MCQ
$\int_0^{\pi /4} {{{\tan }^6}x \, {{\sec }^2}x\,dx = } $
- ✓$\frac{1}{7}$
- B$\frac{2}{7}$
- C$1$
- DNone of these
Now $\int_0^{\pi /4} {{{\tan }^6}x{{\sec }^2}xdx = \int_0^1 {{t^6}dt = \frac{1}{7}[{t^7}]_0^1 = \frac{1}{7}} } $.
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$f(x)=\left\{\begin{array}{l}\max \left\{t^{3}-3 t\right\} ; x \leq 2 \\ t \leq x \\ x^{2}+2 x-6 ; 2 < x < 3 \\ {[x-3]+9 ; 3 \leq x \leq 5} \\ 2 x+1 \quad ; \quad x > 5\end{array}\right\}$
Where $[t]$ is the greatest integer less than or equal to $t$. Let $m$ be the number of points where $f$ is not differentiable and $I =\int\limits_{-2}^{2} f( x ) dx$. Then the ordered pair $( m , I )$ is equal to