CBSE 12th Standard Maths Subject Continuity and Differentiability HOT Questions 4 Mark Questions 2021
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CBSE 12th Standard Maths Subject Continuity and Differentiability HOT Questions 4 Mark Questions 2021
12th Standard CBSE
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Reg.No. :
Maths
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If x sin (a+y) + sin a cos (a+y)=0, prove that \(\frac { dy }{ dx } =\frac { { sin }^{ 2 }(a+y) }{ sin\quad a } \)
(a) -
If \(y={ \left\{ x+\sqrt { { x }^{ 2 }+{ a }^{ 2 } } \right\} }^{ n }\)prove that \(\frac { dy }{ dx } =\frac { ny }{ \sqrt { { x }^{ 2 }+{ a }^{ 2 } } } \)
(a) -
If \(y={ \left\{ x+\sqrt { { x }^{ 2 }+{ a }^{ 2 } } \right\} }^{ n }\)prove that \(\frac { dy }{ dx } =\frac { ny }{ \sqrt { { x }^{ 2 }+{ a }^{ 2 } } } \)
(a) -
If \({ x }^{ 2 }+{ y }^{ 2 }=t-\frac { 1 }{ t } \)and \({ x }^{ 4 }+{ y }^{ 4 }={ t }^{ 2 }+\frac { 1 }{ { t }^{ 2 } } \) show that \(\frac { dy }{ dx } =\frac { 1 }{ { x }^{ 3 }y } \)
(a) -
If \({ x }^{ 2 }+{ y }^{ 2 }=t-\frac { 1 }{ t } \)and \({ x }^{ 4 }+{ y }^{ 4 }={ t }^{ 2 }+\frac { 1 }{ { t }^{ 2 } } \)show that \(\frac { dy }{ dx } =\frac { 1 }{ { x }^{ 3 }y } \)
(a)
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CBSE 12th Standard Maths Subject Continuity and Differentiability HOT Questions 4 Mark Questions 2021 Answer Keys
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\(\frac { dy }{ dx } =\frac { { sin }^{ 2 }(a+y) }{ sin\quad a } \)
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We have \(y={ \left\{ x+\sqrt { { x }^{ 2 }+{ a }^{ 2 } } \right\} }^{ n }\) ....(1)
\(\frac { dy }{ dx } =n{ \left\{ x+\sqrt { { x }^{ 2 }+{ a }^{ 2 } } \right\} }^{ n-1 }\frac { d }{ dx } \left\{ x+\sqrt { { x }^{ 2 }+{ a }^{ 2 } } \right\}\)
\( =n{ \left\{ x+\sqrt { { x }^{ 2 }+{ a }^{ 2 } } \right\} }^{ n-1 }\)
\(\left[ 1+\frac { 1 }{ 2\sqrt { { x }^{ 2 }+{ a }^{ 2 } } } (2x+0) \right] \)
\(=n{ \left\{ x+\sqrt { { x }^{ 2 }+{ a }^{ 2 } } \right\} }^{ n-1 }\left\{ \frac { \sqrt { { x }^{ 2 }+{ a }^{ 2 } } +x }{ \sqrt { { x }^{ 2 }+{ a }^{ 2 } } } \right\} \)
\(=\frac { n{ \left\{ x+\sqrt { { x }^{ 2 }+{ a }^{ 2 } } \right\} }^{ n } }{ \sqrt { { x }^{ 2 }+{ a }^{ 2 } } } =\frac { ny }{ \sqrt { { x }^{ 2 }+{ a }^{ 2 } } } \)
which is true. -
We have \(y={ \left\{ x+\sqrt { { x }^{ 2 }+{ a }^{ 2 } } \right\} }^{ n }\) ....(1)
\(\frac { dy }{ dx } =n{ \left\{ x+\sqrt { { x }^{ 2 }+{ a }^{ 2 } } \right\} }^{ n-1 }\frac { d }{ dx } \left\{ x+\sqrt { { x }^{ 2 }+{ a }^{ 2 } } \right\} \)
\(=n{ \left\{ x+\sqrt { { x }^{ 2 }+{ a }^{ 2 } } \right\} }^{ n-1 } \)
\(\left[ 1+\frac { 1 }{ 2\sqrt { { x }^{ 2 }+{ a }^{ 2 } } } (2x+0) \right] \)
\(=n{ \left\{ x+\sqrt { { x }^{ 2 }+{ a }^{ 2 } } \right\} }^{ n-1 }\left\{ \frac { \sqrt { { x }^{ 2 }+{ a }^{ 2 } } +x }{ \sqrt { { x }^{ 2 }+{ a }^{ 2 } } } \right\} \)
\(=\frac { n{ \left\{ x+\sqrt { { x }^{ 2 }+{ a }^{ 2 } } \right\} }^{ n } }{ \sqrt { { x }^{ 2 }+{ a }^{ 2 } } } =\frac { ny }{ \sqrt { { x }^{ 2 }+{ a }^{ 2 } } } \)
which is true. -
We have \({ x }^{ 2 }+{ y }^{ 2 }=t-\frac { 1 }{ t } \)...(1)
\({ x }^{ 4 }+{ y }^{ 4 }={ t }^{ 2 }+\frac { 1 }{ { t }^{ 2 } } \).....(2)
squaring (1) \({ \left( { x }^{ 2 }+{ y }^{ 2 } \right) }^{ 2 }={ \left( t-\frac { 1 }{ t } \right) }^{ 2 }\)
\({ x }^{ 4 }+{ y }^{ 4 }+2{ x }^{ 2 }{ y }^{ 2 }={ t }^{ 2 }+\frac { 1 }{ { t }^{ 2 } } -2\)
\(=2{ x }^{ 2 }{ y }^{ 2 }=-2\)
\( { y }^{ 2 }=-\frac { 1 }{ { x }^{ 2 } } \)
Differentiating w.r.t.x, \(2y\frac { dy }{ dx } =\frac { -2 }{ { x }^{ 3 } } \)
\(\frac { dy }{ dx } =\frac { 1 }{ { x }^{ 3 }y } \)
which is true. -
We have \({ x }^{ 2 }+{ y }^{ 2 }=t-\frac { 1 }{ t } \)...(1)
\({ x }^{ 4 }+{ y }^{ 4 }={ t }^{ 2 }+\frac { 1 }{ { t }^{ 2 } } \).....(2)
squaring (1) \({ \left( { x }^{ 2 }+{ y }^{ 2 } \right) }^{ 2 }={ \left( t-\frac { 1 }{ t } \right) }^{ 2 }\)
\({ x }^{ 4 }+{ y }^{ 4 }+2{ x }^{ 2 }{ y }^{ 2 }={ t }^{ 2 }+\frac { 1 }{ { t }^{ 2 } } -2\)
\(=2{ x }^{ 2 }{ y }^{ 2 }=-2\)
\({ y }^{ 2 }=-\frac { 1 }{ { x }^{ 2 } } \)
Differentiating w.r.t.x, \(2y\frac { dy }{ dx } =\frac { -2 }{ { x }^{ 3 } } \)
\(\frac { dy }{ dx } =\frac { 1 }{ { x }^{ 3 }y } \)
which is true.
