MATRIX: Definition: A matrix is defined as an ordered rectangular array of numbers. Elementary Transformation 3. R i → R i + kR j means multiply each element of j th row by k and add it to the corresponding elements of i th row. 4. In applying one or more row operations while finding A -1 by elementary row operations, we obtain all zeros in one or more, then A -1 does not exist. NOTE If A is symm. As well as skew- symm., then A is a null matrix.( if A = A T then A is Symm. And if A = - A T then A is skew- symm.)
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MATRIX:
Definition: A matrix is defined as an ordered rectangular array of numbers.
Elementary Transformation
3. Ri → Ri + kRj means multiply each element of jth
row by k and add it to the corresponding elements of ith
row.
4. In applying one or more row operations while finding A-1 by elementary row operations, we obtain all zeros in one or more, then A-1 does not exist.
NOTE If A is symm. As well as skew-symm., then A is a null matrix.( if A = AT then A is Symm. And if A = - AT then A is skew- symm.)
A =[1 0 00 2 00 0 4] is symmetric and B[ 0 −5 8
5 0 12−8 −12 0 ] is skew-
symmetric.
NOTE: (i) If A and B are symmetric matrices, then BA-2AB is neither symm. nor skew-symm.
(ii) If A is symm. matrix then BTAB is symm. (iii) If A and B are symmetric matrices of same order, then AB is symm. iff AB=BA (iv) Zero matrix is both symm. and skew-
symm. (v) Sum of two skew-symm. matrices is always skew-symm (vi) If A is a symm., then A3 is a symm. and if A is skew-symm., then A2 is a symm.
The Determinant of a Matrix
DEFINITION: Determinants play an important role in finding the inverse of a matrix and also in solving systems of linear equations. In the following we assume we have a square matrix (m = n). The determinant of a matrix A will be denoted by det(A) or |A|. Firstly the determinant of a 2×2 and 3×3 matrix will be introduced, then the n×n case will be shown.
Consistent and Inconsistent SolutionsConsistent system : A system of equation is said to be consistent if its solution ( one or more ) exists.Inconsistent system : A system of equation is said to be inconsistent if its solution does not exist. Working rule to check consistency : Case I When A0 System is consistence and has uniquesolution. Case II When A=0 .Find Adj(A) and then find Adj(A) .B. If Adj(A) .B 0 then system is inconsistence .Case III If Adj(A).B=0 Then it may have infinite solutions then it is consistence or have no solution then it is inconsistence.Properties of Determinants:Property 1. If each element of a row (column)ofdeterminant is zero , then value of determinant is zero.Property 2. Value of a determinant is not changed bychanging the rows into columns and columns into rows.
Property 3. If two adjacent rows (columns)of a determinant are interchanged , then the sign of the determinant is changed but its numerical value is unchanged.
Property 4. If two rows (columns) are identical, then the value of the determinant is zero.Property 5. If every element of a row (column) is multiplied by some constant k,the value of the determinant is multiplied by k.Property 6 .If each element in any row (column) consist of two terms , then the determinant can be expressed as the sum of the determinants of same order. Property 7 . The value of a determinant remain unchanged if to each element of a row (column) be add ( or subtracted) equimultiplies of the corresponding elements of one or more rows (columns) of the determinant.Property 8. The value of the determinant of a diagonal matrix is equal to the product of the diagonal elements.Property 9. The value of the determinant of a skew-symmetric matrix of odd order is always zero.Property 10. The determinant of a symmetric matrix of even order is always a perfect square.ASSIGNMENT(matrices)
Qoestion.1 Using matrices, solve the following system of equations
(i) x+2y+z = 1 , 2x – y+z = 5 , 3x+y – z = 0.
[Hint use AX = B ⇨ X = A-1 B, |A|=15≠0 means A is invertible. Adj(A) =
[0 3 35 −4 15 5 −5] ,A-1 = adj(A)
¿ A∨¿¿Ans. x =1, y=-1, z=2]
(ii) 2x+y – 3z = 13, x + y – z = 6 , 2x – y+4z = -12.
Question.2 Use the product [−4 4 4−7 1 35 −3 −1] [1 −1 1
1 −2 −22 1 3 ] to solve the
equations x – y+z = 4, x – 2y – 2z = 9, 2x+y+3z = 1.
[Hint take product of above two matrices, we get identity matrix, then use AB=BA = I means B is the inverse of A
Or A is the inverse of B.
⇨ [−4 4 4−7 1 35 −3 −1] [1 −1 1
1 −2 −22 1 3 ] = 8I3 ,
according to above equation let A
let B (1/8) B is the inverse of A. Ans. x=3, y=-2, z=-1.]
Question.3 Solve the following system of homogenous equations: 2x+3y – z = 0, x – y – 2z = 0, 3x+y+3z = 0.
Solution: system of homogenous equations can be written as AX = O
, A = [2 3 −11 −1 −23 1 3 ] ,|A| = -33
So, the system has only the trivial solution given by x=y=z=0. If |A| = 0 then system has non-trivial solution.]
Question.4 Show that system of equations x+y – z = 0, x – 2y+z = 0, 3x+6y – 5z = 0 has non-trivial solution. Find sol.Answer: |A| = 0, it has infinitely many solutions ∴ let z = k a arbitrary x+y = z = k , x – 2y = -z = -k , 3y = 2k i.e, y = 2k/3 ⇨ x = k/3 from first equation by putting the values of x, y & z in third equation, we get 0 which is true. The required solution is z = k, y = 2k/3, x = k/3 where k is arbitrary.
Question.5 Show that system of equations3x+2y +7 z = 0, 4x – 3y - 2z = 0, 5x+9y +23z = 0 has non-trivial solution. Find the solution. [Hint x = -k, y = -2k, z = k]
Question.6 The system of equations 2x+3y = 7 , 14x+21y = 49 has (a) only one solution (b) finitely many solution (c) no solution (d) infinitely many solution . [give reason]
Question.6 Find the inverse (using elementary transformations) of
Question. If A is singular matrix then underwhatcondition set of equations AX = B may beconsistent. [answer if (adjA)B = O ,then eqns. Will have infinitly many sols. Hence consistent.]
Question. If A is a square matrix of order 3 such that |adjA| = 289, find |A|. [ |A| = ±17 ∵ |adjA| = |A|n-1.]
ASSIGNMENT ( WITH HINTS)(determinant)
Question: (i) Let ∆ = Ax x ² 1By y ² 1Cz z ² 1
and ∆₁ = A B Cx y zzy zx xy
, then ∆ - ∆₁ = 0
[Hint ∆₁ = xyzxyz
Ax By Czx ² y ² z ²1 1 1
]
(ii) If f(x) = 0 x−a x−b
x+a 0 x−cx+b x+c 0
, then which is correctf(a)=0 , f(b)=0,
f(0)=0 and f(1)=0 [ Hint f(0)=0 det.(skewsymm.matrix)=0].
**(iii) Let f(t) = cost t 1
2 sint t 2 tsint t t
, then limt →0¿
f ( t)t ² is equal to 0,1,2,3.
[Hint 0, f ( t)t ² =
cost t 12 sintt
1 2
sintt
1 1
→ 1 0 12 1 21 1 1
as t→∞ ].
(iv) There are two values of a which makes determinant ∆ = 1 −2 52 a −10 4 2a
= 86, then sum of these numbers is 4,5,-4,9. [Hint a=-4, operate R2
– 2R1]
Question.1 Prove that the points P (a, b+c), Q(b, c+a), R(c, a+b) are collinear.
Answer : If P,Q and R are collinear then a b+c 1b c+a 1C a+b 1
= 0
By applying C2 → C2+C1 a a+b+c 1b a+b+c 1c a+b+c 1
= (a+b+c) a 1 1b 1 1c 1 1
=0 ( ∵ C2, C3 are identical)
Question.2 Find the value of k if the area of the triangle with vertices (-2,0),(0,4) and (0,k) is 4 square units.
Answer: Area of ∆ = ½ −2 0 10 4 10 k 1
= 4
⇨ the absolute value of ½ (-2)(4 – k) = 4 ⇨ the absolute value of (k – 4) = 4 ⇨ |k – 4| = 4 ⇨ k – 4 = 4, -4 ⇨ k = 8, 0.Question.3 Without expanding, show that
(i) b−c c−a a−bc−a a−b b−ca−b b−c c−a
= 0
Operating C1 → C1+C2+C3, we get 0 c−a a−b0 a−b b−c0 b−c c−a
since x = -1 does not satisfy the equation ,the equation of the L.H.S is negative, so x = 1/6 is the only solution of the given equation.
ASSIGNMENT:
Question.1 Evaluate: (i) sin-1(sin10) (ii) cos-1 (cos10) (iii) tan-1(tan(-6))
Answer: (i) if –π/2 ≤x ≤π/2, then sin-1(sinx)=x but x= 10 radians does not lie between –π/2 and π/2
3π – 10 lies between –π/2 and π/2 ∴ sin-
1(sin(3π-10)) = 3π-10. Similarly for (ii) cos-1 (cos10) = cos-1 (cos(4π-10)) = 4π-10. [10 radians does not lie between 0 and π. ∴ 0≤4π-10≤π]
For (iii) tan-1(tan(-6)) = tan-1(tan(2π-6)) = 2π-6 . { -6 radians does not lie in [ –π/2 , π/2]}
Question.2 If x = cos-1(cos4) and y = sin-1(sin3), then which holds? (give reason)
(i) x=y=1 (ii) x+y+1=0 (iii) x+2y=2 (iv) tan(x+y) = -tan7.
Question.3 if cos−1 xa + cos−1 y
b = θ, then prove that x ²a ² -
2xyab cosθ
+ y ²a ² = sin2θ
[Hint: cos−1 xa + cos−1 y
b = cos−1 ¿ - √1− x ²a ² √1− y ²
b ² ] = θ ⇨ ¿cosθ)2 = ¿)2
Simplify it]
Question.4 *(i) sin-1x + sin-1y + sin-1z = π, then prove that
X4+y4+z4+4x2y2z2 = 2(x2y2+y2z2+z2x2)
(ii) If tan−1 x + tan−1 y + tan−1 z = π/2 ; prove that xy+yz+xz = 1.
(iii) If tan−1 x + tan−1 y + tan−1 z = π , prove that x+y+z = xyz. [Hint: for (i) sin-1x + sin-1y = π - sin-1z ⇨ cos(sin-1x + sin-
1y) =cos( π - sin-1z)
Use cos(A-B) = cosAcosB – sinAsinB and cos(π – 𝛂)= -cos𝛂
It becomes √(1−x ²)(1− y ²) - xy = - √1−z ² and simply it.
[Hint: for (ii) tan-1 x + tan-1y = tan−1 X+Y1−XY ]
Question.5 Write the following functions in the simplest form:
(i)tan−1 ¿ ) (ii) tan−1 ¿) (iii) tan−1 √ a−xa+ x , -a<x<a
[Hint: for (i) write cosx = cos2x/2 – sin2x/2 and 1+sinx =(cosx/2 +sinx/2)2 , then use tan(A-B), answer is π/4 – x/2 ]
[ Hint: for (ii) write cosx = sin(π/2 – x) and sinx = cos(π/2 – x), then use formula of 1-cos(π/2 – x)= 2sin2(π/4 – x/2) and sin(π/2 – x) = 2 sin(π/4 – x/2) cos(π/4 – x/2)
Same method can be applied for (i) part also. Answer is π/4 + x/2]
[ for (iii) put x=a cos𝛂, then answer will be ½ cos−1 xa ]
Question.6 If y = cot−1¿¿) - tan−1(√cosx ), prove that siny =
tan2(x/2). [Hint: y = π2 - 2 tan−1 ¿¿ , use formula 2tan−1 x = cos−1 ¿)]
= -cos[sin−1¿¿ ] = -cos[cos−1 √1−108x ² ] = - √1−108 x ² etc.
5. 1/2(2tan−1 ¿tanx2 )) , use formula 2tan−1 A = cos−1 1−A ²
1+A ² and
tan2x/2 = 1−cosx1+cosx .
6. Put tan𝛂 = t and use sin2𝛂 = 2 tanα
1+ tan ²α and cos2𝛂 = 1−tan ²α1+tan ² α
then put t/3 = T,answer is 𝛂 = nπ, nπ+π/4. tan−1 2t ² - tan−1 t = ½ sin−1 6 t
9+t ² ⇨ tan−1 2 t ²−t1+2 t ³ = ½ sin−1 2 t /3
1+(t /3)² = ½ sin−1 2T
1+(T ) ²
tan−1 2 t ²−t1+2 t ³ = (2T), then tan𝛂 = 0 ,1.½
ASSESSMENT OF Relations & functions for class—XII Level—1
Q.1 Let f(x) = { x+3 , if x<14 x−2, if 1≤x ≤4.x ²+5 , if x>4
Find f(-1) ,f(4) and f(5).
Q.2 If f(x) = x2 - 1x ² , then find the value of f(x) + f ( 1
x ²¿.
Q.3 Let Q be the set all rational numbers and relation on Q defined by R = {(X, Y): 1+XY > 0}. Prove then R is reflexive and symmetric but not transitive.
Q.4 Write the identity element for the binary operation *defined on set R by a*b = 3ab/8 ∀ a, b ЄR.
Q.5 Show that the function f: R → R defined by f(x) = sin x is neither 1-1 nor onto.
Answers (Level—1)
Ans.1 f (-1) = 2, f (4) = 14, f(5)= 30. Ans.2 0. Ans.3 Consider any x, y Є Q, since 1+x.x =1+x2 ≥ 1 ⇨ (x,x)Є R ⇨ reflexive
Let (x,y) Є R ⇨ 1+xy > 0 ⇨ 1+yx > 0 ⇨ (y,x) ЄR ⇨ symmetric.
But not transitive . Since (-1, 0) and (0, 2) ЄR, because 1 > 0 by putting values. But (-1, 2) ∉ R because -1<0. Ans.4 Let e be the identity element in R. Then a *e =a =e*a ∀ aЄR ⇨ a*e =a ∀ a ЄR ⇨ e = 8/3 in R. Ans.5 f is not 1-1 because sin 0 = 0 =sin π,so the different elements o, π have same images. f is not onto because -1 ≤ sin x ≤ 1 for all x ЄR ∴ the range of f =[-1,1], which is a proper subset of R.
LevelQ.1 If f: R→ R is given by f(x) = (3 – x3)1/3 show that fof =Ig where Ig is the identity map on R.
Q.2 Show that the function f: [-1, 1] →R defined by f(x) = x2+ x is 1-1 .
Find the range of f. Also find the inverse of the function f: [-1, 1] → range of f.
Q.3 Show that the function f: R → R defined by f(x) = cos (5x+2) is neither 1-1 nor onto?
Q.4 If f: R → R be given by f(x) = sin2x +sin2(x+π/3) +cosx .cos(x+π/3) ∀ x Є R, and g: R → R be a function such that g(5/4) =1 , then prove that (gof) : R → R is a constant function.
Q.5 Let R1=R – {-1} and an operation * is defined on R1 by a*b = a + b + ab ∀ a, b Є R1 .
Find the identity element and inverse of an element. ANSWERS OF Level—2
Ans.1 As f: R → R, fof exists and fof : R → R is given by (fof) (x) = f(f(x)) = f(3 – x3)1/3 = (3 – ((3 – x3)1/3 )3 )1/3 = (3 – (3 – x3))1/3 =x ∀ x ЄR Ans.2 f is 1-
1, as consider any x1, x2 Є [-1, 1] such that f(x1) = f(x2) ⇨ x₁2+X ₁= X ₂
2+X ₂⇨ x1x2+2x1 = x1x2 +2x2 ⇨ x1 = x2 For the range of f
Let y = f(x) ⇨ y = x2+ x ⇨ xy +2y =x ⇨ (y – 1) x= -2y ⇨ x = −2 y
y−1
As x Є [-1, 1], so -1 ≤ −2 yy−1 ≤1 , but (y – 1)2 >0 , y ≠ 1⇨ -(y – 1)2 ≤ −2 y
y−1
(y-1)2≤ (y – 1)2 ⇨ -(y2 – 2y +1) ≤ -2y2+2y ≤ y2 – 2y +1 ,y≠ 1⇨ Y2 – 1 ≤ 0 and 0 ≤ 3y2 – 4y +1 ⇨ y ε [-1,1] and (y – 1/3) (y – 1) ≥ 0 ,y ≠ 1 ⇨y Є [-1,1] and y ε (-∞ ,1/3] U [1,∞) , y ≠ 1 ⇨ y Є [-1,1] and y ε (-∞ ,1/3] U (1,∞)⇨ y Є[-1,1/3].∴ its inverse exists as f is 1-1 and onto, to find f-1
x2+ x= y ⇨ xy +2y =x ⇨ 2y = x (1 – y) ⇨ x= 2 y
1− y f-1(y) = x = 2 y1− y .
Ans. 3 For f is not 1-1, 5x+2 = π/2 ⇨ x = (π – 4)/10 ∴ 5x+2 =π/2, again 5x+2 = 3π/2 ⇨ x = (3π – 4)/10, Now f ((π – 4)/10)) = cos[5((π – 4)/10) +2] = cosπ/2 =0
For f is not onto, as -1 ≤ cos (5x+2) ≤1, then -1≤ y ≤1, range of f = [-1, 1] = {y : -1≤ y ≤1 } ≠ co-domain R.
Ans. 4 ½[ 2sin2x +2sin2(x+π/3) +2cosx cos(x+π/3)]
f (x)= ½[ 1 – cos2x +1 – cos (2x+2π/3)+ cos (2x+π/3)+cosπ/3]
( As we know that 2sin2x= 1 – cos2x and 2cosA cosB= Cos(A+B) + cos(A-B).) ½[5/2 – {cos2x + cos(2x+2π/3)} + cos(2x+π/3) ⇨ ½[5/2 – 2cos(2x+π/3) cos π/3 + cos(2x+π/3)] = 5/4 ∀ x ЄR∴ for any x Є R , we have (gof)(x) = g(f(x)) = g(5/4)=1 ,so it is constant function.
Ans. 5 * can be shown to be a binary operation on R1 as let a ≠ -1, b ≠ -1 . a*b = a+b+ab Є R – {-1} ⇨ a+b+ab ≠ -1⇨ a(1+b)+(b+1) ≠0 ⇨ (a+1) (1+b) ≠0 ⇨ a ≠ -1 and b ≠ -1 which is true.Now if e is the identity element, then a*e =a ⇨ a+e+ae =a ⇨ e (1+a) = 0 ⇨ e =0 or a = -1 ⇨e =0 , 0 is the identity w.r.t. *
Let a’ be inverse of a, then a*a’ =0 ⇨ a+a’+aa’ = 0 ⇨ a’(1+a) = - a ∴ a’ = - a/(1+a) , is the inverse of a w.r.t. *.
ASSIGNMENT(continuity & differentiability) (XII)
**Question 1 Determine a and b so that the function f given by
f(x) = 1−sin ² x3 cos ² x , x<п/2
=a, x=п/2
= b(1−sinx)
(п−2x ) ² , x>п/2 Is continuous at x=п/2.
Answer [a = 1/3 , b = 8/3] **Question 2 Find k such that following functions are continuous at indicated point
(i) f(x) ={1−cos4 x8x ²
, x ≠0
k , x=0 at x=0
(ii) f(x) = (2x+2 - 16)/(4x – 16) , x≠2
= k, x = 0 at x=2. Answer [ (i) k=1,(ii) k=1/2]
**Question 3 The function f is defined as {x ²+ax+b ,0≤x<23 x+2 ,2≤ x≤4
2ax+5b ,4< x≤8 If
f(x) is continuous on [0,8], find the values of a and b. Answer [a=3,b=-2]
** Question 4 If f(x) = {√1+ px−√1−pxx
,−1≤ x<0
2 x+1x−1
,0≤ x ≤1 is continuous in the [-
1,1], find p. Answer [p=-1]
**Question 5 Find the value of a and b such that the f(x) defined as
[ Hint: Nr. Can be written as tanx(tanx-1)(tanx+1) =- [tanx(cosx-sinx)(tanx+1)]/cosx Cosx-sinx = √ 2 cos¿) ]
**Question 7 Prove that (i) limx→ 1
√ 2
x−cos (sin−1 x )1−tan (sin−1 x ) = −1
√2 [ Hint: put x=
sin ]Ѳ
(ii) limx→∞x ¿¿ ) = -3/2. [Hint: π4 = tan−11 & use formula of tan−1 x−tan−1 y
]
Question 8 f(x) = a x2+b
x2+1 , limx→0
f (x ) =1 & limx→∞f (x) =1, then p.t. f(-
2)=f(2)=1. [ Hint: limx→∞
1
x2 =0]
Question 9 limx→0
ex−1√1−cosx
[Dr. = √2|sinx/2| &limx→0
e x−1x
=1
|sinx/2| =+ve & -ve as x→0+ & x→0- , ⇨ limit does not exist] Question 10 Show that the function
f(x)¿ {sin 3 xtan 3 x
, x<0
32, x=0
log(1+3 x)e2x−1
, x>0
is continuous at x=0.
[Hint: use limx→0
e x−1x
=1 , limx→0
log (1+x)x =1]
Question11 Show that f(x) = |x-3|,x∊R is cts. But not diff. at x=3.
[Hint:show L.H.lt=R.H.lt by |x-3| = x-3, if x ≥3 and –x+3, if x<3, L.hd=-
1≠1(R.h.d) Question 12 Discuss the continuity of the fn. f(x) = |x+1|
+|x+2|, at x = -1 & -2 [Hint:f(x) = {−2x−3 ,when x←21 ,when−2≤x←1
2 x+3 ,when x ≥−1
yes cts. At x=-1,-2
Question 13 Find the values of p and q so that f(x) ={x ²+3x+ p , if x≤12x+2 , if x>1 is
diff. at x = 1. [ answer is p=3 , q=5]
Question 14 For what choice of a, b, c if any , does the function
F(x)={ax ²+bx+c ,0≤ x≤1bx−c ,1<x ≤2
c , x>2becomes diff at x=1,2 & show that a=b=c=0.
Question15For what values a,b f(x)={ e2x−1 ,when x ≤0
ax+ bx ²2
,when x>0 is diff.at x=0
[Hint: L.H.d= 2by using limx→ 0
ex−1x
=1& R.H.d=a, since f‘(x)=0exists, a=2,b∊R]
ASSIGMENT OF DIFFERENTITION
Question 1 Show that y = aex and y = be –x cut at right angles
aab=1 [ by equating , we get ex = √ ba ⇨ x= ½ log ( b/a) , find
slopes(dy/dx) at pt. of intersection is (½ log ( b/a , √ab).
Question 2 (i) If y√1−x ² + x√1− y ² = 1, prove that
dy/dx= (-1)√ 1− y ²1−x ²
[Hint: put y=sinѲ & x= sinφ , use formula of sin(θ+φ ¿¿
(ii) If cos-1¿ ) = tan-1a , find dy/dx. [let cos(tan-1a )= k(constant), then assume c= 1-k/1+k , dy/dx= y/x]
(iii) If y x = e y−x, prove that dy/dx = (1+ logy ) ²logy
(iv) If xm.yn = (x+y)m+n, then find dy/dx. [ y/x]Question 3 Differentiate w.r.t. x : **(i) Using logarithmic differentiation, differentiate:
Solution:
(ii) x tanx + √ x ²+1x
(iii) (iogx)x + xlogx
Question 4 (i) If y x = e y−x, prove that dy/dx = (1+ logy ) ²logy
(ii) If f(1)= 4,f’(1)=2,find d/dx{logf(ex)} at the point x =0.[1/2](iii)If y = √ x+√x+√ x+…….∞ ,show that (2y – 1)dy/dx =1.(iv) If x = (t+1/t)a , y= a(t+1/t) where a>0,a≠1,t≠0, find dy/dx.[Hint: take dy/dt & dx/dt , then find dy/dx = ylogy/ax. ]Question5(i)differentiate: Sec-1(1/(2x2 – 1)),w.r.t.sin-1(3x –4x3).[Hint: let u=1st fn. & v= 2nd fn. , find du/dv = 1]
[ du/dv= ¼, put x=tan ⇨ u= /2, v=2 , u&v as assumed Ѳ Ѳ Ѳabove](iii) If y = e(msin-1x) , show that (1-x2)y2 – xy1 – m2y= 0.Question 6 Water is driping out from a conical funnel, at the uniform rate of 2cm3/sec. through a tiny hole at the vertex at the bottom. When the slant height of the water is 4cm.,find the rate of decrease of the slant height of the water given that the vertical angle of the funnel is 1200 .[Hint: Let l is slant height ,V = 1/3.π .l(√ 3/2)2.l/2=πl3/8(vertical angle will be 600 (half cone), take dv/dt=-2cm3/sec. l=-1/3⇨ π cm/s.]**Question 7(i) Let f be differentiable for all x. If f(1)=-2 and if f `(x) ≥2 ∀ x∊[1, 6], then prove f(6) ≥8.[ use L.M.V.Thm.,f`(c)≥2,c∊[1, 6]](ii) If the function f(x)= x3 – 6x2+ax+b defined on [1, 3] satisfies the rolle’s theorem for c = (2√ 3 +i)/√ 3 , then p.t. a = 11 & b∊R.[Hint: Take f(1)=f(3) , use rolle’s thm. f`(c)=0⇨ a=11]Question 8 (i) Show that f(x)= x/sinx is increasing in (0, п/2)[HINT: f’(x)>0 , tanx >x](ii) Find the intervals of increase and decrease for f(x) = x3 + 2x2 – 1.[Answer is increasing in (-∞, -4/3)U(0, ∞) & decreasing in (-4/3, 0)](iii) Find the interval of increase&decrease for f(x) =log(1+x)-(x/1+x) ORProve that x/1+x < log(1+x) < x for x > 0.[ Hint: f(x)strictly ↑ in [0, ∞) , x>0 ⇨f(x)>f(0), let g(x)=x-log(1+x)g(x)>0 ↑ in [0,∞) & f(x) ↓ in (-∞, 0].](iv) For which value of a , f(x)=a(x+sinx)+a is increasing. [Hint: f’(x) a(1+cosx) ≥0 ⇨ a>0 ∵ -1≤cosx≤1]**Question 9 Problem: Using differentials, approximate the expression
Solution: We let
Hence, x = 0.05 and y = /4.
Differentiating, we obtain
Substituting, we get
Question 10 For the curve y = 4x3 − 2x5, find all the points at which the tangents passes through the origin. [Hint: eqn. Of tangent at (x0,y0) , put x,y=0,(x0,y0)lies on given curve]Question 11 Find the stationary points of the function f(x) = 3x4 – 8x3+6x2 and distinguish b/w them. Also find the local max. And local mini. Values, if they exist.[ f’(x)=0⇨ x=0,1 f has local mini. At x=0∵f’’>0 & f’’(1)=0, f has point of inflexion at x=1,f(1)=1] Question 12 Show that the semi – vertical angle of right circular cone of given total surface area and max. Volume is sin-1 1/3.[Hint: take S=Пr(l+r) ⇨ l= S/пr – r , take derivative of V² OR can use trigonometric functions for l & h]Question 13 A window has the shape of a rectangle surmounted by an equilateral ∆. If the perimeter of the window is 12 m., find the dimensions of the rectangle so that it may produce the largest area of the window.[Hint: let x=length, y=breadth, then y=6 – 3y/2, A= XY+√3X2 /4, take derivative of A & it is max. ,x=4(6+√3)/11 ,y=6(5−√3)/11]
ASSIGNMENT OF INTEGRATION
Question 1 Evaluate: (i)** Integrate .[ Use the power substitution
Put ]
** (iii) Integrate . [ Use the power substitution Put ]
(iii) ∫0
π /4
secx tan3 x dx [answer is (2 - √2)/3 ]
(iv) ∫ dx [multiply÷ by sin(a-b)] (v)∫ √ 1−√ x
1+√x dx
[multiply & divide by √1−√ x ] (Vi)∫ x
x3−1dx [by partial fraction]
(v)∫ (x−4)ex
(x−2)3 dx [ use ∫ex(f(x)+f’(x))dx] (vi)∫o
π /2dx
3+2 sinx+cosx dx [put sinx= 2 tanx /2
1+ tan ² x /2, cosx =1−tan ² x /2
1+tan ² x /2 , then put t=tanx/2. Answer is tan−1 2– п/2]
f ( x )dx ∵f(2a-x) = f(x) , then put t=tanx, answer is /2√2п² ] (xii) ∫
−5
0
f ( x ) dx , where f(x) =|x|+|x+2|+|x+5|. [∫−5
−2
(−x+3)dx + ∫−2
0
( x+7) dx , answer is 31.5 ] (xiii) Evaluate ∫ ex (1−x ) ²
( x ²+1 ) ² dx [use ∫ ex(f(x)+f’(x))dx
Question 2 Using integration, find the area of the regions: (i) { (x,y): |x-1| ≤y ≤√5−x ² } (ii) {(x,y):0≤y≤x2+3; 0≤y≤2x+3; 0≤x≤3}
[(i) A= ∫−1
2
√5−x ²dx- ∫−1
1
(−x+1 )dx - ∫1
2
(x−1) dx = 5/2 [ sin−1( 2
√5) +sin−1( 1
√5) ] –
] [(ii) ½ A=∫0
2
(x ²+3¿)¿ dx +∫2
3
(2 x+3)dx , answer is 50/3](iii) Find the area bounded by the curve x 2 = 4y & the line x = 4y – 2.[A = ∫
−1
2x+24
dx - ∫−1
2x ²4
dx = 9/8 sq. Unit.]**(iv) Sketch the graph of f(x) = {|x−2|+2 , x≤2
x ²−2 , x>2 ,evaluate∫0
4
f (x ) dx[hint: ∫
0
4
f (x )dx = ∫0
2
(4−x )dx + ∫2
4
(x ²−2) dx = 62/3.]**Question 3 evaluate ∫ √1+ x
√ x dx [ mult. & divide by √1+x , put 1+x =A.(d/dx)(x2+x)+B ,find A=B=1/2, integrate]
Definite integral as the limit of a sum , use formula : ∫a
b
f ( x )dx limh→0
∑r=1
n
f (a+rh), where
nh=b-a & n→∞ Question 4 Evaluate ( i )∫0
4
¿¿) dx (ii) ∫0
3
(x ²−2 x+2) dx[ use lim
h→ 0
eh−1h
= 1 for part (i) , use formulas of special sequences, answer is 6]
Some special case :
(1) Evaluate: ∫ dx(x−3)√ x+1
[ put x+1=t²] (2) ∫ dx(x ²−4)√ x+1
[ put x+1 = t² ]
(3) Evaluate: ∫ dx(x+1)√x ²−1
(4) Evaluate: ∫ dx
x ²√x ²+1 [ put x=1/t for both]
(5) Evaluate: ∫ (x ²+1)dxx4+1
[ divide Nr. & Dr. By x2 , then write x²+1/x²=(x-1/x)² +2
according to Nr. , let x-1/x=t]
(6) Evaluate ∫ x √1+x−x ² dx [ let x=A(d/dx) ( 1+x-x²) +B]
(7) Integrating by parts evaluate ∫ x ²( xsinx+cosx ) ² = ∫ ( xsecx ) . xcosx
( xsinx+cosx )²
(8) Evaluate ∫ 11+cotxdx =∫ sinx
sinx+cosxdx [ put sinx=Ad/dx(sinx+cosx)+B(sinx+cosx)
+C
If Nr. Is constant term then use formulas of sinx,cosx as Ques. No. 1 (vi) part]
ASSESSMENT OF DIFFERENTIAL EQUATIONS FOR CLASS—
XI Level--1 Q.1 Find the order and degree of the following differential equations. State also whether they are linear or non-linear.
(i) X2( d ² ydx ² )3 + y ( dydx )4 y4 =0. (ii) d ² y
dx ² = 3√1+( dydx
¿)² ¿ .
Q.2 Form the differential equation corresponding to y2 = a (b – x)(b+ x) by eliminating parameters a and b. Q.3 Solve the differential equation (1+e2x) dy + (1+y2) ex dx = 0, when x= 0, y =1.
Q.4 Solve the differential equation: dydx = 1−cosx1+cosx .
Q.5 Verify that y = A cosx – Bsinx is a solution of the differential
equation d ² ydx ² + y = 0.
Answers of Level—11. (i) order =2 , degree=3 , non linear( because degree is more than 1 ) , (ii) order 2 , degree 3 , non-linear .
2. y2 = a(b – x)(b+ x) = a (b2 – x2), 2y dydx =-2ax ⇨ ydydx = -ax, again
differentiate Y d ² ydx ² + ¿ )2 = -a , by using the value of a from above step ,
we will get , x{ Y d ² ydx ² + ¿ )2 } = ydydx .
3. dy1+ y ² = - e
xdx1+e2 x , Integrating both sides, we get
tan−1 y = - ∫ exdx
1+e2 x , put ex = t⇨ tan−1 y = - tan−1 t +c
Using x=0, y=1, we have y = 1/ex.
4. y = 2 tan(x/2) – x +c , put tan(x/2) = 1−cosx1+cosx
5. dydx = - A sinx – B cosx , d ² ydx ² = - A cosx + B sinx = -y.
Level---2
Q.1 Solve: y dx + x log ( yx ) dy – 2x dy = 0 .
Q. 2 Which of following transformations reduce the differential
equation dzdx + zx log z = zx ² ( log z ) ² into the form
dudx + P(x) u = Q(x) ?
(i) u = log x (ii) u = ex (iii) u = ¿-1 (iv) u = ( log z )2
Q.3 Solve: dydx +xy = xy3
Q.4 Solve: dydx = cos (x+ y) + sin( x+ y )
Q.5 Solve: dydx + x sin 2 y = x3 cos2y
Answers of Level ---2 1. put x = vy , answer = 1+ log (yx ) = ky .
2. (iii) differentiate w.r.t. x dudx = - 1¿¿ .1
z dzdx , put the value of dzdx in the
given differential equation. 3. put 1y ² = t, answer is 1
y ² = 1 + cex ². 4.
put x+y = v, answer is log (1+tan ( x+ y )
2) = x + c.
5. put tan y = v, I.F. = ex ² , also use sin 2 y = 2siny cosy.
ASSESSMENT OF PROBABILITY FOR CLASS –XII Level—1 Q. 1 If the mean and variance of a binomial distribution are 4 and 4/3 respectively, find P(X≥1). Q. 2 If P(A) = 3/8 , P(B) = ½ and P(A∩B) = ¼ , find P(A/B). Q.3 A bag contains 4 white and 2 black balls. Another bag contains 3 white and 5 black balls.
If one ball is drawn from each bag, find the probability that (i) Both are white balls. (ii) One is white and one is black.
Q.4 If A and B are independent events and P(A∩B) = 1/8, P( A’ ∩ B’) =3/8 , find P(A) and P(B).
Q.5 The probabilities of P, Q and R solving a problem are ½, 1/3 and ¼ respectively. If the problem is attempted by
all simultaneously, find the probability of exactly one of them solving it.
Answers of Level—1 Ans.1 np = 4, npq = 4/3 ⇨ q=1/3 ⇨ p = 1 - 1/3
Ans.3 (i) P(A ∩ B) = P(A).P(B) = (2/3).(3/8) [A,B are independent events] (ii) P(A’ ∩ B) + P(A ∩ B’) = P(A’).P(B)+P(A).P(B’) =( 1/3).(3/8)+ (2/3).(5/8)=13/24. [A’, B are indep. Events, B’ A are indep. events], where A = drawing a white ball from first bag. B= drawing a same ball from second bag.A’ = drawing a black ball from first bag and B’ =drawing from second bag. Ans.4 P(A∩B) = P(A).P(B) = 1/8 let x=P(A), y= P(B), P( A’ ∩ B’) =3/8 = P( A’) .P( B’) =(1- X)(1 – Y) ⇨ X+Y – XY = 5/8 ⇨ X=1/2 , Y= ¼.Ans. 5 P(A’)=1/2 ,P(B’) = 1-1/3=2/3 , P(C’)=3/4 ∴ Req. Prob. = P(A)P(B’)P(C’) = P(A’)P(B)P(C’)+ P(A’)P(B’)P(C) [A,B,C are indep. events] = (1/2)(2/3)(3/4)+(1/2)(1/3)(3/4)+(1/2)(2/3)(1/4)=11/24.
Level---2Q.1 If A ∩ B = ф, show that P(A/B) =0, where A and B are possible events.Q.2 A pair of dice is thrown if the sum is even, find the probability that at least one of the dice Shows three. Q.3 Let X denotes the number of hours you study during a randomly selected school day.The probability that X can take the value x, has the following form, where k is some unknown constant P(X=0)=0.1 and P(X=x) = { kx if x=1∨2
k (5−x ) if x=3∨4o ,otherwise
(i) Find k. (ii) What is the probability that you study at least two hour? Exactly two hour? At most two hours?Q.4 Six dice are thrown 729 times. How many do you expect at least three dice to show a 5 or 6? Q.5 In a class; 5% of the boys and 10% of the girls have an I.Q. of more than 150. In this class 60% of the students are boys. If a student is selected at random and is formed and is found To have an I.Q. of more than 150, find the probability that the student is a boy. Answers of Level—2Ans.1 A and B are possible events ⇨A ≠ ⇨ P(A)≠ 0 , P(B) ≠0 фBut A∩B = ф ⇨ P(A∩B) = P( ) = ф P(A/B) = P (A ∩B)
P(B) =0. Ans. 2 n(S)=36, n(A)=18 Out of these 18, the cases which at least one die shows up 3 are (1, 3),(3,1),(3,3),(3,5),(5,3) Required probability=5/18. Ans.3 X 0 1 2 3 4P(X) 0.1 K 2K 2K K(i) k=0.15 (ii) 0.75, 0.3, o.55. Ans.4 P(success)= 2/6=1/3 ∴ q=2/3 P(x success i.e., getting a 5 or 6)= C(6, x) Px q6-x P(at least three successes in six trials) = P(x≥3)=1 – [p(0)+p(1)+p(2)]By using above result we get 1 – (16/81)(31/9) = 233/729 ∴ required answer is 233/729x729=233. Ans.5 Let E1: The student chosen is a boy. P(E1)=60/100 ∴ E2: ........................................girl. P(E2) = 40/100 E1, E2 are mutually exclusive. A: a student has an I.Q. of more than