Question 1 Find the minimum point on the curve $$y = x^2 - 6x + 5$$ Options (a) (-3,4) (b) (3,-4) (c) (1,5) (d) (2,3)

Topic: Calculus (Differentiation) Correct Option: (b)

Explanation: To find the minimum or maximum point, we compute the first derivative: $$ \frac{dy}{dx} = 2x - 6 $$ >Set $$\frac{dy}{dx} = 0$$:

$$ 2x - 6 = 0 \implies x = 3 $$

Substitute $$x = 3$$ into the original equation to find $$y$$: $$ y = (3)^2 - 6(3) + 5 = 9 - 18 + 5 = -4 $$ The coordinate of the minimum point is $$(3,-4)$$.

Question 2

An A.P has 14 as its second term and 38 as its fifth term. Find the tenth term and the sum of the first five terms.

Options

(a) 75,110 (b) 65,124 (c) 78,110 (d) 141,33

Topic: Sequence and Series (Arithmetic Progression)

Correct Option: (c)

Explanation:

Let the first term be $$a$$ and common difference $$d$$.

Second term: $$a + d = 14 \quad \text{(i)}$$

Fifth term: $$a + 4d = 38 \quad \text{(ii)}$$

Subtract (i) from (ii):

$$ (a + 4d) - (a + d) = 38 - 14 \implies 3d = 24 \implies d = 8 $$

Substitute $$d = 8$$ into (i):

$$ a + 8 = 14 \implies a = 6 $$

Find the tenth term:

$$ T_{10} = a + 9d = 6 + 9(8) = 6 + 72 = 78 $$

Sum of the first five terms:

$$ S_5 = \frac{5}{2}[2a + (5-1)d] = \frac{5}{2}[12 + 32] = \frac{5}{2}[44] = 110 $$

The values are 78 and 110.

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Question 3

Given $$\sec^2\theta + \tan^2\theta = 3$$, find the value of $$\theta$$.

Options

(a) 30° (b) 60° (c) 45° (d) 90°

Topic: Trigonometry

Correct Option: (c)

Explanation:

Recall identity:

$$ \sec^2\theta = 1 + \tan^2\theta $$

Substitute into the equation:

$$ (1 + \tan^2\theta) + \tan^2\theta = 3 \implies 1 + 2\tan^2\theta = 3 $$ $$ 2\tan^2\theta = 2 \implies \tan^2\theta = 1 \implies \tan\theta = 1 $$ $$ \theta = \tan^{-1}(1) = 45^\circ $$

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Question 4

The circumference of a circle is 100 cm. Find the area of the circle.

Options

(a) 142 cm² (b) 795 cm² (c) 830 cm² (d) 419 cm²

Topic: Mensuration

Correct Option: (b)

Explanation:

Circumference: $$C = 2\pi r = 100 \implies r = \frac{100}{2\pi} = \frac{50}{\pi}$$

Area:

$$ A = \pi r^2 = \pi \left(\frac{50}{\pi}\right)^2 = \pi \cdot \frac{2500}{\pi^2} = \frac{2500}{\pi} \approx 795.77 \, \text{cm}^2 $$

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Question 5

Find the value of $$p$$ so that the roots of $$4x^2 + px + 9 = 0$$ are equal.

Options

(a) ±4 (b) ±6 (c) ±12 (d) ±18

Topic: Quadratic Equations

Correct Option: (c)

Explanation:

For equal roots, the discriminant must be zero:

$$ b^2 - 4ac = 0 $$

Here $$a = 4, b = p, c = 9$$

$$ p^2 - 4(4)(9) = 0 \implies p^2 - 144 = 0 \implies p^2 = 144 \implies p = \pm 12 $$>