MCQ
Greatest value of the function, $f(x) = - 1 + \frac{2}{{{2^x}^2 + 1}}$ is
- A$1$
- B$3/2$
- C$2/3$
- ✓$0$
Clearly $f(x)$ in an even function and $f(x)$ is greatest
when $\frac{2}{2^{x^{2}}+1}$ is greatest.
(given)
Also, $\frac{2}{2^{x^{2}}+1}$ is greatest when $2^{x^{2}}+1$ is least,
which occurs when $x=0.$
Hence greatest value is $f(0)=0.$
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| Column-$I$ | Column-$II$ |
| $(A)$ In $R ^2$, if the magnitude of the projection vector of the vector $\alpha \hat{i}+\beta \hat{j}$ on $\sqrt{3} \hat{i}+\hat{j}$ is $\sqrt{3}$ and if $\alpha=2+\sqrt{3} \beta$, then possible value(s) of $|\alpha|$ is (are) | $(P)$ $1$ |
| $(B)$ Let $a$ and $b$ be real numbers such that the function $f(x)=\left\{\begin{array}{cc}-3 a x^2-2, & x < 1 \\ b x+a^2, & x \geq 1\end{array}\right.$ is differentiable for all $x \in R$. Then possible value(s) of a is (are) | $(Q)$ $2$ |
| $(C)$ Let $\omega \neq 1$ be a complex cube root of unity. If $\left(3-3 \omega+2 \omega^2\right)^{4 n+3}$ $+\left(2+3 \omega-3 \omega^2\right)^{4 \omega+3}+\left(-3+2 \omega+3 \omega^2\right)^{4 \omega+3}=0$, then possible value(s) of $n$ is (are) | $(R)$ $3$ |
| $(D)$ Let the harmonic mean of two positive real numbers $a$ and $b$ be $4$. If $q$ is a positive real number such that $a, 5, q, b$ is an arithmetic progression, then the value$(s)$ of $|q-a|$ is (are) | $(S)$ $4$ |
| $(T)$ $5$ |