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Quadratic Equations and Expressions 2.31<br />

COMPREHENSIVE LINK PASSAGES<br />

Passage I: 1. (b) 2. (c) 3. (a)<br />

Passage II: 1. (c) 2. (c) 3. (c) 4. (d)<br />

Passage III: 1. (c) 2. (b) 3. (c) 4. (c)<br />

Passage IV: 1. (b) 2. (c) 3. (c)<br />

Passage V: 1. (d) 2. (b) 3. (c)<br />

MATCH MATRIX<br />

1. A Æ (P, T); B Æ (Q, S); C Æ (R)<br />

2. A Æ (R, S, T); B Æ (P, Q, R); C Æ (R, S, T)<br />

3. A Æ (P, Q) ; B Æ (S, T) ; C((P, R)<br />

4. A Æ (P, Q, R, S, T); B Æ (P, Q); C Æ (R)<br />

5. (B)<br />

6. (C)<br />

7. (D)<br />

8. (A)<br />

ASSERTION AND REASON<br />

1. (b) 2. (d) 3. (c) 4. (b) 5. (a)<br />

6. (d) 7. (b) 8. (b) 9. (a) 10. (b)<br />

HINTS AND SOLUTIONS<br />

LEVEL I<br />

2. Let x 2 2 2 2 ...... to<br />

x= 2 + x<br />

x 2 = 2 + x<br />

x 2 – x – 2 = 0<br />

(x – 2)(x + 1) = 0<br />

x = 2, –1<br />

since square always provide us non negative values, so<br />

x = 2<br />

3. Given x = 2 + 2 2/3 + 2 1/3<br />

x – 2 = 2 2/3 + 2 1/3<br />

(x – 2) 3 = (2 2/3 + 2 1/3 ) 3<br />

x 3 – 6x 2 + 12x – 8<br />

= (2 2/3 ) 3 + (2 1/3 ) 3 + 3.2 2/3 2 1/3 (2 2/3 + 2 1/3 )<br />

= 4 + 2 + 3.2(x – 2)<br />

= 6 + 6x – 12<br />

= 6x – 6<br />

x 3 – 6x 2 + 6x = 8 – 6 = 3<br />

4. Divide by x 2 , we get<br />

Ê 1 ˆ Ê 1ˆ<br />

+ - + + 2=<br />

0<br />

2<br />

Áx<br />

Ë 2 ˜<br />

x ¯<br />

Áx<br />

Ë<br />

˜<br />

x¯<br />

2<br />

Ê 1ˆ Ê 1ˆ<br />

Áx+ - 2- x+ + 2=<br />

0<br />

Ë<br />

˜<br />

x¯ Á<br />

Ë<br />

˜<br />

x¯<br />

2<br />

Ê 1ˆ Ê 1ˆ<br />

Áx+ - x+ = 0<br />

Ë<br />

˜ Á ˜<br />

x¯ Ë x¯<br />

Ê 1ˆ<br />

Put Áx+ = t<br />

Ë<br />

˜<br />

x¯<br />

Then t 2 – t = 0<br />

t = 0 and t = 1<br />

1 1<br />

x + 0 and x 1<br />

x<br />

= + x<br />

=<br />

x 2 + 1 = 0 and x 2 – x + 1 = 0<br />

1±<br />

i 3<br />

x=± iand<br />

x=<br />

2<br />

Hence, the solutions are<br />

Ï 1±<br />

i 3¸<br />

̱<br />

i,<br />

˝<br />

Ó 2 ˛<br />

5. (x + 1)(x + 2)(x + 3)(x + 4) = 120<br />

{(x + 1)(x + 4)}{(x + 2)(x + 3)} = 120<br />

{(x 2 + 5x + 4)}{(x 2 + 5x + 6)} = 120<br />

Put x 2 + 5x = a<br />

(a + 4)(a + 6) = 120<br />

a 2 + 10a – 96 = 0<br />

(a + 16)(a – 6) = 0<br />

a = 6, –16<br />

x 2 + 5x = 6, –16<br />

x 2 + 5x – 6 = 0, x 2 + 5x + 16 – 0<br />

(x + 6)(x – 1) = 0, x 2 + 5x + 16 – 0<br />

- 5±<br />

i 39<br />

x = 1, – 6,<br />

2<br />

Hence, the solutions are<br />

Ï - 5±<br />

i 39¸<br />

Ì1, – 6, ˝<br />

Ó 2 ˛<br />

6. It is true for all real values of x.<br />

So it is an identity in x<br />

Thus the number of real solution is infinite.<br />

1 1<br />

7. x - = 2 -<br />

2 2<br />

x -4 x -4<br />

No real values of x satisfying the given equation<br />

Thus, x = f<br />

8. (x – 1) 2 + (x – 1) 2 + (x – 3) 2 = 0<br />

3x 2 – 12x + 14 = 0<br />

Clearly, D < 0<br />

Number of real solutions is zero<br />

x -ab x -bc x -ca<br />

9. + + = a+ b+<br />

c<br />

a + b b+ c c+<br />

a

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