Your Answer : (Not Answered)

Correct Answer : B

The correct match is (P) → (3), (Q) → (2), (R) → (1), (S) → (5).

(P) The radius is the distance from (1, 2) to the line: \(\frac{|3 + 8 - 1|}{5} = 2\). The circle \((x - 1)^{2} + (y - 2)^{2} = 4\) passes through (3, 2).

(Q) A tangent to \(y^{2} = 8x\) is \(y = mx + \frac{2}{m}\). Tangency to \(x^{2} + y^{2} = 2\) needs its distance from the origin to be \(\sqrt{2}\):

\[ \begin{aligned} \frac{4}{m^{2}} &= 2(1 + m^{2}) \\ m^{4} + m^{2} - 2 &= 0 \Rightarrow m^{2} = 1 \end{aligned} \]

With positive slope, \(m = 1\) and the tangent is \(y = x + 2\), which passes through (7, 9).

(R) The ellipse is \(\frac{x^{2}}{16} + \frac{y^{2}}{12} = 1\), with \(e^{2} = 1 - \frac{12}{16} = \frac{1}{4}\), focus \((2, 0)\) and semi-latus rectum \(\frac{12}{4} = 3\), so \(M = (2, 3)\). The normal at \((x_{1}, y_{1})\) is \(\frac{16x}{x_{1}} - \frac{12y}{y_{1}} = 16 - 12\), here \(8x - 4y = 4\), or \(2x - y = 1\), which passes through (1, 1).

(S) \(ae = 5\) and \(\frac{a}{e} = \frac{16}{5}\) give \(a^{2} = 16\), \(e = \frac{5}{4}\) and \(b^{2} = a^{2}(e^{2} - 1) = 9\). On \(\frac{x^{2}}{16} - \frac{y^{2}}{9} = 1\), the point \((8, 3\sqrt{3})\) gives \(4 - 3 = 1\), so it lies on \(H\).

Only the second option has all four; the third sends R to (2, 5), which gives \(2x - y = -1\), not 1.

Your Answer : (Not Answered)

Correct Answer : B

\(\cos\beta = \frac{3}{7}\).

The direction ratios of the line are the differences of the coordinates:

\[ \begin{aligned} (3 - 1,\ 2 + 1,\ 8 - 2) &= (2, 3, 6) \\ \sqrt{2^{2} + 3^{2} + 6^{2}} &= \sqrt{49} = 7 \end{aligned} \]

The direction cosines are \(\frac{2}{7}, \frac{3}{7}, \frac{6}{7}\). The angle with the y-axis uses the second one, so \(\cos\beta = \frac{3}{7}\). The option \(\frac{2}{7}\) is the cosine of the angle with the x-axis.

Your Answer : (Not Answered)

Correct Answer : C

a = \(-\frac{1}{2}\).

The general term of \((4x + a)^{7}\) is \(\binom{7}{r}(4x)^{7 - r}a^{r}\). The \(x^{4}\) term has 7 − r = 4, so r = 3.

\[ \begin{aligned} \binom{7}{3}4^{4}a^{3} &= -1120 \\ 35 \times 256\,a^{3} &= -1120 \\ a^{3} &= -\frac{1}{8} \\ a &= -\frac{1}{2} \end{aligned} \]

The negative coefficient needs a negative a, which rules out the positive options; \(-\frac{1}{8}\) is \(a^{3}\), not a.

Your Answer : (Not Answered)

Correct Answer : E

The common ratio is \(\dfrac{1}{100}\).

Rewrite the sum in the standard form \(S_{n} = \frac{a(1 - r^{n})}{1 - r}\):

\[ \begin{aligned} s_{n} &= 100 - 100 \cdot 100^{-n} \\ &= 100\left(1 - \left(\tfrac{1}{100}\right)^{n}\right) \end{aligned} \]

The term raised to the power \(n\) is the common ratio, so \(r = \frac{1}{100}\). Check: \(a = s_{1} = 100 - 1 = 99\), and \(\frac{a}{1 - r} = \frac{99}{0.99} = 100\), matching the factor in front. The other fractions do not reproduce \(s_{1} = 99\).

Your Answer : (Not Answered)

Correct Answer : E

The determinant is 0.

In a G.P., \(a_{2} = ra_{1}\), \(a_{4} = ra_{3}\) and \(a_{6} = ra_{5}\). So the second column is \(r\) times the first column, and a determinant with two proportional columns is zero:

\[ \begin{vmatrix} a_{1} & ra_{1} & 1 \\ a_{3} & ra_{3} & 1 \\ a_{5} & ra_{5} & 1 \end{vmatrix} = r\begin{vmatrix} a_{1} & a_{1} & 1 \\ a_{3} & a_{3} & 1 \\ a_{5} & a_{5} & 1 \end{vmatrix} = 0 \]

The values \(a_{1} = 2\) and \(r = \frac{1}{2}\) do not matter; any G.P. gives 0, so the non-zero options are wrong.

Your Answer : (Not Answered)

Correct Answer : B

The area is \(9\sqrt{3}\) square units.

The two vertices are \(-3 - 3i\) and \(3 - 3i\), the points (-3, -3) and (3, -3). They lie on the same horizontal line, so the side is 6.

\[ \begin{aligned} \text{Area} &= \frac{\sqrt{3}}{4}a^{2} \\ &= \frac{\sqrt{3}}{4} \times 36 = 9\sqrt{3} \end{aligned} \]

18 is \(\frac{1}{2} \times 6 \times 6\), the area of a right isosceles triangle with legs 6, not of an equilateral triangle of side 6. The height of the equilateral triangle is \(3\sqrt{3}\), and \(\frac{1}{2} \times 6 \times 3\sqrt{3} = 9\sqrt{3}\) confirms the answer.

Your Answer : (Not Answered)

Correct Answer : C

The maximum value is 385.

\(8\sin x + 15\cos x\) ranges between \(\pm\sqrt{8^{2} + 15^{2}} = \pm 17\). Adding 3 gives the range [-14, 20] for the expression inside the bracket.

\[ \begin{aligned} \max (8\sin x + 15\cos x + 3)^{2} &= 20^{2} = 400 \\ \max f &= 400 - 15 = 385 \end{aligned} \]

The square is largest at the end of the range farther from zero, which is 20, not -14 (whose square, 196, is smaller). Taking 17 without the 3 gives 274, and \(17^{2} + 3^{2}\) or similar slips give the other options.

Your Answer : (Not Answered)

Correct Answer : E

The value is \(\frac{\pi}{2}\).

Use \(2\tan^{-1}x = \tan^{-1}\frac{2x}{1 - x^{2}}\) for \(|x| \lt 1\):

\[ \begin{aligned} 2\tan^{-1}\frac{1}{3} &= \tan^{-1}\frac{2/3}{8/9} = \tan^{-1}\frac{3}{4} \\ \cot^{-1}\frac{3}{4} &= \tan^{-1}\frac{4}{3} \end{aligned} \]

Then \(\tan^{-1}\frac{3}{4} + \tan^{-1}\frac{4}{3} = \frac{\pi}{2}\), because \(\tan^{-1}y + \tan^{-1}\frac{1}{y} = \frac{\pi}{2}\) for \(y \gt 0\).

The common slip is to treat \(\cot^{-1}\frac{3}{4}\) as \(\tan^{-1}\frac{3}{4}\); it is \(\tan^{-1}\frac{4}{3}\), the complementary angle, which is what makes the sum exactly a right angle.

Your Answer : (Not Answered)

Correct Answer : C

The expression equals 1.

The denominator is a difference of squares, \((1 + \sin\alpha)^{2} - \cos^{2}\alpha\).

\[ \begin{aligned} (1 + \sin\alpha)^{2} - \cos^{2}\alpha &= 1 + 2\sin\alpha + \sin^{2}\alpha - \cos^{2}\alpha \\ &= 2\sin\alpha + 2\sin^{2}\alpha \\ &= 2\sin\alpha(1 + \sin\alpha) \end{aligned} \]

Here \(1 - \cos^{2}\alpha = \sin^{2}\alpha\) was used. The denominator equals the numerator, so the value is 1 for every admissible \(\alpha\). tan \(\alpha\) and the two quotients are what partial simplifications leave behind.

Your Answer : (Not Answered)

Correct Answer : C

The eccentricity is \(\frac{6}{5}\).

For \(\frac{X^{2}}{a^{2}} - \frac{Y^{2}}{b^{2}} = 1\), the eccentricity satisfies \(e^{2} = 1 + \frac{b^{2}}{a^{2}}\). The shift of centre to (1, -2) does not change it.

\[ \begin{aligned} e^{2} &= 1 + \frac{11}{25} = \frac{36}{25} \\ e &= \frac{6}{5} \end{aligned} \]

\(\frac{25}{11}\) and \(\frac{5}{11}\) are ratios of the denominators, not eccentricities, and \(\frac{7}{5}\) or \(\frac{5}{3}\) would need \(b^{2}\) = 24 or \(\frac{400}{9}\). A hyperbola always has e greater than 1, which \(\frac{5}{11}\) is not.

Your Answer : (Not Answered)

Correct Answer : D

The maximum value is 11.

For a bounded feasible region the maximum of a linear objective occurs at a corner point. Evaluate z = 2x + 3y at each.

\[ \begin{aligned} (0, 0)&: 0 \qquad (5, 0): 10 \\ (4, 1)&: 8 + 3 = 11 \qquad (0, 2): 6 \end{aligned} \]

The largest is 11 at (4, 1). 10 and 6 are values at other corners, 0 is the minimum, and 16 does not occur at any corner.

Your Answer : (Not Answered)

Correct Answer : C

The distance is \(\frac{17}{4}\).

For \(y^{2} = 16x\), \(4a = 16\), so \(a = 4\) and the focus is \((4, 0)\). Write points as \((at^{2}, 2at)\); the tangent at parameter \(t\) has slope \(1/t\). The point \((64, 32)\) has \(8t = 32\), so \(t = 4\).

Perpendicular tangents need \(\frac{1}{t_{1}}\cdot\frac{1}{4} = -1\), so \(t_{1} = -\frac{1}{4}\):

\[ \begin{aligned} x_{1} &= at_{1}^{2} = 4 \cdot \tfrac{1}{16} = \tfrac{1}{4} \\ y_{1} &= 2at_{1} = 8 \cdot \left(-\tfrac{1}{4}\right) = -2 \end{aligned} \]

For a parabola, the distance of a point from the focus equals its distance from the directrix \(x = -4\), so it is \(x_{1} + a = \frac{1}{4} + 4 = \frac{17}{4}\).

\(\frac{15}{4}\) is what you get by subtracting, \(a - x_{1}\), instead of adding. 4 is just \(a\), the distance from the vertex to the focus, and 5 does not correspond to this point.

Your Answer : (Not Answered)

Correct Answer : D

The value is \(\pi - 2\).

Write \(\frac{\sin x}{1 + \sin x} = 1 - \frac{1}{1 + \sin x}\).

\[ \begin{aligned} \int_{0}^{\pi}\frac{dx}{1 + \sin x} &= \int_{0}^{\pi}\frac{1 - \sin x}{\cos^{2}x}dx \\ &= \left[\tan x - \sec x\right]_{0}^{\pi} = 2 \\ \text{value} &= \pi - 2 \end{aligned} \]

The antiderivative \(\tan x - \sec x\) equals 1 at \(\pi\) and -1 at 0 (it is continuous across \(\frac{\pi}{2}\), with limit 0 there). \(\pi + 2\) adds instead of subtracting the second part.

Your Answer : (Not Answered)

Correct Answer : A

The integral is \(\log|\sin x| + C\).

Simplify the integrand first. \(\tan x + \cot x = \frac{1}{\sin x\cos x}\) and \(1 + \tan^{2}x = \sec^{2}x\).

\[ \begin{aligned} \frac{\tan x + \cot x}{\sec^{2}x} &= \frac{\cos^{2}x}{\sin x\cos x} = \cot x \\ \int \cot x\,dx &= \log|\sin x| + C \end{aligned} \]

\(\log|\cos x|\) would come from integrating \(-\tan x\), and \(\log|\tan x|\) from integrating \(\frac{1}{\sin x\cos x}\) without the \(\sec^{2}x\) in the denominator.

Your Answer : (Not Answered)

Correct Answer : D

\(f(2)\) is 8.

Differentiate the right-hand side and match it with the integrand:

\[ \begin{aligned} &\frac{d}{dt}\left[f(t)\sin\tfrac{1}{t}\right] \\ &= f'(t)\sin\tfrac{1}{t} - \frac{f(t)}{t^{2}}\cos\tfrac{1}{t} \end{aligned} \]

Comparing with \(3t^{2}\sin\frac{1}{t} - t\cos\frac{1}{t}\) gives \(f'(t) = 3t^{2}\) and \(\frac{f(t)}{t^{2}} = t\). Both are satisfied by \(f(t) = t^{3}\), so \(f(2) = 8\).

4 is \(t^{2}\) at \(t = 2\) and 16 is \(t^{4}\); neither function satisfies both conditions.

Your Answer : (Not Answered)

Correct Answer : B

The integral is \(\cos(\cot^{-1}x) + C\).

Substitute \(u = \cot^{-1}x\). Then \(du = -\frac{dx}{1 + x^{2}}\):

\[ \begin{aligned} \int \frac{\sin(\cot^{-1}x)}{1 + x^{2}}\,dx &= -\int \sin u\,du \\ &= \cos u + C = \cos(\cot^{-1}x) + C \end{aligned} \]

The option \(-\cos(\cot^{-1}x)\) forgets that the derivative of \(\cot^{-1}x\) is negative. The options with \(1 + x^{2}\) or 2 in the denominator keep a factor that the substitution has already absorbed.

Your Answer : (Not Answered)

Correct Answer : C

The area is 8 sq.units.

For \(x \ge 0\), \(y = |x|\) is the line \(y = x\). Together with the line \(x = 4\) and the \(x\)-axis it closes a right triangle with vertices \((0, 0)\), \((4, 0)\) and \((4, 4)\), which is the region the key takes:

\[ A = \int_{0}^{4}x\,dx = \frac{1}{2}(4)(4) = 8 \]

The \(x\)-axis is the implied third boundary, as in the usual 'area under the curve' reading; the other branch \(y = -x\) lies at \(x \le 0\), on the other side of the origin from the line \(x = 4\). 4 halves the triangle again, and 12 and 13 do not correspond to any region bounded by these lines.

Your Answer : (Not Answered)

Correct Answer : D

\(f(x) = x^{3} + \frac{1}{x^{2}} - 2\).

Integrate \(f'(x) = 3x^{2} - 2x^{-3}\) term by term:

\[ \begin{aligned} f(x) &= x^{3} - 2 \cdot \frac{x^{-2}}{-2} + C \\ &= x^{3} + \frac{1}{x^{2}} + C \\ f(1) &= 1 + 1 + C = 0, \quad C = -2 \end{aligned} \]

The options \(+1\), \(+2\) and \(-1\) have the right variable part but the wrong constant; only \(-2\) makes \(f(1) = 0\). The first option, \(x^{2} + \frac{1}{x^{3}} + 1\), has the powers wrong as well.

Your Answer : (Not Answered)

Correct Answer : B

The value is \(p = \frac{-1}{3}\).

Substitute \(u = \frac{1}{x^{6}} + 1\), so \(du = -\frac{6}{x^{7}}dx\) and \(\frac{dx}{x^{7}} = -\frac{du}{6}\).

\[ \begin{aligned} \int \frac{dx}{x^{7}u^{2/3}} &= -\frac{1}{6}\int u^{-2/3}du \\ &= -\frac{1}{6}\cdot 3u^{1/3} = -\frac{1}{2}u^{1/3} \end{aligned} \]

The answer is written in terms of \(\frac{1}{u}\), and \(u^{1/3} = \left(\frac{1}{u}\right)^{-1/3}\), so \(p = -\frac{1}{3}\). The value \(\frac{1}{3}\) is the power of u itself, before rewriting in terms of the reciprocal.

Your Answer : (Not Answered)

Correct Answer : C

The number of words is 60.

VIRUS has five different letters. A three-letter word without repetition is an ordered arrangement of 3 of them, a permutation:

\[ ^{5}P_{3} = 5 \times 4 \times 3 = 60 \]

30 halves this, and 40 is not a count of arrangements of these letters. 120 is \(5!\), the number of five-letter arrangements. Counting combinations \(\binom{5}{3} = 10\) would ignore order, which words do not.