Logarithms - 6

If log_0.3(x-1)0.09(x-1), then x lies in the interval:

a.(2,∞)
b. (1,2)
c. (-2,-1)
d. none of these

(JEE 1983)

Answer (a)

log_0.3(x-1)0.09(x-1),

=> log (x-1)/log 0.3 < log (x-1)/log 0.09

=> log (x-1)/log 0.3 < log (x-1)/2log 0.3
=> log (x-1)>1/2 log (x-1) (as log 0.3<0)
=> long (x-1)>0
=> x-1>1
=> x>2
thus x Є (2,∞)

Straight Line - Model Problems - 6

Problems with complexity

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)

Answer (3/4, ½)

3a+2b+4c = 0
Dividing the equation by 4

3a/4+2b/4+4c/4 = 0
=> 3a/4 + b/2+c = 0
x = ¾ and y = ½ satisfied this relation

Hence the set of lines are concurrent at (3/4, ½)

2. The straight lines x+y = 0, 3x+y-4 = 0, and x+3y-4 = 0 form a triangle which is

a. isosceles
b. equilateral
c. right angled
d. none of these

(JEE 1984)
Answer: (a)

Solution:

x+y = 0 … (1) slope = -1
3x+y = 4 … (2) slope = -3/1
x+3y = 4 … (3) slope = -1/3

So no two lines are perpendicular to each other as m1*m2 is not equal to -1 for any two lines.

From (1) and (2) intersection point is

A(2,-2)

From (2) and (3) intersection point is

B(1,1)

From (3) and (1) intersection point is
C(-2,2)

AC² = [(2-(-2)] ² + [-2-2] ² = 32
BC² = [1-(-2)] ²+[1-2] ² = 10
AB²= [2-1] ² + [-2-1] ² = 10
As BC² = AB² which means BC = AB and AC is different, the triangle is a isosceles triangle.

3. Three lines px+qy+r = 0, qx+ry+p =0, and rx+py+q +0 are concurrent if:

a. p+q+r + 0
b. p²+q²+r² = pq+rq+rp
c. p³+q³+r³ = 3pqr
d. none of these

Answer: (a), (b),©

Solution

The condition for concurrency of three lines
a1x +b1y+c1 = 0,
a2x+b2y+c2 = 0 and
a3x+b3y+c3=0 is
The determinant

|a1 b1 c1|
|a2 b2 c2|
|a3 b3 c3|

= 0

Hence the condition for given lines to be concurrent is
|p q r|
|q r p|
|r p q|

= 0
=> p(rq-p²) –q(q²-rp) +r(pq-r²) = 0
=> prq - p³ - q³+qrp +rpq - r³ = 0
= > -p³-q³-r³ = -3pqr
=> p³+q³+r³ = 3pqr
=> p³+q³+r³ - 3pqr = 0
=> (p+q+r)(p²+q²+r²-pq-qr-rp) = 0
=> p+q+r = 0 or
=> p²+q²+r² = pq+qr+rp

4. The points (0,8/3), (1,3) and (82,30) are vertices of:

a. an obtuse angled triangle
b. an acute angled triangle
c. a right angled triangle
d. an isosceles triangle
e. none of these

(JEE 1986)

Answer (e)

Solution

Concepts involed:

The slope of a line passing through (x1,y1) and (x2,y2) is (y2-y1)/(x2-x1), provided x1≠x2.

The equation of a line having slope m and passing through (x1,y1)and (x2,y2) is

(y-y1)/(x-x1) = (y1-y2)/(x1-x2)

The equation of the line passing through (0,8/3) and (1,3) is

(y - 8/3)/(x) = (8/3 – 3)/(0-1)

y- 8/3 = x/3
=> x/3 –y +8/3 = 0
=> x – 3y +8 = 0

Checking whether (82,30) is on this line

82 – 3(30)+8 = 90-90 turns out to be zero.

Hence given points are collinear.

5. State whether the following statement is true or false
The lines 2x +3y +19 = 0 and 9x+6y-17 = 0 cut the coordinate axes in concyclic points.
(JEE 1988)

Answer: True

Concyclic points are points on a circle.

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

x² + y² -(a+c)x – (b+d) y +k = 0
In this case the points are A(-19/2, 0), B(0,-19/3), C(17/9,0), D (0,17/6)

ac = -19/2*17/9 = -323/18

bd = -19/3*17/6 = -323/18

Hence the given lines cut the coordinate axes in concyclic points.

(Note: This problem belongs to “circle” topic. But as it is worded it seems to be a problem under straight line topic)

Parabola - 1

1. Find the focus, directrix, and vertex for the parabola x² = -16y

For x² = 4ay

Focus is (0,a), directrix is y = -a, and vertex is (0,0)

Write the given equation as x² = 4 (-4)y

Hence focus is (0,-4), directrix is y = 4, and vertex is (0,0).
The parabola opens downward.

2. Find the equation of the parabola that satisfied the following conditions.

Vertex (0,0), latus rectum = 16, opens to the right.

Latus rectum for standard parabola = |4a|
Hence |a| for the given parabola = 4
As vertex is at (0,0) and the parabola open to the right focus is (4,0)
The standard equation is y² = 4ax.

The equation of the given parabola = y² = 4*4x = 16x

Hyperbola - 1

1. Find the following properties of hyperbola for the given equation

(i) Points where hyperbola intersects the x-axis
(ii) eccentricity
(iii) foci
(iv) distance between the foci
(v) length of the latus rectum

Equation of the hyperbola is x2. Equation of hyperbola in its standard form

x²/16 - y²/9² = 1

Concepts to be used

For the hyperbola x²/a² - y²/b² = 1


The hyperbola intersects x axis at (a,0) and (-a,0).


b² = a²(e²-1)
Where e = eccentricity

Focus is (ae,0)


Solution:
Hyperbola given is x²/16 - y²/9 = 1

a =4, b= 3

Intersecting points on the x axis = (4,0) and (-4,0)
Eccentricity 9 = 16(e²-1)
=> 9/16 + 1 = e²
=> 25/16 = e²
=> e = 5/4

foci are (±ae,) = (5,0) and (-5,0)

distance between the foci = 10

Length of the latus rectum = 2b²/a = 18/4 = 9/2

Hyperbola - Past JEE

If x = 9 is the chord of contact of the hyperbola x²-y² = 9, then the equation of the corresponding pair of tangents is

a. 9x² -8y² +18x-9 = 0
b. 9x² -8y² -18x+9 = 0
c. 9x² -8y² -18x-9 = 0
d. 9x² -8y² +18x+9 = 0

(JEE 1999)

Answer: b.

The chord of contact x =9 meets the hyperbola at (9,6√2) and (9,-6√2).
the equations of tangent at these points are 3x-2√(2y)-3 = 0 and 3x+2√(2y)-3 = 0.
the combined equation of the two is:

9x² -8y² -18x+9 = 0

Definite Integrals - 6

1. The value of the integral ∫₀ ^{π /2} [√(cot x)/[ √(cot x) +√(tan x)]]dx is

a. π/4
b. π/2
c. π
d. none of these
(JEE 1983)

Answer: (a)

Solution:

Concept from definite integration to be used:

∫₀^af(x) = ∫₀ ^af(x-a)


I = ∫₀ ^{π /2} [√(cot x)/[ √(cot x) +√(tan x)]]dx ….(i)

= ∫₀ ^{π /2} [√(cot(π /2- x))/[ √(cot (π /2 -x)) +√(tan(π /2- x))]]dx

= ∫₀ ^{π /2} [√(tan x)/[ √(tan x) +√(cot x)]]dx … (ii)

Adding (i) and (ii)

2I = ∫₀ ^{π /2} [[√(cot x) +√(tan x)]/[ √(tan x) +√(cot x)]]dx

2I = ∫₀ ^{π /2}dx = π /2
I = π /4

Differential Equations - 1

Prob: Find the solution of the differential equation

log (dy/dx) = 3x + 4y, y(0) = 0


Solution:

log (dy/dx) = 3x + 4y

=> dy/dx = e^3x+4y = e^3xe^4y

=> e^-4ydy = e^3xdx

Integrating both sides

e^-4y/-4 = e^3x/3 +C

Using the condition y(0) = 0

-1/4 = 1/3 +C

=> C = -1/4 - 1/3 = -7/12

=>e^-4y/-4 = e^3x/3 -7/12

=> -(12/4)e^-4y = (12/3)e^3x - 7

=>3e^-4y + 4e^3x = 7

Limits-1

Limit - Definition

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.

Limits-5

Prob: Find limit of (sin^-1x - tan^-1x)/x³ when x→0

when x → 0 both numerator and denominator become zero

So L'Hospital rule becomes applicable

L'Hospital rule

(i). If f and g are differentiable functions on (0,δ) such that
(ii) g'(x)≠ 0 or any x Є (0,δ)
(iii) lim x→0⁺ f(x) = 0 = lim x→0⁴ g(x)

(iv) lim x→0⁺ f'(x)/g'(x) = L, then lim x→0⁺ f(x)/g(x) = L


lim x→0(sin^-1x - tan^-1x)/x³ =

lim x→0 [1/√(1-x²) - 1/(1+x²)]/3x² =

1/3 lim x→0 (1/x²)[(1+x²) -√(1-x²)]/[(1+x²)*√(1-x²)] =

1/3 lim x→0 (1/x²)[(1+x²) -√(1-x²)]*[(1+x²) +√(1-x²)]/[(1+x²)*√(1-x²)]*[(1+x²) +√(1-x²)]

1/3 lim x→0 (1/x²)[(1+x²)² -(1-x²)]/ [(1+x²)*√(1-x²)]*[(1+x²) +√(1-x²)]=

1/3 lim x→0 (1/x²)[ 1 + 2x²+x⁴ -1 +x²]/[(1+x²)*√(1-x²)]*[(1+x²) +√(1-x²)]=

1/3 lim x→0 (1/x²)[ 3x²+x⁴]/[(1+x²)*√(1-x²)]*[(1+x²) +√(1-x²)]=

1/3 lim x→0 (1/x²)[ x²(3+x²)]/[(1+x²)*√(1-x²)]*[(1+x²) +√(1-x²)]=

1/3 lim x→0 (3+x²)/[(1+x²)*√(1-x²)]*[(1+x²) +√(1-x²)]= (1/3)(3/2) = 1/2

Limits-10 - Problems with Complexity

Lim x→∞ [√(x²+1) - ³√(x³+1)]/[⁴√(x⁴+1) - ⁵√(x³+1)

a. 0
b. -1
c. 1
d. limit does not exist

Answer: (a)

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.

Lim x→∞ [√(x²+1) - ³√(x³+1)]/[⁴√(x⁴+1) - ⁵√(x³+1)]

= Lim x→∞ [√(1+(1/x²)) - ³√(1+(1/x³))]/[⁴√(1+(1/x⁴)) - ⁵√((1/x²+(1/x⁵))]

= (1-1/(1-0) = 0

1/x and Other terms with highest powers of x in the denominator approach zero as x→∞

Differentiation - 5 - Inverse Trigonometric Functions

d/dx of sin^-1x = 1/√(1-x²)

d/dx of cos^-1x = -1/√(1-x²)

d/dx of tan^-1x = 1/(1+x²)

d/dx of cot^-1x = -1/(1+x²)

d/dx of sec^-1x = 1/|x|√(x²-1)

d/dx of cosec^-1x = -1/|x|√(x²-1)

Differentiation - Model Problems - 6

Problems having some complexity and requiring use multiple concepts from this chapter as well as other chapters of algebra, trigonometry and geometry

1. Prob:

If y = tan^-1[(√(1+x²) -1)/x]
find y'(1)

To solve the problem, transform the given expression by putting x = tan θ or θ = tan^-1x

[(√(1+x²) -1)/x] = [(√(1+tan²θ) - 1)/tan θ)
=> (√(sec²θ)-1)/tan θ = (sec θ - 1)/tan θ

writing sec θ = 1/cos θ and tan θ = sin θ/cos θ

(sec θ - 1)/tan θ = (1- cos θ)/sin θ
=> tan θ/2

Therefore y tan^-1(tan θ/2) = θ/2
=> y = (tan^-1x)/2

As d/dx of tan^-1x = 1(1+x²)

dy/dx = (1/2)* 1(1+x²)

Therefore y'(1) = (1/2)* 1(1+1) = (1/2)*(1/2) = 1/4



2. Prob: if f(x) = log_x(ln x), then find value of f'(x) at e. (JEE 1985)

Solution:
f(x) = log_x(ln x),
=> x^f(x) = ln(x)

Taking logaritm on both sides
f(x)ln x = ln (ln x)
Write y = f(x)
y (ln x) = ln (ln x)

Differentiating both sides with respect to x

y*(1/x)+(ln x)*(dy/dx) = (1/ln x)* (1/x)

substitute x = e

y*(1/e) + (ln e)*(dy/dx) = (1/ln e)*(1/e)
=> dy/dx = (1/e) - (y/e)

value of y at e = log_x(ln e) = log_x(1) = 0

Therefore dy/dx = 1/e

Differentiation - Model Problems - 7

1. There is exists a function f(x) satisfying f(0) = 1, f’(0) = -1, f(x)>0 for all x , and

a. f’’(x)>0 for all x
b. -1c. -2≤f’’(x) ≤-1 or all x
d. f’’(x)<-2 for all x

(JEE, 1982)

Answer: (a)

Reason:

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

f’’(x) = 2 and hence f’’(x)>0 for all x.

We may assume f(x) = e^-x and get a similar conclusion. But e^-x may tend to zero as x tends to infinity. Hence f(x) = x² -x+1 is more appropriate function.

2. If y = f[(2x-1)/( x²+1)] and f’(x) = sin x², then dy/dx = --------------.
(JEE 1982)

Answer: dy/dx = sin [(2x-1)/( x²+1)] ² * [2 +2x–2x²]/ ( x²+1) ²

Solution: f’(x) = sin x²
=> f’(t) = sin t²
Given problem is visualized as
y = f(t)
t = [(2x-1)/( x²+1)]
dy/dx = dy/dt*dt/dx
=> sin t² [( x²+1)(2) – (2x-1)(2x)]/ ( x²+1) ²
=> sin [(2x-1)/( x²+1)] ² * [( x²+1)(2) – (2x-1)(2x)]/ ( x²+1) ²
=> sin [(2x-1)/( x²+1)] ² * [2x²+2 –(4x²-2x)]/ ( x²+1) ²
=> sin [(2x-1)/( x²+1)] ² * [2 +2x–2x²]/ ( x²+1) ²

Progressions - Model Problems - 1

Chapter: Sequences, Series and Progressions

Arithmetic Progressions

Finding nth term of arithmetic progessions
nth term = a +(n-1)d

Prob: Find the 20th term of the A.P. starting with 4 and common difference 5.

a = 4
d = 5
n = 20

20th term = 4 + (20-1)*5 = 4+19*5 = 4 + 95 = 99.

Finding sum to n terms of AP

Sn = ½ n(a + l)
Sn = ½ n{2a+(n-1)d}


Prob: Find the sum to 20th term of the A.P. starting with 4 and common difference 5.

We found in the earlier problem the 20th term.

a = 4
n = 20
l = 99

Sn = ½ *20(4+99) = 10*103 = 1030.

Insertion of arithmetic means

Prob: Insert 4 arithmetic means between 4 and 19.

We have a = 4, l = 19 and 4 arithmetic means are to be inserted. It means the last term l is the 6th term.

19 = 4+5d
d = 3
the firm A.M. = second term = 4+3 = 7
second A.M. = 7+3 = 10
third A.M. = 10+3 = 13
fourth A.M. = 13+3 = 16

Progressions - Model Problems - 6

Complex problems in progressions

Prob 1: Find the sum of the series

S = 1²-2²+3²-4²+…-2002²+2003²

We need transform the given series into

S = (1-2)(1+2) + (3-4)(3+4)+…+(2001-2002)(2001+2002)+2003²
=> (-1)(3) + (-1)(7) + (-1)(11) +(-1)(15)+…+(-1)(4003) + 2003²

=> (-1)[3+7+11+…+4003]+2003²
So we have an arithmetic progression in the brackets with a =3, d = 4 and l = 4003 and n = 1001 terms.

Sum to n terms in AP is
Sn = ½ n(a + l)

Hence the sum of 1001 terms in the brackets = 1001*[3+4003]/2
= 1001*4006/2 = 1001*2003

So S = -1001*2003 + 2003²
= 2003(2003-1001) = 2003*1002 = 2007006


Prob 2: Find the value of n for which
704 + ½ *(704)+1/4 * (704) +… up to n terms =
1984 – ½ *(1984) + ¼* (1984)- … up to n terms

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.
For LHS a = 704 and r = ½.
For RHS a = 1984 and r = -1/2

sum of n terms of GP
Sn = a(1-rⁿ)/(1-r)

704[1 – (1/2)ⁿ]/(1 – ½ ) = 1984 [1 – (-1/2)ⁿ]/(1 – (-½ ))
704*2(1 – 1/2ⁿ] = 1984*2/3*[1 – (-1) ⁿ/2ⁿ]
704*6(1 – 1/2ⁿ] = 1984*2[1 – (-1) ⁿ/2ⁿ]
4224 – 4224/2ⁿ = 3968 - 3968(-1) ⁿ/2ⁿ
4224-3968 = 4224/2ⁿ - 3968(-1) ⁿ/2ⁿ
256 = 4224/2ⁿ - 3968(-1) ⁿ/2ⁿ
128 = 2112/2ⁿ - 1984(-1) ⁿ/2ⁿ

If n is assumed as odd
128 = (2112+1984)/ 2ⁿ
=> 2ⁿ = 4096/128 = 1024/32 = 128/4 = 32
=> 2ⁿ = 32
n = 5

If n is assumed as even

128 = (2112 – 1984)/ 2ⁿ = 128/2ⁿ
=> 2ⁿ = 128/128 = 1
Implies n = 0

Hence the answer is n =5 or n = 0. The logical answer is 5 as n = 0 is a trivial solution..

Progressions - Model Problems - 7

1. The third term of a geometric progression is 4. The product of the first five terms is:

a. 4³
b. 4⁴
c. 4⁵
d. none of these

(JEE 1982)

Answer ©

Select the five terms as a/r², a/r,a, ar,ar².

Product of the five terms is a/r² * a/r * a *ar*ar² = a⁵

As the third term is 4, a = 4
Hence product is 4⁵

2. The sum of integers from 1 to 100 that are divisible by 2 or 5 is ---------------------.
(JEE 1984)

Answer: 3050

Integers divisible by 2 are 2,4,6,…,100 ….(50 numbers)
Integers divisible by 5 are 5, 10, 15,…100 (20 numbers)

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

So integers for which sum is to be taken 2,4,6…,100 and
5,15,25,…,95

First progression sum 2(sum of 1,2,3,…,50)
= 2*50*51/2 = 2550

Second progression is an A.P. with d = 10 and a = 5

Hence sum = n/2(a+l) = 10/2(100) = 500

Total sum = 2550 +500 = 3050

Complex Numbers -Model Problems - 1

To find iⁿ divide n by 4 to get 4m+r where m is the quotient and r is the remainder.

iⁿ will be equal to i^r

Prob: The value of i⁵³/i¹²¹ is

a. 2i
b. i
c. -2i
d. 2

53 = 13*4 + 1

So i⁵³ = i

121 = 30*4 + 1

So i¹²¹ = i

i⁵³/i¹²¹ is equal to i/i = 1

Answer (b)





Multiplication of complex numbers

(a1+ib1) (a2+ib2) by multiplying and simplifying we get

(a1a2 – b1b2) + i(a1b2+a2b1)


Prob: Multiply (2+i) by (2+i)

(2+i)(2+i) = 2*2 - 1*1 + i(2*1+2*1) = 3+4i


Division of complex numbers

z1/z2 = z1* Multiplicative inverse of z2

Multiplicative inverse of a+ib = a/(a² + b²) - ib/(a² + b²)

Prob: Find the result of (7+i)/(1+3i)

First step; Find multiplicative inverse of 1+3i = 1/(1²+3²) - i 3/(1²+3²)
= 1/10 - i 3/10

Therefore (7+i)/(1+3i) = (7+i)(1/10 - 3i/10)
= 7*1/10 - (1)(-3/10) + i(7*(-3/10)+1(1/10))
= 7/10 + 3/10 + i(-21/10+1/10)
= 10/10 + i(-20/10)
= 1 - 2i