CBSE 11th Standard Maths Subject HOT Questions 4 Mark Questions With Solution 2021
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QB365 Provides the HOT Question Papers for Class 11 Maths, and also provide the detail solution for each and every HOT Questions. HOT Questions will help to get more idea about question pattern in every exams and also will help to get more marks in Exams
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CBSE 11th Standard Maths Subject HOT Questions 4 Mark Questions With Solution 2021
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 Mathematics and Science but not in English?
(a) -
If \(cos\left( \alpha +\beta \right) =\frac { 4 }{ 5 } ,\quad sin(\alpha -\beta )=\frac { 5 }{ 13 } \) and \(\alpha , \beta \) lie between 0 and \(\frac { \pi }{ 4 } \) then prove that \(tan 2\alpha =\frac { 56 }{ 33 } .\)
(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) -
If z is a complex number such that \(|z-1|=|z+1|\) then show that Re(z) = 0.
(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 With Solution 2021 Answer Keys
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\(n(E\cap S\cap \overline { E } )=n(E\cap S)-n(E\cap M\cap S)\)
=74=3 -
since, \(\alpha \),\(\beta \) lie between 0 and \(\frac { \pi }{ 4 } \)
\(\therefore -\frac { \pi }{ 4 } <\alpha -\beta \frac { \pi }{ 4 } and\quad 0<\alpha +\beta <\frac { \pi }{ 2 } \)
\(\Rightarrow cos\quad (\alpha -\beta )\quad and\quad sin\quad (\alpha +\beta )\quad are\quad positive.\)
\(Now,\quad sin\quad (\alpha +\beta )=\sqrt { 1-{ cos }^{ 2 }\left( \alpha +\beta \right) } =\frac { 3 }{ 5 } \)
\(and\quad \quad cos\quad (\alpha -\beta )=\sqrt { 1-{ sin }^{ 2 }\left( \alpha -\beta \right) } =\frac { 12 }{ 13 } \)
\(\therefore \quad tan\quad (\alpha +\beta )=\frac { sin\left( \alpha +\beta \right) }{ cos\left( \alpha +\beta \right) } =\frac { 3/5 }{ 4/5 } =\frac { 3 }{ 4 } \)
\(and\quad tan\quad (\alpha -\beta )=\frac { sin\quad \left( \alpha -\beta \right) }{ cos\quad \left( \alpha -\beta \right) } =\frac { 5/13 }{ 12/13 } =\frac { 5 }{ 12 }\)
\(now,\quad tan\quad 2\alpha =tan\quad [\left( \alpha +\beta \right) +\left( \alpha -\beta \right) ]\)
\(=\frac { tan\quad (\alpha +\beta )+tan\quad (\alpha -\beta ) }{ 1-tan\quad (\alpha +\beta )tan\left( \alpha -\beta \right) } =\frac { \frac { 3 }{ 4 } +\frac { 5 }{ 12 } }{ 1-\frac { 3 }{ 4 } \times \frac { 5 }{ 12 } } =\frac { 56 }{ 33 } \)
Hence proved. -
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 \(\theta \). -
Let z = x + iy, then \(|z-1|=|z+1|\)
\(\Rightarrow \left| x+iy-1 \right| =\left| x+iy+1 \right| \)
\(\Rightarrow \left| (x-1)+iy \right| =\left| (x+1)-iy \right| \)
\(\Rightarrow\sqrt { (x-1)^{ 2 }+{ y }^{ 2 } } =\sqrt { (x+1)^{ 2 }+{ y }^{ 2 } } \)
\(\Rightarrow (x-1)^{ 2 }+{ y }^{ 2 }=(x+1)^{ 2 }+{ y }^{ 2 }\)
\(\Rightarrow { x }^{ 2 }+1-2x={ x }^{ 2 }+1+2x\Rightarrow 4x=0\Rightarrow x=0\)
\(\therefore \quad Re(z)=0\) -
\(\left| (1+i)(1+2i)(1+3i)....(1+ni) \right| =\left| x+iy \right| \)
\(\Rightarrow \left| 1+i \right| \left| 1+2i \right| \left| 1+3i \right| ....\left| 1+ni \right| =\left| x+iy \right| \)
\(\Rightarrow \sqrt { 1+1 } \sqrt { 1+4 } \sqrt { 1+9 } ....\sqrt { 1+{ n }^{ 2 } } =\sqrt { { x }^{ 2 }+{ y }^{ 2 } }\)
\( \Rightarrow \sqrt { 2 } \sqrt { 5 } \sqrt { 10 } ....\sqrt { 1+{ n }^{ 2 } } =\sqrt { { x }^{ 2 }+{ y }^{ 2 } } \)
