<p><strong>Let <math><msubsup><mrow><mrow><mo>⋃</mo></mrow></mrow><mrow><mrow><mi>i</mi><mo>=</mo><mn>1</mn></mrow></mrow><mrow><mrow><mn>50</mn></mrow></mrow></msubsup><mo> </mo><msub><mrow><mrow><mi>X</mi></mrow></mrow><mrow><mrow><mi>i</mi></mrow></mrow></msub><mo>=</mo><msubsup><mrow><mrow><mo>⋃</mo></mrow></mrow><mrow><mrow><mi>i</mi><mo>=</mo><mn>1</mn></mrow></mrow><mrow><mrow><mi>n</mi></mrow></mrow></msubsup><mo> </mo><msub><mrow><mrow><mi>Y</mi></mrow></mrow><mrow><mrow><mi>i</mi></mrow></mrow></msub><mo>=</mo><mi>T</mi></math> , where each <math><msub><mrow><mrow><mi>X</mi></mrow></mrow><mrow><mrow><mi>i</mi></mrow></mrow></msub></math> contains 10 elements and each <math><mi>Y</mi><mi>i</mi></math> contains 5 elements. If each element of the set <math><mi>T</mi></math> is an element of exactly 20 of sets <math><msub><mrow><mrow><mi>X</mi></mrow></mrow><mrow><mrow><mi>i</mi></mrow></mrow></msub></math> 's and exactly 6 of sets <math><msub><mrow><mrow><mi>Y</mi></mrow></mrow><mrow><mrow><mi>i</mi></mrow></mrow></msub></math> 's then <math><mi>n</mi></math> is equal to:</strong></p>

f(x)=∫13?xdx(1+x)2=∫13?t.2tdt1+t22 (put x=t )=-t1+t213+tan-1?t13 [Applying by parts]=-34-12+π3-π4=12-34+π12

Maths NCERT Exemplar Solutions Class 12th Chapter Twelve Maths Class 12th Sets Maths NCERT Exemplar Solutions Class 12th Chapter One

Let $⋃_{i = 1}^{50} X_{i} = ⋃_{i = 1}^{n} Y_{i} = T$ , where each $X_{i}$ contains 10 elements and each $Y i$ contains 5 elements. If each element of the set $T$ is an element of exactly 20 of sets $X_{i}$ 's and exactly 6 of sets $Y_{i}$ 's then $n$ is equal to:

Option 1 -

Option 2 -

Option 3 -

Option 4 -

3 Views | Posted 2 months ago

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1 Answer

Answered by

alok kumar singh | Contributor-Level 10

2 months ago

Correct Option - 1

Detailed Solution:

$f (x) = \int_{1}^{3} ? \frac{\sqrt[]{x} d x}{(1 + x)^{2}} = \int_{1}^{\sqrt[]{3}} ? \frac{t . 2 t d t}{{(1 + t^{2})}^{2}}$ (put $\sqrt[]{x} = t$ )

$= {(- \frac{t}{1 + t^{2}})}_{1}^{\sqrt[]{3}} + {({t a n}^{- 1} ? t)}_{1}^{\sqrt[]{3}}$ [Applying by parts]

$\begin{array}{r} = - (\frac{\sqrt[]{3}}{4} - \frac{1}{2}) + \frac{π}{3} - \frac{π}{4} \\ = \frac{1}{2} - \frac{\sqrt[]{3}}{4} + \frac{π}{12} \end{array}$

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Let $⋃_{i = 1}^{50} X_{i} = ⋃_{i = 1}^{n} Y_{i} = T$ , where each $X_{i}$ contains 10 elements and each $Y i$ contains 5 elements. If each element of the set $T$ is an element of exactly 20 of sets $X_{i}$ 's and exactly 6 of sets $Y_{i}$ 's then $n$ is equal to:

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Let ⋃i=150 Xi=⋃i=1n Yi=T , where each Xi contains 10 elements and each Yi contains 5 elements. If each element of the set T is an element of exactly 20 of sets Xi 's and exactly 6 of sets Yi 's then n is equal to:

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Let $⋃_{i = 1}^{50} X_{i} = ⋃_{i = 1}^{n} Y_{i} = T$ , where each $X_{i}$ contains 10 elements and each $Y i$ contains 5 elements. If each element of the set $T$ is an element of exactly 20 of sets $X_{i}$ 's and exactly 6 of sets $Y_{i}$ 's then $n$ is equal to: