Solution - Career Point

MATHSStudents' ForumExpert’s Solution for Question asked by IIT-JEE Aspirants1. Let f : R → R and f(x) = g(x) + h(x), where g(x) is apolynomial and h(x) is a continuous anddifferentiable bounded function on both sides, thenf(x) is onto if g(x) of odd degree and f(x) is into ifg(x) is of even degree. Then check whether f(x) isone one, many one, onto or into.(i) f(x) = a 1 x + a 3 x 3 + a 5 x 5 + .... + a 2n + 1 x 2n + 1 – cot –1 xwhere 0 < a 1 < a 2 < a 3 ....... < a 2n + 144x(x + 1)(x + 1) + x + 2(ii) f(x) =2x + x + 1Sol. (i) f(x) = odd degree polynomial + bounded functioncot –1 x also f´(x) > 0⇒ y = f(x) will be one one and onto42(x + 1)(x + x + 1)(ii) f(x) == x 4 1+ 1 +22x + x + 1x + x + 1= even degree polynomial⎛ 4 ⎞+ bounded function ∈ ⎜0, ⎟⎝ 3 ⎠also f´(x) = 0 has at least one root real which is notrepeated since f´(x) is a polynomial of degree 7.⇒ f(x) = 0 has at least one point of extrema.⇒ many one & Into2. A rectangle ABCD of dimensions r and 2r is foldedalong the diagonal BD such that planes ABD andCBD are perpendicular to each other. Let the positionof the vertex A remains unchanged and C 0 is the newposition of C, then find the distance of C 0 from Aand shortest distance between the edges AB & C 0 D.(0, r) DN(2r, r)CA(0, 0)B (2r, 0)Sol. Let the rectangle ABCD lies on the plane xy. Afterfolding the rectangle along the BD co-ordinates ofpoints in 3-D are-A : (0, 0, 0), B : (2r, 0, 0), C : (2r, r, 0), D(0, r, 0)⎛ 2r r ⎞ ⎛ 2rand N : ⎜ , ,0⎟ and C 0⎝ 5 5⎜ ,⎠ ⎝ 5r52r ⎞, ⎟ ,5 ⎠Now AC 0 =85 r5and shortest distance =AC0.(AB×C0D)=| AB×C D |05 r3unit.3. Let f(x) is a polynomial one-one function such thatf(x).f(y) + 2 = f(x) + f(y) + f(xy) ∀ x, y ∈ R–{0},xf(1) ≠ 1, f´(1) = 3. Let g(x) = (f(x) + 3) – 4 ∫ xf (x) dx,then prove that g(x) is an identity for all givenx ∈ R –{0}.Sol. putting x = y = 1 in given condition we getf(1) 2 + 2 = f(1) + f(1) + f(1)⇒ {f(1)} 2 – 3{f(1)} + 2 = 0⇒ f(1) = 1 or 2 ⇒ f(1) = 2Now put y = 1/x,⎛ 1 ⎞⎛ 1 ⎞f(x) . f ⎜ ⎟ + 2 = f(x) + f ⎜ ⎟ + f(1)⎝ x ⎠ ⎝ x ⎠⇒ f(x) = 1 ± x nAccording to given conditions, f(x) = 1 + x 3xNow, g(x) = [1 + x 3 + 3] – 4 ∫ x(1 + x 3 ) dx = 0⇒ g(x) = 0 for ∀ x ∈ R –{0}4. Let three normals are drawn to the parabola y 2 = 4axat three points P, Q and R, from a fixed point A. Twocircles S 1 and S 2 are drawn on AP and AQ asdiameter. If slope of the common chord of the circlesS 1 and S 2 be m 1 and the slope of the tangent to theparabola at the point R be m 2 , then prove thatm 1 . m 2 = 2.Sol. Let A(h, k) be a fixed pointat 3 + (2a – h)t – k = 0{Q three normals are drawn from (h, k)}Let feet of normals P,Q and R are three points withparameters t 1 , t 2 and t 3 .Common chord of S 1 and S 2 = S 1 – S 2 = (t 1 + t 2 )x +2y – h(t 1 + t 2 ) – 2k = 02Tangent to the parabola at R = t 3 y = x + at 3⎛ t1 + t 2 ⎞ 1 1m 1 . m 2 = – ⎜ ⎟ . =⎝ 2 ⎠ t 3 2(Q t 1 + t 2 + t 3 = 0 for co-normal points).00XtraEdge for IIT-JEE 44 APRIL 2010