General Engineering Test 1 Nddc Scholarship

Correct Answer D. (2x + 5)(6x – 11)

\[ \begin{aligned}
y &= (2x + 5)^2(x - 4) \\
\dfrac{dy}{dx} &= \dfrac{d}{dx} \left[ (2x + 5)^2 \cdot (x - 4) \right] \\
&= \dfrac{d}{dx}(2x + 5)^2 \cdot (x - 4) + (2x + 5)^2 \cdot \dfrac{d}{dx}(x - 4) \\
&= 2(2x + 5) \cdot 2 \cdot (x - 4) + (2x + 5)^2 \cdot 1 \\
&= 4(2x + 5)(x - 4) + (2x + 5)^2 \\
&= (2x + 5)\left[4(x - 4) + (2x + 5)\right] \\
&= (2x + 5)\left(4x - 16 + 2x + 5\right) \\
&= (2x + 5)(6x - 11)
\end{aligned}\]  

Correct answer C. 2

\[ \begin{aligned}
y &= 2k x^2 + x + 1 \\
\dfrac{dy}{dx} &= \dfrac{d}{dx}(2k x^2 + x + 1) = 4k x + 1 \\
\text{At } x &= 1,\ \dfrac{dy}{dx} = 9 \\
4k(1) + 1 &= 9 \\
4k + 1 &= 9 \\
4k &= 8 \\
k &= \boxed{2}
\end{aligned}\]

C. i

\[ \begin{aligned}
x^2 + 1 &= 0 \\
x^2 &= -1 \\
x &= \sqrt{-1} = \boxed{i}
\end{aligned}\]  

Correct answer D. 1

\[ \begin{aligned}
y &= -3 - 2x + x^2 \\
  &= x^2 - 2x - 3 \\
\text{Minimum at } x &= \dfrac{-b}{2a} = \dfrac{-(-2)}{2(1)} = \dfrac{2}{2} = \boxed{1}
\end{aligned}\]  

Correct answer A. 60

There are 5 distinct digits, and we are to choose 3 of them to form 3-digit numbers.

This is a permutation problem (order matters in number formation):

\[ \text{Number of 3-digit numbers} = P(5, 3) = \dfrac{5!}{(5 - 3)!} = 5 \times 4 \times 3 = \boxed{60}\]  

\[ \text{Given: } T_3 = a + 2d = -9\]

\[ T_7 = a + 6d = -29\]

\[ \text{Subtracting: } (a + 6d) - (a + 2d) = -29 - (-9)\]

\[ 4d = -20 \Rightarrow d = -5\]

\[ \text{Substitute } d \text{ into } T_3: \quad a + 2(-5) = -9\]

\[ a - 10 = -9 \Rightarrow a = 1\]

\[ T_{10} = a + 9d = 1 + 9(-5) = 1 - 45 = -44\]  

Correct Answer D. 10 sin5x cos5x

\[ \begin{aligned}
y &= \sin^2(5x) = \left( \sin(5x) \right)^2 \\
\dfrac{dy}{dx} &= 2\sin(5x) \cdot \dfrac{d}{dx}[\sin(5x)] \\
&= 2\sin(5x) \cdot 5\cos(5x) \\
&= 10\sin(5x)\cos(5x)
\end{aligned}\]

Correct Answer A. 720 minutes

Correct option D. -25

\[ T_n = n^2 - 6n - 4\]\[ T_3 = 3^2 - 6(3) - 4 = -13,\quad T_4 = 4^2 - 6(4) - 4 = -12\]\[ T_3 + T_4 = -13 + (-12) = -25\]  

Correct answer A. ¹/₃

\[ \text{Total fruits} = 9 + 8 + 7 = 24\]\[ \text{Favorable outcomes (bananas only)} = 8\]\[ P(\text{neither apple nor orange}) = \dfrac{8}{24} = \dfrac{1}{3}\]