Educational

If 3x + 2/x = 7, then (9x² – 4/x²) = Question: If \[ 3x+\frac{2}{x}=7, \] then \[ \left(9x^2-\frac{4}{x^2}\right)= \] (a) 25 (b) 35 (c) 49 (d) 30 Solution: Using identity: \[ (a+b)(a-b)=a^2-b^2 \] Here, \[ a=3x, \quad b=\frac{2}{x} \] So, \[ 9x^2-\frac{4}{x^2} = \left(3x+\frac{2}{x}\right) \left(3x-\frac{2}{x}\right) \] Given: \[ 3x+\frac{2}{x}=7 \] Now find: \[ 3x-\frac{2}{x} […]

If x – 1/x = 15/4, then x + 1/x = Question: If \[ x-\frac{1}{x}=\frac{15}{4}, \] then \[ x+\frac{1}{x}= \] (a) 4 (b) \[ \frac{17}{4} \] (c) \[ \frac{13}{4} \] (d) \[ \frac{1}{4} \] Solution: Using identity: \[ \left(x+\frac{1}{x}\right)^2 = \left(x-\frac{1}{x}\right)^2+4 \] Substituting the given value: \[ \left(x+\frac{1}{x}\right)^2 = \left(\frac{15}{4}\right)^2+4 \] \[ = \frac{225}{16}+\frac{64}{16} \]

If x⁴ + 1/x⁴ = 194, then x³ + 1/x³ = Question: If \[ x^4+\frac{1}{x^4}=194, \] then \[ x^3+\frac{1}{x^3}= \] (a) 76 (b) 52 (c) 64 (d) none of these Solution: Using identity: \[ x^4+\frac{1}{x^4} = \left(x^2+\frac{1}{x^2}\right)^2-2 \] Substituting the given value: \[ 194 = \left(x^2+\frac{1}{x^2}\right)^2-2 \] \[ \left(x^2+\frac{1}{x^2}\right)^2 = 196 \] \[ x^2+\frac{1}{x^2} =

If x⁴ + 1/x⁴ = 623, then x + 1/x = Question: If \[ x^4+\frac{1}{x^4}=623, \] then \[ x+\frac{1}{x}= \] (a) 27 (b) 25 (c) \[ 3\sqrt{3} \] (d) \[ -3\sqrt{3} \] Solution: Using identity: \[ x^4+\frac{1}{x^4} = \left(x^2+\frac{1}{x^2}\right)^2-2 \] Substituting the given value: \[ 623 = \left(x^2+\frac{1}{x^2}\right)^2-2 \] \[ \left(x^2+\frac{1}{x^2}\right)^2 = 625 \] \[

(x – y)(x + y)(x² + y²)(x⁴ + y⁴) is equal to Question: \[ (x-y)(x+y)(x^2+y^2)(x^4+y^4) \] is equal to (a) \[ x^{16}-y^{16} \] (b) \[ x^8-y^8 \] (c) \[ x^8+y^8 \] (d) \[ x^{16}+y^{16} \] Solution: Using identity: \[ (x-y)(x+y)=x^2-y^2 \] Therefore, \[ (x^2-y^2)(x^2+y^2) = x^4-y^4 \] Now, \[ (x^4-y^4)(x^4+y^4) = x^8-y^8 \] Hence, \[

75 × 75 + 2 × 75 × 25 + 25 × 25 is equal to Question: \[ 75 \times 75 + 2 \times 75 \times 25 + 25 \times 25 \] is equal to (a) 10000 (b) 6250 (c) 7500 (d) 3750 Solution: Using identity: \[ a^2+2ab+b^2=(a+b)^2 \] Here, \[ a=75,\quad b=25 \] Therefore,

If the volume of a cuboid is 3x² – 27, then its possible dimensions are Question: If the volume of a cuboid is \[ 3x^2-27, \] then its possible dimensions are (a) \[ 3,\ x^2,\ -27x \] (b) \[ 3,\ x-3,\ x+3 \] (c) \[ 3,\ x^2,\ 27x \] (d) 3, 3, 3 Solution: Volume

If a – b = -8 and ab = -12, then a³ – b³ = Question: If \[ a-b=-8 \] and \[ ab=-12, \] then \[ a^3-b^3= \] (a) -244 (b) -240 (c) -224 (d) -260 Solution: Using identity: \[ a^3-b^3 = (a-b)(a^2+ab+b^2) \] Now, \[ (a-b)^2 = a^2+b^2-2ab \] Substituting the given values: \[

If a + b = 3 and ab = 2, then a³ + b³ = Question: If \[ a+b=3 \] and \[ ab=2, \] then \[ a^3+b^3= \] (a) 6 (b) 4 (c) 9 (d) 12 Solution: Using identity: \[ a^3+b^3 = (a+b)^3-3ab(a+b) \] Substituting the given values: \[ a^3+b^3 = 3^3-3(2)(3) \] \[ =27-18

(a – b)³ + (b – c)³ + (c – a)³ = Question: \[ (a-b)^3+(b-c)^3+(c-a)^3= \] (a) \[ (a+b+c)(a^2+b^2+c^2-ab-bc-ca) \] (b) \[ (a-b)(b-c)(c-a) \] (c) \[ 3(a-b)(b-c)(c-a) \] (d) none of these Solution: Let \[ x=a-b \] \[ y=b-c \] \[ z=c-a \] Then \[ x+y+z=0 \] Using identity: \[ x^3+y^3+z^3=3xyz \quad \text{when} \quad x+y+z=0