May 2026

If a² + b² + c² = 29 and ab + bc + ca = 26, then a + b + c = Question: If \[ a^2+b^2+c^2=29 \] and \[ ab+bc+ca=26, \] then \[ a+b+c= \] (a) 9 (b) 6 (c) 7 (d) 10 Solution: \[ (a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca) \] \[ =(29)+2(26) \] \[ =29+52 \] \[ […]

If x² + y² + xy = 1 and x + y = 2, then xy = Question: If \[ x^2+y^2+xy=1 \] and \[ x+y=2, \] then \[ xy= \] (a) \(-3\) (b) \(3\) (c) \(-\frac{3}{2}\) (d) \(0\) Solution: \[ (x+y)^2=x^2+y^2+2xy \] \[ 2^2=x^2+y^2+2xy \] \[ 4=(x^2+y^2+xy)+xy \] \[ 4=1+xy \] \[ xy=3 \] \[

If x + y = 2 and xy = 1, then x⁴ + y⁴ = Question: If \[ x+y=2 \] and \[ xy=1, \] then \[ x^4+y^4= \] (a) 6 (b) 4 (c) 8 (d) 2 Solution: \[ x^2+y^2=(x+y)^2-2xy \] \[ =2^2-2(1) \] \[ =4-2 \] \[ =2 \] \[ x^4+y^4=(x^2+y^2)^2-2x^2y^2 \] \[ =2^2-2(1)^2 \]

If x/y + y/x = -1, then the value of x³ – y³ is Question: If \[ \frac{x}{y}+\frac{y}{x}=-1 \quad (x,y \ne 0), \] then the value of \[ x^3+y^3 \] is (a) 1 (b) -1 (c) 0 (d) \(\frac{1}{2}\) Solution: \[ \frac{x}{y}+\frac{y}{x}=-1 \] \[ \frac{x^2+y^2}{xy}=-1 \] \[ x^2+y^2=-xy \] \[ x^2+xy+y^2=0 \] Using identity \[

Factor of (x + y)³ – (x³ + y³) Question: Which of the following is a factor of \[ (x+y)^3-(x^3+y^3)? \] (a) \(x^2+2xy+y^2\) (b) \(x^2-xy+y^2\) (c) \(xy^2\) (d) \(3xy\) Solution: \[ (x+y)^3-(x^3+y^3) \] \[ =x^3+3x^2y+3xy^2+y^3-x^3-y^3 \] \[ =3x^2y+3xy^2 \] \[ =3xy(x+y) \] Hence, one factor is \[ \boxed{3xy} \] Next Question / Full Exercise

The Value of 249² – 248² is Question: The value of \[ 249^2-248^2 \] is (a) 1 (b) 477 (c) 487 (d) 497 Solution: \[ 249^2-248^2 \] \[ =(249-248)(249+248) \] \[ =1 \times 497 \] \[ =497 \] \[ \boxed{497} \] Next Question / Full Exercise

The Coefficient of x in (x + 3)³ is Question: The coefficient of \(x\) in \[ (x+3)^3 \] is (a) 1 (b) 9 (c) 18 (d) 27 Solution: \[ (x+3)^3 \] \[ =x^3+3x^2(3)+3x(3)^2+3^3 \] \[ =x^3+9x^2+27x+27 \] The coefficient of \(x\) is \[ \boxed{27} \] Next Question / Full Exercise

If 9x² – b = (3x + 1/2)(3x – 1/2), then the value of b is Question: \[ 9x^2-b=(3x+\frac{1}{2})(3x-\frac{1}{2}) \] (a) 0 (b) \(\frac{1}{\sqrt{2}}\) (c) \(\frac{1}{4}\) (d) \(\frac{1}{2}\) Solution: \[ (3x+\frac{1}{2})(3x-\frac{1}{2}) \] \[ =(3x)^2-(\frac{1}{2})^2 \] \[ =9x^2-\frac{1}{4} \] \[ 9x^2-b=9x^2-\frac{1}{4} \] \[ b=\frac{1}{4} \] \[ \boxed{\frac{1}{4}} \] Next Question / Full Exercise

If 49a² – b = (7a + 1/2)(7a – 1/2), then the value of b is Question: \[ 49a^2-b=(7a+\frac{1}{2})(7a-\frac{1}{2}) \] (a) 0 (b) \(\frac{1}{4}\) (c) \(\frac{1}{\sqrt{2}}\) (d) \(\frac{1}{2}\) Solution: \[ (7a+\frac{1}{2})(7a-\frac{1}{2}) \] \[ =(7a)^2-(\frac{1}{2})^2 \] \[ =49a^2-\frac{1}{4} \] \[ 49a^2-b=49a^2-\frac{1}{4} \] \[ b=\frac{1}{4} \] \[ \boxed{\frac{1}{4}} \] Next Question / Full Exercise

If a/b + b/a = 1, then a² + b² = Question: \[ \frac{a}{b}+\frac{b}{a}=1, \] then \[ a^2+b^2= \] (a) 1 (b) -1 (c) 1/2 (d) 0 Solution: \[ \frac{a}{b}+\frac{b}{a}=1 \] \[ \frac{a^2+b^2}{ab}=1 \] \[ a^2+b^2=ab \] \[ \boxed{ab} \] Next Question / Full Exercise