Solve: \(\frac { 2 }{ x } +\frac { 3 }{ y } +\frac { 10 }{ z } =4,\frac { 4 }{ x } -\frac { 6 }{ y } +\frac { 5 }{ z } =1,\frac { 6 }{ x } +\frac { 9 }{ y } -\frac { 20 }{ z } \) = 2

Verify that arg(1+i) + arg(1-i) = arg[(1+i) (1-i)]

If a, b, c, d and p are distinct non-zero real numbers such that (a²+b²+c²) p²-2 (ab+bc+cd) p+(b²+c²+d²)≤ 0 then prove that a, b, c, d are in G.P and ad = bc

If \({ tan }^{ -1 }\left( \frac { \sqrt { 1+{ x }^{ 2 } } -\sqrt { 1-{ x }^{ 2 } } }{ \sqrt { 1+{ x }^{ 2 } } +\sqrt { 1-{ x }^{ 2 } } } \right) =a\) than prove that x² = sin 2a

A kho-kho player In a practice session while running realises that the sum of tne distances from the two kho-kho poles from him is always 8m. Find the equation of the path traced by him of the distance between the poles is 6m.

ABCD is a quadrilateral with \(\overset { \rightarrow }{ AB } =\overset { \rightarrow }{ \alpha } \) and \(\overset { \rightarrow }{ AD } =\overset { \rightarrow }{ \beta } \) and \(\overset { \rightarrow }{ AC } =2\overset { \rightarrow }{ \alpha } +3\overset { \rightarrow }{ \beta } \). If the area of the quadrilateral is λ times the area of the parallelogram with \(\overset { \rightarrow }{ AB } \) and \(\overset { \rightarrow }{ AD } \) as adjacent sides, then prove that \(\lambda =\frac { 5 }{ 2 } \)

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plane

Show that the curves 4x = y² and 4xy = k cut at right angles if k² = 512.

If the curves 4x=y² and 4xy=k cut at right angles show that k²=512.

Find the intervals for which the function f(x)=2x²-9x²-12x+1 is increasing or decfreasing and find the local extermems.

Find the local maximum and local minimum values of f(x)=x⁴-3x+3x²-x.

Using differential find the approximate value of cos 61; if it is given that sin 60° = 0.86603 and 1⁰ = 0.01745 radians.

Find \(\frac { \partial w }{ \partial u } ,\frac { \partial w }{ \partial v } \) if w=sin^-1(x,y) where x=u+v,y=u-v

Using integration, find the area of the triangle with sides y = 2x + 1, y = 3x + 1 and x = 4.

Find the area of the loop of the curve 3ay²=x(x-a)²

Find the area of the region bounded by a²y²=a²(a²-x²)

Find the area bounded by the curve y2(2a-x)=x² and the line x=2a.

Solve: \(\frac { dy }{ dx } \) = (3x+2y+1)²

Solve :(x²+xy)dy=(x²+y²)dx

Verify (p ∧ ~p) ∧ (~q ∧ p) is a tautlogy, contradiction or contingency.