Class 12th

Let f : R ® R be a function defined as

$f (x) = a s i n (\frac{π [x]}{2}) + [2 - x], a \in R,$ where [t] is the greatest integer less than or equal to t, if $\underset{x \to - 1}{l i m} f (x)$ exists, then the value of $\int_{0}^{4} f (x) d x$ is equal to

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Contributor-Level 10

Find RHL, x = -1 + h

$\underset{h \to 0}{l i m} a . s i n (\frac{π [- 1 + h]}{2}) + [2 + 1 - h] = - a + 2$

Find LHL, x = -1 – h

$\underset{h \to 0}{l i m} a s i n \frac{(π [- 1 - h])}{2} = [2 + 1 + h] = 3$

$\int_{0}^{4} f (x) d x = \int_{0}^{1} 1 d x + \int_{1}^{2} (- 1) d x + \int_{2}^{3} (- 1) d x + \int_{3}^{4} (- 1) d x = - 2$

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11 months ago

A 1m long wire is broken into two unequal parts X and Y. The X part of the wire is stretched into another wire W. Length of W is twice the length of X and the resistance of W is twice that of Y. Find the ratio of length of X and Y.

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Contributor-Level 10

R_w = 2R_y

$\Rightarrow ρ (\frac{2 x}{\frac{A}{2}}) = \frac{2 ρ (1 - x)}{A}$

$\frac{4 ρ x}{A} = \frac{2 ρ (1 - x)}{A}$

4x = 2 – 2x 6x = 2

x = $\frac{2}{6} = \frac{1}{3}$ $∴ \frac{L_{x}}{L_{y}} = \frac{\frac{1}{3}}{1 - \frac{1}{3}} = \frac{1}{3} * \frac{3}{2}$

$= \frac{1}{2}$

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11 months ago

Let A = $(\begin{matrix} 1 & 2 \\ - 2 & - 5 \end{matrix}) . L e t α, β \in R$ such that aA² + bA = 2I. Then a + b is equal to

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Contributor-Level 10

Kindly go through the solution

Use characteristic equation = 0

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11 months ago

Let the minimum value v₀ of $v = {| z |}^{2} + {| z - 3 |}^{2} + {| z - 6 i |}^{2}, z \in C$ is attained at z = z₀. Then ${| 2 z_{0}^{2} - {\bar{z}}_{0}^{3} + 3 |}^{2} + v_{0}^{2}$ is equal to

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Contributor-Level 10

$V = {| z |}^{2} + {| z - 3 |}^{2} + {| z - 6 i |}^{2}$

Put z = x + iy

$V = 3 [{(x - 1)}^{2} + {(y - 2)}^{2} + 10]$

For minimum x = 1, y = 2, V₀ = 30

So, z₀ = 1 + 2i

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11 months ago

Given below are two statements: one is labeled as Assertion A and the other is labeled as Reason R.

Assertion A : Alloys such as constantan and manganin are used in making standard resistance coils.

Reason R : Constantan and maganin have very small value of temperature coefficient of resistance.

In the light of the above statements, choose the correct answer from the options given below.

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Contributor-Level 10

Based on theoretical data.

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11 months ago

Let $f, g : N - {1}$ N be functions defined by f(a) = a, where a is the maximum of the powers of those primes p such that p^a divides a, and g(a) = a + 1, for all $a \in N$ - {1}. Then, the function f + g is

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Contributor-Level 10

f (a) = a, a is max of powers of prime P such that P^a divides a.

f (2) = 1; g (2) = 3

f (3) = 1; g (3) = 4

f (4) = 2; g (4) = 5

f (5) = 1; g (5) = 6

f (2) + g (2) = 4

f (3) + g (3) = 5

f (4) + g (4) = 7

f (5) + g (5) = 7

So, f (x) + g (x) is many-one-and an into function. We won't get 'I' in the range.

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11 months ago

Two identical thin metal plates has charge q₁ and q₂ respectively such that q₁ > q₂. The plates were brought close to each other to from a parallel plate capacitor of capacitance C. The potential difference between them is

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Contributor-Level 10

V = $\frac{q_{(c a p a c i t a n c e)}}{C} = \frac{q_{1} - q_{2}}{2 c}$

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11 months ago

Two identical metallic spheres A and B when placed at certain distance in air repel each other with force of F. Another identical uncharged sphere C is first place in contact with A and then in contact with B and finally placed at midpoint between sphere A and B. The force experienced by sphere C will be

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Contributor-Level 10

For sphere 'C' after contacting with 'A'. q_A = q_C = $\frac{q}{2} .$

offer contacting with 'B'. q_B = q_C = $\frac{3 q}{4}$

$F_{N e t} = | F_{1} - F_{2} |$

F₂ = $\frac{K \cdot 3 q^{2} * 4}{r^{2} 8} = \frac{3}{2} \frac{k q^{2}}{r^{2}}$

$F_{1} = \frac{9 k q^{2} * 4}{16 r^{2}} = \frac{9 k q^{2}}{4 r^{2}} = \frac{9}{4} F$

$∴ F_{N e t} = | \frac{9}{4} F - \frac{3}{2} F |$

= $\frac{9 - 6}{4} F = \frac{3}{4} F$

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11 months ago

Le the mirror image of the point (a, b, c), with respect to the plane 3x – 4y + 12z + 19 = 0 be $(a - 6, β, γ)$ . If a + b + c = 5, then $7 β - 9 γ$ is equal to……….s

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Contributor-Level 10

$\frac{x - a}{3} = \frac{y - b}{- 4} = \frac{z - c}{12} = \frac{- 2 (3 a - 4 b + 12 c + 19)}{3^{2} + {(- 4)}^{2} + 1 2^{2}}$

$\frac{x - a}{3} = \frac{y - b}{- 4} = \frac{z - c}{12} = \frac{- 6 a + 8 b - 24 c - 38}{169}$

$(x, y, z) \equiv (a - 6, β, γ)$

$\frac{(a - b) - a}{3} = \frac{β - b}{- 4} = \frac{γ - c}{12} = \frac{- 6 a + 8 b - 24 c - 38}{169}$

$\frac{β - b}{- 4} = - 2$

$\Rightarrow β = 8 + b$

$\Rightarrow 3 a - 4 b + 12 c = 150$ ….(i)

a + b + c = 5

$\Rightarrow 3 a + 3 b + 3 c = 15$ ….(ii)

Applying (i) – (ii), we get :

= 56 + 216 + 7b – 9c = 56 + 216 – 135 = 137

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11 months ago

Class 12th Chemistry

Some drugs bind to a site other than the active site of an enzyme. This site is known as

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