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Concept -

Range = highest value - lowest value

\(Average = \frac{Sum \ of\ observation } {total\ no\ of\ observation}\)

Calculation - 

Range of C = 250 - 80 = 170

The average of participants from E will be

\( \frac{112+105+180+80+183 } {5}=132\)

The difference is 170 - 132 = 38.

∴ the difference between 'the range of C and average of E participants is 38.

Calculation-

\(\rm \begin {vmatrix}\frac{2}{x-4} \end{vmatrix}\) > 1 ⇒\(\rm \begin {vmatrix}\frac{x-4}{2} \end{vmatrix}\) < 1 ⇒|x - 4| < 2

⇒ -2 < x - 4 < 2 ⇒ 2 < x < 6 

As x ≠ 4 hence x ϵ (2, 4) ∪ (4, 6) 

Since, (4, 6) is lying in the interval for all the acceptable values of x ϵ (2, 4) ∪ (4, 6) so this will be the best option.  

∴ The solution of the equality \(\rm \begin {vmatrix}\frac{2}{x-4} \end{vmatrix}\) > 1 is x ϵ (4, 6).

Concept-

log a - log b = \(log\frac{a}{b}\)

Calculation-

\(\int_{2}^{3}\frac{1}{x}dx\)

⇒\(\left [ logx \right ]_{2}^{3}\)

⇒ log 3 - log 2 

⇒ \(log\left| \frac{3}{2} \right|\)

∴ \(\int_{2}^{3}\frac{1}{x}dx = log\left| \frac{3}{2} \right| \)

Concept-

\(\int x^{n} = \frac{x^{n+1}}{n+1} \)

Calculation-

\(\int_{1}^{2}x^{2}dx\)

⇒\(\left[\frac{x^{3}}{3} \right]_{1}^{2}\)

⇒\(\left[\frac{2^{3}}{3}-\frac{1^{3}}{3} \right]\)

⇒\( {7} \over {3}\)

Concept-

(x + y)ⁿ + (x - y)ⁿ

1. \( {n+1} \over {2}\) if n is odd
2. \( {n} \over {2} \)+1 if n is even.

Calculation-

(x + a)100 + (x − a)¹⁰⁰

⇒\( {100} \over {2}\) + 1

⇒ 51

∴The total number of terms in the expansion of (x + a)100 + (x - a)100 is 51

Concept-

Sum of the angles of a triangle is 180° 

If a side of a triangle is produced, then the exterior angle so formed is equal to the sum of the two interior opposite angles.

Calculation-

By linear angle property

∠ ACB + 110° = 180° 

∠ ACB = 70° 

Now in Δ ABC

∠ BAC + ∠ ABC + ∠ ACB = 180°

∠ BAC + 62° + 70° = 180° 

∠ BAC = 48° 

Alternate Method

By exterior angle property in ΔABC

∠ BAC + 62° = 110° 

⇒ ∠ BAC = 48° 

∴ the measure of ∠ BAC is 48°.

Concept-

• (a + b)² = a² + b² + 2ab

Calculation-

We have 3x + 5y = 14 and xy = 6

Now 3x + 5y = 14

On squaring both sides we get

⇒ 9x² + 25y² + 30xy = 196

⇒ 9x2 + 25y² + 30(6) = 196  

⇒ 9x2 + 25y² = 16

∴ the value of 9x2 + 25y2 is 16

Concept-

Use the method of Variable separable-

The general form of separable differential equation will be \(\frac{dy}{dx} = g(x)\times h(y)\)

Write the equation in a form such that one side only involves x and the other only has y ( \(g(x)dx = \frac{dy}{h(y)}\) )

Integrate each side with respect to its own variable. Combine results to get the solution.

1. 1 - cos x = 2 sin² x/2.
2. 1 + cos x = 2 cos² x/2
3. 1 + tan² x = sec² x

Calculation-

(dy - dx) + cos x(dy + dx) = 0 

⇒ dy - dx + cos x dy + cos x dx = 0  ⇒ dy(1 + cos x) - dx(1 - cos x) = 0 

⇒ dy = \( {1-cosx} \over {1+cosx}\) dx ⇒ dy = \( {2sin^{2}\frac{x}{2} \over 2cos^{2}\frac{x}{2}}dx\)

⇒ dy = \( tan^{2} \frac{x} {2}\) dx ⇒ dy = \(sec^{2}\frac{x}{2}-1\)

⇒ \(\int dy = \int \left ( sec^{2}\frac{x}{2}-1 \right )dx\)  ⇒ y = 2 \(tan{x\over {2}} \) - x + c

∴solution of the differential equation (dy - dx) + cos x (dy + dx) = 0 is y = 2 \(tan{x\over {2}} \) - x + c .

Hence option (3) is correct.

Concept-

• (a + b)² = a² + b² + 2ab

Calculation-

\(\rm 5x +\frac{1}{3x}=4\)

Multiply the equation by 3/5 we get 

\(\rm 3x +\frac{1}{5x}= \frac{12}{5}\)

Now squaring both sides 

\(\rm 9x^{2} +\frac{1}{25x^{2}}+ 2 \times 3x \times \frac{1}{5x} =\frac{144}{25}\)

⇒\(\rm 9x^{2} +\frac{1}{25x^{2}} + \frac{6}{5}=\frac{144}{25}\)

⇒ \(\rm 9x^{2} +\frac{1}{25x^{2}} =\frac{114}{25}\)

∴ The value of \(\rm 9x^{2} +\frac{1}{25x^{2}} \ is\ \frac{114}{25}\)

Concept-

Use the method of Variable separable-

The general form of separable differential equation will be \(\frac{dy}{dx} = g(x)\times h(y)\)

Write the equation in a form such that one side only involves x and the other only has y ( \(g(x)dx = \frac{dy}{h(y)}\) )

Integrate each side with respect to its own variable. Combine results to get the solution.

• \(\int {{dx} \over {1+x^{2}}} \) = tan^-1x

Calculation-

dy = (1 + y2)dx  ⇒ \( {dy} \over {1+y^{2}} \) = dx

on integrating both side we get 

⇒ tan^-1y = x + c ⇒ y = tan(x + c)

∴ The solution of the equation dy = (1 + y2) dx is y = tan(x + c).