Mathematics · Functions

JEE Main 2025 — 4 April, Evening Shift — Question 36

Let the domains of the functions f(x)=log⁡4log⁡3log⁡7(8f(x)=\log _{4} \log _{3} \log _{7}(8 −log⁡2(x2+4x+5))\left.-\log _{2}\left(x^{2}+4 x+5\right)\right) and

g(x)=sin⁡−1(7x+10x−2)g(x)=\sin ^{-1}\left(\frac{7 x+10}{x-2}\right) be (α,β)(\alpha, \beta) and [γ,δ][\gamma, \delta], respectively. Then α2+β2+γ2+δ2\alpha^{2}+\beta^{2}+\gamma^{2}+\delta^{2} is equal to:

  1. Option A:

    14

  2. Option B:

    16

  3. Option C:

    13

  4. Option D:

    15

    Correct

Answer: D

Step-by-step solution

f(x)=log⁡4(log⁡3(log⁡7(8−log⁡2(x2+4x+5)))f(x)=\log _{4}\left(\log _{3}\left(\log _{7}\left(8-\log _{2}\left(x^{2}+4 x+5\right)\right)\right)\right.

log⁡3(log⁡1(8−log⁡2(x2+4x+5)))>0\log _{3}\left(\log _{1}\left(8-\log _{2}\left(x^{2}+4 x+5\right)\right)\right)>0

log⁡7(8−log⁡2(x2+4x+5))>1\log _{7}\left(8-\log _{2}\left(x^{2}+4 x+5\right)\right)>1

8−log⁡2(x2+4x+5)>78-\log _{2}\left(x^{2}+4 x+5\right)>7

−log⁡2(x2+4x+5)>−1-\log _{2}\left(x^{2}+4 x+5\right)>-1

log⁡2(x2+4x+5)<1\log _{2}\left(x^{2}+4 x+5\right)<1

x2+4x+5<2x^{2}+4 x+5<2

x2+4x+3<0x^{2}+4 x+3<0

⇒(x+3)(x+1)<0…(1)\Rightarrow(x+3)(x+1)<0 …(1)

log⁡7(8−log⁡2(x2+4x+5))>0\log _{7}\left(8-\log _{2}\left(x^{2}+4 x+5\right)\right)>0

8−log⁡2(x2+4x+5)>18-\log _{2}\left(x^{2}+4 x+5\right)>1

log⁡2(x2+4x+5)<9\log _{2}\left(x^{2}+4 x+5\right)<9

x2+4x+5<29x^{2}+4 x+5<2^{9}

x2+4x+5<512x^{2}+4 x+5<512

⇒x2+4x−507<0\Rightarrow x^{2}+4 x-507<0

⇒x=−4±16+2028\Rightarrow x=-4 \pm \sqrt{16+2028}

x=−4±20442…(2)x=\frac{-4 \pm \sqrt{2044}}{2} …(2)

⇒(x−(−4+20442))(x−(−4−20442))<0\Rightarrow\left(x-\left(\frac{-4+\sqrt{2044}}{2}\right)\right)\left(x-\left(\frac{-4-\sqrt{2044}}{2}\right)\right)<0

x2+4x+5>0x^{2}+4 x+5>0

D>0D>0

x∈Rx \in R

Also, 8−log⁡2(x2+4x+5)>08-\log _{2}\left(x^{2}+4 x+5\right)>0

log⁡2(x2+4x+5)<8\log _{2}\left(x^{2}+4 x+5\right)<8

x2+4x+5<256x^{2}+4 x+5<256

⇒x2+4x−251<0\Rightarrow x^{2}+4 x-251<0

⇒x=−4±16+1004\Rightarrow x=-4 \pm \sqrt{16+1004}

⇒x=−4±10202\Rightarrow x=\frac{-4 \pm \sqrt{1020}}{2}

⇒(x−(−4+10202))(x−(−4−10202))<0\Rightarrow\left(x-\left(\frac{-4+\sqrt{1020}}{2}\right)\right)\left(x-\left(\frac{-4-\sqrt{1020}}{2}\right)\right)<0

∴\therefore Intersection of (1), (2) and (3)

∴x∈(−3,−1)\therefore x \in(-3,-1)

−1≤7x+10x−2≤1-1 \leq \frac{7 x+10}{x-2} \leq 1

⇒x∈[−2,−1]\Rightarrow x \in[-2,-1]

∴α2+β2+γ2+δ2=(−3)2+(−1)2+(−2)−2+(−1)2\therefore \alpha^{2}+\beta^{2}+\gamma^{2}+\delta^{2}=(-3)^{2}+(-1)^{2}+(-2)^{-2}+(-1)^{2}

=9+1+4+1=9+1+4+1

=15=15

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Subject
Mathematics
Chapter
Functions
Topic
Domain & range of functions
Let the domains of the functions f(x)=log 4 log 3 log 7 (8 .-log 2 (x… | JEE Main 2025 PYQ with Solution · DhiX AI