CBSE 11th Standard Maths Subject HOT Questions 4 Mark Questions 2021 Part - II
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CBSE 11th Standard Maths Subject HOT Questions 4 Mark Questions 2021 Part - II
11th Standard CBSE
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Reg.No. :
Mathematics
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Out of 100 students, 15 passed in English, 12 passed in Mathematics, 8 in Science, 4 in English and Science, 4 in all the three.Find how many students passed in English and Mathematics but not in Science?
(a) -
If tan(πcosθ)=cot(πsinθ) then prove that \(cos\left( \theta -\frac { \pi }{ 4 } \right) =\pm \frac { 1 }{ 2\sqrt { 2 } } .\)
(a) -
If \(cos \left( \theta +\phi \right) = m\quad cos\quad \left( \theta -\phi \right) \)then find the value of \(\frac { 1-m }{ 1+m } cot\quad \phi .\)
(a) -
Find the modulus and argument of \(z=\frac { (1+i)^{ 13 } }{ (1+i)^{ 7 } }\)
(a) -
If (1+i)(1+2i)(1+3i)...(1+ni)=(x+iy), then show that 2.5.10....(1+n2)=x2+y2
(a)
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CBSE 11th Standard Maths Subject HOT Questions 4 Mark Questions 2021 Part - II Answer Keys
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n(E)=15,n(M)=12,n(s)=8,\(n(E\cap M)=6\)
\(n(M\cap S)=7,n(E\cap S)=4,n(E\cap M\cap S)=4\)
(i)\(n(E\cap M\cap \overline { S } )=n(E\cap M)-n(E\cap M\cap S)\)
=6-4=2 -
\(\frac { sin\quad (\pi \quad cos\quad \theta ) }{ cos\quad (\pi \quad sin\quad \theta ) } =\frac { cos\quad (\pi \quad sin\quad \theta ) }{ sin\quad \left( \pi \quad sin\quad \theta \right) }\)
\(\Rightarrow cos\quad \left( \pi \quad cos\quad \theta \quad +\quad \pi \quad sin\quad \theta \right) =0\)
\(\Rightarrow cos\quad \theta \quad +\quad sin\quad \theta \quad =\quad \pm \frac { 1 }{ 2 }\)
\( \Rightarrow cos\quad \theta \quad cos\quad \frac { \pi }{ 4 } +sin\quad \theta \quad sin\quad \frac { \pi }{ 4 } =\pm \frac { 1 }{ 2\sqrt { 2 } } \)
\(\therefore \quad cos\left( \theta -\frac { \pi }{ 4 } \right) =\pm \frac { 1 }{ 2\sqrt { 2 } } \) -
We have, \(cos \left( \theta +\phi \right) = m\quad cos\left( \theta -\phi \right) \)
\(\Rightarrow \frac { 1 }{ m } =\frac { cos\left( \theta -\phi \right) }{ cos\left( \theta +\phi \right) } \)
Now, \(\frac { 1-m }{ 1+m } =\frac { cos\quad \left( \theta -\phi \right) -cos\quad \left( \theta +\phi \right) }{ cos\quad \left( \theta -\phi \right) +cos\left( \theta +\phi \right) } =tan\quad \theta \quad tan\quad \phi \\ \\ \)
Ans. tan \(\theta \). -
We have, \(z=\frac { (1+i)^{ 13 } }{ (1+i)^{ 7 } }\)
\(z=\frac { (1+i)^{ 13 } }{ (1+i)^{ 7 } } x\frac { (1+i{ ) }^{ 7 } }{ (1+i{ ) }^{ 7 } } \)
\([by\quad rationalising\quad the\quad denominator]\)
\(=\frac { (1+i{ ) }^{ 20 } }{ (1-{ i }^{ 2 }{ ) }^{ 7 } } =\frac { (1+i{ ) }^{ 20 } }{ (1+1{ ) }^{ 7 } } =\frac { (1+i{ ) }^{ 20 } }{ 2^{ 7 } } =\frac { 1 }{ 2^{ 7 } } (1+i{ ) }^{ 20 }\)
\( \frac { 1 }{ 2^{ 7 } } [(1+i{ )^{ 2 }] }^{ 10 }=\frac { 1 }{ 2^{ 7 } } (1+2i+{ i }^{ 2 }{ ) }^{ 10 }\)
\(\frac { 1 }{ 2^{ 7 } } (2i{ ) }^{ 10 }=\frac { (2{ ) }^{ 10 } }{ 2^{ 7 } } (i{ ) }^{ 10 }=(2{ ) }^{ 3 }({ i }^{ 2 })^{ 5 }=8(-1{ ) }^{ 5 }\)
\(z=-8\)
\(|z=\sqrt { (-8{ ) }^{ 2 } } =\sqrt { { 8 }^{ 2 } } =8\)
and arg(z) = \(\pi \) [z lies on negative X-axis] -
\(\left| (1+i)(1+2i)(1+3i)....(1+ni) \right| =\left| x+iy \right| \)
\(\Rightarrow \left| 1+i \right| \left| 1+2i \right| \left| 1+3i \right| ....\left| 1+ni \right| =\left| x+iy \right| \)
\(\Rightarrow \sqrt { 1+1 } \sqrt { 1+4 } \sqrt { 1+9 } ....\sqrt { 1+{ n }^{ 2 } } =\sqrt { { x }^{ 2 }+{ y }^{ 2 } }\)
\( \Rightarrow \sqrt { 2 } \sqrt { 5 } \sqrt { 10 } ....\sqrt { 1+{ n }^{ 2 } } =\sqrt { { x }^{ 2 }+{ y }^{ 2 } } \)
