General Engineering Test 1 Nddc Scholarship

Correct answer D. 20

Total students} = 40
Students offering neither} = 4
Students offering at least one subject} = 40 - 4 = 36
Let x = number offering both Mathematics and Physics
Using inclusion-exclusion
32 + 24 - x = 36
56 - x = 36
x = 56 - 36 = 20

Correct answer B. –12 cos (-4x)

Given:

\[y = 3 \sin(-4x)\]

Differentiate using the chain rule:

\[\dfrac{dy}{dx} = 3 \cdot \cos(-4x) \cdot \dfrac{d}{dx}(-4x)\]

\[\dfrac{dy}{dx} = 3 \cdot \cos(-4x) \cdot (-4)\]

\[\dfrac{dy}{dx} = -12 \cos(-4x)\]  

Correct option A. 1

We are given the polynomial:

\[f(x) = 3x^3 + 5x^2 - 11x + 4\]

To find the remainder when this polynomial is divided by \( x + 3 \), we apply the Remainder Theorem.  
This means we evaluate \( f(-3) \):

\[f(-3) = 3(-3)^3 + 5(-3)^2 - 11(-3) + 4\]

\[= 3(-27) + 5(9) + 33 + 4\]

\[= -81 + 45 + 33 + 4\]

\[= 1\]

Therefore, the remainder is:

\[\boxed{1}\]  

Correct option B. 54

Let the total number of oranges be \( x \).

The first boy received:\[\dfrac{1}{3}x\]

Remaining oranges after the first boy:\[x - \dfrac{1}{3}x = \dfrac{2}{3}x\]

The second boy received:\[\dfrac{2}{3} \times \dfrac{2}{3}x = \dfrac{4}{9}x\]

The third boy received the rest:\[x - \left( \dfrac{1}{3}x + \dfrac{4}{9}x \right) = x - \left( \dfrac{3}{9}x + \dfrac{4}{9}x \right) = x - \dfrac{7}{9}x = \dfrac{2}{9}x\]

We are told the third boy received 12 oranges:

\[\dfrac{2}{9}x = 12\]

Multiply both sides by 9:

\[2x = 108 \Rightarrow x = 54\]

Therefore, the total number of oranges shared is:

\[\boxed{54}\]  

We are to evaluate the definite integral:

The integral you posted is: $$ \int_0^{\pi/2} \sin^2 x \, dx $$ Use the identity: $$ \sin^2 x = \dfrac{1 - \cos 2x}{2} $$ So the integral becomes: $$ \int_0^{\pi/2} \dfrac{1 - \cos 2x}{2} \, dx = \dfrac{1}{2} \int_0^{\pi/2} (1 - \cos 2x) \, dx $$ Integrate term by term: $$ \dfrac{1}{2} \left[ \int_0^{\pi/2} 1 \, dx - \int_0^{\pi/2} \cos 2x \, dx \right] = \dfrac{1}{2} \left[ x - \dfrac{\sin 2x}{2} \right]_0^{\pi/2} $$ Apply the limits: $$ \dfrac{1}{2} \left[ \left(\dfrac{\pi}{2} - \dfrac{\sin \pi}{2}\right) - \left(0 - \dfrac{\sin 0}{2}\right) \right] = \dfrac{1}{2} \left( \dfrac{\pi}{2} - 0 \right) = \dfrac{\pi}{4} $$ $$ \int_0^{\pi/2} \sin^2 x \, dx = \dfrac{\pi}{4} $$  

Correct Answer: A. 513

The given series is:

\[3, 6, 9, \ldots, 36\]

This is an arithmetic progression (AP) with:

- First term \( a = 3 \)  
- Common difference \( d = 6 - 3 = 3 \)

We are to find the sum of the first 18 terms:

\[S_n = \dfrac{n}{2} \left[ 2a + (n - 1)d \right]\]

Substitute \( n = 18, a = 3, d = 3 \):

\[S_{18} = \dfrac{18}{2} \left[ 2(3) + (18 - 1)(3) \right]\]

\[= 9 \left[ 6 + 17 \cdot 3 \right]\]

\[= 9 \left[ 6 + 51 \right] = 9 \cdot 57 = 513\]

Therefore, the sum of the first 18 terms is:

\[\boxed{513}\]

Correct option:B. ¹/₃​

We are given the curve:

\[y = x^3 + x^2 - x + 1\]

To find the minimum point, we first compute the first derivative:

\[\dfrac{dy}{dx} = 3x^2 + 2x - 1\]

Set the derivative to zero:

\[3x^2 + 2x - 1 = 0\]

Solve using the quadratic formula:

\[x = \dfrac{-2 \pm \sqrt{2^2 - 4(3)(-1)}}{2(3)} = \dfrac{-2 \pm \sqrt{4 + 12}}{6} = \dfrac{-2 \pm \sqrt{16}}{6}\]

\[x = \dfrac{-2 \pm 4}{6}\]

\[x = \dfrac{2}{6} = \dfrac{1}{3} \quad \text{or} \quad x = \dfrac{-6}{6} = -1\]

Now determine which of these is the minimum point using the second derivative:

\[\dfrac{d^2y}{dx^2} = 6x + 2\]

At \( x = -1 \):

\[\dfrac{d^2y}{dx^2} = 6(-1) + 2 = -4 \quad (\text{negative} \Rightarrow \text{maximum})\]

At \( x = \dfrac{1}{3} \):

\[\dfrac{d^2y}{dx^2} = 6\left( \dfrac{1}{3} \right) + 2 = 2 + 2 = 4 \quad (\text{positive} \Rightarrow \text{minimum})\]

Thus, the minimum occurs at:

\[\boxed{\dfrac{1}{3}}\]