Exams Joint Entrance Examination Mathematics Limits, Continuity and Differentiability Question 54 Question 54 of 401 Medium MCQ
If α > β > 0 \alpha >\beta >0 α > β > 0 are the roots of the equation a x 2 + b x + 1 = 0 a{x}^{2}+bx+1=0 a x 2 + b x + 1 = 0 , and l i m x → 1 α ( 1 − cos ( x 2 + b x + a ) 2 ( 1 − α x ) 2 ) 1 2 = 1 k ( 1 β − 1 α ) , k \underset{x\to \frac{1}{\alpha }}{lim}{(\frac{1-\cos ({x}^{2}+bx+a)}{2(1-\alpha x{)}^{2}})}^{\frac{1}{2}}=\frac{1}{k}(\frac{1}{\beta }-\frac{1}{\alpha }),k x → α 1 l im ( 2 ( 1 − α x ) 2 1 − c o s ( x 2 + b x + a ) ) 2 1 = k 1 ( β 1 − α 1 ) , k :
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Since, α , β \alpha ,\beta α , β are roots of a x 2 + b x + 1 = 0 a{x}^{2}+bx+1=0 a x 2 + b x + 1 = 0
Replace x → 1 x x\to \frac{1}{x} x → x 1
a x 2 + b x + 1 = 0 ⇒ x 2 + b x + a = 0 \frac{a}{{x}^{2}}+\frac{b}{x}+1=0\Rightarrow {x}^{2}+bx+a=0 x 2 a + x b + 1 = 0 ⇒ x 2 + b x + a = 0
So, 1 α , 1 β \frac{1}{\alpha },\frac{1}{\beta } α 1 , β 1 are the roots.
Now, l i m x → 1 α [ 1 − cos ( x 2 + b x + a ) 2 ( 1 − α x ) 2 ] 1 2 \underset{x\to \frac{1}{\alpha }}{lim}{[\frac{1-\cos ({x}^{2}+bx+a)}{2(1-\alpha x{)}^{2}}]}^{\frac{1}{2}} x → α 1 l im [ 2 ( 1 − α x ) 2 1 − c o s ( x 2 + b x + a ) ] 2 1
= l i m x → 1 α [ 2 sin 2 ( x 2 + b x + a 2 ) 2 ( 1 − α x ) 2 ] 1 2 =\underset{x\to \frac{1}{\alpha }}{lim}{[\frac{2{\sin }^{2}(\frac{{x}^{2}+bx+a}{2})}{2(1-\alpha x{)}^{2}}]}^{\frac{1}{2}} = x → α 1 l im [ 2 ( 1 − α x ) 2 2 s i n 2 ( 2 x 2 + b x + a ) ] 2 1
= l i m x → 1 α [ [ ∣ 2 sin 2 ( x − 1 α ) ( x − 1 β ) 2 4 × 2 α 2 ( x − 1 α ) 2 ( x − 1 β ) 2 4 ( x − 1 β ) 2 ] ] ∣ 1 2 = l i m x → 1 α [ [ ∣ ± 1 2 sin ( x − 1 α ) ( x − 1 β ) 2 α ( x − 1 α ) ( x − 1 β ) 2 ( x − 1 β ) ] ] ∣ = 1 2 α [ − 1 α + 1 β ) ⇒ 1 k [ 1 β − 1 α ] = 1 2 α [ 1 β − 1 α ] ⇒ k = 2 α \begin{gathered} =\underset{x\to \frac{1}{\alpha }}{lim}{[[|\frac{2{\sin }^{2}\frac{(x-\frac{1}{\alpha })(x-\frac{1}{\beta })}{2}}{4\times 2{\alpha }^{2}\frac{{(x-\frac{1}{\alpha })}^{2}{(x-\frac{1}{\beta })}^{2}}{4}{(x-\frac{1}{\beta })}^{2}}]]|}^{\frac{1}{2}} \\ =\underset{x\to \frac{1}{\alpha }}{lim}[[|\pm \frac{1}{2}\frac{\sin \frac{(x-\frac{1}{\alpha })(x-\frac{1}{\beta })}{2}}{\alpha \frac{(x-\frac{1}{\alpha })(x-\frac{1}{\beta })}{2}}(x-\frac{1}{\beta })]]| \\ =\frac{1}{2\alpha }[\frac{-1}{\alpha }+\frac{1}{\beta }) \\ \Rightarrow \frac{1}{k}[\frac{1}{\beta }-\frac{1}{\alpha }]=\frac{1}{2\alpha }[\frac{1}{\beta }-\frac{1}{\alpha }] \\ \Rightarrow k=2\alpha \end{gathered} = x → α 1 l im [[ ∣ 4 × 2 α 2 4 ( x − α 1 ) 2 ( x − β 1 ) 2 ( x − β 1 ) 2 2 sin 2 2 ( x − α 1 ) ( x − β 1 ) ]] ∣ 2 1 = x → α 1 l im [[ ∣ ± 2 1 α 2 ( x − α 1 ) ( x − β 1 ) sin 2 ( x − α 1 ) ( x − β 1 ) ( x − β 1 )]] ∣ = 2 α 1 [ α − 1 + β 1 ) ⇒ k 1 [ β 1 − α 1 ] = 2 α 1 [ β 1 − α 1 ] ⇒ k = 2 α Answer the question to unlock