Magnus Academy (Maths Specialist): December 2021

Sunday, 26 December 2021

INTEGRATION (COMPETITION)

EXERCISE-1

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1) ∫ (2ˣ + 3ˣ)/5ˣ. (2/5)ˣ/log (2/5) + (3/5)ˣ)/log(3/5)

2) ∫ (aˣ + b ˣ)²/(aˣ bˣ). (a/b)ˣ/log(a/b) + (b/a(ˣ/log(b/a) + 2x

3) ∫ (e ⁵ˡᵒᵍˣ - e⁴ˡᵒᵍˣ)/(e³ ˡᵒᵍˣ - e²ˡᵒᵍˣ). x³/3

4) ∫ (cosx - cos 2x)/(1- cosx). 2 sinx + x

5) ∫ (sin⁶x + cos⁶x)/(sin³x cos²x). tanx - cotx - 3x

6) ∫ dx/{√(3x+4) - √(3x+1)}. 2/27 {√(3x+4)³ + √(3x+1)³}

7) ∫ (8x+13)/√(4x+ 7). 1/3 √(4x +7)³ - 1/2 ³√(4x+7)

8) ∫ (7x-2) √(3x+2). 14/45 √(3x+2)⁵ - 40/27 √(3x+2)³

9) ∫ x²/(a+ bx)². 1/b³[bx - 2a log|bx +a| - a²/(a+ bx)]

10) ∫ sin³x cos³x dx. 1/31(- 3/2 cos 2x + 1/6 cos 6x)

11) ∫ sin⁴x dx. 1/8(3x - 2 sin2x + 1/4 sin 4x)

12) ∫ cos⁴x dx. 1/8(3x + 2 sin2x + 1/4 sin 4x)

13) ∫ sin⁴x cos⁴x dx. 1/128 (3x - sin 4x + 1/8 sin 8x)

14) ∫ dx/{sin(x-a) cos(x-b)}. 1/cos(a-b) log|sin(x-a)/cos(x-b)|

15) ∫ dx/{cos(x-a) cos(x-b)}. 1/sin(a-b) log|cos(x-a)/cos(x-b)|

17) ∫ tanx /(a+ b tan²x). 1/{2(b-a)} log|a cos²x + b sin²x|

18) ∫ sin(x+a)/sin(x+b). (x+b) cos(a- b) + sin(a- b) log|sin(x+ b)|

19) ∫√{1+ 2 tanx (tanx + secx)}. log|secx+ tanx| + log|Secx|

20) ∫ dx/(x²- x+1). 2/√3 tan⁻¹{(2x-1)/√3}

21) ∫ dx/(2x²+x -1). 1/3 log|{(2x-1)/2(x+1)}|

22) ∫ dx/(3+ 2x -x²). 1/4 log|{(x+1)/(3-x)}|

23) ∫ dx/[x{6(log x)²+ 7 logx +2}]. Log|(2 logx +1)/(3 logx +2)|

24) ∫ e⁻ˣ/(16+ 9e⁻²ˣ). 1/12 tan⁻¹(3e⁻ˣ/4)

25) ∫ aˣ/√(1- a²ˣ). 1/log a. Sin⁻¹(aˣ)

26) ∫ 2x/(√(1- x² - x⁴). Sin⁻¹{(2x²+1)/√5}

27) ∫ eˣ/√(5 - 4eˣ- e²ˣ). Sin⁻¹{(eˣ+2)/3}

28) ∫ cosx/√(sin²x - 2 sinx -3). Log|(sinx -1) + √(sin²x - 2 sinx -3|

29) ∫ √{x/(a³ - x³)}. 2/3 Sin⁻¹{√x³/√a³}

30) ∫(secx -1)dx. - log|(cosx + 1/2)+ √(cos²x + cosx)|

31) ∫ dx/√(1 - e²ˣ). Log|e⁻ˣ + √(e⁻²ˣ -1)|

32) ∫ √{sin(x- a)/sin(x+a)}. - Cos a sin⁻¹(cosx/cos a) - sin a. log|sinx + √(sin²x - sin²a)|

Continue.........

EXERCISE -

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1) If ¥(x)= ∫cot⁴x dx+ 1/3 cot³x - cot x and ¥(π/2)=π/2 then ¥(x) is

A) π -x B) x - π C) π/2 - x D) n

2) Integral of f(x)= √(1+ x²) with respect to x² is..

A) 2/3 √(1+x²)³/x + k

B) 2/3 √(1+x²)³ + k

C) 2x/3 √(1+x²)³ + k D) n

3) ∫ d(x²+1)/√(x²+2) is..

A) 2√(x²+2) + k B) 2√(x²+2) + k

C) 1√(x²+2)³+ k D) n

4) ∫cos{2 tan⁻¹√{(1-x)/(1+x)}} dx is.

A) 1/8 (x² -1) + k B) x²/2 + k

C) x/2 + k D) n

5) ∫xˣ(1+ logx)dx is

A) xˣ logx + k B) ₑxˣ + k

C) xˣ + k D) none

6) Let the equation of a curve passing through the point (0,1) be given by y= ∫x² ₑxˣ dx. If the equation of the curve is written in the form x = f(y) then f(y) is..

A) √log(3y-2) B) ³√log(3y-2)

C) ³√log(2- 3y) D) none

7) ∫x dx/(1+ x⁴) is

A) tan⁻¹x²+ k B) 1/2 tan⁻¹x²+ k

C) log(1+ x⁴)+ k D) n

8) The antiderivatives of ∫ 2ˣ/(1- 4ˣ) w.r.z.x is

A) logₑ 2. sin⁻¹(2ˣ) + k

B) sin⁻¹(2ˣ) + k

C) cos⁻¹(2ˣ)/logₑ 2 + k. D) n

9) ∫(1+ x)²/(x+ x³) dx is

A) logₑ x + logₑ (1+x²)+ k

B) logₑ x + tan⁻¹x+ k

C) logₑ x + 2tan⁻¹x+ k. D) n

10) ∫ √x⁵/√(1+ x⁷) dx is

A) 2/7 log(√x⁷ + √(1+ x⁷) + k

B) 1/2 log {(x⁷+1)/(x⁷- 1)}+ c

C) 2 √(1+ x⁷) + c. D) n

11) dx/√{⁵√x(1+ ⁵√x⁴)} is

A) √(1+ ⁵√x⁴)+ k

B) 5/2 √(1+ ⁵√x⁴)+ k

C) ⁵√x⁴ √(1+ ⁵√x⁴)+ k. D) none

12) The primitive of the function x|cosx| when π/2 < x <π is given by

A) cosx + x sinx B) - cosx - x sinx

C) x sinx - cosx D) n

13) ∫ x sec x² dx is..

A) 1/2 log(secx²+ tanx²)+ k

B) x²/2 log(secx²+ tanx²)+ k

C) 2 log(secx²+ tanx²)+ k D) n

14) ∫{f(x). ¥'(x) - f'(x). ¥(x)}{log ¥(x) - log f(x)}/(f(x). ¥(x)) dx is.

A) log{¥(x)/f(x)} + k

B) 1/2 {log{¥(x)/f(x)}² + k

C) ¥(x)/f(x) log{¥(x)/f(x)} + k D)N

15) ∫ sin 2x. Log cosx dx is..

A) cos²x(1/2 + log cos x) + k

B) cos²x. Log cosx + k

C) cos²x(1/2 - log cos x) + k D) n

16) ∫ e⁻ˣ(1- tan x)secx dx is

A) e⁻ˣ secx + c. B) e⁻ˣ tanx + c

C) - e⁻ˣ tanx + c D) n

17) ∫ (1+ sinx).eˣ(1/(1+ cosx) is

A) eˣ tan(x/2) + c

B) eˣ tan x + c

C) 1/2 eˣ tan(x/2) + c

D) eˣ sec²(x/2) + c

18) ∫ eˣ{f(x)- f'(x)}dx= ¥(x). Then ∫ eˣ f(x) dx is.

A) ¥(x) + eˣ f(x)

B) ¥(x) - eˣ f(x)

C) 1/2 {¥(x)+ eˣ f(x)}

D) 1/2 {¥(x)+ eˣ f'(x)}

19) If f(0) = f'(0)= 0 and f"(x)= tan²x then f(x) is.

A) log secx - x²/2

B) log cos x + x²/2

C) log secx + x²/2. D) none

20) Let f(x)= ∫ x²dx/{(1+ x²)(1+ √(1+x²))} and f(0)=0, then f(1) is.

A) log(1+ √2) B) log(1+√2) - π/4

C) log(1+√2) + π/4. D) n

21) ∫ dx/(cosx + √(3sinx)) is.

A) log tan(x/2 + π/3)+ k

B) log tan(x/2 - π/3)+ k

C) 1/2 log tan(x/2 + π/3)+ k. D) n

22) If ∫ tan⁴x dx= a tan³x + b tanx + k(x) then

A) a= 1/3 B) b=1

C) k(x)= x+ c D) b = -1

23) If ∫ sinx(sin(x -a) dx= Ax + B logsin(x -a) + C then

A) A= sin a B) B= cos a

C) A= cos a D) B= sin a

24) If ∫ (4eˣ + 6e⁻ˣ)/(9eˣ - 4e⁻ˣ) dx = Ax + B log(9e²ˣ -4)+ C then

A) A= 3/2 B) 35/36

C) C is undefinite D) A+ B= - 19/36

25) If ∫ x log(1+ x²) dx = k(x). Log(1+ x²) + m(x) + c then

A) k(x)= (1+x²)/2

B) m(x)= (1+x²)/2

C) m(x)= - (1+x²)/2

D) k(x)= - (1+x²)/2

26) ∫ dx/{x+1)(x -2)} = A log(x+1) + B log(x -2) + C, where

A) A+ B= 0 B) AB = -1

C) A : B =-1 D) n

ANSWER:

1) D 2) B 3) A 4) B. 5) C

6) B 7) B 8) A 9) C 10) A

11) B 12) B 13) A 14) B 15) C

16) D 17) A 18) C 19) A 20) B

21) C 22) A,C,D 23) C, D

24) B, C, D 25) A, C 26) A, C

Tuesday, 21 December 2021

SPECIAL INTEGRAL(XII)

Expression. Substitution

1) a² + x². x= a tan k or a cot k

2) a² - x². x= a sin k or a cos k

3) x² + a². x= a sec k ora cosec k

4) √{(a-x)/(a+x)} or √{(a+x)/(a- x) x = a cos 2k

5) √{(x- ¢)/(¥-x)} or √{(x-¢)(x- ¥). x = ¢ cos² k + ¥ sin² k

FORMULA;

1) dx/(x²+ a²)= 1/a tan⁻¹(x/a)

2) dx/(x²- a²)= 1/2a log|(x-a)/(x+a)|

3) dx/(a²- x²)= 1/2a log|(a+x)/(a - x)|

4) dx/√(a²- x²)= sin⁻¹(x/a)

5) dx/√(a²+ x²)= log|x + √(a² +x²)|

6) dx/√(x²- a²)= log|x + √(x² - a²)|

EXERCISE --1

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1) ∫ dx/(x²+25). 1/5 tan⁻¹(x/5)

2) ∫ dx/(4+ 9x²). 1/6 tan⁻¹(3x/2)

3) ∫ dx/(4x²+9). 1/6 tan⁻¹(2x/3)

4) ∫ dx/(9x² -1). 1/6 log|(3x-1)/(3x+1)|

5) ∫ dx/(9x² -4). 1/12 log|(3x-2)/(3x+2)|

6) ∫ dx/(16 - 9x²). 1/24 log|(4+3x)/(4 - 3x)|

7) ∫ dx/(a² - m²x²). 1/2am log|(a+ mx)/(a - mx)|

8) ∫ dx/(a²x² - b²). 1/2ab log|(ax- b)/(ax + b)|

9) ∫ dx/(a²x² + b²). 1/ab tan⁻¹(ax/b)

10) ∫ dx/√(4 + x²). log|x+ √(x²+4)|

11) ∫ dx/√(1 + 4x²). 1/2 log|2x + √(4x²+1)|

12) ∫ dx/√(9 - 25x²). 1/5 sin⁻¹(5x/3)

13) ∫ dx/√(16x²+ 25). 1/4 log|x + √(x² +25/16)

14) ∫ dx/√(4x² -9). 1/2 log(x+ √(x²- 9/4)

15) ∫ dx/√(a²+ b²x²). 1/b log|(bx+ √(a² + b²x²|

16) ∫ dx/√(a² - b²x²). 1/b sin⁻¹(bx/a)

17) ∫ dx/√((2- x)²+1). - log|(2- x) + √((2- x)² +1)|

18) ∫ dx/√((2- x)²-1). - log|(2- x) + √((2- x)² -1)|

19) ∫ dx/((x+2)²+1). tan⁻¹{(x+2)/1}

20) ∫ (x²-1)/(x²+4). x - 5/2 tan⁻¹(x/2)

21) ∫ (x⁴+1)/(x²+1). x³/3 - x + 2 tan⁻¹x

EXERCISE --2

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1) ∫ dx/(x² - x +1). 2/√3 tan⁻¹{(2x-1)/√3}

2) ∫ dx/(x² +8x +20). 1/2 tan⁻¹{(x+4)/2}

3) ∫ dx/(x² +2x +5). 1/2 tan⁻¹{(x+1)/2}

4) ∫ dx/(x² +4x +8). 1/2 tan⁻¹{(x+2)/2}

5) ∫ dx/(x² - 10x +34). 1/3 tan⁻¹{(x-5)/3}

6) ∫ 3/(x² - 8x +25). tan⁻¹{(x-4)/3}

7) ∫ dx/(2x² + x -1). 1/3 log|+2x-1)/(2(x+1)|

8) ∫ dx/(3x² +13x -10). 1/17 log|(3x-2)/3(x+5)|

9) ∫ dx/(4x² - 4x +3). 1/2√2 tan⁻¹{(2x-2)/√2

10) ∫ dx/(9x² +6x +10). 1/9 tan⁻¹{(3x+1)/3}

11) ∫ dx/(3+ 2x -x²). 1/4 log|(x+1)/(3-x)|

12) ∫ dx/(1+ x -x²). 1/√5 log|(√5 - 1 + 2x)/(√5 + 1- 2x)|

13) ∫ dx/(3- 2x -x²). 1/4 log|(3+ x)/(1 -x)|

14) ∫ dx/(7 - 6x -x²). 1/8 log|(7+x)/(1 -x)|

EXERCISE -- 3

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1) ∫ x/(x⁴+ x²+1). 1/√3 tan⁻¹{(2x²+1)/3

2) ∫ x/(x⁴+ 2x²+3). 1/2√2 tan⁻¹{(x²+1)/√2}

3) ∫ dx/{x(xⁿ+1)}. 1/n log{xⁿ/(xⁿ+1)}

4) ∫ dx/{x(x⁵+1)}. 1/5 log |x⁵/(x⁵+1)|

5) ∫ 3x⁵/(1+ x¹²)}. 1/2 tan⁻¹(x⁶)

6) ∫ x²/(x⁶ - a⁶). 1/(6a³) log|(x³ - a³)/(x³ + a³)|

7) ∫ x²/(x⁶ + a⁶). 1/(3a³) tan⁻¹(x³/a³)

8) ∫ dx/{x(x⁶ + 1)}. 1/6 log|x⁶/(x⁶+1)|

9) ∫ dx/{x(x³+ 1)}. 1/3 log|x³/(x³+1)|

10) ∫ dx/{x(x⁴+ 1)}. 1/4 log|x⁴/(x⁴+1)|

11) ∫ x/(x⁴ - x²+1). 1/√3 tan⁻¹{(2x²-1)/√3}

12) ∫ x/(3x⁴ - 18x² + 11). √3/48 log |(x² - 3 - 4/√3)/(x² - 3 + 4/√3)|

13) ∫ eˣ/(e²ˣ + 6eˣ+5). 1/4 log|(eˣ+1)/(eˣ+5)|

14) ∫ eˣ/(e²ˣ + 5eˣ+6). log|(eˣ+2)/(eˣ+3)|

15) ∫ eˣ/(1+ e²ˣ). tan⁻¹(eˣ/1)

16) ∫ e³ˣ/(4e⁶ˣ - 9). 1/36 log|(2e³ˣ- 3)/(2e³ˣ+3)|

17) ∫ e⁻ˣ/(16+ 9e⁻²ˣ). -1/12 tan⁻¹(3e⁻ˣ/4)

18) ∫ dx/(eˣ +e⁻ˣ). tan⁻¹(e⁻ˣ)

19) ∫ dx/x{(6 log x)² + 7 log x +2}. Log |(2 log x +1)/(3 log x+2)|

20) ∫ sinx/(1+ cos²x). tan⁻¹(cosx)

21) ∫ cosx/(sin²x+ 4 sinx +5). tan⁻¹(sinx+2)

21) ∫ sec²x/(1 - tan²x). 1/2 log|(1+ tanx)/(1- tanx)|

** FORMULA

1) dx/√(a² + x²)= log|x+ √(a²+x²)|

2) dx/√(x² - a²)= log|x+ √(x²- a²)|

3) dx/√(a² - x²)= sin⁻¹(x/a)

EXERCISE-4

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1) ∫dx/√(2- 4x+ x²). log|(x-2)+ √(x² - 4x +2)|

2) ∫ dx/√(x² - 4x +2). log|(x-2)+ √(x² - 4x +2)|

3) ∫ dx/√(2x- x²). sin⁻¹(x- 1)

4) ∫ dx/√(8+ 3x- x²). sin⁻¹{(2x- 3)/√41}

5) ∫ dx/√(2- 4x- x²). sin⁻¹{(x +2)/√6}

6) ∫ dx/√(5+ 4x- x²). sin⁻¹{(x- 2)/3}

7) ∫ dx/√(1+ 2x- 3x²). 1/√3 sin⁻¹{(3x- 1)/2}

8) ∫ dx/√{(2- x)²+1}. - log|2-x + √(x² - 4x +5)|

9) ∫ dx/√{(2- x)² -1}. - log|2- x + √(x² - 4x +3)|

10) ∫ dx/√(x²+ 4x+2). log|x+2 + √(x² + 4x+ 2)|

11) ∫ dx/√(8+ 4x- 4x²). 1/2 sin⁻¹{(2x- 1)/3}

12) ∫ dx/√(3x²+ 5x+7). 1/√3 log{x + 5/6 + √(x²+5x/3 +7/3}}

13) ∫ dx/√(1+ x- x²). sin⁻¹{(2x- 1)/√5}

14) ∫ dx/√(8+3x- x²). sin⁻¹{(2x- 3)/√41

15) ∫ dx/√(7- 3x- 2x²). 1/√2 sin⁻¹{(4x +3)/√65}

16) ∫ dx/√(6 - x- x²). sin⁻¹{(2x+ 1)/5}

17) ∫ dx/√(6+ x- x²). sin⁻¹{(2x- 1)/5}

18) ∫ dx/√(4- 2x- x²). sin⁻¹{(x+ 1)/√5}

19) ∫ dx/√(7- 6x- x²). sin⁻¹{(x +3)/4}

20) ∫ dx/√(2+2x- x²). sin⁻¹{(x- 1)/√3}

21) ∫ dx/√(x²+ 12x+ 11). log|x + 6 + √(x² +12x +11)|

22) ∫ dx/√{(x+5)(x+1)}. Log|(x+3) + √(x²+ 6x+ 5)|

23) ∫ dx/√{(x-1)(x-2)}. Log|(x -3)/2 + √(x²- 3x +2)|

24) ∫ dx/√{x(1- 2x)}. 1/√2 sin⁻¹{(4x -1)}

25) ∫ dx/√{(x- a)(b -x)}. 2 log|√(x- a) + √(x-b)|

EXERCISE-5

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1) ∫ x²/√(1- x⁶). 1/3 sin⁻¹(x³)

2) ∫ x/√(x⁴ + a⁴). 1/2 log |x² + √(x⁴+ a⁴)|

3) ∫ x/√(4 - x⁴). 1/2 sin⁻¹(x²/2)

4) ∫ 2x/√(1- x²- x⁴). sin⁻¹{(2x²+1)/√5}

5) ∫ √{x/(a³ - x³)}. 2/3 sin⁻¹(x³⁾²/a³⁾²)

6) ∫ dx/{x²⁾³ √(x²⁾³ - 4)}. 3 log|x¹⁾³ + √(x²⁾³ - 4)|

7) ∫ sec²x/(√(16 + tan²x). log|tan x + √(16+ tan²x)|

8) ∫ sec²x/(√(4 + tan²x). log|tan x + √(4+ tan²x)|

9) ∫ cosx/√(sin²x - 2 sinx -3). log|(sinx -1) + √(sin²x - 2 sinx -3)|

10) ∫ cosx/(√(4 + sin²x). log|sin x + √(4+ sin²x)|

11) ∫ sinx/(√(4cos²x -1). - 1/2 log|2 cos x + √(4cos²x - 1)|

12) ∫ sin 8x/√{8+ sin⁴4x). 1/4 log|sin²4x + √(9+ sin⁴4x)|

13) ∫ cos 2x/√(sin²2x +8). 1/2 log|sin2x + √(sin²2x +8)|

14) ∫ sin 2x/√(sin⁴x + 4 sin²x -2).

log|sin²x + 2+ √(sin⁴x + 4 sin²x - 2)|

15) ∫ sin 2x/√(cos⁴x- sin²x+2). - log|(cos²x + 1/2) + √(cos⁴x + cos²x +1)|

16) ∫ cosx/√{4 - sin²x). sin⁻¹ {(sinx)/2}

17) ∫ cosx/√(sin²x- 2sinx -3). log|(sinx - 1) + 1 √(sin²x - 2 sinx -3)|

18) ∫ √(secx - 1) dx. - log|(cosx + 1/2) + √(cos²x+ cosx)|

19) ∫ √(cosecx - 1) dx. log|(sinx + 1/2) + √(sin²x + sin x)|.

20) ∫ √{sin(x- a)/sin(x+ a)}. - cos a sin⁻¹(cosx/cos a) - sin a. log|sinx + √(sin²x - sin²a)|

21) ∫ dx/{√(1- x²)(9+ (sin⁻¹x)²)}. Log|sin⁻¹x + √(9+ (sin⁻¹x)²)|

22) ∫ dx/{x √(log x)² -5}. log|log x +√(log x)² - 5|

23) ∫ dx/{x √(4 - (9log x)²}. 1/3 sin⁻¹{(3 log x)/2}

24) ∫ eˣ/√(4 - e²ˣ). sin⁻¹(eˣ/2)

25) ∫ eˣ/√(16 - e²ˣ). sin⁻¹(eˣ/4)

26) ∫ eˣ/√(5 - 4eˣ - e²ˣ). sin⁻¹ {(eˣ+2)/3}

27) ∫ aˣ/√(1 - a²ˣ). 1/(log a) sin⁻¹(aˣ)

INTEGRAL OF THE FORM :-

∫(px +q)/(ax²+ bx+ c)

Step 1: Write the numerator px + q in the following form:

px+ q = K{d/dx (ax²+ bx+c}+ M

i.e. px + q= K(2ax + b) + M

Step 2: Obtain the values of K and M by equating the coefficient of like powers of x on both sides.

Step 3: Replace px+q by K(2ax+b)+ M in the given integral to get

∫ (px +q)/(ax²+ bx+ c)

= K ∫(2ax +b)/(ax²+ bx+ c) + M ∫ dx/(ax²+ bx+ c).

Step 4: Integrate RHS in step 3 and put the values of K and M obtained in step 2.

EXERCISE-6

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1) ∫ x/(x²+ x+ 1). 1/2 log|x²+x+1| - 1/√3 tan⁻¹{(2x+1)/√3}

2) ∫ (4x+1)/(x²+ 3x+ 2). 2 log|x²+3x+2| - 5 log|{(x+1)/(x+2)}

3) ∫ (2x- 3)/(x²+ 3x -18). log|x²+3x -18| - 2/3 log|{(x- 3)/(x+6)}

4) ∫ x/(x²+ 3x+ 2). 1/2 log|x²+3x+2| - 3/2 log|{(x+1)/(x+2)}

5) ∫ (x+1)/(x²+ x+ 3). 1/2 log|x²+x+3| + 1/√11 tan⁻¹{( {(2x+1)/√11}

6) ∫ (x -3)/(x²+ 2x-4). 1/2 log|x²+2x-4| - 2/√5 log|{(x+1-√5)/(x+1+ √5)}

7) ∫ (2x -3)/(x²+ 6x +13). log|x²+6x+ 13| - 9/2 tan⁻¹{(x+3)/2}

8) ∫ (x -1)/(3x² -4x +3). 1/6 log|3x²- 4x +3| - √5/15 tan⁻¹{(3x-2)/√5}

9) ∫ 2x/(2+x -x²). - log|2+ x-x²| + 1/3 log|{(1+x)/(2-x)}

10) ∫ (1 - 3x)/(3x² +4x +2). - 1/2 log|3x²+ 4x +2| + 3/√2 tan⁻¹{(3x+2)/√2}

11) ∫ (2x +5)/(x² -x -2). log|x²- x-2| + 2 log|(x-2)/(x+1)|

12) ∫ (x³ +x)/(x⁴ - 9). 1/4 log|x⁴ - 9| + 1/12 log|{(x²- 3)/(x²+3)}

13) ∫ (ax³ +bx)/(x⁴ + c²). a/4 log|x⁴ + c²| + b/2c tan⁻¹{x²/c}

14) ∫ (x +2)/(2x² +6x +5). 1/4 log|2x²+ 6 x +5| + 1/2 tan⁻¹(2x+3)

15) ∫ (2 sin2x - cosx)/(6 - cos²x - 4 sinx). 2 log|sin²x - 4 sinx +5| + 7 tan⁻¹(sinx - 2)

16) ∫ (3 sin2x - 2) cosx)/(5 - cos²x - 4 sinx). 3 log|2 - sinx|+ 4/(2- sinx)

17) ∫ dx/(2eˣ + 3eˣ+1). -1/2 log|e⁻ˣ + 3e⁻ˣ+2)| + 3/2 log|(e⁻ˣ +1)/(e⁻ˣ+2)|

** INTEGRAL OF THE FORM ::

∫ P(x)/(ax²+ bx + c) where P(x) is a polynomial of degree greater than or equal to 2.

To evaluate this type of integral we divided the numerator by the denominator and express the integrand as

Q(x) + R(x)/(ax²+ bx + c). Where R(x) is a linear function of x.

∫ P(x)/(ax²+ bx + c) = ∫ Q(x) + ∫ R(x)/(ax²+ bx + c

Now to evaluate the second integral on RHS apply the method as Exerciss -6.

EXERCISE --7

-------------------

1) ∫ (x³ + x+1)/(x² -1). x²/2 + log|x² - 1| + 1/2 log|(x-1)/(x+1)|

2) ∫ (x² + x+1)/(x² -x). x + log|x² - x| + 2 log|(x-1)/x|

3) ∫ (x² + 5x+3)/(x²+3x +2). x + log|x² +3x+2| - 2 log|(x+1)/(x+2)|

4) ∫ (x² + x-1)/(x²+x -6). x + log|(x -2)/(x+3)|

5) ∫ (x² + 1)/(x²- 5x +6). x - 5 log|x - 2| + 10 log|(x - 3)|

6) ∫ (1- x²)/{x(1 - 2x)}. x/2 + log|x| - 3/4 log|(1- 2x)|

7) ∫ x²/(x²+7x +10). x - 7/2 log|x² +7x+ 10| + 29/6 log|(x+2)/(x+5)|

8) ∫ (x² + x+1)/(x²-x +1). x + log|x² - x+1| + 2/√3 tan⁻¹{(2x -1)/√3}

9) ∫ (x -1)²/(x²+ 2x +2). x - 2 log|x² + 2x+2| + 3 tan⁻¹{(x +1)}

10) ∫ (x³+x² + 2x+1)/(x²-x +1). x²/2 + 2x+ 3/2 log|x² - x+1| + 1/√3 tan⁻¹{(2x -1)/√3}

11) ∫ {x²(x⁴ +4)}/(x² +4). x⁵/5 - 4x³/3 + 20x - 40 tan⁻¹{(x/2}

12) ∫ x²/(x²+ 6x +12). x - 3 log|x² + 6x+12| + 2√3 tan⁻¹{(x+3)/√3}

*** INTEGRAL OF THE FORM

∫ (px +q)/√(ax²+ bx + c)

For Solution:

Step -1: Write the numerator px + q in the following form:

px +q = K{d/dx (ax² +bx +c)} + M

i. e., px + q= K(2ax+ b)+ M

Step-2: Obtain the values of K and M by equating the coefficient of like powers of x on both sides.

Step- 3: Replace px + q by K(2ax + b) + M in the given integral to get

∫ (px+q)/(ax²+ bx + c) =

K ∫ (2ax+b)/√(ax²+ bx + c) + M∫ dx/√(ax²+ bx + c)

Step-4: Integrate RHS in step 3 and put the values of K and M obtained in step 2

EXERCISE--8

----------------

1) ∫ (2x+ 3)/√(x²+ 4x + 1). 2√(x²+ 4x +1) - log|x + 2 + √(x²+ 4x +1)|

2) ∫ (2x+ 1)/√(x²+ 2x -1). 2√(x²+ 2x -1) - log|x + 1+ √(x²+ 2x -1)|

3) ∫ (6x -5)/√(3x²- 5x + 1). 2√(3x²-5x +1)|

4) ∫ (x+ 2)/√(x²+ 2x - 1). √(x²+ 2x -1) + log|(x + 1) + √(x²+ 2x -1)|

5) ∫ (2x+ 5)/√(x²+ 2x + 5). 2√(x²+ 2x +5) + 3 log|x + 1 + √(x²+ 2x +5)|

6) ∫ (2x+ 1)/√(x²+ 4x + 3). 2√(x²+ 4x +3) - 3log|x + 2 + √(x²+ 4x +3)|

7) ∫ (2x+ 3)/√(x²+ 4x + 5). 2√(x²+ 4x +5) - log|x + 2 + √(x²+ 4x +5)|

8) ∫ (x+ 2)/√(x²+ 5x + 6). √(x²+ 5x +6) - 1/2 log|(x + 5/2) + √(x²+ 5x +6)|

9) ∫ x/√(x²+ x + 1). √(x²+ x +1) - 1/2 log|((2x + 1)/2 + √(x²+ x +1)|

10) ∫ (x+ 1)/√(4+ 5x- x²). - 4√(4 + 5x -x²) + 7/2 sin⁻¹{(2x-5)/√41}

11) ∫ (3x+ 1)/√(5 - 2x - x²). - 3√(5 - 2x -x²) - 2 sin⁻¹{(x +1)/√6}

12) ∫ x/√(8+ x - x²). - 8 √(8 +x -x²) + 1/2 sin⁻¹{(2x-1)/√33}

13) ∫ √{(1+x)/x}. √(x²+ x) + 1/2 log|(x+ 1/2) + √(x²+ x)|

14) ∫ √{(1- x)/(1+x)}. sin⁻¹x + √(1 - x²)

15) ∫ √{(a- x)/(a+x)}. a sin⁻¹{x/a} + √(a² - x²)

16) ∫ x. √{(a²- x²)/(a²+x²)}. a²/2 sin⁻¹{x²/a²} + 1/2 √(a⁴ - x⁴)

17) ∫ (x+2)/√(x² - 1)}. √(x²-1) + 2 log|x + √(x² -1)|

18) ∫ (x +1)/√(x² +1)}. √(x²+1) + log|x + √(x² +1)|

19) ∫ (x-1)/√(x² + 1)}. √(x²+1) - log|x + √(x² +1)|

20) ∫√{(1+x)/x}. √(x²+x) 1/2 log|(x+ 1/2)+ √(x²+x)|

*** INTEGRAL OF THE FORM

∫ dx/(a sin²x + b cos²x), OR

∫ dx/(a + bsin²x), OR

∫ dx/(a + b cos²x), OR

∫ dx/(a sinx + b cosx)² OR

∫ dx/(a + b sin²x + c cos²x)

For Solution::::

Step-1: Divide numerator and denominator both by cos²x.

Step-2: Replace sec²x, if any, in denominator by 1 + tan²x.

Step-3: Put tan x= t

so that sec²x dx = dt.

This substitution reduces the integral in the form of

∫ dt/(at²+ bt + c)

Step-4: Evaluate the integral obtained in step 3 by using the method discussed earlier.

EXERCISE--9

-------------------

1) ∫ dx/(a² sin²x+ b² cos²x. 1/ab

tan⁻¹{(a tanx)/b}

2) ∫ dx/(1+ 3sin²x+ 8cos²x. 1/6 tan⁻¹{(2 tanx)/3}

3) ∫ dx/(9sin²x+ 4cos²x. 1/6

tan⁻¹{(3tanx)/2}

4) ∫ dx/(4sin²x+ 5 cos²x. 1/2√5

tan⁻¹{(2 tanx)/√5}

5) ∫ dx/(2sinx+ 3cosx)². -1/2{(2tanx +3)}

6) ∫ dx/(3+ sin 2x). 1/2√2 tan⁻¹{(3 tanx +1)/2√2}

7) ∫ 2 dx/(2+ sin 2x). 2/√3 tan⁻¹{(2 tanx +1)/√3}

8) ∫ dx/(2 - 3 cos 2x). 1/2√5 log|{(√5 tanx -1)/(√5 tan x +1)}|

9) ∫ dx/(1+ 3 sin²x). 1/2 tan⁻¹{ (2tanx)}

10) ∫ dx/(3+ 2 cos²x). 1/√15 tan⁻¹{(√3 tanx)/√5}

11) ∫ dx/{(sin x- 2 cosx)(2 sinx+ cosx) 1/5 log|tanx -2))/(2 tanx +1)}|

12) ∫ sin 2x/(sin⁴x + cos⁴x). tan⁻¹{ tan²x}

13) ∫ dx/{cosx(sin x + 2 cosx). log|tan x +2|

14) ∫ dx/(sin²x + sin2x). 1/2 log|tanx/(tanx +2)|

15) ∫ dx/(cos2x + 3 sin²x). 1/√2 tan⁻¹{(√2 tanx}

16) ∫ sinx/sin3x. 1/2√3 log|(√3+ tanx)/(√3 - tanx)|

17) ∫ cosx/cos3x. 1/2√3 log|(1+ √3 tanx)/(1 - √3 tanx)|

**INTEGRAL OF THE FORM

∫ dx/(a sinx + b cos x),. OR

∫ dx/(a + b sin x),. OR

∫ dx/(a + b cos x),. OR

∫ dx/(a sinx + b cos x + c),.

For Solution:

Step-1: Put sinx = 2tan x/2/(1+ tan² x/2),

cosx= (1- tan² x/2)/(1+ tan² x/2)

Step-2: Replace 1+ tan²x/2 in the numerator by sec² x/2.

Step-3 : Put tan x/2 = t so that

1/2 sec² x/2 dx = dt.

This substitution reduces the integral in the form

∫ dt/(at² + bt +c),.

Step-4 : Evaluate the integral in STEP-3 by using method discussed earlier.

EXERCISE- 10

--------------------

1) ∫ dx/(2+ cos x). 2/√3 tan⁻¹{( tan x/2)/√3}

2) ∫ dx/(1 - 2sin x). 1/√3 log|(tan x/2 - 2 - √3)/(tan x/2 - 2 + √3)|

3) ∫ dx/(5+ 4cos x). 2/3 tan⁻¹{( tan x/2)/3}

4) ∫ dx/(5 - 4 sin x). 2/3 tan⁻¹{(5tan x/2 - 4)/3}

5) ∫ dx/(1- 2 sinx). 1/√3 log|( tan x/2 - 2 - √3)/(tan x/2 - 2 +√3)}|

6) ∫ dx/(4cos x -1). 1/√15 log|{(√3+ √5 tan x/2)/(√3 - √5 tan x/2)}|

7) ∫ dx/(1 + sinx + cos x). log|{( tan x/2) +1}

8) ∫ dx/(1 - sinx + cos x). - log|{(1- tan x/2)}|

9) ∫ dx/(3 + 2sinx + cos x). tan⁻¹{1+ tan x/2}

10) ∫ dx/(13 + 3sinx + 4cos x). 1/6 tan⁻¹{(5tan x/2 +2)}/6

11) ∫ dx/(cosx - sinx). 1/√2 log|(√2+ tan x/2 +1)/(√2 - tan x/2 -1)}|

12) ∫ dx/(sinx + cos x). 1/√2 log{(√2+ tan x/2 - 1)/(√2 - tan x/2 +1)}

13) ∫ dx/(2 + sinx + cos x). √2 tan⁻¹{1+ tan x/2}/√2

14) ∫ dx/(sinx + √3 cos x). 1/2 log|(1+ √3 tan x/2)/(3- √3 tan x/2)|

15) ∫ dx/(√3sinx + cos x). 1/2 log|{tan (x/2 + π/12)}

16) ∫ dx/(sinx - √3 cos x). 1/2 log|{tan (x/2 - π/6)}

17) ∫ dx/(5+ sinx + 7cos x). 1/5 log|(tan (x/2 +2)/(tan x/2 -3)}

18) ∫ (1+ sinx)/{sinx(1 + cos x)}. 1/2{ log|{tan x/2| + (tan² x/2)/2 + 2 tan x/2}.

ALTERNATIVE METHOD TO EVALUATE INTEGRAL OF THE FORM:

) ∫ dx/(asinx + bcos x).

For Solution, We substitute

a= r cos k, b= r sin k and so

r² = √(a² + b²), k= tan⁻¹(b/a)

So, a sink + b cos k

= r cos k sin k + r sin k cos k

= r sin(x+ k)

So, ∫ dx/(asinx + bcos x).

= 1/r ∫ dx/sin(x+k).

= 1/r ∫ cosec(x+k) dx.

= 1/r log|tan(x/2 + k/2)|+ C

= 1/√(a²+ b²) log|tan(x/2 + 1/2 tan⁻¹(b/a)|+ C.

EXERCISE-11

-------------------

1) ∫ dx/(√3 sinx + cos x). 1/2 log|tan(x/2 + π/12)|

2) ∫ dx/(sinx + √3 cos x). 1/2 log|tan(x/2 + π/6)|

*** INTEGRALS OF THE FORM:

∫ (a sinx + b cosx)/(c sinx + d cos x).

For Solution:

Step-1: Write

Numerator= K(Diff. of denominator) + M(Denominator).

i.e., a sin x + b cos x= K(c cos x - d sinx)+ M(c sinx + d cosx).

Step-2: Obtain the values of K and M by equating the coefficient of sin x and cosx on both the sides.

Step-3: Replace numerator in the integrand by K(c cosx - d sinx) + M(c sinx+ d cosx) to obtain

∫ (a sinx + b cosx)/(c sinx + d cos x). = K∫ (c cosx - d sinx)/(c sinx + d cos x) dx + M∫ (c sinx + d cosx)/(c sinx + d cos x).

= K log|c sinx + d cosx| + Mx + C

EXERCISE--12

--------------------

1) ∫ (3 sinx + 2 cosx)/(3 cosx + 2 sin x). -5x/13 + 12/13 log|3 cosx + 2 sinx |

2) ∫ dx/(1+ tanx). x/2 + 1/2 log|sinx + cosx|

3) ∫ dx/(1+ cotx). -1/2 log|sinx + cosx| + x/2

*** INTEGRALS OF THE FORM:

∫ (a sinx +b cosx + c)/(p sinx + q cos x + r).

For Solution:

Step-1: Write

Numerator= K(Diff. of denominator) + M(Denominator) + v.

i.e., a sin x + b cos x + c= K(p cos x - q sinx)+ M(p sinx + q cosx +r)+ v.

Step-2: Obtain the values of K and M by equating the coefficient of sin x and cosx and constant on both the sides.

Step-3: Replace numerator in the integrand by K(p cosx - q sinx) + M(p sinx+ q cosx + r)+ v to obtain

∫ (a sinx + b cosx+ c)/(p sinx + q cos x +r).

= K∫ (p cosx - q sinx)/(p sinx + q cos x + r) dx + M ∫ (p sinx + q cosx +r)/(p sinx + q cos x +r)+ v + ∫ dx/(p sinx + q cosx +r).

= K log|p sinx + q cosx + r| + Mx + v ∫ dx/(p sinx + q cosx +r).

Step-4: Evaluate the integral on RHS in step 3 by using the method discussed earlier.

EXERCISE- 13

-----------------

1) ∫ dx/(1- cotx). x/2 + 1/2 log|sinx - cosx|

2) ∫ dx/(1- tanx). x/2 - 1/2 log|sinx - cosx|

3) ∫ dx/(p + q tanx). px/(p²+ q²) + q/(p²+ q²) log|p cosx + q sinx|

4) ∫ dx/(4 cotx + 3). 3x/25 - 4/25 log|3sinx + 4 cosx|

5) ∫(2 tanx + 3)/(3 tanx+ 4). 18x/25 + 1/25 log|3 sinx + 4 cosx|

6) ∫ dx/(4 + 3 tan x). 4x/25 + 3/25 log|3sinx + 4 cosx|

7) ∫(8 cotx + 1)/(3cot x + 2). 2x + log|2sinx + 3 cosx|

8) ∫(3 cosx + 2)/(sinx + 2 cos x + 3). 6x/5 + 3/5 log|sinx + 2 cosx +3| + -8/5 tan⁻¹{(tan x/2 +1)/2}

9) ∫(3 + 2 cosx + 4)/(2sinx + cos x + 3). 2x - 3 tan⁻¹{(tan x/2 +1)}

10) ∫(5cosx + 6)/(sinx + 2 cos x + 3). 2x + log|sinx + 2 cosx +3|

11) ∫(3 cosx + 2 sinx)/(3sinx + 4 cos x). 18x/25 + 1/25 log|3sinx + 4 cosx|

12) ∫(5cosx + 4 sinx)/(5sinx + 4 cos x). 40x/41 + 9/41 log|5sinx + 4 cosx|

Friday, 17 December 2021

LOGARITHM (A- Z) C

EXERCISE - A

**Convert in to logarithmic form:

1) 6⁻¹ = ⅙ is

A) log₆(1/6)= 1 B) log₆(1/6)= -1

C) log₆6= 1 D) none

2) ³√(27) =3 is

A) log 27¹⁾³= 3 B) log₂₇3 = 1/3

C) log₂₇3= 1/3 D) log₂₇(1/3)= 3

3) 2⁵= 32 is..

A) log₂32= 5 B) log₂5= 32

C) log₅32= 2 D) log₅2= 32

4) 2⁰= 1 is..

A) log₁2= 0 B) log₂1= 0

C) log₁0 =2 D) log₂0=2

5) ³√64= 4 is..

A) log₆₄4= 1/3 B) log₄64= 1/3

C) log₃64= 1/4 D) log₃4= 64

6) 8⁻²⁾³ = 1/4 is...

A) log₈(1/4)= -2/3

B) log₈(-1/4)= 2/3

C) log₈(2/3)= -1/4

D) log₈(-2/3)= 1/4

7) 10⁻² = 0.01 is..

A) log₁₀(0.01) = 2 B) log₁₀(0.01) =-2

C) log₂(0.01)=10 D) log₂(0.01) = 0.1

8) 4⁻¹= 1/4 is..

A) log₄(1/4)= 1

B) log₄(1/4)= -1

C) log₁(1/4)= 4

D) log₁4= 1/4

EXERCISE - B

**Convert into Exponential form:

1) log₅(625) = 4 is..

A) 5⁴= 625 B) 4⁵= 625 C) 5/4= 625 D) n

2) log√₃ 27= 6 is...

A) (√3)²⁷=6 B) (√3)⁶=27 C) 3²⁷=2 D) 3²⁷= √2

3) log₄(4) = 1

A) 1⁴= 4 B) 4⁴= 0 C) 4⁴= 1 D) 4⁰=1

4) log√₅ 625= 8 is

A) √5= 625 B)(√5)⁸= 625 C) 5=√625 D)n

5) log₂(1/32)= -5 is..

a) 2⁵=32 b) 2⁻⁵=1/32 v) 1/2⁵= 1/32 d) n

EXERCISE - C

** Find the Value of:

1) log₃(81) is

A) 1 B) 2 C) 3 D) 4

2) log₁₀³√(100) is

A) 1/3 B) 2/3 C) 10 D) 1

3) log₂(1/32) is..

A) 5 B) 1/5 C) 1/2 D) 2

4) log₉(27) is..

A) 1/2 B) 2/3 C) 1/3 D) √3

5) log₇343 is .

A) 7 B) 3 C) 7/3 D) 3/7

6) Log₂64 is..

A) 2 B) 4 C) 6 D) 8

7) log₈32 is..

A) 5 B) 5/2 C) 5/3 D) 5/4

8) log ₃(1/9) is...

A) 1 B) -1 C) 2 D) -2

9) log₀·₅(16)

A) 2 B) -2 C) 4 D) -4

10) log₂(0.125) is..

A) -1 B) -2 C) -3 D) -4

11) log₇7 is.

A) 0 B) 1 C) -1 D) none

12) log₅√₅125

A) 2 B) 4 C) 6 D) 8

EXERCISE - D

** Find the value of x if..

1) log₃(x)= 4

A) 3 B) 9 C) 17 D) 81

2) log₂₅(x)=-1/2

A) 5 B) -5 C) 1/5 D) -1/5

3) log₁/₂(x)= -3

A) 8 B) 1/8 C) 4 D) 2

4) x=log₁₀(0.001)

A) 3 B) -3 C) 1/3 D) -1/3

5) log₂₅(x) = -½

A) 5 B) 1/5 C) -5 D) -1/5

6) logₓ(243) = 5

A) 3 B) 5 C) -3 D) -5

7) log₂x=-2

A) 4 B) 1/4 C) -4 D) -1/4

8) logₓ9 = 1

A) 1 B) 9 C) 3 D) 1/9

9) log₉243= x

A) 2 B) 5 C) 2/5 D) 5/2

10) log₃x= 0

A) 0 B) 1 C) 3 D) 1/3

11) log√₃(x-1)= 2

A) 1 B) -2 C) 3 D) 4

12) log₅ (x²-19)= 3

A) 12 B) -12 C) ± 12 D) none

13) logₓ64 = 3/2

A) 4 B) 8 C) 16 D) 2

14) log ₂ (x²- 9)= 4

A) 2 B) 16 C) 5 D) ±5

15) logₓ(0.008)= -3

A) 3 B) 5 C) ±5 D) -3

EXERCISE - E

1) The base when 2 is the logarithm of 9

A) 2 B) 9 C) 1 D) ±3

2) The base when 3 is the logarithm of 216 is.

A) 6 B) √6 C) 2 D) 3

3) The base when 6 is the logarithm of 49 is.

A) 7 B) √7 C) ³√7 D) 1/7

4) The base when (-1/6) is the logarithm of √5 is..

A) 0.8 B) 0.08 C) 0.008 D) 0.0008

5) the base, when 4 is the logarithm of 1296 is..

A) 6 B) -6 C) 1/6 D) 1

6) The base, when 6 is the logarithm of 343 is .

A) 7 B) -7 C) ±7 D) √7

7) The logarithms of 243 to the base 3 is..

A) 2 B) 3 C) 4 D) 5

8) The logarithms of 16 to the base 32 is :

A) 4 B) 5 C) 4/5 D) 5/4

9) The logarithms of 81 to the base ³√9 is..

A) 3 B) 4 C) 5 C) 6

10) If log 2= 0.31 then log 8 is.

A) 0.31 B) 0.62 C) 0 D) 0.93

11) If log 2= 0.642 then log 5 is

A) 0.1284 B) 0.358 C) 0.853 D) n

12) What is the base if the logarithm of 144 is 4 ?

A) 2 B) 3 C) √2 D) 2√3 E) none

EXERCISE - F

****Find the value of:

1) log 5+ log 20+ log 24 + log 25 - log 60 is

A) 0 B) 1 C) 2 D) 3 E) n

2) log 6+ 2log 5+ log 4 - log 3 - log 2 is

A) 0 B) 1 C) 2 D) 3 E) n

3) 2log 5+ log 8- 1/2 log 4 is

A) 0 B) 1 C) 2 D) 3 E) n

4) log 8+ log 25+2 log 3 - log 18 is

A) 0 B) 1 C) 2 D) 3 E) n

5) 5log 2+ 3/2log 25+ 1/2 log 49 - log 28 is

A) 0 B) 1 C) 2 D) 3 E) n

6)1/2 log 25 - 2log 3 +log 18 is

A) 0 B) 1 C) 2 D) 3 E) n

7) log 2 + 16 log(16/15) + 12 log(25/24) + 7 log(81/80) is..

A) 0 B) 1 C) 2 D) 3 E) n

8) log(81/8) - 2log(3/2) + 3log(⅔) + log(3/4) is

A) 0 B) 1 C) 2 D) 3 E) n

9) 7log(16/15)+ 5log(25/24) + 3 log(81/80) is

A) 0 B) 1 C) 2 D) 3 E) n

10) 2 log 2+ log 5 - 1/2 log 36 - log (1/30) is

A) 0 B) 1 C) 2 D) 3 E) n

11) log(1.2)+ 2 log (0.75) - log(6.75) is

A) 0 B) 1 C) 2 D) 3 E) n

12) log 5+ 16 log(625/6) + 12 log(4/375) + 7 log(81/1250) is..

A) 0 B) 1 C) 2 D) 3 E) n

13) 2log₁₀8 + log₁₀36 - log₁₀(1.5)

- 3 log₁₀ 2 is

A) 0 B) 1 C) log₁₀ 2 D) log₁₀ 32 E) n

14) 2 log₁₀5+ 2 log₁₀3 - log₁₀2 + 1 is..

A) 0 B) 1 C) log₁₀ 25 D) log₁₀ 1125

15) 2+ 1/2 log₁₀9 - 2 log₁₀5 is..

A) 0 B) 1 C) log₁₀ 2 D) log₁₀ 12 E) n

16) 1/2 log₁₀9 + 1/4 log₁₀81 + 2 log₁₀6 - log₁₀12 is .

A) 0 B) 1 C) log₁₀ 2 D) log₁₀ 7 E) log₁₀ 27

17) 2log₁₀(11/13) + log₁₀(130/77) - log₁₀(55/91) is .

A) 0 B) 1 C) log₁₀ 2 D) log₁₀ 20 E)n

18) 1 - 1/3 log₁₀64 is..

A) 0 B) 1 C) log₁₀ 2 D) log₁₀ 5 E) log₁₀ (2.5)

19) log(32/243) - log(16/75) - 2 log(5/9) is..

A) 0 B) 1 C) 2 D) 3 E) log 2

20) 7 log (15/16) + 6 log (8/3) + 5 log(2/5) + log (32/25)

A) 0 B) 1 C) log 2 D) log 3 E) log 5

21) {log√(27)+log(8)+log√(1000)}/ log(120)

A) 0 B) 1 C) 2 D) 3 E) 3/2

22) {log√(27) + log√(8) + log√(125)} /{log(6) + log(5)} is

A) 0 B) 1 C) 2 D) 3 E) 3/2

23) value of log₂√6+ log₂√(2/3) is

A) 0 B) 1 C) 10 D) 2 E) none

24) value of log 144 - log 90 + log (0.0625) is.

A) 0 B) 1 C) -1 D) 2 E) -2

EXERCISE - G

** Solve::

1) 1/2 log (11+4√7)= Log(2+x)

A) 7 B) √ 7 C) 0 D) 1 E) none

2) log(x+2) + log(x-2)= log 5.

A) 1 B) 2 C) 3 D) 0 E) none

3) log(x+4) - log(x- 4)= log 2

A) 0 B) 1 C) 2 D) 12 E) none

4) log(x+3) - log(x- 3)= 1.

A) 0 B) 1 C) 3 D) 11 E) 11/3

5) log(x² - 21) = 2.

A) 0 B) 1 C) ±2 D) ±11 E) 11/3

6) 2 logx + 1 = log 250.

A) 1 B) 2 C) 5 D) ±6 E) none

7) logx/log 5 = log 9/log(1/3).

A) 1 B) 2 C) 25 D) 1/25

8) log 7 - log 2 + log 16 - 2log 3 - log (7/45) = 1+ log x.

A) 0 B) 1 C) 3 D) 4.

9) log₁₀x - log(2x -1)= 1.

A) 1 B) 0 10 D) 19 E) 10/19

10) log₅(x²+x) - log₅ (x+1)= 2

A) ±1 B) ±3 C) ± 5 D) ± 6 E) 25

11) log₁₀5+log₁₀ (5x+1)=log₁₀(x+5) +1

A) 0 B) 1 C) 2 D) 3

12) log₂ log₃ log₂ x= 1.

A) 5 B) 1 C) 2 D) 512

13) logₓ (8x-3) - logₓ 4 = 2

A) 3/2 B) 1/2 C) 1 D) 4

14) (log₁₀x - 5)/2 + (13 - log₁₀x)/3 = 2

A) 0 B) 1 C) 2 D) 10

15) log₃ (3+x)+ log₃(8-x) - log₃(9x-8) = 2 - log ₃ 9.

A) 0 B) 2 C) 4 D) 6 D) 8

16) log₂x + log₂(x+4)= 5

A) 0 B) 1 C) 2 D) 3 E) 4

17) ₓˡᵒᵍ10ˣ = 100x.

A) 0 B) 1 C) 10 D) 100 E) none

EXERCISE - H

A) If log₁₀2=0.3010, and log₁₀3 =0.7781, log₁₀7= 0.8451 then

1) log₅(5) is

A) 0.6990 B) 0.6099 C) 9069 D) n

2) log₁₀(45)

A) 0.6532 B)1.6532 C) -0.6532 D) n

3) log₁₀(2.4)

A) 1.6811 B) 0.6811 C) 0.7781 D) n

4) log₁₀(6)

A) 2.7781 B) 0.7781 C) 1.7781 D) n

5) log₁₀(108)

A) 0.333 B) 1.333 C) 2.0333 D) n

6) log₁₀³√(5)

A) 0.2330 B) 1.2330 C) 2.2330 D)n

7) log₁₀(70)

A) 0.8451 B) 1.8451 C) 2.8451 D)n

8) log 84

A) 0.9242783 B) 1.9242793

C) 2.9242783 D) none

9) log (21.6)

A) 0.3344539 B) 1.3344539

C) 2.3343539 D) none

10) log(0.00693)

A) 0.840733 B) 1.840733

C) 2.840733 D) 3.840733

11) log 294

A) 0.4683473 B) 1.4683473

C) 2.4683473 D) 4.4683473

12) log√(4.5)

A) 0.3266063 B) 1.3266063

C) 2.3266063 D) 2.3266063

13) If log₁₀2=0.3010, log₁₀3=0.4771, then log₁₂40 is

A) 0.485 B) 1.485 C) 2.485 D) 3.485

14) If log₁₀3= 0.4771 then log₂₅125 is

A) 2 B) 3 C) 3/2 D) 2/3

15) If log₁₀3= 0.4771 then log₁₀3000 is..

A) 0.4771 B) 1.4771 C) 2.4771 D) 3.4771

B) if log 2 =0.3010 and log(3)= 0.4771 then

16) a) log 8 is..

A) 0.3010 B) 0.6020 C) 0.9030 D) n

17) log 24 is

A) 0.4771 B) 0.9030 C) 1.3802 D) n

18) log 108 is..

A) 0.6020 B) 1.4313 C) 1.0333 D) 2.0333 E) n

19) log 25 is.

A) 2.0000 B)0.6990 C) 1.3980 D) n

20) log (0.405)¹/²

A) 1.9084 B) 3.9084 C) 0.3010 D) 3.6074 E) none

21) value of log₃log₄log₃81

A) 1 B) 2 C) 0 D) 3

22) log₃log₂log₂(256)

A) 0 B) 1 C) 2 D) 3 E) n

23) log₂log√₂log₃(81)

A) 0 B) 1 C) 2 D) 3 E) n

24) log₂[log₂{log₃(log₃27³}]

A) 0 B) 1 C) 2 D) 3 E) n

EXERCISE- I

Find the value of:

1) log₂(10) - log₁₆(625)

A) 0 B) 1 C) 2 D) 1/2 E) none

2) log₃log₂log₂(2⁸)

A) 0 B) 1 C) 2 D) 1/2 E) none

3) logᵤa . logᵥx . logₐv

A) 0 B) 1 C) 2 D) 1/2 E) none

4) log₃5 x log₂₅27 is...

A) 0 B) 1 C) 2 D) 3 E) 3/2

5) log₄5 x log₅3 is..

A) log₂3 B)2log₂3 C)log3 D)log2 E) n

6) log₄2 x log₂3 is..

A)log₂3 B)2log₂3 C)log3 D)log2 E) n

7) log₂10 - log₈125 is

A) 0 B) 1 C) 2 D) 10 E) none

8) logₐx . logₓc. log꜀a is..

A) 0 B) 1 C) 2 D) 10 E) none

9) value of log₆ log√₂ 8 is

A) 0 B) 1 C) 2 D) none

EXERCISE- J

Evaluate :

1) log₈√[8 {√8√(8)...∞}]

A) 0 B) 1 C) 2 D) 10 E) none

2) log₄√[4{√4√4...∞}]

A) 0 B) 1 C) 2 D) 10 E) none

EXERCISE - K

**SOLVE:

1)5ˡᵒᵍ ˣ + 3ˡᵒᵍ ˣ= 3ˡᵒᵍ ˣ⁺¹ - 5ˡᵒᵍ ˣ ⁻¹

A) 0 B) 1 C) 10 D) 100 E) none

2) log₅ (5¹⁾ˣ+ 125)= log₅6 + 1 + 1/2x

A) 1 B) 1/2 C) 1/3 D) 1/4 E) none

3) 1/(logₓ 10) + 2= 2/(log₀·₅2)

A) 0.25 B) 0.025 C) 0.0025 E) none

4) logₓ 2. Logₓ/₁₆ 2= log ₓ/₆₄2

A) 0 B) 1 C) 2 D) none

5) log₂x + log₄ x+ log₁₆ x= 5.25.

A) 1 B) 1/10 C) 8 D) none

6) logₓ5 log ₓ/₁₂₅5= log ₓ/₆₂₅5.

A) 25 B) 1/25 C) 50 D)1/50 E) n

7) Log₇ log₅{√(x+5) + √x}= 0

A) 0 B) 1 C) 2 D) 4 E) none

8) a²ˣ⁻³. b²ˣ = a⁶⁻ˣ. b ⁵ˣ then x log(a/b) is

A) 3 B) log a C) 3 log a D) none

9) If logₐ b= 6 and log₁₄ₐ(8b)= 3, then the value of a is..

A) 5 B) 6 C) 7 D) none

9) If the logarithm of y² to the base x³ is equal to the logarithm of x⁸ to the base y¹², then the value of each logarithm is.

A) ±1/3 B) 2/3 C) ±1/5 D) none

10) If log(x²y³)= a and log(x/y)= b, then values of log x and log y is

A) 1/5(a+3b), 1/5(a-2b)

B) 1/5(a-3b), 1/5(a+2b)

C) 1/5(a+3b), 1/5(a+2b) D) none

11) If log (x³y³)= 6, log(x²/y)= -1/2 then the value of x and y is

A) √10, 10√10 B) 10√10, 10√10

C) 10√10, √10 D) none

11) If log x +1= 0, then x is

A) 3 B) 4 C) 5 D) 1/10 E) none

11) if log√₈ b= 10/3 then b is..

A) 32 B) 33 C) 34 D) 35

13) If logₓ(1/2)= 1/2 then x is

A) 1/2 B) 1/3 C) 1/4 D) 1/5

EXERCISE - L

1) If x²+y² = 6xy, then the value of 1/2(l ogx + log y +3 log 2) is

A) log(x+y) B) log x C) log y D) n

2) If a² +b² =23ab Then the value of 1/2(log a+ log b)

A) log a B) log b C) log(a+b) D) n

3) If a³⁻ˣ.b⁵ˣ=aˣ⁺⁵.b³ˣ, then the value of x log(b/a) is

A) log a B) log b C) 1 D) none

4) If a²+b²=14ab, then the value of 1/2(log a + log b) is

A) log a B) log b C) log(a+b) D) n

5) If a²+b² = 27ab, then the value of 1/2 (log a + log b) is

A) log a B) log b C) log(a+b) D) n

6) If log{(a+b)/3} =1/2(log a+ log b) then the value of a/b + b/a is

A)7 B) 6 C) 5 D) 1 E) none

7) If log{(a-b)/4} =1/2 (lig a+ log b) then the value of (a² + b²) is

A)18ab B) 18a C) 18b D) none

8) If a²+b²= 7ab, then the value of ½(loga +logb) is

A) log a B) log b C) log(a+b) D) n

9) If x² + y² =11xy, then the value of 2log3 + log(x) +log(y) is

A) 2log(x-y) B) log(x-y) C) 2xy D) n

10) If a² =b³=c⁵ =x⁶ then the value of logₓ(abc) is

A) 31 B) 5 C) 155 D) 31/5

11) If log(a+b)/7= 1/2 {log a + log b) then the value of a/b +b/a is

A) 47 B) 7 C) 4 D) none

12) The value of log (aˣ/bˣ) + log (bˣ/cˣ) + log ( cˣ/aˣ) is..

A) 0 B) 1 C) log a D) log b E) log ab

Mg. A- R.1

1) If p log a= q and q log b = p, then the value of log (aᑫbᵖ).

A) p+q B) p³+ q³ C) (p³+q³)/p D) (p³+q³)/pq E) none

2) logₐ m+ logₐ n= logₐ (m+n), then value of m in terms of n is

A) n B) n/(n-1) C) 1/(n-1) D) (n-1)/n

3) If (log p)/m= (log q)/n = (log r)/l = log x express p²/qr as a power of x.

A) x²⁻ᵐ⁻ⁿ⁻ˡ B) x²⁺ᵐ⁻ⁿ C) x1⁺ᵐ⁻ⁿ D) n

4) If 3 + log x = 2 log y, express x in terms of y

A) y/1000 B) y²/1000 C) y³/1000 D) n

5) If log₁₀ a= r, then value of (a)²⁾ʳ is

A) 1 B) 10 C) 100 D) 1000 E) 10000

6) If a= b²= c³= d⁴, then the value of log ₐ abcd is

A) 2 B) 1 C) 25 D) 25/2

7) Value of ₐlogₐx is

A) a B) x C) ax D) a/x E) none

8) If logₐ log₂ Log₂ 256=2 then a is

A) 0 B) 1 C) 2 D) 3

9) If log₃ log₂logₐ81=1, then a is

A) √2 B) √3 C) √5 D) √6 E) none

10) value of log ₘn . logₙm is

A) 0 B) 1 C) 2 D) √3

11) If logₘA= log ₙA . P then P is

A) logₘn B) log n C) log m D) none

12) logₘₙm= x,then log ₘₙn is

A) logₘn B) logₙm C) logₘₙm D) n

13) value oflog₁₂(log₁₉16.log₁₆19)

A) 0 B) 1 C) 2 D) none

14) value of log₇9. Log₅7. log₃5

A) 0 B) 1 C) 2 D) none

15) value of log₁₀25(1+ log₂₅40) is

A) 0 B) 1 C) 2 D) none

16) If log₂ 3= a, then₈27 is

A) 0 B) 1 C) 2 D) a E) 3

17) If log ₂₀2= a, then log₂₀10 is

A) 0 B) 1 C) a D) 1-a E) 1+a

18) If log₁₀2= p, then log₂₀5 is

A) 1-p B) 1+p C) (1-p)/(1+p) D) n

19) If logₑ2 logₘ625=log₁₀16. logₑ10 then the value of m is

A) 2 B) 3 C) 4 D) 5 E) none

19) If log₅k. log₃5. logₖx= k then the value of x. If k= 3.

A) 3 B) 3ᵏ C) k³ D) none

20) If logₓ4+ logₓ8+ logₓ32= 5, then the value of x is..

A) 1 B) 2 C) 3 D) 4 E) none

21) value of log 1+ log 2 + log 3 is

A) log 5 B) log(1+2+3) C) 1 D) none

22) Value of (log 81)/(log 27) is

A) 3 B) 4 C) 3/4 D) 4/3 E) none

23) Value of (log128)/(log 32) is

A) 5 B) 7 C) 5/7 D) 7/5 E) none

24) Value of (log 27)/(log √3) is

A) 2 B) 3 C) 6 D) none

25) Value of (log 9 - log 3)/(log 27) is

A) 1 B) 3 C) 1/3 D) none

26) If log(m+n)= log m+ log n then n/(n-1) is

A) m B) mn C) 1 D) 0 E) none

27) Value of (log 32)/(log 4) is

A) 2 B) 5 C) 5/2 D) 2/5 E) none

Mg. A- R. 2

1) Value of (log 27)/(log 9) is

A) 2 B) 3 C) 2/3 D) 3/2 E) none

2) If log 2=x, log 3= y and log 7= z, express log (4 ³√(63)) in terms of x,y ,z.

A) 2x+ 3z+ 2y/3

B) 2x + z/3 + 2y/3

C) 2x + z/3 - 2y/3

D) 2x - z/3 + 2y/3 E) none

3) Value of 2 log(11/13)+ log(130/77) - log(55/91) is

a) 0 b) log 2 c) log 3 d) 2log 2

4) If log (m+n)= log m + log n then the value m/(m -1) is

a) 1 b) n c) n² d) n/2

5) If log {(x+y)/2}= 1/2 (logx + log y) then the value of x is

a) x b) xy c) x² d) y²

6) If log{(a - b)/2}= 1/2(log a + log b) then the value of a²+ b² is

a) 27 b) 27a c) 27b d) 27ab

7) If logₓ (1/49)=-2 then x is

a) 5 b) 6 c) 7d) 8 e) 9

8) If logₓ(1/4√2)=-5 then x is:

a) 1 b) 2 c) √2 d) -1 e)-√2

9) If logₓ(1/243)=10 then x is

a) 1 b) 3 c) √3 d) -3 e) 1/√3

10) If log₄ (32)= x -4 then x is

a) 1 b) 3 c) 13 d) 13/2 e) 2/13

11) If log ₇(2x²-1)=2 then x is

a) 5 b) -5 c) ±5 d) 1/5

12) if log(x²-21)= 2 then x is

a) ±1 b) ±2 c) ±9 d) ±11 e) ±13

13) If log₆{(x-2)(x+3)}= 1 then x is

a) 3 b) 4 c) -3 d) -4 e) 3,-4

14) if log₆(x-2)+ log₆(x+3)=1 then x is:

a) 1 b) 2 c) 3 d) 4 e) 5

15) log(x+1)+ log(x-1)= log11+ 2 log3 then x is

a) 9 b) 10 c) 11 d) 12 e) 13

16) If 1/3 log x= 1/2 log y and log(xy)= 5 then the value of x and y are

a) 1000,100 b) 100,10 c) 10,1 d) 10,100

17) If log 27 base √3= x then x is

a) 3 b) 4 c) 6 d) 9

18)

Tuesday, 14 December 2021

QUICK REVISION (THEORY OF QUADRATICS EQUATION)

Theory of quadratic equation

--------------------------------------------

Exercise --1

----------------

A) SHORT ANSWER TYPE:

1) If m, n are the roots of the equation x²+ x+1= 0, then find the value of m⁴+ n⁴+ 1/mn. 0

2) For what value of p(≠0) sum of the root of px²+2x+3p= 0 is equal to their product? -2/3

3) Form a quadratic equation whose one root is 2-√5. x²-4x-1=0

4) If 2 +i√3 is a root of x²+ px+q= 0, find p and q. -4,7

5) If one root of 2x²- 5x+k= 0 be double the other, find k. 25/9

6) If one root of x²+ (2-i)x - c= 0 be i. Find the value of c and other root of the equation. 2i, -2

7) Form a quadratic equation whose one root is 2 - 3i. x²-4x+13=0

8) If the roots of the equation qx²+ px+ q= 0 are imaginary, find the nature of the roots of the equation px²-4qx+ p=0. Real, unequal

9) If one root of x²+ px+8= 0 is 4 and two roots of x²+ px+q= 0 are equal, find q. 9

10) Construct a quadratic in x such that AM of its roots is A and GM is G. x²-2Ax+ G²= 0

11) if 5p²- 7p+4= 0 and 5q²- 7q+4= 0, but p≠ q, find pq. 4/5

12) if the equation x²+px+6= 0 and x²+4x+4=0 have a common root, find p. 5

13) if x is a real, show that the expression is always positive. Find its minimum value and the value of x for which it will be minimum. 14/5, 4/5

14) If c, d are the roots of (x-a)(x-b) - K= 0 show that a, b are the roots of (x- c)(x- d)+ K= 0.

15) If the roots of the equation x²- 4x - log₂a=0 are real, find the minimum value of a. 1/16

16) Given that m, n are the roots of x² -(a -2)x - a+1= 0. If a be real, Find the least value of m²+n². 1

17) If m, n are the roots of x²- 4x+5 = 0, form an equation whose roots are m/n +1 and n/m +1. 5x²-16x+16=0

B) CHOOSE THE CORRECT:

1) The sum of their reciprocals of the roots of 4x²+3x+7= 0 is

A) 7/4 B) -7/4 C) -3/7 D) 3/7

2) If one root of 5x²-6x+K= 0 be reciprocal of the other, then

A) K= 6 B) K= 5 C) K= -5 D) K= 1/5

3) If x be real, the maximum value of 5+ 4x- 4x² will be

A) 5 B) 6 C) 1 D) 2

4) The roots of x²+ 2(3m,+5)x+ 2(9m²+25) = 0 will be real if

A)m>5/3 B)m=5/3 C)m<5/3 D) m=0

5) The equation (4-n)x²+(2n+4)x +8n +1= 0 has equal integral roots, if

A) n= 0 B) n=1 C) n=3 D) none

6) The equation whose roots are reciprocal of the roots of ax²+ bx+c= 0, is

A) bx²+ cx+a= 0 B)cx²+ bx+a= 0

C) bx²+ ax+c= 0 D) cx²+ ax+b= 0

7) The value of the expression (ax)²+ bx+c, for any real x, will be always positive, if

A) b²- 4ac>0 B) b² - 4ac< 0

C) b²- 4a²c> 0 D) b² - 4a²c< 0

8) The value of m for which the equation x²-x+m²= 0, has no real roots, can satisfy

A) m>1/2 B) m>-1/2 C) m<-1/2 D) m<1/2

9) If x be real and a> 0, the least value of ax²+ bx+c will be

A) -b/a B) -b/2a C) -(b²-4ac)/2a D) -(b²- 4ac)/4a

10) The roots of ax²+ bx+c= 0 will be both negative, if

A) a>0, b> 0, c< 0

B) a>0, c> 0 ,b< 0

C) a>0, b> 0, c>0

D) b>0, c> 0 a< 0

11) If a, b are the roots of x² -2x +2= 0, the least integer n(>0) for which aⁿ/bⁿ = 1, is

A) 2 B) 3 C) 4.D) none

C) GENERAL QUESTIONS:

1) If the roots of 2x²+ x+1= 0 are p and q, from an equation whose roots are p²/q and q²/p. 4x²-5x+2=0

2) the equation x² - cx+d= 0 and x²- ax+b= have one root common and the second equation has equal roots.

Prove that ac= 2(b+d).

3) If the roots x²+ 3x+4= 0 are m,n, form an equation whose roots are (m-n)² and (m+n)². x² - 2x -63= 0.

4) If the roots of x²- px+q=0 are in the ratio 2:3, show that 6p²=25q.

5) If the roots of ax²+ bx+c=0 are m, n, form an equation whose roots are 1/(m+n), and 1/m + 1/n. bcx²+ (ac+b²)x + ab= 0.

6) If m, n are the roots of ax²+ 2b x+c= 0 and m+ + K, n+ K those of Ax²+ 2Bx+C= 0, prove that (b²- ac)/(B² - AC)= (a/A)².

7) Show that if one root of ax²+ bx+c=0 be the square of the other, than b³ + a²c + ac²= 3abc.

8) If m, n are the roots of the equation x²+ px - q= 0 and a, b those of the equation x²+ px+q=0, prove that (m- a)(m - b)= (n- a)(n- b)= 2q.

9) If the ratio of the roots of ax²+ cx+c= 0 be p: q, show that, √(p/q) + √(q/p)+ √(c/d)= 0.

10) if m be a root of equation 4x²+ 2x-1=0, prove that its other root is 4m³ - 3m.

11) If the sum of the roots of 1/(x+p) + 1/(x+ q) = 1/r be equal to zero, show that the product of root is 1/2 (p²+ q²).

12) If a, b are the roots of x²+ px+1= 0 and c, d are the roots of x²+ qx+1=0, show that q²- p²= (A-- c)(b - c)(a+ d)(b+ d).

13) Show that if x is real, the expression (x²- bc)/(2x- b - c) has no real values between b and c.

14) If one root of the equation ax²+ bx+c= 0 be the cube of the Other, show that ac(a+ c)²= (b² - 2ac)².

15) If a²= 5a - 3, b² = 5b - 3 but a≠ b, then find the equation roots are a/b and b/a. 3x²- 19x+3= 0

16) the coefficient of x in x²+ px+q= 0 is misprinted 17 for 13 and the roots of the original equation. -3, -10

17) if b³ + a²c + ac²= 3abc, then what relation may exist between the roots of the equation ax²+ bx+c= 0 ? One root is the square of the other.

18) find the maximum and minimum value of: x/(x²-5x+9). 1, -1/11

19) If m, n are the roots of ax²+ 2bx+c= 0, form an equation, whose roots are mw + nw² and mw² + nw (w= omega). (ax - b)²= 3(ac - b²)

20) If √m ± √n denote the roots of x² - px+q= 0, show that the equation, whose roots are m± n is (4x - p²)²= (p² - 4q)².

21) prove that for all real value of x, the value of p²/(1+x) - q²/(1- x) is real.

22) if x be real, prove that 4(a - x)(x - a + √(a²+ b²)) can never be greater than (a²+ b²).

23) If the quadratics x²+ px+q=0 and x²+ qx+p= 0 have a common root, prove that their other roots will satisfy the equation x²+ x+pq = 0

24) Show that if a, b, c are real, the roots of the equation (b - c) x²+ (c - a)x+(a - b)= 0 are real and they are equal if a, b, c are in AP.

25) If the the roots of the equation ax²+ 2bx+b =0 are Complex, show that the roots of the equation bx²+ (b - c)x- (a+ c - b)= 0 are real and cannot be equal unless a =b =c.

26) If a, b, c are real, show that the roots of the equation 1/(x+a) + 1/(x+ b) + 1/(x- c) = 3/x are real.

27) Show that the equation (b - c)x²+ (c - a)x+(a - b)= 0, (c - a)x²+ (a - b)x+(b - c)= 0, have a common root, find it and the remaining roots of the equations. 1, (a-b)/(b- c) and (b-c)/(c-a)

28) Prove that the roots of the equation (a - b)x²+ 2(a + b - 2c)x++ 1= 0, are real or complex according as c does not or lie between a and b.

29) prove that if the equation ax²+ bx+ c= 0 and bx²+ cx+ a= 0 have a common root, then neither a+ b+ c= 0 or a= b= c.

30) If the equation ax+ by =1 and cx²+ dy² = 1 have only one solution, prove that, a²/c + b²/d = 1 and x= a/c, y= b/d.

31) if (a - K)x²+ b(b - K)y²+ (c - K)z²+ 2fyz+ 2gzx + 2hxy is a perfect square, show that a - gh/f = b - hf/g = c - fg/h = K

32) Prove that x²+ y²+ z² + 2ayz + 2bzx + 2cxy can be resolved into two rational factors if if a² + b² + c² - 2abc = 1.

33) find K so that the value of x given by K/2x = a/(x+ c) + b/(x- c) may be equal. If m, n are two values of K and l, p the corresponding values of x, show that m. n = (a - b)² and l² p²= c².

a+ b± 2√(ab)

Monday, 13 December 2021

MOMENTS, SKEWNESS, KURTOSIS

SKEWNESS:
Karl Pearson: (Mn - Mod)/S. D
: 3(Mn - Med)/S. D
: (Q₃+ Q₁ - 2 Med)/(Q₃ - Q₁)

1) Find the coefficient of skewness from the following:
a) Value: 6 12 18 24 30 36 42

F: 4 7 9 18 15 10 5. +0.139

b) variable frequency

20.5 - 23.5 17

23.5 - 26.5 193

26.5 - 29.5 399

29.5 - 32.5 194

32.5 - 35.5 27

35.5 - 38.5 10

With the help of mode. 0.068

c) Marks          No. Of students
Above 0               150
    ,,      10               140
    ,,      20               100
    ,,      30                 80
    ,,      40                 80
    ,,      50                 70
    ,,      60                 30
    ,,      70                 14
    ,,      80                   0
With the help of Median - 0.754

d) Marks          No. Of students
Below 80               12
    ,,       90                30
    ,,      100              65
    ,,      110             107
    ,,      120             157
    ,,      130             202
    ,,      140             222
    ,,      150             330 - 0.332

e) Variable         Frequency
    10 - 20               358
    20 - 30              2417
    30 - 40               976
    40 - 50               129
    50 - 60                 62
    60 - 70                 18
    70 - 80                 10
With the help of Quartiles. 0.131

f) Convert the following into an ordinary frequency table and obtain the values of Quartiles Deviation and Coefficient of Skewness.
Marks below.    No of students
    80                         240
    70                         190
    60                         125
    50                           95
    40                           75
    30                           60
    20                           40
    10                           25 -0.473

2) In a certain distribution the following results were obtained:
Mean = 45.00
Median= 48.00
Coefficient of skewness= - 0.4. find the standard deviation. 22.5

3) For a moderately skewed data, the arithmetic mean is 100, the coefficient of variation is 35 and the Karl Pearson's coefficient of Skewness is 0.2. find the mode and the median. 97.7

4) Karl Pearson's coefficient of Skewness of a distribution is +0.40. Its standard deviations is 8 and mean is 30. Find the mode and median of the distribution. 26.8, 28.93

5) For a group of 10 items, ∑x= 452, ∑x² =24270, Mode= 43.7. Find the Pearson's coefficient of skewness.
A) +0.08 B) 0.80 C) -0.08 D) -0.80

6) For a moderately skewed distribution, mean= 160, Mode=157, S. D= 50. What is the value of coefficient of variation?
A) 31.52 B) 31.25 C) 31.35 D) 31.58

7) For a moderately skewed distribution, mean= 160, mode= 157, SD= 50. What is the value of Pearson coefficient of skewness.
A) -0.06 B)0.06 C)+0.60 D)+0.60

8) For a moderately skewed distribution, mean =160, Mode= 157, S D= 50. what is the value of Median ?
A) 195 B)159 C)169 D) 191

9) For a moderately skewed distribution, mean= 172, median= 167, SD= 60. what is the value of coefficient of skewness ?
A) -0.25 B) 0.52 C) -0.52 D)0.25

10) For a moderately skewed distribution, mean=172, median= 167, SD= 60. What is value of mode ?
A) 175 B)157 C)159 D) 169

11) The Karl Pearson's coefficient of skewness of a distribution is 0.32. its SD is 6.5, and the mean is 29.6. find the mode.
A)27.25 B)27.45 C)27.54 D) 27.52

12) the Karl Pearson's coefficient of skewness of a distribution is 0.32. Its SD is 6.5, and the mean is 29.6. find the median.
A) 28.9 B)25.9 C)29.8 D) 29.2

13) Given the coefficient of skewness= -0.475, mean= 64, median= 66; find the value of standard deviation.
A)12.36 B)12.53 C)12.63 D) 12.68

14) the measure of skewness for a certain distribution is -0.8. if the lower and upper quartiles are 44.1 and 56.6 respectively, find the median.
A) 55.35 B) 55.53 C) 55.85 D) 55.58

15) For a moderately skewed distribution, the mean is 100, the coefficient of variation is 35, and Karl Pearson Coefficient of Skewness is 0.2. find the mode.
A) 91 B) 93 C) 92 D) 94

16) For a moderately skewed distribution, the mean is 100, the coefficient of variation is 35, and Karl Pearson's coefficient of skewness is 0.2. Find the median.
A) 97.27 B) 97.77 C) 97.57 D) 97.67

17) In a certain distribution mean = 45, median =48, coefficient of skewness = - 0.4. what is the value of standard deviation ?
A) 22.3 B) 22.7 C) 22.5 D) 22.9

18) For a frequency distribution with coefficient of variation= 5, standard deviations= 2, Karl Pearson's coefficient of skewness =0.5. find the mean and standard deviation.
A) 39,40 B) 38, 40 C) 40, 39 D) 40, 42

19) the median, mode and coefficient of skewness for a certain distribution are respectively 17.4 ,15.3, 0.35. Calculate the coefficient of variation.

A) 47 B) 49. C) 48 D) 50

*** For a particular distribution, let mean= 50, Coefficient of Vari6= 40%, Coefficient of skewness =- 0.4 find

20) Find the Variance

A) 399 B)400. C) 401 D) 402

21) find the median.

A) 52.76 B)53.76 C) 53.67 D)52.67.

22) find the mode

A) 58. B) 59 C) 60 D) 61

23) Find the coefficient of variation of a frequency distribution, given that its mean is 120, Mode= 123, coefficient of skewness= - 0.3.

A)8.33%. B)8.53% C)8.34% D)8.53%

24) The median, mode and the coefficient of skewness for a certain distribution are respectively 17.4, 15.3 and 0.35. Find the value of coefficient of variation.

A) 47 B) 49 C) 51. D) 53

** It is given that mean, median and coefficient variation of a set of variable are 45, 42 and 40 respectively.

25) find the mode.

A) 33 B) 34 C) 35 D) 36.

26) find SD

A) 17 B) 19. C) 18 D) 20

27) find the coefficient of skewness

A) 0.3 B) 0.4 C) 0.5. D) 0.6

28) The measure of skewness for a certain distribution is - 0.8. if the lower and upper quartiles are 44.1, and 56.6 respectively. find the median.

A) 56.35 B) 55.36

C) 56.56 D) 55.35.

Saturday, 11 December 2021

QUICK REVISION (COMPLEX NUMBER)

A) SHORT ANSWER TYPE & OBJECTIVE QUESTIONS:

1) Find the conjugate of the complex number (2+3i)/(2-3i). 5/13 - 12i/13

2) find the modulus of (1-i)³/(1-i³). 2

3) Find the amplitude of -3- 3i. -3π/4

4) Resolve into factors: a²+ ab+ b². (a- bw)(a- bw²)

5) Find the square root of -i. ±1/√2(1- i)

6) Evaluate : (1- w²)(1- w⁴)(1- w⁸)(1- w¹⁶). 9

7) If z= x+ iy and |z- 2| = |2z - 1|, Prove that, x² + y² = 1.

8) Find the smallest positive integers n for which {(1+i)/(1 - i)}ⁿ = 1. 4

9) Find the square root of q+ √(q² -1), 0< q <1. ±1/√2 {√(1+q)+ i √(2-q)}

10) If x= a+ b, y= aw + bw², z= aw² + be, find xyz. a³ + b³

11) Solve: |z| - z = 1+ 2i, (z= x+ i y). 3/2 - 2i

12) The sum and the product of two complex numbers are respectively 6 and 25. Find the numbers. 3±4i

13) Determine the three cube roots of i. i, (i+√3)/2

14) If z= x + i y and (z -1)/(z+1) is purely imaginary, prove that, |z| = 1.

15) The absolute value of (i/2) is.

A) i/2. B) 1/2. C) 1/√2 D) none

16) The value of √4 . √9 is

A) 6 B) -6. C) -6 or 6 D) none

17) The value of w⁴+ w⁸+ 1/w .1/w² is

A) w B) w². C) -1 D) 0.

18) The argument of ia(a< 0), is.

A) 0 B) π/2 C) 3π/2. D) π

19) The value of (1- w +w²)⁵ + (1+ w -w²)⁵ is..

A) -1 B) 1 C) -32 D) 32.

20) The amplitude of (a+ i b)² is

A) tan⁻¹(b/a) B) 2tan⁻¹(b/a).

C) 2tan⁻¹(a/b) D) tan⁻¹(a/b)

21) If z= x + i y, the value of (amp z + amp z') is.

A) 0 B) π/2 C) π D) none.

22) the quantity, whose cube root is 1/2 (√3 + i), is.

A) -1 B) 1 C) -i D) i.

23) The value of (1+ w)(1+ w²)(1+ w⁴)(1- w⁸)....2n factors, is

A) 1. B) 2ⁿ C) -1 D) wⁿ

24) For any complex number z, the minimum value of |z | + |z - 1| is...

A) 0 B) 1/2 C) 1. D) 3/2

25) The real part of (2- i)²/(2+ i) is

A) -2/5 B) -6/5 C) -11/5 D) none

B) GENERAL QUESTIONS:

1) Show that (1- w+ w²) (1- w²+w⁴) (1- w⁴+ w⁸)....2n factors = 2²ⁿ.

2) If X+ i Y be one of the cube root of x + i y, prove that, 4(X² - Y²)= x/X + y/Y.

3) If x= a + i b, y= a¢+ b$, z= a$ + b¢, where ¢, $ are complex cube roots of unity, show that x³ + y³ + z³ = 3(a³ + b³).

4) If x= w² - w - 2, evaluate x⁴+ 3x³ + 2x² -11 x -4. 3

5) If x= 2 - i √3, Find the value of k from the equation 2x⁴ - 5x³ - 3x² + 41x + k= 0. -35

6) If cos a + i sin a, b= cos b+ I sin b, c= cos d + i sin d and a+ b+c= 0, prove that a²+ b²+ c²= 0.

7) If z= x + i y and (z- i)/(z + i) = i b, show that (x - 1/2)² + (y - 1/2)² = 1/2.

8) Prove that (a + bw + cw²) + (a + bw² + cw)³ = (2a- b - c)(2b - c - a)(2c - a- b) and 27abc, if a+ b+ c= 0.

9) If z= x + i y and arg{(z -1)/(z+ 1)} = π/4, Show that the locus of (x,y) is a circle.

10) If w be a Complex cube root of unity, find the simplified value of (a +bw + cw²)/(c + aw+ bw²) + (a +bw + cw²)/(b + cw+ aw²). -1

11) Show that the points 2+ 3i , 0 and 1/(-2 + 3i) are collinear.

12) Express a+ i b in the form pw + qw². (b/√3 - a)w + (-b/√3 - a)w² or (-b/√3 - a)w + (b/√3 - a)w²

13) solve: z² + z'= 0 (z= x+ i y). 0, -1, 1/2 ± √3 i/2.

14) If a²+ b²+ c²= 1 and b+ ic= (1+ a)z, then prove that (1+ i z)/(1- i z) = (a + i b)/(1+ c).

Saturday, 4 December 2021

LINEAR INEQUATION(XI)

* SOLVING AN INEQUATION:

It is the process of obtaining all the possible solution of an inequation.

* SOLUTION SET:

The set of all possible solutions of an inequation is known as its solution set.

=> SOLVING LINEAR INEQUATION IN ONE VARIABLE:

• RULE-1: Same number may be added to (or subtracted from) both sides of an inequation without changing the sign of inequality.

•RULE-2: Both sides of an inequation can be multiplied(or divided) by the same positive real number without changing the sign of inequality. However, the sign of inequality is reversed when both sides of an inequation are multiplied or divided by a negative number.

• RULE-3: Any term of an inequation may be taken to the other side with its sign changed without affecting the sign of inequality.

A linear inequation in one variable is of the form

ax+b<0 or, ax+b≤0 or, ac+b> 0 or, ax+ b ≥ 0.

We follow the following STEP to solve a linear inequation in one variable.

• STEP-1: Obtain the linear inequation.

• STEP -2 : Collect all terms involving the variable on one side of the inequation and the constant terms on the other side.

• STEP -3: Simplified both sides of inequality in their simplest form to reduce the inequation in the form

ax <b, or ax≤b, or ax>b, or ax≥b

• STEP -4: Solve the inequation obtained in step - 3 by dividing both sides of the inequation by the coefficient of the variable.

• STEP -5: Write the solution set obtained in step -4 in the form of an interval on the real line.

EXERCISE - 1

😀 ------------------ 😀

* Solve the following inequation:

1) 2x - 4≤ 0. (-∞, 2]

2) - 3x +12< 0. (4,∞)

3) 4x - 12 ≥ 0. [3,∞)

4) 12x < 50, when

a) x ∈ R. (-∞,25/6)

b) x ∈ Z. {....,-3,-2,-1,0,1,2,3,4}

c) x ∈ N. {1,2,3,4}

5) - 4x > 30, when

a) x ∈ R. (-∞,-15/2)

b) x ∈ Z. {....,-9,-8}

c) x ∈ N. null set

6) 7x +9> 30. (3,∞)

7) 4x -2 < 8, when

a) x ∈ R. (-∞,5/2)

b) x ∈ Z. {....-2,-1,0,1,2}

c) x ∈ N. {1,2}

8) 5x -3< 3x+1 when

a) x is a real number. (-∞, 2)

b) x is integer number. {..,-4,-3, -2, -1, 0,1}

c) x is a natural number. {1}

8) 3x - 7 > x+1. (4,∞)

9) x+ 5 > 4x -10. (-∞, 5)

10) 3x+ 9 ≥ - x+19. (5/2,∞)

11) 3x+17≤ 2(1-x). (-∞, 3]

12) 2(2x+3)-10≤6(x-2). [4,∞)

13) 2(3-x)≥ x/5 + 4. (-∞, 10/11]

14) -(x-3)+4< 5- 2x. (-∞,-2)

15) (2x-3)/4 +9≥ 3+ 4x/3. (-∞,63/10]

16) (3x-2)/5 ≤ (4x-3)/2. [11/14,∞)

17) (5x-2)/3 -(7x-3)/5 > x/4. (4,∞)

18) 1/2(3x/5 +4) ≥ (x- 6)/3. (-∞,120]

19) 3(x- 2)/5 ≥ 5(2 - x)/3. [2,∞)

20) x/5< (3x-2)/4 - (5x-3)/5. (-∞,2/9)

21) 2(x-1)/5 ≤ 3(2+x)/7. [-44,∞)

22) 5x/2 + 3x/4 ≥ 39/4. [3,∞)

23) (x-1)/3 +4< (x -5)/5 - 2. (-∞,50)

24) (2x+3)/4 - 3 < (x-4))3 - 2. (-∞, - 13/2)

25) (5-2x)/3 < x/6 - 2. (8,∞)

26) (4+2x)/3 ≥x/2 - 3. (-26,∞)

27) (2x+3)/5 - 2 < 3(x-3)/5. (-1,∞)

28) (x-2)≤(5x+8)/3. (-7,∞)

29) 1/(x -2)< 0. (-∞,2)

30) (x+1)/(x +2)≥ 1. (-∞,-2)

------------------------------------------------------

Type ::2

EQUATION OF THE FORM

* (ax+b)/(cx+d)> k, OR

* (ax+b)/(cx+d)≥ k, OR

* (ax+b)/(cx+d)< k, OR

* (ax+b)/(cx+d) ≤ k.

STEP -1: Obtain the inequation.

STEP-2:Transpose all terms on LHS

STEP-3: Simplify LHS of the inequation obtained in STEP-2 to obtain an inequation of the form

(px+q)/(rx+s)> 0 OR

(px+q)/(rx+s)≥0 OR

(px+q)/(rx+s)< 0, OR

(px+q)/(rx+s) ≤ 0

STEP-4: Make coefficient x positive in numerator and denominator if they are not.

STEP-5: Equate numerator and denominator separately to zero and obtain the values of x. These values of x are generally called critical points.

STEP-6: Plot the critical points obtained in STEP-5 on real line. These points will divide the real line in three regions.

STEP-7: In the right most region the expression on LHS of the inequation obtained in STEP-4 will be positive and in other regions it will be alternatively negative and positive. So, mark positive signs in the right most region and then mark alternatively negative and positive signs in other regions.

STEP-8: Select appropriate region on the basis of the sign of the inequation obtained in STEP-4 , Write these region in the form of interval to obtain the desired solution sets of the given inequation.

EXERCISE-2

-----------------

Solve the inequation of followings:

1) (2x+4)/(x-1) ≥ 5. (1,3]

2) (x+3)/(x-2) ≤2. (-∞,2)U(7,∞)

3) (2x-3)/(3x-7) ≤2. (-∞,3/2) U (7/3,∞)

4) 3/(x-2) < 1 (-∞,2)U(5,∞)

5) 1/(x-1) ≤2. (-∞,2)U(3/2,∞)

6) (4x+3)/(2x-5) ≤2. (-∞,5/2) U (33/8,∞)

7) (5x-6)/(x+6) <1. (-6,3)

8) (5x+8)/(4-x) <2. (-∞,0) U (4,∞)

9) (x-1)/(x+3) >2. (-7,-3)

10) (7x-5)/(8x+3)>4 (-17/25, -3/8)

11) x/(x-5) > 1/2. (-∞,-5)U(5,∞)

12) (x-3)/(x+4) >0, x ∈R {x∈ R: x < -4}U {x ∈ R: x> 3}

13) (x+5)/(x-2) >0, x ∈R. {x∈ R: x ≤ -5}U {x ∈ R: x> 2}

14) (2x+5)/(x+3) >1, x ∈R. {x∈ R: x < -3} U {x∈ R: x > -2}

15) (x+7)/(x+4) >1, x ∈R. {x∈ R: x > - 4}

16) (x+4)/(x+6) >1, x ∈R. {x∈ R: x < -6}

17) 3/(x -2) >2, x ∈R. {x∈ R: 2< x < 7/2}

18) (x-3)/(x+1) < 0, x ∈R. {x∈ R: -1 < x < 3}

-----------------------------------------------------

TYPE -3:

-------------

* If a is a positive real number, then

1) |x| < a <=> -a<x< a i.e. x∈ (-a, a).

2) |x|≤a <=> -a≤x≤a i.e. x x∈ [-a, a].

3) | x|> a <=> x<-a or x> a.

4) | x|≥ a <=> x≤-a or x≥ a.

* Let r be a positive real number and a be a fixed real number. Then,

1) |x - a| < r <=> a-r <x< a+r i.e. x∈ (a - r, a+ r).

2) |x -a|≤r <=> a-r ≤x≤a+r i.e.x∈ [a - r, a +r].

3) | x - a|> r <=> x<a -r or x> a +r

4) | x -a|≥ r <=> x≤-a- r or x≥ a+r.

EXERCISE -3

------------------

Solve the inequation of followings:

1) | x | < 5, x ∈R. {x∈ R: -5< x < 5}

2) |x |≥ 5, x ∈R. {x∈ R: x < -5} U {x∈ R: x ≥ 5}

3) |3x -2|≤ 1/2. {x∈ R: x < -5}

4) | 4x - 5 | ≤ 1/3, x∈[1/2,5/7}

5) | x - 2|≥5 (-∞,-3)U[7,∞)

6) |5 - 2x | ≤ 3, x ∈R. {x∈ R: 1≤x< 4}

7) | 3x - 7| > 4, x ∈R. {x∈ R: x < 1} U {x∈ R: x > 11/3}

8) |2(3-x)/5| < 9/5, x ∈R. {x∈ R: -3/2 < x < 15/2}

9) |x +1/3| > 8/3. (-∞,-3)U(7/3,∞)

10) |4 - x| + 1<3. (2,6)

11)|(3x-4)/2|≤ 5/12. (19/18,29/18)

12) |x -1|≤ 5, |x|≥ 2. (-∞,-2]U[2,∞)

13) 1≤| x-2|≤ 3. [-1,1]U[3,5]

14) |x- 1|≤5, | x |≥2. (-∞,-2]U[2,∞)

15) |x-2|/(x-2) > 0. (2,∞)

16) (|x| -1)/(|x| -2) ≥0, x∈R, x≠ ±2. [-1,1]U(-∞,-2]U(2,∞)

17) -1/(|x| -2) ≥ 1, where x∈R, x≠ ±2. [-2,-1]U[1,2)

18) |2/(x-4)|> 1, x ≠4. (2,4)U(4,6)

19) (|x+3| + x)/(x+2) > 1. (-5,-2) U(-1,∞)

20) 1/(|x| -3)< 1/2. (-∞,-5) U(-3, 3) U(5,∞)

21) (|x +2| - x)/x < 2. (-∞,2]U(1,∞)

22) |x -1| + |x-2|≥ 4. (-∞,-1/2]U [7/2,∞)

23) |x-1|+ |x-2| + |x -3|≥6. (-∞,0] U [4,∞).

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TYPE:: 4

STEP-1: Convert the given inequation, say ax + by≤ c, into the equation ax+ by = c which represents a straight line in xy-plane put.

STEP-2: Put y= 0 in the equation obtained in STEP- 1 to get the point where the line meets with x-axis. Similarly, put x= 0 to obtain a point where the line meets y-axis.

STEP-3: Join the points obtained in STEP-2 to obtain the graph of the line obtained from the given inequation. In case of strict inequalities like ax+ by< c or ax + by > c, draw the dotted line, otherwise mark it thick line.

STEP-4: Choose a point, if possible (0,0), not lying on this line: substitute its coordinates in the inequation. If the inequation is satisfied, then shade the portion of the plane which contains the chosen point; otherwise shade the portion which does not contain the chosen point.

STEP-5: The shaded regions obtained in STEP-4 represents the desired solution set.

*** NOTE::

In case of inequalities ax+ by≤ c and ax + by ≥ c points on the line are also a part of the shaded region while in the case of inequalities ax + by < c and ax + by > c points on the line ax + by = c are not shaded region.

EXERCISE-4

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** Solve graphically in a two- dimensional plane.

1) 5x - 8 ≥ 0.

2) 2y -3 < 0.

3) 2x - 3 ≥ 0.

4) 4x - 3 ≤ 0.

5) x - 1 < 0.

6) x - 2 > 0.

7) 2y - 3≥ 0.

8) 5y - 4 ≤ 0.

9) y - 1< 0.

10) 2y +5> 0.

11) 2x+3y ≤ 6.

12) 2x - y≥ 1

13) x ≥ 2

14) y ≤ -3

15) x+ 2y - y ≤ 0.

16) x+ 2y ≥ 6.

17) - 3x + 2y≤ 6.

18) 0≤ 2x - 5y +10.

19) 3x - 2y≤ x+y - 8.

20) |x|≤ 3.

21) |y - x|≤ 3.

22) |x - y|≥ 1.

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LINEAR INEQUATIONS(In Two Variables)

TYPE: 5 (A)

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STEP-1: Consider the equation ax+ by+ c= 0.

Draw the graph of this equation, which is a line.

In case of strict inequation > or < draw the line dotted, otherwise make it thick.

This line divides the plane into two equal parts.

STEP-2: Choose a point [if possible (0,0)], not lying on this line. If this point satisfies the given inequation then shade the part of the plane containing this point, otherwise shade the other part.

The shaded portion represents the solution set of the given inequation. The dotted line is not a part of the sulution, while thick line is a part of it.

EXAMPLE:

Draw the graph of the solution set of the inequation 2x -y≥ 1.

Solution: consider the equation 2x- y= 1.

The value of (x,y) satisfying 2x - y= 1 are:

x: 2 0

y: 3 -1

Plot the points A(2,3) and B (0,-1) on a graph paper.

Join A and B by a thick line.

This line divides the plane of the paper into two equal parts. Consider the point (0,0). It does not lie on 2x - y= 1.

Clearly, (0,0) doesn't satisfy 2x - y≥ 1.

So shade that part of the plane divided by line AB which does not contain (0,0).

The shaded part of the plane together with all points on line AB constitute the solution set of the inequation 2x -y ≥ 1.

EXERCISE-5

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A) Draw the graph of the solution of each of the followings:

1) x+ y ≥ 4.

2) x - y ≤ 3.

3) x+ 2y > 1.

4) 2x - 3y < 4.

5) x ≥ y - 2.

6) y - 2 ≤ 3x.

7) 3x+ 5y < 15.

8) 4x - 3y > 12.

9) 3x+ 2y > 6.

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TYPE:5(B)

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STEP-1:

B) Solve each of the following systems of inequation graphically:

1) 3x +2y ≤ 18, x+ 2y ≤10, x ≥0, y ≥ 0.

2) 2x+3y≤6,x+y≥2,x≥0,y≥0.

3) 3x+4y≥12,4x+ 7y≤28, y≥1, x≥0, y≥0.

4) x- 2y≤2, x+y≥3, -2x+y≤4, x≥0,y≥0.

5) x+2y≤100, 2x+y≤120, x+y≤70, x≥ 0, y≥ 0.

6) x+ 2y ≤2000, x+ y≤1500, y≤ 600, x≥ 0, y≥ 0.