Graph of function f contains point P (1, 2) and Q(s, r) . equation of secant line through P and Q is

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1.Relation and Function

normal

The graph of function $f$ contains the point $P (1, 2)$ and $Q(s, r)$. The equation of the secant line through $P$ and $Q$ is $y = \left( {\frac{{{s^2} + 2s - 3}}{{s - 1}}} \right)$ $x - 1 - s$. The value of $f ‘ (1)$, is

$2$

$3$

$4$

non existent

Solution

$(I)$ By definition $f ‘(1)$ is the limit of the slope of the secant line when $s \to 1$.
Thus $f ‘(1) =$$\mathop {Lim}\limits_{s \to 1} \,\frac{{{s^2} + 2s – 3}}{{s – 1}}$
=$\mathop {Lim}\limits_{s \to 1} \,\frac{{(s – 1)(s + 3)}}{{s – 1}}$
=$\mathop {Lim}\limits_{s \to 1} \,(s + 3)$ $= 4 \to \,\,(D)$
$(II)$ By substituting $x = s$ into the equation of the secant line, and cancelling by $s – 1$ again,

we get $y = s^2 + 2s – 1$.

This is $f (s)$, and its derivative is $f ‘(s) = 2s + 2, so f ‘ (1) = 4.$

Standard 12

Mathematics

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Let $P(x)$ be a polynomial with real coefficients such that $P\left(\sin ^2 x\right)=P\left(\cos ^2 x\right)$ for all $x \in[0, \pi / 2)$. Consider the following statements:

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$II.$ $P(x)$ can be expressed as a polynomial in $(2 x-1)^2$

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(IIT-1990)

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The graph of function $f$ contains the point $P (1, 2)$ and $Q(s, r)$. The equation of the secant line through $P$ and $Q$ is $y = \left( {\frac{{{s^2} + 2s - 3}}{{s - 1}}} \right)$ $x - 1 - s$. The value of $f ‘ (1)$, is

Solution

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