Exercise: 13-D
Competency Focused Questions
Q1: \(\left(3x+2y\right)^2-\left(3x-2y\right)^2=\)
Step 1: Use the identity \((a+b)^2 – (a-b)^2 = 4ab\)
Here, \(a = 3x\) and \(b = 2y\)
Step 2: Substitute values into the formula:
\((3x+2y)^2 – (3x-2y)^2 = 4 \cdot (3x) \cdot (2y)\)
Step 3: Multiply:
\(4 \cdot 3x \cdot 2y = 24xy\)
Answer: b. 24xy
Q2: \(\left(9.7\right)^2-\left(0.3\right)^2=\)
Step 1: Use the identity \(a^2 – b^2 = (a+b)(a-b)\)
Here, \(a = 9.7\) and \(b = 0.3\)
Step 2: Apply the formula:
\((9.7)^2 – (0.3)^2 = (9.7 + 0.3)(9.7 – 0.3)\)
Step 3: Simplify the sums and differences:
\(9.7 + 0.3 = 10\)
\(9.7 – 0.3 = 9.4\)
Step 4: Multiply:
\(10 \cdot 9.4 = 94\)
Answer: d. 94
Q3: If \(apq=\left(3p+q\right)^2-\left(39-q\right)^2\), then the value of a is:
Step 1: Use the identity \(a^2 – b^2 = (a+b)(a-b)\)
Here, \(a = (3p+q)\) and \(b = (3p-q)\)
Step 2: Apply the formula:
\((3p+q)^2 – (3p-q)^2 = [(3p+q) + (3p-q)] \cdot [(3p+q) – (3p-q)]\)
Step 3: Simplify each bracket:
\((3p+q) + (3p-q) = 6p\)
\((3p+q) – (3p-q) = 2q\)
Step 4: Multiply the simplified brackets:
\(6p \cdot 2q = 12pq\)
Answer: c. 12
Q4: If \(x-\frac{1}{x}=3\), then the value of \(x^4+\frac{1}{x^4}\) is:
Step 1: Use the identity:
\((x – \frac{1}{x})^2 = x^2 + \frac{1}{x^2} – 2\)
Step 2: Substitute the given value:
\((3)^2 = x^2 + \frac{1}{x^2} – 2\)
\(9 = x^2 + \frac{1}{x^2} – 2\)
\(x^2 + \frac{1}{x^2} = 9 + 2 = 11\)
Step 3: Use the square formula again to find \(x^4+\frac{1}{x^4}\):
\((x^2 + \frac{1}{x^2})^2 = x^4 + \frac{1}{x^4} + 2\)
\((11)^2 = x^4 + \frac{1}{x^4} + 2\)
\(121 = x^4 + \frac{1}{x^4} + 2\)
\(x^4 + \frac{1}{x^4} = 121 – 2 = 119\)
Answer: b. 119
Q5: If \(x^2+\frac{1}{x^2}=627\), then the value of \(\left(x-\frac{1}{x}\right)\) is:
Step 1: Use the identity:
\((x – \frac{1}{x})^2 = x^2 + \frac{1}{x^2} – 2\)
Step 2: Substitute the given value:
\((x – \frac{1}{x})^2 = 627 – 2\)
\((x – \frac{1}{x})^2 = 625\)
Step 3: Take square root on both sides:
\(x – \frac{1}{x} = \sqrt{625}\)
\(x – \frac{1}{x} = 25\)
Answer: a. 25
Q6: \(\left(x^4-1\right)+\left(x-1\right)\) is:
Step 1: Recall the identity for difference of squares:
\(a^2 – b^2 = (a-b)(a+b)\)
Step 2: Apply difference of squares to \(x^4-1\):
\(x^4-1 = (x^2)^2 – 1^2 = (x^2-1)(x^2+1)\)
Now factor \(x^2-1\) again using the same identity:
\(x^2-1 = (x-1)(x+1)\)
So, \(x^4-1 = (x-1)(x+1)(x^2+1)\)
Step 3: Divide by \((x-1)\):
\(\frac{(x-1)(x+1)(x^2+1)}{x-1} = (x+1)(x^2+1)\)
Answer: d. \((x+1)(x^2+1)\)
Q7: Which of the following will be equal to \(98 \times 102\)?
Step 1: Recall the identity for product of numbers in the form \((a-b)(a+b) = a^2 – b^2\)
Step 2: Express \(98\) and \(102\) as \(100-2\) and \(100+2\) respectively:
\(98 \times 102 = (100-2)(100+2)\)
Step 3: Apply the identity \((a-b)(a+b) = a^2 – b^2\):
\((100-2)(100+2) = 100^2 – 2^2\)
Answer: b. \(100^2 – 2^2\)
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