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JEE Main 202623 JanuaryMorning Shift
Exam
JEE Main
Year
2026
Date
23 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Hyperbola

JEE Main 2026 · 23 January · Morning Shift · Mathematics · Hyperbola

Let the domain of the function f(x)= _ 3 _ 5 _ 7 (9 x-x^ 2 -13) be...

Let the domain of the function f(x)=log⁡3log⁡5log⁡7f(x)=\log _{3} \log _{5} \log _{7}\left(9 x-x^{2}-13\right)betheintervalbe the interval(\mathrm{m}, \mathrm{n}).Letthehyperbola. Let the hyperbola\frac{x^{2}}{\mathrm{a}^{2}}-\frac{y^{2}}{\mathrm{~b}^{2}}=1haveeccentricityhave eccentricity\frac{\mathrm{n}}{3}andthelengthofthelatusrectumand the length of the latus rectum\frac{8 \mathrm{~m}}{3}.Then. Then\mathrm{b}^{2}-\mathrm{a}^{2} is equal to :

Options

  1. A77
  2. B1111
  3. C5
  4. D9

Correct answer

A. 77

Step-by-step solution

For f(x)=log⁡3log⁡5log⁡7(9x−x2−13)f(x) = \log_3\log_5\log_7(9x-x^2-13):
log⁡7(⋅)>0⇒9x−x2−13>1⇒(x−2)(x−7)<0⇒x∈(2,7)\log_7(\cdot) > 0 \Rightarrow 9x-x^2-13 > 1 \Rightarrow (x-2)(x-7) < 0 \Rightarrow x \in (2,7).
log⁡5(⋅)>0⇒log⁡7(9x−x2−13)>1⇒9x−x2−13>7⇒(x−4)(x−5)<0⇒x∈(4,5)\log_5(\cdot) > 0 \Rightarrow \log_7(9x-x^2-13) > 1 \Rightarrow 9x-x^2-13 > 7 \Rightarrow (x-4)(x-5) < 0 \Rightarrow x \in (4,5).
Domain =(4,5)= (4, 5), so m=4,n=5m = 4, n = 5.
Eccentricity e=n/3=5/3e = n/3 = 5/3, latus rectum =8m/3=32/3= 8m/3 = 32/3.
e2=1+b2/a2⇒b2/a2=16/9e^2 = 1 + b^2/a^2 \Rightarrow b^2/a^2 = 16/9.
2b2/a=32/3⇒32a/9=32/3⇒a=32b^2/a = 32/3 \Rightarrow 32a/9 = 32/3 \Rightarrow a = 3, so a2=9a^2 = 9, b2=16b^2 = 16.
b2−a2=16−9=7b^2 - a^2 = 16 - 9 = 7.
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