Basics & LogarithmsmediumJEE Main 2021Free

Basics & Logarithms — JEE Maths practice question (JEE Main 2021)

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

Question
The number of real roots of the equation (x+1)2+∣x−5∣=274(x+1)^{2}+|x-5|=\dfrac{27}{4} is
Solution
Answer: 2
Step 1: Case I: x≥5x\ge 5. Then ∣x−5∣=x−5|x-5|=x-5:
(x+1)2+(x−5)=274(x+1)^{2}+(x-5)=\dfrac{27}{4}
x2+2x+1+x−5=274x^{2}+2x+1+x-5=\dfrac{27}{4}
x2+3x−4=274x^{2}+3x-4=\dfrac{27}{4}
4x2+12x−16=27  ⇒  4x2+12x−43=04x^{2}+12x-16=27\;\Rightarrow\;4x^{2}+12x-43=0
 Step 2: Solve:
x=−12±144+6888=−12±8328=−12±4528=−3±2132x=\dfrac{-12\pm\sqrt{144+688}}{8}=\dfrac{-12\pm\sqrt{832}}{8}=\dfrac{-12\pm 4\sqrt{52}}{8}=\dfrac{-3\pm 2\sqrt{13}}{2}
 −3+2132≈−3+7.22≈2.1\dfrac{-3+2\sqrt{13}}{2}\approx\dfrac{-3+7.2}{2}\approx 2.1 (not ≥5\ge 5, rejected). −3−2132<0\dfrac{-3-2\sqrt{13}}{2}<0 (rejected). Step 3: Case II: x<5x<5. Then ∣x−5∣=5−x|x-5|=5-x:
(x+1)2+(5−x)=274(x+1)^{2}+(5-x)=\dfrac{27}{4}
x2+2x+1+5−x=274x^{2}+2x+1+5-x=\dfrac{27}{4}
x2+x+6=274x^{2}+x+6=\dfrac{27}{4}
4x2+4x+24=27  ⇒  4x2+4x−3=04x^{2}+4x+24=27\;\Rightarrow\;4x^{2}+4x-3=0
 Step 4: Solve:
x=−4±16+488=−4±88x=\dfrac{-4\pm\sqrt{16+48}}{8}=\dfrac{-4\pm 8}{8}
x=12 or x=−32x=\dfrac{1}{2}\text{ or }x=-\dfrac{3}{2}
 Both satisfy x<5x<5. Accepted. Step 5: Total real roots: 22. Answer: 22
Still stuck on this question?Ask your doubt on WhatsApp
Similar questions
Basics & Logarithms · easy
The product of all solutions of the equation e 5( e x) 2 +3 =x 8 , x>0 , is
Basics & Logarithms · easy
The number of real solutions of the equation x (x 2 +3|x|+5|x-1|+6|x-2| )=0 is
Basics & Logarithms · easy
If a+b+c=1 , ab+bc+ca=2 and abc=3 , then the value of a 4 +b 4 +c 4 is equal to
Basics & Logarithms · easy
Sum of all the values of x satisfying the equation 17 11 ( x+11 + x ) = 0 is:

Solve more, learn faster

Sign up free to solve more JEE Maths questions and explore doMath — timed drills, mastery sprints, bookmarks, and chapter-wise progress tracking.

Sign up freeExplore doMath