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SSC CGL 2023 PYQ — Question 63

SSC CGL - 18 July 2023 - Shift 4 • Quantitative Aptitude • Q63

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If p = 2+12−1\frac{\sqrt 2 +1}{\sqrt 2-1} and q = 2−12+1\frac{\sqrt 2-1}{\sqrt 2+1} then find the value of p2q+q2p\frac{p^2}{q}+\frac{q^2}{p}.

Show answer & explanation

Correct answer: C

Explanation

The correct answer is Option 3 i.e. 198.

p = 2+12−1\frac{\sqrt 2 +1}{\sqrt 2-1}

On rationalize 

⇒ p = 2+12−1×2+12+1\frac{\sqrt 2 +1}{\sqrt 2-1} \times{\frac{\sqrt 2 +1}{\sqrt 2+1}}

⇒ p = (2+1)222−12\frac{({\sqrt 2 +1})^2}{{\sqrt 2}^2-1^2} = (2+1+22)2−1\frac{({ 2 +1 + 2\sqrt2})}{{2}-1} = (3+22)({ 3 + 2\sqrt2})

Now, solve for q

⇒ q = 2−12+1\frac{\sqrt 2-1}{\sqrt 2+1}

On rationalize 

⇒ q = 2−12+1×2−12−1\frac{\sqrt 2 -1}{\sqrt 2+1} \times{\frac{\sqrt 2 -1}{\sqrt 2-1}}

⇒ q = (2−1)222−12\frac{({\sqrt 2 -1})^2}{{\sqrt 2}^2-1^2} = (2+1−22)2−1\frac{({ 2 +1 - 2\sqrt2})}{{2}-1} = (3−22)({ 3 - 2\sqrt2})

The value of p2q+q2p\frac{p^2}{q}+\frac{q^2}{p}  = p3+q3pq{p^3}+{q^3}\over{pq} = (p+q)(p2+q2−pq)pq({p}+{q})(p^2+q^2-pq)\over{pq}

Let's find out individual values

⇒ (p + q) = (3+22)({ 3 + 2\sqrt2}) + (3−22)({ 3 - 2\sqrt2}) = 6

⇒ pq = (3+22)({ 3 + 2\sqrt2})(3−22)({ 3 - 2\sqrt2}) = (9 - 62\sqrt{2} + 62\sqrt{2} - 8) = 1

⇒ p2 + q2 = (3+22)({ 3 + 2\sqrt2})2 + (3−22)({ 3 - 2\sqrt2})2

⇒ 9 + 8 + 122\sqrt{2} + 9 + 8 - 122\sqrt{2}

⇒ 18 + 16 = 34

On putting these values in the given question

⇒ p2q+q2p\frac{p^2}{q}+\frac{q^2}{p} = (p+q)(p2+q2−pq)pq({p}+{q})(p^2+q^2-pq)\over{pq}

⇒ (6)(34−1)1(6)(34-1)\over{1} = 6×33=1986\times 33 = 198