Correct Answer: A. \(\frac{7}{6}\)

Explanation

3x\(^2\) - 7x + 6 = 0

α + β = \(\frac{7}{3}\);

αβ = 2;

\(\frac{1}{α}\) + \(\frac{1}{β}\) = \(\frac{α +β}{αβ}\)

\(\frac{7}{3}\) ÷ 2

= \(\frac{7}{6}\)

Explanation

\((\log_{2} m)^{2} - \log_{2} m^{3} = 10\)

Let \(\log_{2} m\) = x.

\((\log_{2} m)^{2} - 3\log_{2} m - 10 = 0\)

\(\implies x^{2} - 3x - 10 = 0\)

\(x^{2} + 2x - 5x - 10 = 0 \)

\(x(x + 2) - 5(x + 2) = 0\)

\(\implies x = \text{-2 or 5}\)

When \(\log_{2} m = -2 \implies m = 2^{-2} = \frac{1}{4}\)

When \(\log_{2} m = 5 \implies m = 2^{5} = 32\).

Correct Answer: C. \(\begin{bmatrix} 4&0\\0&4\end{bmatrix}\)

Explanation

\(P : (x, y) → (2x + y, -2y)\)

\(p\begin{bmatrix} x\\y\end{bmatrix}=\begin{bmatrix} 2x & y\\0 &-2y\end{bmatrix}\)

\(\therefore p = \begin{bmatrix} 2 & 1\\0 &-2\end{bmatrix}\)

\(\therefore p^2 = \begin{bmatrix} 2&1\\0&-2\end{bmatrix}\) \(\begin{bmatrix} 2&1\\0&-2\end{bmatrix}\) = \(\begin{bmatrix} 4&0\\0&4\end{bmatrix}\)

Correct Answer: A. 12

Explanation

\(Mean : \frac{x + 4 + z + t}{4} = 5 \implies x + 4 + z + t = 20\)

\(\implies x + z + t = 16 ... (1)\)

\(Median : \frac{4 + z}{2} = z \implies 4 + z = 2z\)

\(4 = z\)

From 1,

\(\implies x + 4 + t = 16 \)

\(x + t = 12\)

Correct Answer: B. 84

Explanation

y 3x\(^2\) (2x +1) = 6x\(^3\) + 3x\(^2\)

dy/dx = 18x\(^2\) + 6x

At x = 2, 18(2)\(^2\) + 6(2)

= 72 + 12 = 84

Correct Answer: D. -1

Explanation

Midpoint between \((x_{1}, y_{1})\) and \((x_{2}, y_{2})\) = \((\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})\).

\((-k, k) = (\frac{2 + (1 + k)}{2}, \frac{-4 + (k + 1)}{2})\)

\(-k = \frac{k + 3}{2} \implies -2k = k + 3\)

\(-3k = 3 \implies k = -1\)

Correct Answer: B. \(1+ \frac{1}{2}\sqrt{3}\)

Explanation

\(\frac{1}{(1-\sqrt{3})^{2}}\)

\((1-\sqrt{3})^{2} = (1-\sqrt{3})(1-\sqrt{3})\)

\(1 - 2\sqrt{3} + 3 = 4 - 2\sqrt{3}\)

\(\frac{1}{4-2\sqrt{3}}\)

After rationalising (multiplying the denominator and numerator with \(4+2\sqrt{3}\), we have

\(\frac{4+2\sqrt{3}}{4} = 1 + \frac{1}{2}\sqrt{3}\)

Correct Answer: B. \(4(2x + \sqrt{x})(2 + \frac{1}{2\sqrt{x}})\)

Explanation

\(y = 2(2x + \sqrt{x})^{2}\)

Let \(u = 2x + \sqrt{x}\)

\(y = 2u^{2}\)

\(\frac{\mathrm d y}{\mathrm d u} = 4u\)

\(\frac{\mathrm d u}{\mathrm d x} = 2 + \frac{1}{2\sqrt{x}}\)

\(\therefore \frac{\mathrm d y}{\mathrm d x} = (\frac{\mathrm d y}{\mathrm d u})(\frac{\mathrm d u}{\mathrm d x})\)

= \(4u(2 + \frac{1}{2\sqrt{x}}) \)

= \(4(2x + \sqrt{x})(2 + \frac{1}{2\sqrt{x}})\)

Explanation

volume of rectangular block(cuboid) = l × b × h

V = 3x × 2x × y

72 = 6x\(^2\)y

y = \(\frac{72}{6x^2}\)

y = \(\frac{12}{x^2}\)

ii.

Area of cuboid = 2(lb + lh + bh)

l = 3x, b = 2x, h = y

A = 2(3x(2x) + 3x(y) + 2x(y))

A = 2(6x\(^2\) + 3xy + 2xy)

A = 12x\(^2\) + 6xy + 4xy

A = 12x\(^2\) + 10xy

where y = \(\frac{12}{x^2}\)

A = 12x\(^2\) + \(\frac{120}{x}\)

iii.

At stationary point \(\frac{dy}{dx}\) = 0

\(\frac{dA}{dx} = 2(12x^2-1) + -120x^{-1-1}\)

24x - 120x\(^{-2}\)

\(\frac{dA}{dx}\) = 0

24x- \(\frac{120}{x^2}\) = 0

multiply through by x\(^2\)

24x\(^3\) - 120 = 0

24x\(^3\) = 120

x\(^3\) = 5

x = ?5

x = 1.71

Explanation

If we let \(\beta\) be the angle between the forces 10N and 16N,

then \(\beta\) = (180\(^o\) - \(\theta\))

Applying the cosine rule; cos\(\theta\) = \(\frac{10^2 + 16^2 - 14^2}{2 \times 10 \times 16}\) = 0.5

If \(\theta\) = cos\(^{-1}\)(0.5) = 60\(^o\)

\(\beta\) = 180\(^o\) - 60\(^o\) = 120\(^o\)