tag:blogger.com,1999:blog-42466292249224254672024-03-08T08:40:03.169-08:00IIT JEE Mathematics - Model ProblemsKVSSNraohttp://www.blogger.com/profile/05748254811752425330noreply@blogger.comBlogger20125tag:blogger.com,1999:blog-4246629224922425467.post-34288228510965961502009-01-01T01:55:00.000-08:002009-01-01T01:57:49.635-08:00Ask questions and answer questions about IIT JEE SubjectsKNOWLEDGE QUESTION AND ANSWER BOARD<br /><a href="http://knol.google.com/k/narayana-rao-kvss/-/2utb2lsm2k7a/654#">http://knol.google.com/k/narayana-rao-kvss/-/2utb2lsm2k7a/654#</a>KVSSNraohttp://www.blogger.com/profile/05748254811752425330noreply@blogger.com3tag:blogger.com,1999:blog-4246629224922425467.post-5890740960323784032008-09-27T21:48:00.000-07:002008-09-27T21:50:25.535-07:00Analytical geometry - Combination problemsProblems that require concepts from two or more chapters<br /><br />The radius of the circle passing through the foci of the ellipse<br /><br />x²/16 + y²/9 = 1 and having its centre at (0,3) is<br /><br />a. 4<br />b. 3<br />c. √12<br />d. 7/2<br /><br />JEE 1995KVSSNraohttp://www.blogger.com/profile/05748254811752425330noreply@blogger.com5tag:blogger.com,1999:blog-4246629224922425467.post-18649864380223424632008-08-30T23:53:00.010-07:002008-08-31T01:36:53.117-07:00Binomial Theorem - Model Problems - Past JEE1. The coefficient of x<sup>4</sup> in [(x/2)- (3/x²)]<sup>10</sup> is<br /><br />a. 405/256<br />b. 504/259<br />c. 450/263<br />d. none of these<br /><br />(JEE 1983)<br /><br />2. Let n be a positive integer. If the coefficient of 2nd, 3rd, and 4th terms in the expansion of (1 + x)<sup>n</sup> are in A.P. then the value of n is ________________ .<br />(JEE 1994)<br /><br /><br />3. If the binomial expansion of (a-b)<sup>n</sup>, n≥5, the sum of the 5th and 6th terms is zero. The a/b equals<br /><br />a. (n-5)/6<br />b. (n-4)/5<br />c. 5/(n-4)<br />d. 6/(n-5)<br /><br />(JEE 2001)<br /><br /><br />4. The coefficient of t<sup>24</sup> in (1+t²)<sup>12</sup>(1+t<sup>12</sup>)(1+<sup>24</sup>) is<br /><br />a. <sup>12</sup>C<sub>6</sub> + 3<br />b. <sup>12</sup>C<sub>6</sub> + 1<br />c. <sup>12</sup>C<sub>6</sub><br />d. <sup>12</sup>C<sub>6</sub> + 2<br /><br />(JEE 2003)<br /><br /><br />Answers and Solutions<br /><br />1. The coefficient of x<sup>4</sup> in [(x/2)- (3/x²)]<sup>10</sup> is<br /><br />a. 405/256<br />b. 504/259<br />c. 450/263<br />d. none of these<br /><br />(JEE 1983)<br /><br />Answer (a)<br /><br />Solution:<br /> (r+1) th term of the expansion is given by<br /><sup>10</sup>C<sub>r</sub> (x/2) <sup>10-r</sup>(-3/x²) <sup>r</sup><br /><br />= <sup>10</sup>C<sub>r</sub> (x) <sup>10-r</sup>(1/ x²)<sup>r</sup>(1/2) <sup>10-r</sup>(-3) <sup>r</sup><br /><br />= <sup>10</sup>C<sub>r</sub>(-3) <sup>r</sup>(1/2) <sup>10-r</sup>(x) <sup>10-3r</sup><br /><br />In this term (x) <sup>10-3r</sup> is equal to x<sup>4</sup><br /> 10-3r = 4 or r = 2<br /><br />So coefficient = <br /><sup>10</sup>C<sub>r</sub>(-3) <sup>r</sup>(1/2) <sup>10-r</sup><br /><br /><sup>10</sup>C<sub>2</sub>(-3) <sup>2</sup>(1/2) <sup>10-2</sup><br /><br />= [10*9/2]*9*(1/2<sup>8</sup>)<br /><br />= (45*9)/256<br />= 405/256<br /><br /><br /><br /><br />2. Let n be a positive integer. If the coefficients of 2nd, 3rd, and 4th terms in the expansion of (1 + x)<sup>n</sup> are in A.P. then the value of n is ________________ .<br />(JEE 1994)<br /><br /><br />Answer: 7<br /><br />Solution:<br /><br />The question requires concepts from binomial theorem and concept from arithmetic progression.<br /><br />Coefficients of 2nd, 3rd, and 4th terms in the expansion of (1 + x)<sup>n</sup> are <br /><sup>n</sup>C<sub>1</sub>, <sup>n</sup>C<sub>2</sub>, <sup>n</sup>C<sub>3</sub><br /><br />According to arithmetic progression properties, three terms can be taken as a-d, a, and a+d. Hence sum of first and third terms = 2*second term<br /><br />=> 2(<sup>n</sup>C<sub>2</sub>) = <sup>n</sup>C<sub>1</sub> + <sup>n</sup>C<sub>3</sub><br /><br />=> 2 n(n-1)/2 = n + [n(n-1)(n-2)/6]<br /><br />=> n-1 = 1 +[(n² – 3n+2)/6]<br /><br />=> n² – 3n+2 = 6n-12<br />=> n² – 9n+14 = 0<br />=> (n-2)(n-7) = 0<br /><br />Since 4th term will be there only when n>2, n is equal to 7.<br /><br /><br /><br /><br /><br />3. If the binomial expansion of (a-b)<sup>n</sup>, n≥5, the sum of the 5th and 6th terms is zero. The a/b equals<br /><br />a. (n-5)/6<br />b. (n-4)/5<br />c. 5/(n-4)<br />d. 6/(n-5)<br /><br />(JEE 2001)<br /><br />Answer: (b)<br /><br />Solution<br /><br />5th term = 4+1th term = <sup>n</sup>C<sub>4</sub> a <sup>n-4</sup>(-b) <sup>4</sup><br /><br />6th term = 5+1th term = <sup>n</sup>C<sub>5</sub> a <sup>n-5</sup>(-b) <sup>5</sup><br /><br />As sum of 5th terms and 6th terms is zero<br /> <br /><sup>n</sup>C<sub>4</sub> a <sup>n-4</sup>(-b) <sup>4</sup> + <sup>n</sup>C<sub>5</sub> a <sup>n-5</sup>(-b) <sup>5</sup> = 0<br /><br /><sup>n</sup>C<sub>4</sub> a <sup>n-4</sup>(b) <sup>4</sup> = <sup>n</sup>C<sub>5</sub> a <sup>n-5</sup>(b) <sup>5</sup><br /><br />a <sup>n-4</sup>(b) <sup>4</sup>/ a <sup>n-5</sup>(b) <sup>5</sup> = <sup>n</sup>C<sub>5</sub>/<sup>n</sup>C<sub>4</sub><br /><br />a/b = [n!/(n-5)!(5!)]/[n!/(n-4)!(4!)<br /><br />= (n-4)!/(5(n-5)!)<br />= (n-4)/5<br /><br /><br /><br />4. The coefficient of t<sup>24</sup> in (1+t²)<sup>12</sup>(1+t<sup>12</sup>)(1+<sup>24</sup>) is<br /><br />a. <sup>12</sup>C<sub>6</sub> + 3<br />b. <sup>12</sup>C<sub>6</sub> + 1<br />c. <sup>12</sup>C<sub>6</sub><br />d. <sup>12</sup>C<sub>6</sub> + 2<br /><br />(JEE 2003)<br /><br />Answer: (d)<br /><br />Solution:<br />Multiplying the last two expressions we have to find coefficient of t<sup>24</sup> in (1+t²)<sup>12</sup>(1+t<sup>12</sup>+<sup>24</sup>) <br /><br />t<sup>24</sup> term will come as a sum of three terms in the expanded expression<br /><br />1st term is t<sup>24</sup> term in 1st expression multiplied by 1 in the second expression.<br />2nd term is t<sup>12</sup> term in 1st expression multiplied by t<sup>12</sup> term in the second expression.<br />3rd term is 1 in the first expression multiplied by t<sup>24</sup> term in the second expression.<br /><br />Coefficients of 1 and 3 terms are 1 and 1.<br /><br />Coefficient of 2 term = coefficient t<sup>12</sup> term in 1 st expression which is (1+t²)<sup>12</sup><br /><br />7th term or 6+1th term in the expansion will have t<sup>12</sup> term<br /><br />Hence coefficient of 7th term = <sup>12</sup>C<sub>6</sub><br /><br />Therefore coefficient of t<sup>24</sup> term will be<br /><br /><sup>12</sup>C<sub>6</sub> + 2KVSSNraohttp://www.blogger.com/profile/05748254811752425330noreply@blogger.com1tag:blogger.com,1999:blog-4246629224922425467.post-52515218378753599472008-06-25T09:48:00.005-07:002008-06-25T10:04:59.863-07:00Logarithms - 6If log<sub>0.3</sub>(x-1)<log<sub>0.09</sub>(x-1), then x lies in the interval:
<br />
<br />a.(2,∞)
<br />b. (1,2)
<br />c. (-2,-1)
<br />d. none of these
<br />
<br />(JEE 1983)
<br />
<br />Answer (a)
<br />
<br />log<sub>0.3</sub>(x-1)<log<sub>0.09</sub>(x-1),
<br />
<br />=> log (x-1)/log 0.3 < log (x-1)/log 0.09
<br />
<br />=> log (x-1)/log 0.3 < log (x-1)/2log 0.3
<br />=> log (x-1)>1/2 log (x-1) (as log 0.3<0)
<br />=> long (x-1)>0
<br />=> x-1>1
<br />=> x>2
<br />thus x Є (2,∞)KVSSNraohttp://www.blogger.com/profile/05748254811752425330noreply@blogger.com0tag:blogger.com,1999:blog-4246629224922425467.post-46901075466281998722008-06-21T06:32:00.011-07:002008-06-21T21:11:55.485-07:00Straight Line - Model Problems - 6Problems with complexity<br /><br />1. The set of lines ax+by+c = 0 where a,b,c satisfy the relation 3a+2b+4c = 0 is concurrent at the point --------------. (JEE, 1982)<br /><br />Answer (3/4, ½)<br /><br />3a+2b+4c = 0<br />Dividing the equation by 4<br /><br />3a/4+2b/4+4c/4 = 0<br />=> 3a/4 + b/2+c = 0<br />x = ¾ and y = ½ satisfied this relation<br /><br />Hence the set of lines are concurrent at (3/4, ½)<br /><br />2. The straight lines x+y = 0, 3x+y-4 = 0, and x+3y-4 = 0 form a triangle which is <br /><br />a. isosceles<br />b. equilateral<br />c. right angled<br />d. none of these<br /><br />(JEE 1984)<br />Answer: (a)<br /><br />Solution:<br /><br />x+y = 0 … (1) slope = -1<br />3x+y = 4 … (2) slope = -3/1<br />x+3y = 4 … (3) slope = -1/3<br /><br />So no two lines are perpendicular to each other as m1*m2 is not equal to -1 for any two lines.<br /><br />From (1) and (2) intersection point is<br /><br />A(2,-2)<br /><br />From (2) and (3) intersection point is<br /><br />B(1,1)<br /><br />From (3) and (1) intersection point is<br />C(-2,2)<br /><br />AC² = [(2-(-2)] ² + [-2-2] ² = 32<br />BC² = [1-(-2)] ²+[1-2] ² = 10<br />AB²= [2-1] ² + [-2-1] ² = 10<br />As BC² = AB² which means BC = AB and AC is different, the triangle is a isosceles triangle.<br /><br />3. Three lines px+qy+r = 0, qx+ry+p =0, and rx+py+q +0 are concurrent if:<br /><br />a. p+q+r + 0<br />b. p²+q²+r² = pq+rq+rp<br />c. p³+q³+r³ = 3pqr<br />d. none of these<br /><br />Answer: (a), (b),©<br /><br />Solution<br /><br />The condition for concurrency of three lines<br />a1x +b1y+c1 = 0,<br />a2x+b2y+c2 = 0 and<br />a3x+b3y+c3=0 is<br />The determinant<br /><br />|a1 b1 c1|<br />|a2 b2 c2|<br />|a3 b3 c3|<br /><br />= 0<br /><br />Hence the condition for given lines to be concurrent is<br />|p q r|<br />|q r p|<br />|r p q|<br /><br />= 0<br />=> p(rq-p²) –q(q²-rp) +r(pq-r²) = 0<br />=> prq - p³ - q³+qrp +rpq - r³ = 0<br />= > -p³-q³-r³ = -3pqr<br />=> p³+q³+r³ = 3pqr<br />=> p³+q³+r³ - 3pqr = 0<br />=> (p+q+r)(p²+q²+r²-pq-qr-rp) = 0<br />=> p+q+r = 0 or<br />=> p²+q²+r² = pq+qr+rp<br /><br />4. The points (0,8/3), (1,3) and (82,30) are vertices of:<br /><br />a. an obtuse angled triangle<br />b. an acute angled triangle<br />c. a right angled triangle<br />d. an isosceles triangle<br />e. none of these<br /><br />(JEE 1986)<br /><br />Answer (e)<br /><br />Solution<br /><br />Concepts involed:<br /><br />The slope of a line passing through (x1,y1) and (x2,y2) is (y2-y1)/(x2-x1), provided x1≠x2.<br /><br />The equation of a line having slope m and passing through (x1,y1)and (x2,y2) is<br /><br />(y-y1)/(x-x1) = (y1-y2)/(x1-x2)<br /><br />The equation of the line passing through (0,8/3) and (1,3) is<br /><br />(y - 8/3)/(x) = (8/3 – 3)/(0-1)<br /><br />y- 8/3 = x/3<br />=> x/3 –y +8/3 = 0<br />=> x – 3y +8 = 0<br /><br />Checking whether (82,30) is on this line<br /><br />82 – 3(30)+8 = 90-90 turns out to be zero.<br /><br />Hence given points are collinear.<br /><br />5. State whether the following statement is true or false<br />The lines 2x +3y +19 = 0 and 9x+6y-17 = 0 cut the coordinate axes in concyclic points. <br />(JEE 1988)<br /><br />Answer: True<br /><br />Concyclic points are points on a circle.<br /><br />If ac = bd = k, then the equation of the circle through A(a,0), B(0,b), C(c,0)) and D(0,d) is<br /><br />x² + y² -(a+c)x – (b+d) y +k = 0<br />In this case the points are A(-19/2, 0), B(0,-19/3), C(17/9,0), D (0,17/6)<br /><br />ac = -19/2*17/9 = -323/18<br /><br />bd = -19/3*17/6 = -323/18<br /><br />Hence the given lines cut the coordinate axes in concyclic points. <br /><br />(Note: This problem belongs to “circle” topic. But as it is worded it seems to be a problem under straight line topic)KVSSNraohttp://www.blogger.com/profile/05748254811752425330noreply@blogger.com6tag:blogger.com,1999:blog-4246629224922425467.post-53279884573710103202008-06-21T06:29:00.005-07:002008-06-23T21:18:09.146-07:00Parabola - 11. Find the focus, directrix, and vertex for the parabola x² = -16y<br /><br />For x² = 4ay<br /><br />Focus is (0,a), directrix is y = -a, and vertex is (0,0)<br /><br />Write the given equation as x² = 4 (-4)y<br /><br />Hence focus is (0,-4), directrix is y = 4, and vertex is (0,0).<br />The parabola opens downward.<br /><br />2. Find the equation of the parabola that satisfied the following conditions.<br /><br />Vertex (0,0), latus rectum = 16, opens to the right.<br /><br />Latus rectum for standard parabola = |4a|<br />Hence |a| for the given parabola = 4<br />As vertex is at (0,0) and the parabola open to the right focus is (4,0)<br />The standard equation is y² = 4ax.<br /><br />The equation of the given parabola = y² = 4*4x = 16xKVSSNraohttp://www.blogger.com/profile/05748254811752425330noreply@blogger.com0tag:blogger.com,1999:blog-4246629224922425467.post-24532186548956800882008-06-19T23:27:00.005-07:002008-06-26T01:36:07.993-07:00Hyperbola - 11. Find the following properties of hyperbola for the given equation<br /><br />(i) Points where hyperbola intersects the x-axis<br />(ii) eccentricity<br />(iii) foci<br />(iv) distance between the foci<br />(v) length of the latus rectum<br /><br />Equation of the hyperbola is x2. Equation of hyperbola in its standard form<br /><br />x²/16 - y²/9² = 1<br /><br />Concepts to be used<br /><br />For the hyperbola x²/a² - y²/b² = 1<br /><br /><br />The hyperbola intersects x axis at (a,0) and (-a,0).<br /><br /><br />b² = a²(e²-1)<br />Where e = eccentricity<br /><br />Focus is (ae,0)<br /><br /><br />Solution:<br />Hyperbola given is x²/16 - y²/9 = 1<br /><br />a =4, b= 3<br /><br />Intersecting points on the x axis = (4,0) and (-4,0)<br />Eccentricity 9 = 16(e²-1)<br />=> 9/16 + 1 = e²<br />=> 25/16 = e²<br />=> e = 5/4<br /><br />foci are (±ae,) = (5,0) and (-5,0)<br /><br />distance between the foci = 10<br /><br />Length of the latus rectum = 2b²/a = 18/4 = 9/2KVSSNraohttp://www.blogger.com/profile/05748254811752425330noreply@blogger.com0tag:blogger.com,1999:blog-4246629224922425467.post-13246226583466533272008-06-19T23:26:00.001-07:002008-12-06T09:23:29.701-08:00Hyperbola - Past JEEIf x = 9 is the chord of contact of the hyperbola x²-y² = 9, then the equation of the corresponding pair of tangents is<br /><br />a. 9x² -8y² +18x-9 = 0<br />b. 9x² -8y² -18x+9 = 0<br />c. 9x² -8y² -18x-9 = 0<br />d. 9x² -8y² +18x+9 = 0<br /><br />(JEE 1999)<br /><br />Answer: b.<br /><br />The chord of contact x =9 meets the hyperbola at (9,6√2) and (9,-6√2).<br />the equations of tangent at these points are 3x-2√(2y)-3 = 0 and 3x+2√(2y)-3 = 0.<br />the combined equation of the two is:<br /><br />9x² -8y² -18x+9 = 0KVSSNraohttp://www.blogger.com/profile/05748254811752425330noreply@blogger.com0tag:blogger.com,1999:blog-4246629224922425467.post-33117984799974831762008-06-17T23:28:00.007-07:002008-06-22T08:20:43.086-07:00Definite Integrals - 61. The value of the integral ∫<sub>0</sub><sup> π /2</sup> [√(cot x)/[ √(cot x) +√(tan x)]]dx is<br /><br />a. π/4<br />b. π/2<br />c. π<br />d. none of these<br />(JEE 1983)<br /><br />Answer: (a)<br /><br />Solution:<br /><br />Concept from definite integration to be used:<br /><br />∫<sub>0</sub><sup>a</sup>f(x) = ∫<sub>0</sub><sup> a</sup>f(x-a)<br /><br /><br />I = ∫<sub>0</sub><sup> π /2</sup> [√(cot x)/[ √(cot x) +√(tan x)]]dx ….(i)<br /><br />= ∫<sub>0</sub><sup> π /2</sup> [√(cot(π /2- x))/[ √(cot (π /2 -x)) +√(tan(π /2- x))]]dx<br /><br />= ∫<sub>0</sub><sup> π /2</sup> [√(tan x)/[ √(tan x) +√(cot x)]]dx … (ii)<br /><br />Adding (i) and (ii)<br /><br />2I = ∫<sub>0</sub><sup> π /2</sup> [[√(cot x) +√(tan x)]/[ √(tan x) +√(cot x)]]dx<br /><br />2I = ∫<sub>0</sub><sup> π /2</sup>dx = π /2<br />I = π /4KVSSNraohttp://www.blogger.com/profile/05748254811752425330noreply@blogger.com1tag:blogger.com,1999:blog-4246629224922425467.post-10184528185143684102008-06-17T22:18:00.010-07:002008-06-17T22:47:52.183-07:00Differential Equations - 1Prob: Find the solution of the differential equation<br /><br />log (dy/dx) = 3x + 4y, y(0) = 0<br /><br /><br />Solution:<br /><br />log (dy/dx) = 3x + 4y<br /><br />=> dy/dx = e<sup>3x+4y</sup> = e<sup>3x</sup>e<sup>4y</sup><br /><br />=> e<sup>-4y</sup>dy = e<sup>3x</sup>dx<br /><br />Integrating both sides<br /><br />e<sup>-4y</sup>/-4 = e<sup>3x</sup>/3 +C<br /><br />Using the condition y(0) = 0<br /><br />-1/4 = 1/3 +C<br /><br />=> C = -1/4 - 1/3 = -7/12<br /><br />=>e<sup>-4y</sup>/-4 = e<sup>3x</sup>/3 -7/12<br /><br />=> -(12/4)e<sup>-4y</sup> = (12/3)e<sup>3x</sup> - 7<br /><br />=>3e<sup>-4y</sup> + 4e<sup>3x</sup> = 7KVSSNraohttp://www.blogger.com/profile/05748254811752425330noreply@blogger.com0tag:blogger.com,1999:blog-4246629224922425467.post-87396366656832504032008-06-16T22:02:00.004-07:002008-06-16T22:10:05.086-07:00Limits-1Limit - Definition<br /><br />a given function f(x) is said to have a limit l as x approaches a, lim x→a f(x) =l uf f(x) is as near to l as we please for all values of x≠a but sufficiently near to a.KVSSNraohttp://www.blogger.com/profile/05748254811752425330noreply@blogger.com0tag:blogger.com,1999:blog-4246629224922425467.post-24530641362199945342008-06-16T22:01:00.007-07:002008-06-16T23:08:23.353-07:00Limits-5Prob: Find limit of (sin<sup>-1</sup>x - tan<sup>-1</sup>x)/x³ when x→0<br /><br />when x → 0 both numerator and denominator become zero<br /><br />So L'Hospital rule becomes applicable<br /><br /> L'Hospital rule<br /><br />(i). If f and g are differentiable functions on (0,δ) such that<br />(ii) g'(x)≠ 0 or any x Є (0,δ)<br />(iii) lim x→0<sup>+</sup> f(x) = 0 = lim x→0<sup>4</sup> g(x)<br /><br />(iv) lim x→0<sup>+</sup> f'(x)/g'(x) = L, then lim x→0<sup>+</sup> f(x)/g(x) = L<br /><br /><br />lim x→0(sin<sup>-1</sup>x - tan<sup>-1</sup>x)/x³ =<br /><br />lim x→0 [1/√(1-x²) - 1/(1+x²)]/3x² =<br /><br />1/3 lim x→0 (1/x²)[(1+x²) -√(1-x²)]/[(1+x²)*√(1-x²)] =<br /><br />1/3 lim x→0 (1/x²)[(1+x²) -√(1-x²)]*[(1+x²) +√(1-x²)]/[(1+x²)*√(1-x²)]*[(1+x²) +√(1-x²)]<br /><br />1/3 lim x→0 (1/x²)[(1+x²)² -(1-x²)]/ [(1+x²)*√(1-x²)]*[(1+x²) +√(1-x²)]=<br /><br />1/3 lim x→0 (1/x²)[ 1 + 2x²+x<sup>4</sup> -1 +x²]/[(1+x²)*√(1-x²)]*[(1+x²) +√(1-x²)]=<br /><br />1/3 lim x→0 (1/x²)[ 3x²+x<sup>4</sup>]/[(1+x²)*√(1-x²)]*[(1+x²) +√(1-x²)]=<br /><br />1/3 lim x→0 (1/x²)[ x²(3+x²)]/[(1+x²)*√(1-x²)]*[(1+x²) +√(1-x²)]=<br /><br />1/3 lim x→0 (3+x²)/[(1+x²)*√(1-x²)]*[(1+x²) +√(1-x²)]= (1/3)(3/2) = 1/2KVSSNraohttp://www.blogger.com/profile/05748254811752425330noreply@blogger.com0tag:blogger.com,1999:blog-4246629224922425467.post-29792921102923149892008-06-15T20:44:00.000-07:002008-06-15T21:01:21.648-07:00Limits-10 - Problems with ComplexityLim x→∞ [√(x²+1) - ³√(x³+1)]/[<sup>4</sup>√(x<sup>4</sup>+1) - <sup>5</sup>√(x<sup>3</sup>+1)<br /><br />a. 0<br />b. -1<br />c. 1<br />d. limit does not exist<br /><br />Answer: (a)<br /><br />When limits x→∞ are to be calculated, polynomials have to divided by their highest power term throughout. In this case, dividing numerator and denominator by x will result in the expression with in the root symbols getting divided by highest power term.<br /><br />Lim x→∞ [√(x²+1) - ³√(x³+1)]/[<sup>4</sup>√(x<sup>4</sup>+1) - <sup>5</sup>√(x<sup>3</sup>+1)]<br /><br />= Lim x→∞ [√(1+(1/x²)) - ³√(1+(1/x³))]/[<sup>4</sup>√(1+(1/x<sup>4</sup>)) - <sup>5</sup>√((1/x<sup>2</sup>+(1/x<sup>5</sup>))]<br /><br />= (1-1/(1-0) = 0<br /><br />1/x and Other terms with highest powers of x in the denominator approach zero as x→∞KVSSNraohttp://www.blogger.com/profile/05748254811752425330noreply@blogger.com0tag:blogger.com,1999:blog-4246629224922425467.post-1866035101035325822008-06-15T20:31:00.000-07:002008-06-16T21:58:11.882-07:00Differentiation - 5 - Inverse Trigonometric Functionsd/dx of sin<sup>-1</sup>x = 1/√(1-x²) <br /><br />d/dx of cos<sup>-1</sup>x = -1/√(1-x²) <br /><br />d/dx of tan<sup>-1</sup>x = 1/(1+x²) <br /><br />d/dx of cot<sup>-1</sup>x = -1/(1+x²)<br /><br />d/dx of sec<sup>-1</sup>x = 1/|x|√(x²-1) <br /><br />d/dx of cosec<sup>-1</sup>x = -1/|x|√(x²-1)KVSSNraohttp://www.blogger.com/profile/05748254811752425330noreply@blogger.com0tag:blogger.com,1999:blog-4246629224922425467.post-8317711329985903962008-06-13T00:20:00.005-07:002008-06-13T01:03:58.991-07:00Differentiation - Model Problems - 6Problems having some complexity and requiring use multiple concepts from this chapter as well as other chapters of algebra, trigonometry and geometry<br /><br />1. <strong>Prob:</strong><br /><br />If y = tan<sup>-1</sup>[(√(1+x²) -1)/x]<br />find y'(1)<br /><br />To solve the problem, transform the given expression by putting x = tan θ or θ = tan<sup>-1</sup>x<br /><br />[(√(1+x²) -1)/x] = [(√(1+tan²θ) - 1)/tan θ) <br />=> (√(sec²θ)-1)/tan θ = (sec θ - 1)/tan θ<br /><br />writing sec θ = 1/cos θ and tan θ = sin θ/cos θ<br /><br /> (sec θ - 1)/tan θ = (1- cos θ)/sin θ <br />=> tan θ/2<br /><br />Therefore y tan<sup>-1</sup>(tan θ/2) = θ/2<br />=> y = (tan<sup>-1</sup>x)/2<br /><br />As d/dx of tan<sup>-1</sup>x = 1(1+x²)<br /><br />dy/dx = (1/2)* 1(1+x²)<br /><br />Therefore y'(1) = (1/2)* 1(1+1) = (1/2)*(1/2) = 1/4<br /><br /><br /><br />2. Prob: if f(x) = log<sub>x</sub>(ln x), then find value of f'(x) at e. (JEE 1985)<br /><br />Solution: <br />f(x) = log<sub>x</sub>(ln x), <br />=> x<sup>f(x)</sup> = ln(x)<br /><br />Taking logaritm on both sides<br />f(x)ln x = ln (ln x)<br />Write y = f(x)<br />y (ln x) = ln (ln x)<br /><br />Differentiating both sides with respect to x<br /><br />y*(1/x)+(ln x)*(dy/dx) = (1/ln x)* (1/x)<br /><br />substitute x = e<br /><br />y*(1/e) + (ln e)*(dy/dx) = (1/ln e)*(1/e)<br />=> dy/dx = (1/e) - (y/e)<br /><br />value of y at e = log<sub>x</sub>(ln e) = log<sub>x</sub>(1) = 0<br /><br />Therefore dy/dx = 1/eKVSSNraohttp://www.blogger.com/profile/05748254811752425330noreply@blogger.com0tag:blogger.com,1999:blog-4246629224922425467.post-48233934186902977772008-06-13T00:20:00.003-07:002008-06-22T08:22:16.048-07:00Differentiation - Model Problems - 71. There is exists a function f(x) satisfying f(0) = 1, f’(0) = -1, f(x)>0 for all x , and
<br />
<br />a. f’’(x)>0 for all x
<br />b. -1<f’’’(x)<0 for all x
<br />c. -2≤f’’(x) ≤-1 or all x
<br />d. f’’(x)<-2 for all x
<br />
<br />(JEE, 1982)
<br />
<br />Answer: (a)
<br />
<br />Reason:
<br />
<br />x² -x+1 = (x – ½)² + ¾ can be a solution to the f(x) satisfying f(0) = 1, f’(0) = -1, f(x)>0 for all x as f(0) = 1 and f’(0) = -1
<br />
<br />f’’(x) = 2 and hence f’’(x)>0 for all x.
<br />
<br />We may assume f(x) = e<sup>-x</sup> and get a similar conclusion. But e<sup>-x</sup> may tend to zero as x tends to infinity. Hence f(x) = x² -x+1 is more appropriate function.
<br />
<br />2. If y = f[(2x-1)/( x²+1)] and f’(x) = sin x², then dy/dx = --------------.
<br />(JEE 1982)
<br />
<br />Answer: dy/dx = sin [(2x-1)/( x²+1)] ² * [2 +2x–2x²]/ ( x²+1) ²
<br />
<br />Solution: f’(x) = sin x²
<br />=> f’(t) = sin t²
<br />Given problem is visualized as
<br />y = f(t)
<br />t = [(2x-1)/( x²+1)]
<br />dy/dx = dy/dt*dt/dx
<br />=> sin t² [( x²+1)(2) – (2x-1)(2x)]/ ( x²+1) ²
<br />=> sin [(2x-1)/( x²+1)] ² * [( x²+1)(2) – (2x-1)(2x)]/ ( x²+1) ²
<br />=> sin [(2x-1)/( x²+1)] ² * [2x²+2 –(4x²-2x)]/ ( x²+1) ²
<br />=> sin [(2x-1)/( x²+1)] ² * [2 +2x–2x²]/ ( x²+1) ²
<br />
<br />
<br />KVSSNraohttp://www.blogger.com/profile/05748254811752425330noreply@blogger.com0tag:blogger.com,1999:blog-4246629224922425467.post-26232136478926007042008-06-10T23:43:00.006-07:002008-06-10T23:52:40.833-07:00Progressions - Model Problems - 1Chapter: Sequences, Series and Progressions<br /><br /><strong>Arithmetic Progressions</strong><br /><br /><strong>Finding nth term of arithmetic progessions</strong><br />nth term = a +(n-1)d<br /><br />Prob: Find the 20th term of the A.P. starting with 4 and common difference 5.<br /><br />a = 4<br />d = 5<br />n = 20<br /><br />20th term = 4 + (20-1)*5 = 4+19*5 = 4 + 95 = 99.<br /><br /><strong>Finding sum to n terms of AP</strong><br /><br />Sn = ½ n(a + l)<br />Sn = ½ n{2a+(n-1)d}<br /><br /><br />Prob: Find the sum to 20th term of the A.P. starting with 4 and common difference 5.<br /><br />We found in the earlier problem the 20th term.<br /><br />a = 4<br />n = 20<br />l = 99<br /><br />Sn = ½ *20(4+99) = 10*103 = 1030.<br /><br /><strong>Insertion of arithmetic means</strong><br /><br />Prob: Insert 4 arithmetic means between 4 and 19.<br /><br />We have a = 4, l = 19 and 4 arithmetic means are to be inserted. It means the last term l is the 6th term.<br /><br />19 = 4+5d<br />d = 3<br />the firm A.M. = second term = 4+3 = 7<br />second A.M. = 7+3 = 10<br />third A.M. = 10+3 = 13<br />fourth A.M. = 13+3 = 16KVSSNraohttp://www.blogger.com/profile/05748254811752425330noreply@blogger.com0tag:blogger.com,1999:blog-4246629224922425467.post-43297206813673354482008-06-10T23:42:00.007-07:002008-06-11T00:48:01.890-07:00Progressions - Model Problems - 6Complex problems in progressions<br /><br /><strong>Prob 1:</strong> Find the sum of the series<br /><br />S = 1²-2²+3²-4²+…-2002²+2003²<br /><br />We need transform the given series into<br /><br />S = (1-2)(1+2) + (3-4)(3+4)+…+(2001-2002)(2001+2002)+2003²<br />=> (-1)(3) + (-1)(7) + (-1)(11) +(-1)(15)+…+(-1)(4003) + 2003²<br /><br />=> (-1)[3+7+11+…+4003]+2003²<br />So we have an arithmetic progression in the brackets with a =3, d = 4 and l = 4003 and n = 1001 terms.<br /><br />Sum to n terms in AP is <br />Sn = ½ n(a + l)<br /><br />Hence the sum of 1001 terms in the brackets = 1001*[3+4003]/2<br />= 1001*4006/2 = 1001*2003<br /><br />So S = -1001*2003 + 2003²<br />= 2003(2003-1001) = 2003*1002 = 2007006 <br /><br /><br />Prob 2: Find the value of n for which <br />704 + ½ *(704)+1/4 * (704) +… up to n terms =<br />1984 – ½ *(1984) + ¼* (1984)- … up to n terms<br /><br />Actually there is no complexity in the problem. But for a first look it looks to be a complex problem. You have to identify both LHS and RHS as geometric progressions.<br />For LHS a = 704 and r = ½.<br />For RHS a = 1984 and r = -1/2<br /><br />sum of n terms of GP<br />Sn = a(1-r<sup>n</sup>)/(1-r)<br /><br />704[1 – (1/2)<sup>n</sup>]/(1 – ½ ) = 1984 [1 – (-1/2)<sup>n</sup>]/(1 – (-½ ))<br />704*2(1 – 1/2<sup>n</sup>] = 1984*2/3*[1 – (-1) <sup>n</sup>/2<sup>n</sup>] <br />704*6(1 – 1/2<sup>n</sup>] = 1984*2[1 – (-1) <sup>n</sup>/2<sup>n</sup>]<br />4224 – 4224/2<sup>n</sup> = 3968 - 3968(-1) <sup>n</sup>/2<sup>n</sup><br />4224-3968 = 4224/2<sup>n</sup> - 3968(-1) <sup>n</sup>/2<sup>n</sup><br />256 = 4224/2<sup>n</sup> - 3968(-1) <sup>n</sup>/2<sup>n</sup><br />128 = 2112/2<sup>n</sup> - 1984(-1) <sup>n</sup>/2<sup>n</sup><br /><br />If n is assumed as odd<br />128 = (2112+1984)/ 2<sup>n</sup><br />=> 2<sup>n</sup> = 4096/128 = 1024/32 = 128/4 = 32<br />=> 2<sup>n</sup> = 32<br />n = 5<br /><br />If n is assumed as even<br /><br />128 = (2112 – 1984)/ 2<sup>n</sup> = 128/2<sup>n</sup><br />=> 2<sup>n</sup> = 128/128 = 1<br />Implies n = 0<br /><br />Hence the answer is n =5 or n = 0. The logical answer is 5 as n = 0 is a trivial solution..KVSSNraohttp://www.blogger.com/profile/05748254811752425330noreply@blogger.com0tag:blogger.com,1999:blog-4246629224922425467.post-50435866000368240062008-06-10T23:42:00.005-07:002008-06-24T08:24:52.730-07:00Progressions - Model Problems - 71. The third term of a geometric progression is 4. The product of the first five terms is:<br /><br />a. 4³<br />b. 4<sup>4</sup><br />c. 4<sup>5</sup><br />d. none of these<br /><br />(JEE 1982)<br /><br />Answer ©<br /><br />Select the five terms as a/r², a/r,a, ar,ar².<br /><br />Product of the five terms is a/r² * a/r * a *ar*ar² = a<sup>5</sup><br /><br />As the third term is 4, a = 4<br />Hence product is 4<sup>5</sup><br /><br />2. The sum of integers from 1 to 100 that are divisible by 2 or 5 is ---------------------.<br />(JEE 1984)<br /><br />Answer: 3050<br /><br />Integers divisible by 2 are 2,4,6,…,100 ….(50 numbers)<br />Integers divisible by 5 are 5, 10, 15,…100 (20 numbers)<br /><br />Integers divisible by 10 will be in integers divisible by 2. We can remove them from integers divisible by 5 to find out the sum required<br /><br />So integers for which sum is to be taken 2,4,6…,100 and<br />5,15,25,…,95<br /><br />First progression sum 2(sum of 1,2,3,…,50)<br />= 2*50*51/2 = 2550<br /><br />Second progression is an A.P. with d = 10 and a = 5<br /><br />Hence sum = n/2(a+l) = 10/2(100) = 500<br /><br />Total sum = 2550 +500 = 3050KVSSNraohttp://www.blogger.com/profile/05748254811752425330noreply@blogger.com0tag:blogger.com,1999:blog-4246629224922425467.post-16434272698470003642008-06-09T23:35:00.000-07:002008-06-10T00:20:35.574-07:00Complex Numbers -Model Problems - 1To find i<sup>n</sup> divide n by 4 to get 4m+r where m is the quotient and r is the remainder.<br /><br />i<sup>n</sup> will be equal to i<sup>r</sup><br /><br />Prob: The value of i<sup>53</sup>/i<sup>121</sup> is<br /><br />a. 2i<br />b. i<br />c. -2i<br />d. 2<br /><br />53 = 13*4 + 1<br /><br />So i<sup>53</sup> = i <br /><br />121 = 30*4 + 1<br /><br />So i<sup>121</sup> = i<br /><br />i<sup>53</sup>/i<sup>121</sup> is equal to i/i = 1<br /><br />Answer (b)<br /><br /><br /><br /><br /><br /><strong>Multiplication of complex numbers</strong><br /><br />(a1+ib1) (a2+ib2) by multiplying and simplifying we get<br /><br />(a1a2 – b1b2) + i(a1b2+a2b1)<br /><br /><br />Prob: Multiply (2+i) by (2+i)<br /><br />(2+i)(2+i) = 2*2 - 1*1 + i(2*1+2*1) = 3+4i<br /><br /><br /><strong>Division of complex numbers</strong> <br /><br />z1/z2 = z1* Multiplicative inverse of z2<br /><br />Multiplicative inverse of a+ib = a/(a² + b²) - ib/(a² + b²)<br /><br />Prob: Find the result of (7+i)/(1+3i)<br /><br />First step; Find multiplicative inverse of 1+3i = 1/(1²+3²) - i 3/(1²+3²)<br /> = 1/10 - i 3/10<br /><br />Therefore (7+i)/(1+3i) = (7+i)(1/10 - 3i/10)<br /> = 7*1/10 - (1)(-3/10) + i(7*(-3/10)+1(1/10))<br /> = 7/10 + 3/10 + i(-21/10+1/10)<br /> = 10/10 + i(-20/10)<br /> = 1 - 2iKVSSNraohttp://www.blogger.com/profile/05748254811752425330noreply@blogger.com1