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

English

Gujarati

Hindi

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

The range of the function $f(x) = \frac{{\sqrt {1 – {x^2}} }}{{1 + \left| x \right|}}$ is

normal

Consider a function $f:\left[ { – 1,1} \right] \to R$ where $f(x) = {\alpha _1}{\sin ^{ – 1}}x + {\alpha _3}\left( {{{\sin }^{ – 1}}{x^3}} \right) + ….. + {\alpha _{(2n + 1)}}{({\sin ^{ – 1}}x)^{(2n + 1)}} – {\cot ^{ – 1}}x$ Where $\alpha _i\ 's$ are positive constants and $n \in N < 100$ , then $f(x)$ is

normal

Consider the identity function $I _{ N }: N \rightarrow N$ defined as $I _{ N }$ $(x)=x$ $\forall $ $x \in N$ Show that although $I _{ N }$ is onto but $I _{ N }+ I _{ N }:$ $ N \rightarrow N$ defined as $\left(I_{N}+I_{N}\right)(x)=$ $I_{N}(x)+I_{N}(x)$ $=x+x=2 x$ is not onto.

easy

Let $\mathrm{f}(\mathrm{x})$ be a polynomial of degree $3$ such that $\mathrm{f}(\mathrm{k})=-\frac{2}{\mathrm{k}}$ for $\mathrm{k}=2,3,4,5 .$ Then the value of $52-10 \mathrm{f}(10)$ is equal to :

hard

(JEE MAIN-2021)

If the domain of the function $f(x)=\log _e\left(4 x^2+11 x+6\right)+\sin ^{-1}$ $(4 x+3)+\cos ^{-1}\left(\frac{10 x+6}{3}\right) \text { is }(\alpha, \beta]$ Then $36|\alpha+\beta|$ is equal to :

hard

(JEE MAIN-2023)

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

Similar Questions

The range of the function $f(x) = \frac{{\sqrt {1 – {x^2}} }}{{1 + \left| x \right|}}$ is

Consider the identity function $I _{ N }: N \rightarrow N$ defined as $I _{ N }$ $(x)=x$ $\forall $ $x \in N$ Show that although $I _{ N }$ is onto but $I _{ N }+ I _{ N }:$ $ N \rightarrow N$ defined as $\left(I_{N}+I_{N}\right)(x)=$ $I_{N}(x)+I_{N}(x)$ $=x+x=2 x$ is not onto.

Let $\mathrm{f}(\mathrm{x})$ be a polynomial of degree $3$ such that $\mathrm{f}(\mathrm{k})=-\frac{2}{\mathrm{k}}$ for $\mathrm{k}=2,3,4,5 .$ Then the value of $52-10 \mathrm{f}(10)$ is equal to :

If the domain of the function $f(x)=\log _e\left(4 x^2+11 x+6\right)+\sin ^{-1}$ $(4 x+3)+\cos ^{-1}\left(\frac{10 x+6}{3}\right) \text { is }(\alpha, \beta]$ Then $36|\alpha+\beta|$ is equal to :

Start a Free Trial Now