Exams Joint Entrance Examination Mathematics Limits, Continuity and Differentiability Question 398 Question 398 of 401 Medium MCQ
For α , β , γ ∈ R \alpha, \beta, \gamma \in \mathbf{R} α , β , γ ∈ R , if \lim _\limits{x \rightarrow 0} \frac{x^2 \sin \alpha x+(\gamma-1) \mathrm{e}^{x^2}}{\sin 2 x-\beta x}=3 , then β + γ − α \beta+\gamma-\alpha β + γ − α is equal to :
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\[ At x → 0 sin 2 x − β x → 0 ⇒ 0 0 form ⇒ ( γ − 1 ) e 0 + 0 sin ( α x ) → 0 ⇒ ( γ − 1 ) = 0 ⇒ γ = 1 ⇒ lim x → 0 x 2 sin ( α x ) ( sin 2 x − β x ) = 3 \begin{aligned}
& \text { At } x \rightarrow 0 \\
& \sin 2 x-\beta x \rightarrow 0 \\
& \Rightarrow \quad \frac{0}{0} \text { form } \\
& \Rightarrow \quad(\gamma-1) e^0+0 \sin (\alpha x) \rightarrow 0 \\
& \Rightarrow \quad(\gamma-1)=0 \\
& \Rightarrow \quad \gamma=1 \\
& \Rightarrow \quad \lim _{x \rightarrow 0} \frac{x^2 \sin (\alpha x)}{(\sin 2 x-\beta x)}=3
\end{aligned} At x → 0 sin 2 x − β x → 0 ⇒ 0 0 form ⇒ ( γ − 1 ) e 0 + 0 sin ( α x ) → 0 ⇒ ( γ − 1 ) = 0 ⇒ γ = 1 ⇒ x → 0 lim ( sin 2 x − β x ) x 2 sin ( α x ) = 3 \]
⇒ lim x → 0 x 2 [ α x − ( α x ) 3 3 ! + ( α x ) 5 5 ! − ⋯ ] [ ( 2 x ) − ( 2 x ) 3 3 ! + ( 2 x ) 5 5 ! ] − β x \Rightarrow \lim _{x \rightarrow 0} \frac{x^2\left[\alpha x-\frac{(\alpha x)^3}{3!}+\frac{(\alpha x)^5}{5!}-\cdots\right]}{\left[(2 x)-\frac{(2 x)^3}{3!}+\frac{(2 x)^5}{5!}\right]-\beta x} ⇒ lim x → 0 [ ( 2 x ) − 3 ! ( 2 x ) 3 + 5 ! ( 2 x ) 5 ] − β x x 2 [ α x − 3 ! ( α x ) 3 + 5 ! ( α x ) 5 − ⋯ ]
⇒ lim x → 0 α x 3 − α 3 x 5 3 ! + α 5 x 7 5 ! − ⋯ x ( 2 − β ) − 8 x 3 6 + 2 5 ⋅ x 5 5 ! − ⋯ = 3 \Rightarrow \lim _{x \rightarrow 0} \frac{\alpha x^3-\frac{\alpha^3 x^5}{3!}+\frac{\alpha^5 x^7}{5!}-\cdots}{x(2-\beta)-\frac{8 x^3}{6}+\frac{2^5 \cdot x^5}{5!}-\cdots}=3 ⇒ lim x → 0 x ( 2 − β ) − 6 8 x 3 + 5 ! 2 5 ⋅ x 5 − ⋯ α x 3 − 3 ! α 3 x 5 + 5 ! α 5 x 7 − ⋯ = 3
\[ ⇒ 2 − β = 0 and α − 8 6 = 3 ⇒ β = 2 α = 3 ( − 8 6 ) = − 4 ⇒ γ = 1 , β = 2 , α = − 4 ⇒ β + γ − α = 7 \begin{aligned}
\Rightarrow & 2-\beta=0 \text { and } \frac{\alpha}{\frac{-8}{6}}=3 \\
\Rightarrow & \beta=2 \\
& \alpha=3\left(-\frac{8}{6}\right)=-4 \\
\Rightarrow & \gamma=1, \beta=2, \alpha=-4 \\
\Rightarrow & \beta+\gamma-\alpha=7
\end{aligned} ⇒ ⇒ ⇒ ⇒ 2 − β = 0 and 6 − 8 α = 3 β = 2 α = 3 ( − 6 8 ) = − 4 γ = 1 , β = 2 , α = − 4 β + γ − α = 7 \] Answer the question to unlock