Option: Answer is D
-7 ≤ 9 – 8x < 16 – x
-7 ≤ 9 – 8x and 9 – 8x < 16 – x
-7 – 9 ≤ -8x and -8x + x < 16 – 9
-16 ≤ -8x and -7x < 7
∴ x ≤ 2 and -1 < x
-1 < x ≤ 2.
29. The nth term of a sequence is given by 2\(^{2n – 1}\). Find the sum of the first four terms.
- A. 74
- B. 32
- C. 42
- D.170
Correct Option: Answer is D
\(T_n = 2^{2n – 1}\)
\(T_1 = 2^{2(1) – 1} \)
= 2
\(T_2 = 2^{2(2) – 1}\)
= 8
\(T_3 = 2^{2(3) – 1}\)
= 32
\(T_4 = 2^{2(4) – 1}\)
= 128
\(T_1 + T_2 + T_3 + T_4 = 2 + 8 + 32 + 128\)
= 170
30. Integrate \(\int_{-1} ^{2} (2x^2 + x) \mathrm {d} x\)
- A.\(4\frac{1}{2}\)
- B.\(3\frac{1}{2}\)
- C.\(7\frac{1}{2}\)
- D.\(5\frac{1}{4}\)
Correct Option: Answer is C
\(\int_{-1} ^{2} (2x^2 + x) \mathrm {d} x\)
= \([\frac{2x^{2 + 1}}{3} + \frac{x^{1 + 1}}{2}]_{-1} ^{2}\)
= \([\frac{2x^{3}}{3} + \frac{x^{2}}{2}]_{-1} ^{2}\)
= \((\frac{2(2)^{3}}{3} + \frac{2^2}{2}) – (\frac{2(-1)^{3}}{3} + \frac{(-1)^{2}}{2})\)
= \((\frac{16}{3} + 2) – (\frac{-2}{3} + \frac{1}{2})\)
= \(\frac{22}{3} – (-\frac{1}{6})\)
= \(\frac{22}{3} + \frac{1}{6}\)
= \(\frac{15}{2}\)
= \(7\frac{1}{2}\)