Exams Joint Entrance Examination Mathematics Limits, Continuity and Differentiability Question 39 Question 39 of 401 Medium MCQ
If the function
f ( x ) = 72 x − 9 x − 8 x + 1 2 − 1 + cos x , x ≠ 0 a log e 2 log e 3 , x = 0 f(x)= \begin{array}{|ccc|}\frac{{72}^{x}-{9}^{x}-{8}^{x}+1}{\sqrt{2}-\sqrt{1+\cos x}}, & x\ne 0 \\ a{\log }_{e}2{\log }_{e}3 & ,x=0\end{array} f ( x ) = 2 − 1 + c o s x 72 x − 9 x − 8 x + 1 , a log e 2 log e 3 x = 0 , x = 0
is continuous at x = 0 x=0 x = 0 , then the value of a 2 {a}^{2} a 2 is equal to
Solutions · 1 Post your solution Official solution
f ( x ) = 72 x − 9 x − 8 x + 1 2 − 1 + cos x , x ≠ 0 a log e 2 log e 3 , x = 0 f(x)= \begin{gathered} \frac{{72}^{x}-{9}^{x}-{8}^{x}+1}{\sqrt{2}-\sqrt{1+\cos x}},\, x\ne 0 \\ a{\log }_{e}2{\log }_{e}3\, \, \, \, \, \, ,\, \, x=0 \end{gathered} f ( x ) = 2 − 1 + cos x 72 x − 9 x − 8 x + 1 , x = 0 a log e 2 log e 3 , x = 0
∵ f ( x ) \because f(x) ∵ f ( x ) is continuous at x = 0 x=0 x = 0
⇒ l i m x → 0 72 x − 9 x − 8 x + 1 2 − 1 + cos x l i m x → 0 ( 9 x − 1 ) ( 8 x − 1 ) ( 2 + 1 + cos x ) ( 1 − cos x ) x 2 × x 2 = ( ln 9 ⋅ ln 8 ) ( 2 2 ) × 2 = 4 2 × 2 × 3 ln 2 ⋅ ln 3 24 2 ⋅ ln 2 ⋅ ln 3 ⇒ a = 24 2 a 2 = 1152 \begin{gathered} \Rightarrow \underset{x\to 0}{lim}\frac{{72}^{x}-{9}^{x}-{8}^{x}+1}{\sqrt{2}-\sqrt{1+\cos x}} \\ \underset{x\to 0}{lim}\frac{({9}^{x}-1)({8}^{x}-1)(\sqrt{2}+\sqrt{1+\cos x})}{\frac{(1-\cos x)}{{x}^{2}}\times {x}^{2}} \\ =(\ln 9\cdot \ln 8)(2\sqrt{2})\times 2 \\ =4\sqrt{2}\times 2\times 3\ln 2\cdot \ln 3 \\ 24\sqrt{2}\cdot \ln 2\cdot \ln 3 \\ \Rightarrow \, a=24\sqrt{2} \\ \, {a}^{2}=1152 \end{gathered} ⇒ x → 0 l im 2 − 1 + cos x 72 x − 9 x − 8 x + 1 x → 0 l im x 2 ( 1 − c o s x ) × x 2 ( 9 x − 1 ) ( 8 x − 1 ) ( 2 + 1 + cos x ) = ( ln 9 ⋅ ln 8 ) ( 2 2 ) × 2 = 4 2 × 2 × 3 ln 2 ⋅ ln 3 24 2 ⋅ ln 2 ⋅ ln 3 ⇒ a = 24 2 a 2 = 1152 Answer the question to unlock