Basics & LogarithmsmediumPYQ 2016 · JEE MainFree

Basics & Logarithms: Sum Real Values Satisfying Equation (JEE Main 2016)

JEE Maths question with a full step-by-step solution.

Question
The sum of all real values of xx satisfying the equation (x2−5x+5)x2+4x−60=1(x^{2}-5x+5)^{x^{2}+4x-60}=1 is
A−4-4
B66
C55
D33correct
Solution
Step 1: ab=1a^{b}=1 when (i) a=1a=1, or (ii) b=0b=0 with a≠0a\neq 0, or (iii) a=−1a=-1 with bb even. Step 2: Case 1: base =1=1, i.e., x2−5x+5=1  ⇒  x2−5x+4=0  ⇒  (x−1)(x−4)=0x^{2}-5x+5=1\;\Rightarrow\;x^{2}-5x+4=0\;\Rightarrow\;(x-1)(x-4)=0. x=1x=1 or x=4x=4. Step 3: Case 2: exponent =0=0, i.e., x2+4x−60=0  ⇒  (x+10)(x−6)=0x^{2}+4x-60=0\;\Rightarrow\;(x+10)(x-6)=0. x=−10x=-10 or x=6x=6. Check base ≠0\neq 0 for each: x=−10x=-10: base =100+50+5=155≠0=100+50+5=155\neq 0. Valid. x=6x=6: base =36−30+5=11≠0=36-30+5=11\neq 0. Valid. Step 4: Case 3: base =−1=-1 with exponent even, i.e., x2−5x+5=−1  ⇒  x2−5x+6=0  ⇒  (x−2)(x−3)=0x^{2}-5x+5=-1\;\Rightarrow\;x^{2}-5x+6=0\;\Rightarrow\;(x-2)(x-3)=0. x=2x=2: exponent =4+8−60=−48=4+8-60=-48 (even). Valid. x=3x=3: exponent =9+12−60=−39=9+12-60=-39 (odd). Invalid. Step 5: Valid solutions: x=1,4,−10,6,2x=1,4,-10,6,2. Step 6: Sum =1+4+(−10)+6+2=3=1+4+(-10)+6+2=3. Answer: (4) 33
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