InterviewSolution
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InterviewSolution
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The slope of the tangent at `(x , y)`to a curvepassing through `(1,pi/4)`is given by`y/x-cos^2(y/x),`then theequation of the curve is(a) `( b ) (c) y=( d ) (e)tan^(( f ) (g)-1( h ))( i )(( j ) (k)log(( l ) (m) (n) e/( o ) x (p) (q) (r))( s ))( t )`(u)(v) `( w ) (x) y=x (y) (z)tan^(( a a ) (bb)-1( c c ))( d d )(( e e ) (ff)log(( g g ) (hh) (ii) x/( j j ) e (kk) (ll) (mm))( n n ))( o o )`(pp)(qq)`( r r ) (ss) y=x (tt) (uu)tan^(( v v ) (ww)-1( x x ))( y y )(( z z ) (aaa)log(( b b b ) (ccc) (ddd) e/( e e e ) x (fff) (ggg) (hhh))( i i i ))( j j j )`(kkk)(d) none of theseA. `y=tan^(-1)[log((e)/(x))]`B. `y=x tan^(-1)[log((x)/(e))]`C. `y=x tan^(-1)[log((e)/(x))]`D. None of these |
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Answer» Correct Answer - C According to the given condition, `(dy)/(dx)=(y)/(x)-cos^(2)((y)/(x))` On putting y=mx `Rightarrow (dy)/(dx)=b+x(dv)/(dx),` we get `v+x(dy)/(dx)=v-cos^(2)v` `Rightarrow (dv)/(cos^(2)v)=-(dx)/(x)` `Rightarrow sec^(2)vdv=(-1)/(x)dx` On integrating both sides, we get `tan v=-log x+log x` `Rightarrow tan((y)/(x))=-log x+log c` Since, this curve is pasing through `(1,pi//4)`, `therefore tan((pi)/(4))=-log1+log cRightarrow log c=1` `therefore tan ((y)/(x))=-log x+1` `Rightarrow tan((y)/(x))=-logx+log x` `Rightarrow y=xtan ^(-1)[log((e)/(x))]` |
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