PUC Science 2nd PUC Class 12 - Karnataka Board PUC Question Bank Solutions

Find the point of local maximum or local minimum, if any, of the following function, using the first derivative test. Also, find the local maximum or local minimum value, as the case may be:

f(x) = x³(2x \[-\] 1)³.

f(x) =\[\frac{x}{2} + \frac{2}{x} , x > 0\] .

Find the projection of \[\vec{b} + \vec{c}  \text { on }\vec{a}\]  where \[\vec{a} = 2 \hat{i} - 2 \hat{j} + \hat{k} , \vec{b} = \hat{i} + 2 \hat{j} - 2 \hat{k} \text{ and } \vec{c} = 2 \hat{i} - \hat{j} + 4 \hat{k} .\]

If \[\vec{a} = 5 \hat{i} - \hat{j} - 3 \hat{k} \text{ and } \vec{b} = \hat{i} + 3 \hat{j} - 5 \hat{k} ,\] then show that the vectors \[\vec{a} + \vec{b} \text{ and } \vec{a} - \vec{b} \] are orthogonal.

f(x) = x⁴ \[-\] 62x² + 120x + 9.

f(x) = x³\[-\] 6x² + 9x + 15

f(x) = (x - 1) (x + 2)².

`f(x) = 2/x - 2/x^2,  x>0`

f(x) = xe^x.

`f(x) = x/2+2/x, x>0 `.

`f(x) = (x+1) (x+2)^(1/3), x>=-2` .

`f(x)=xsqrt(32-x^2),  -5<=x<=5` .

f(x) = \[x^3 - 2a x^2 + a^2 x, a > 0, x \in R\] .

A unit vector \[\vec{a}\] makes angles \[\frac{\pi}{4}\text{ and }\frac{\pi}{3}\] with \[\hat{i}\] and \[\hat{j}\]  respectively and an acute angle θ with \[\hat{k}\] .  Find the angle θ and components of \[\vec{a}\] .

f(x) = \[x + \frac{a2}{x}, a > 0,\] , x ≠ 0 .

f(x) = \[x\sqrt{2 - x^2} - \sqrt{2} \leq x \leq \sqrt{2}\] .

If two vectors \[\vec{a} \text{ and } \vec{b}\] are such that \[\left| \vec{a} \right| = 2, \left| \vec{b} \right| = 1 \text{ and } \vec{a} \cdot \vec{b} = 1,\]  then find the value of \[\left( 3 \vec{a} - 5 \vec{b} \right) \cdot \left( 2 \vec{a} + 7 \vec{b} \right) .\] 

f(x) = \[x + \sqrt{1 - x}, x \leq 1\] .

f(x) = (x \[-\] 1) (x \[-\] 2)².

If \[\vec{a}\] is a unit vector, then find \[\left| \vec{x} \right|\]  in each of the following. 

\[\left( \vec{x} - \vec{a} \right) \cdot \left( \vec{x} + \vec{a} \right) = 8\]