MATHSStudents' ForumExpert’s <strong>Solution</strong> 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
5. Let z 1 , z 2 and z 3 are unimodular complex numbersthen find the greatest value of |z 1 – z 2 | 2 + |z 2 – z 3 | 2 +|z 3 – z 1 | 2 .Sol. |z 1 – z 2 | 2 + |z 2 – z 3 | 2 + |z 3 – z 1 | 2= 2[|z 1 | 2 + |z 2 | 2 + |z 3 | 2 ] – [z 1 z 2 + z1z 2 + z1z 3+ z3z 1 + z 2 z 3 + z2z 3 ]= 6 – [z 1 z 2 + z 2 z 1 + z 1 z 3 + z 3 z 1 +z 2 z 3 + z 3 z 2 ]...(1)Now |z 1 + z 2 + z 3 | 2 ≥ 0⇒ z 1 z 2 + z 2 z 3 + z 2 z 1 + z 3 z 2 + z 1 z 3 + 1From (1) & (2)max m value of |z 1 – z 2 | 2 + |z 2 – z 3 | 2 + |z 3 – z 1 | 2= 6 – (–3) = 9z z 3 ≥ –3...(2)6. Consider following two infinite series in real θ and rC = 1 + r cos θ +22r cos 2θ r 3 cos3θ + + ....2! 3!r 2 sin 2θ r 3 sin 3θ S = r sin θ + + + .....2! 3!If a remains constant and r varies the prove thatdC dS(i) C + S = (C 2 + S 2 ) cos θdr dr⎛ dC ⎞(ii) ⎜ ⎟⎠⎝ drSol. We have2C + iS =C – iS =Now,C 2 + S 2 =⎛ dS ⎞+ ⎜ ⎟⎠⎝ dr2= C 2 + S 2iθree ...(1)−iθree ...(2)re i2e θ =ercosθ.e2i rsin θ= e 2r cos θ ...(3)Differentiating (1) w.r.t.r, we getdC dS + i = e iθ iθr.e. e ....(4)dr dr2⎛ dC ⎞ ⎛ dS ⎞iθiθr.e∴ ⎜ ⎟⎠ + ⎜ ⎟⎠ = e .e = e⎝ dr ⎝ dr= C 2 + S 2 (Form 3)multiply (2) and (4)⎛ dC dS ⎞⎜ + i ⎟ (C – iS) = e iθ iθr.e. e . e⎝ dr dr ⎠= e iθ i ir(e e )[eθ − θ+]2−iθr.e= e iθ 2r cosθ. eNow equating real parts in both sidesdC dSC + S = (C 2 + S 2 ) cosθdr drHence proved.2r cos θScience FactsSkin Deep StorageChip implants that keeptrack of personalinformation seem like anovelty but do they have amore useful future?These days, some people are following theirpets and getting tagged. Radio frequency identification(RFID) chips are the size of a grain of rice and can beloaded up with personal information like passwordsand implanted under the skin. Instead of having toremember a login code, an RFID reader can be set upto automatically detect it and grant you access to arange of things from your computer to your frontdoor. It seems like it could be useful to people withexceptionally poor memories, but right now thesechips are being snapped up by technology geeks likeAmal Graafstra. The 29-year-old businessman fromVancouver, Canada, is one of the first people to havean RFID implant and so far is happy with the results. "Ijust don't want to be without access to the things thatI need to get access to. In his chip, he has stored aunique identification number which can be used to loghim into various electronic devices. It didn't cost himan arm and a leg either: he got the whole set-up onthe internet for about $50 (£30), including the $2 costof the chip itself.The procedure to implant the chip is quitesimple and painless. Amal's chip was implanted underthe skin of his left hand while he was under a localanesthetic. It is possible to inject the chips using alarge enough needle, but in Amal's case the chip wasinserted by simply cutting through his skin with ascalpel. Other than complaining of sensitivity in thearea of the implant, Amal said that it doesn't hurt andhe expects that eventually the chip will be completelyunobtrusive.A hand implanted with anRFID chip and the chip reader.The chip is made of silicon and isdigitally encoded with information..A RFID reader, which is installed in a computer or anelectronic device like a reader by a front door, emits aradio signal of a particular frequency, just like radiostations each broadcast on their own frequency. Thechip acts passively when it is within 3 inches of thereader: the right incoming radio signal induces justenough energy in the antenna of the chip for a circuitin the chip to power up and produce a response. Thereader can then access the information on the chip andpass it on to the computer or device that requires it.XtraEdge for IIT-JEE 45 APRIL 2010
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