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TS EAMCET 6-Aug-2021 Shift 2 Question Paper

Section: Mathematics

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Question : 64 of 160

Marks: +1, -0

If a, b are roots of the equation

px^{2}+qx+r

0

\lim\limits_{x\to b}

\frac{1-\cos 2(px^{2}+qx+r)}{2(px-pb)^{2}}

=

$\frac{1}{2}(b-a)^2$
$(a+b)^2$
$\frac{1}{2}$
$a^2-2ab+b^2$

Solution:

p x^{2}+q x+r=0

has roots

a

and

b

[using L-Hospital rule]

\therefore a+b=\frac{-a}{p}\;\text{and}\;ab=\frac{r}{p}

=\lim\limits_{x\to b}\frac{1-\cos 2(p x^{2}+q x+r)}{2(p x-p b)^{2}}

=\lim\limits_{x\to b}\frac{0+2\sin 2(p x^{2}+q x+r)(2 p x+q)}{4(p x-p b)(p)}

2\cos 2(p x^{2}+q x+r)(2 p x+q)(2 p x+q)

=\lim\limits_{x\to b}\frac{+\sin 2(p x^{2}+q x+r)(2 p)}{2(p)(p)}

\{

using L-Hospital rule again

\}

=\frac{1 \times (2 p b+q)^{2}+0}{p^{2}}=\left(\frac{p b+p b+q}{p}\right)^{2}

. . . (i)

\because a+b=\frac{-q}{p}, ab=\frac{r}{p} \Rightarrow pb=\frac{r}{a}

\Rightarrow pa+pb=-q \Rightarrow pb=-q-pa

\because b

is a root

\Rightarrow pb^{2}+qb+r=0

\Rightarrow b(pb+q)=-r=pb+q=\frac{-r}{b}

\therefore

From Eq. (i), we get

\left(\frac{\frac{r}{a}-\frac{r}{b}}{p}\right)^{2}=\frac{r^{2}(b-a)^{2}}{(ab)^{2}p^{2}}=\frac{r^{2}(b-a)^{2}}{\left(\frac{r}{p}\right)^{2}p^{2}}=(b-a)^{2}

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