MCQ
Let $f(x)=\left\{\begin{array}{ll}x^3-x^2+10 x-7 & ,x \leq 1 \\ -2 x+\log _2\left(b^2-4\right) & , \quad x>1\end{array}\right.$ Then the set of all values of $b$, for which $f(x)$ has maximum value at $x=1$, is
  • A
    $(-6,-2)$
  • B
    $(2,6)$
  • $[-6,-2) \cup(2,6]$
  • D
    $[-\sqrt{6},-2) \cup(2, \sqrt{6}]$

Answer

Correct option: C.
$[-6,-2) \cup(2,6]$
$f(x)=\left\{\begin{array}{ll}x^3-x^2+10 x-7, & x \leq 1 \\ -2 x+\log _2\left(b^2-4\right), & x>1\end{array}\right.$
Since $f(x)$ has maximum value at $x=1$, therefore
$\text{L.H.L.} \geq \text{R.H.L.}$ at $x=1$
$\Rightarrow \lim _{x \rightarrow 1}\left(x^3-x^2+10 x-7\right) \geq \lim _{x \rightarrow 1}\left[-2 x+\log _2\left(b^2-4\right)\right]$
$\Rightarrow 3 \geq-2+\log _2\left(b^2-4\right) $
$\Rightarrow 2^5 \geq b^2-4$
$\Rightarrow b^2 \leq 36 $
$\Rightarrow(b-6)(b+6) \leq 0 $
$\Rightarrow b \in[-6,6]$
$\text { but } b^2-4 \neq 0 $
$\Rightarrow b \neq \pm 2 $
$\Rightarrow b \in[-6,-2) \cup(2,6]$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

If line $\frac{x-3}{2}=\frac{y-4}{3}=\frac{z-5}{4}$ lies in the plane $4 x+4 y- cz - d =0$, then values of $c , d$ are
Every invertible function is:
The diameter of a circle is increasing at the rate of $1\ \text{cm/sec.}$ When its radius is $\pi$ the rate of increase of its area is:
In $\triangle A B C, L, M, N$ are points on $BC , CA , AB$ respectively, dividing them in the ratio $1: 2$, $2: 3,3: 5$. If the point K divides AB in the ratio $5: 3$, then $\frac{|\overline{ AL }+\overline{ BM }+\overline{ CN }|}{|\overline{ CK }|}=$
In ABC, ac cos B – bc cos A = ____________.
If line joining points $A$ and $B$ having position vectors $6 \vec{a}-4 \vec{b}+4 \vec{c}$ and $-4 \vec{c}$ respectively, and the line joining the points $C$ and $D$ having position vectors $-\vec{a}-2 \vec{b}-3 \vec{c}$ and $\vec{a}+2 \vec{b}-5 \vec{c}$ intersect, then their point of intersection is
The area bounded by the curve $y^2= 8x$, the $x-$axis and the lastus rectum is:
Let $\text{X}=\begin{bmatrix}\text{x}_1\\\text{x}_2\\\text{x}_3\end{bmatrix},\text{A}=\begin{bmatrix}1&-1&2\\2&0&1\\3&2&1\end{bmatrix}\text{and }\text{B}=\begin{bmatrix}3\\1\\4\end{bmatrix}$. If $AX = B,$ then $X$ is equal to:
If $\text{f(x)}=\begin{cases}\frac{1-\sin\text{x}}{(\pi-2\text{x}^2)}\times\frac{\log\sin\text{x}}{\log(1+\pi^2-4\pi\text{x}+4\text{x}^2)},&\text{x}\neq\frac{\pi}{2}\\\text{k},&\text{x}=\frac{\pi}{2}\end{cases}$ is continuous at $\text{x}=\frac{\pi}{2},$ then $k =$
If $x, y, z$ are non$-$zero real numbers, then the inverse, then the inverse of the matrix $\begin{bmatrix}\text{x} & 0 & 0\\ 0 & \text{y} & 0 \\ 0 & 0 & \text{z}\end{bmatrix}$, is: