MCQ
The three different face diagonals of a cuboid (rectangular parallelopiped) have lengths $39,40,41$. The length of the main diagonal of the cuboid which joins a pair of opposite corners is
- ✓$49$
- B$49 \sqrt{2}$
- C$60$
- D$60 \sqrt{2}$
Let the length, breadth and height of cuboid is $l, b$ and $h$ respectively.
$Given, l^2+h^2=39^2$
$\Rightarrow b^2+h^2=40^2$
$\Rightarrow \quad l^2+b^2=41^2$
$\Rightarrow \quad 2\left(l^2+b^2+h^2\right)=39^2+40^2+41^2$
$\Rightarrow \quad l^2+b^2+h^2=2401$
$\therefore$ Length of longest diagonal
$=\sqrt{l^2+b^2+h^2}$
$=\sqrt{2401}=49$
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$(ii)$ $f '(-5) = 0 \,; \,f '(2)$ is not defined and $f '(4) = 0$
$(iii)$ $(-5, 12)$ is a point which lies on the graph of $f (x)$
$(iv)$ $f ''(2)$ is undefined, but $f ''(x)$ is negative everywhere else.
$(v)$ the signs of $f '(x)$ is given below
On the possible graph of $y = f (x)$ we have 