Equations

Step 1: Given,

2f(x)-3f%5C!%5Cleft(%5Cfrac%7B1%7D%7Bx%7D%5Cright)%3Dx%5E2%20%5Cqquad%20(1)

Replace

x

%5Cdfrac%7B1%7D%7Bx%7D

2f%5C!%5Cleft(%5Cfrac%7B1%7D%7Bx%7D%5Cright)-3f(x)%3D%5Cfrac%7B1%7D%7Bx%5E2%7D%20%5Cqquad%20(2)

Step 2: Write equations (1) and (2) in standard form:

2f(x)-3f%5C!%5Cleft(%5Cfrac%7B1%7D%7Bx%7D%5Cright)%3Dx%5E2

-3f(x)%2B2f%5C!%5Cleft(%5Cfrac%7B1%7D%7Bx%7D%5Cright)%3D%5Cfrac%7B1%7D%7Bx%5E2%7D

Step 3: Multiply equation (1) by

2

and equation (2) by

3

4f(x)-6f%5C!%5Cleft(%5Cfrac%7B1%7D%7Bx%7D%5Cright)%3D2x%5E2

-9f(x)%2B6f%5C!%5Cleft(%5Cfrac%7B1%7D%7Bx%7D%5Cright)%3D%5Cfrac%7B3%7D%7Bx%5E2%7D

Step 4: Add the above equations:

-5f(x)%3D2x%5E2%2B%5Cfrac%7B3%7D%7Bx%5E2%7D

f(x)%3D-%5Cfrac%7B2x%5E2%2B%5Cfrac%7B3%7D%7Bx%5E2%7D%7D%7B5%7D

Step 5: Substitute

x%3D2

f(2)%3D-%5Cfrac%7B2(2)%5E2%2B%5Cfrac%7B3%7D%7B(2)%5E2%7D%7D%7B5%7D%0A%3D-%5Cfrac%7B8%2B%5Cfrac%7B3%7D%7B4%7D%7D%7B5%7D%0A%3D-%5Cfrac%7B%5Cfrac%7B35%7D%7B4%7D%7D%7B5%7D%0A%3D-%5Cfrac%7B7%7D%7B4%7D

Step 1: Let $t%3D%5Csqrt%7Bx%7D$ , where $t%5Cge%200$ . The given equation becomes: $2%7Ct-3%7C%2Bt(t-6)%2B6%3D0$
Step 2: Simplify the polynomial part: $t(t-6)%2B6%3Dt%5E2-6t%2B6$ So, $2%7Ct-3%7C%2Bt%5E2-6t%2B6%3D0$
Step 3: Consider cases due to the absolute value. Case I: $t%5Cge%203%20%5CRightarrow%20%7Ct-3%7C%3Dt-3$ $2(t-3)%2Bt%5E2-6t%2B6%3D0$ $t%5E2-4t%3D0$ $t(t-4)%3D0$ Valid solution (since $t%5Cge%203$ ): $t%3D4$ Case II: $0%5Cle%20t%3C3%20%5CRightarrow%20%7Ct-3%7C%3D3-t$ $2(3-t)%2Bt%5E2-6t%2B6%3D0$ $t%5E2-8t%2B12%3D0$ $(t-2)(t-6)%3D0$ Valid solution (since $t%3C3$ ): $t%3D2$
Step 4: Convert back to $x$ : $x%3Dt%5E2%20%5CRightarrow%20x%3D16%2C%5C%3B4$
Step 5: Hence, $S%3D%5C%7B4%2C16%5C%7D$ which contains exactly two elements.

High-Yield Trend

Chapter Questions
2 MCQs

Official Solution

Official Solution

Equations

High-Yield Trend

Chapter Questions 2 MCQs

Official Solution

Official Solution

Chapter Questions
2 MCQs