Download App

STD 12 - 1. relation and function — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 1. relation and function1 Mark
MCQ
If $\,\,f(x) = \left\{ {\begin{array}{*{20}{c}}
  {3 + x;\,\,\,\,\,x \geqslant 0} \\ 
  {2 - 3x;\,\,\,\,\,x < 0} 
\end{array}} \right.$ then $\mathop {\lim }\limits_{x \to 0} f(f(x))$ is equal to -
  • A
    $9$
  • B
    $6$
  • C
    $-4$
  • ✓
    does not exist
✓

Answer

Correct option: D.
does not exist
d
$(2^x-1)\,(2^x-7)=7 \Rightarrow x = 3$ only

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

Explore more

More "M.C.Q (1 Marks)" from STD 12 - 1. relation and function→STD 12 - 1. relation and function — all question types→Mathematics STD 12 Science question papers→CBSE Board STD 12 Science question papers→CBSE Board question paper generator→

Similar questions

Choose the correct answer in Exercises:If $\frac{\text{d}}{\text{dx}}\text{f}\text{(x)}=4\text{x}^3-\frac{3}{\text{x}^4}$such that $\text{f}(2)=0.$Then $\text{f}\text{(x)}$ is
→
A curve satisfying the initial condition, $y(1) = 0,$ satisfies the differential equation, $x \frac{{dy}}{{dx}} = y -x^2.$ The area bounded by the curve and the $x-$ axis is
→
Let $n \ge 2$ be a natural number and $0 < \theta  < \frac{\pi }{2}$. Then $\int {\frac{{{{\left( {{{\sin }^n}\,\theta  - \sin \,\theta } \right)}^{\frac{1}{n}}}\,\cos \,\theta }}{{{{\sin }^{n + 1}}\,\theta }}} d\theta $ is equal to
→
Minimum integral value of $\alpha$ for which graph of $f(x) = ||x -2| -\alpha|-5$ has exactly four $x-$intercepts-
→
Let $\vec{a}=\hat{i}+\hat{j}-\hat{k}$ and $\vec{c}=2 \hat{i}-3 \hat{j}+2 \hat{k}$. Then the number of vectors $\vec{b}$ such that $\vec{b} \times \vec{c}=\vec{a}$ and $|\vec{b}| \in\{1,2, \ldots ., 10\}$ is
→
Solution of $\cos x\frac{{dy}}{{dx}} + y\sin x = 1$ is
→
If $f(x) = {\log _x}(\log x),$ then $f'(x)$ at $x = e$ is
→
The derivative of the function $\cot^{-1}\Big|(\cos2\text{x})^{\frac{1}{2}}\Big|$ at $\text{x}=\frac{\pi}{6}$ is:
→
If $tan^{-1} (x+ 2)+ tan^{- 1}( x -2)= tan^{-1} (\frac{1}{2}),$ then sum of value(s) of $x$ is equal to-
→
Let $f(1) = -2$ and $f'(x)  \ge 4.2$ for $1 \le x \le 6$. The possible value of $f(6)$ lies in the interval
→