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Q.1. If f (x) be a twice differentiable function such that f"(x) = f (x) and f'(x) = g (x ). If h'(x) = [f(x)]^{2}+[g(x)]^{2} ’h (1) = 8 and h (0) = 2 then find the value of h (2) .
Ans. h'( X ) = [f ( x )]^{2}+[ g ( x )]^{2}
h'' ( x ) = 2 f (x ) f' ( x ) + 2 g ( x ) g ' (x )
h'' ( x ) = 2 f(x) g (x ) + 2g ( x ) f'' (x )
h'' ( x ) = 2 f(x) g (x ) + 2g ( x ) (f (x )) = 0
Thus, h'(x) = c, a constant for all x.
⇒ h (x) = cx + c_{1}
It is given that h (l) = 8 and h (0) = 2, therefore c_{1} = 2 and c_{2} = 6
∴ h (x ) = 6 x + 2 ⇒ h (2) = 6 x 2 + 2 = 14
Q.2. If y = ax then find the value of dy/dx as a function of x and y ( a is a constant).
Ans. We can write y = a^{xy}
Taking Logrithm of both sides, we obtain ln y = x^{y} ln a.
⇒ ln (ln y) = y ln x + ln (ln a)
Differentiating with respect to x gives
Q.3. Find the coordinates of the point P on the curve y^{2} = 2x^{3} such that the tangent at P is perpendicular to the line 4 x  3 y + 2 = 0.
Ans.
Let, P(x_{0},y_{0}) be the point at which the tangent is perpendicular to the line 4x  3y + 2 = 0.
Thus slope at P is equal to m =
Thus (i)
From the equation of curve (ii)
From equation (i) and (ii)
Thus there are two points and (0,0) at which the tangent line is perpendicular
to the line 4x  3y + 2 = 0.
Q.4. If the tangent to the curve xy + ax + by = 0 at (1,1) is inclined at an angle tan^{1} 2 with x axis then find the values of a and b .
Ans. The point (1,1) lies on the curve xy + ax + by = 0.
Hence, a+b = 1 (i)
Differentiating xy + ax + by = 0 with respect to x we get
Since the tangent at (1,1) makes an angle of tan^{1} 2 with the x  axis hence
Solving equations (i) and (ii) we obtain
a = 1 and b = 2
Q.5. If f (x) = a log x + bx^{2} + x has its extreme values (local maximum or minimum value) at x = 1 and x = 2 , then find the values of a and b .
Ans. The logarithmic function is defined for all x > 0. Hence the domain of f (x) is (0, ∞) . Using this fact we can write
f (x) = a log x + bx^{2} + x
We see that the derivative is defined at all points of the domain of the function. Hence the function can attain extreme values only at points where f'(x) = 0. Since f (x) attains its extreme values at x = 1,2.
∴ f'(1) = 0 and f'(2) = 0
Q.6. If y = sin^{1} where α is a constant then find the value of y'(0) .
Ans.
Q.7. Find the equation of the tangent to the curve x = t cos t and y = t sin t at the origin of xy plane.
Ans.
At the origin x = 0, y = 0
⇒ t cos t = 0 and t sin t = 0
t = 0 and cos t = 0 and t = 0 and sin t = 0
Since there is no value of t for which cos t and sin t are simultaneously 0 , hence t = 0.Thus the slope of the tangent at (0,0) is Thus the equation of the tangent at the origin is y = 0.
Q.8. A curve with equation of the form y = ax^{4} + bx^{3} + c + cx + d has slope 0 at the point (0,1) and also touches the x axis at the point (1,0) then find the values of x for which the curve has negative slope.
Ans:
Hence, 4a • 0+3b • 0+c = 0 ⇒ c = 0
And, 4a (1) + 3b = 0 (i)
Also, the curve passes through (0,1) and (1,0), hence
d = 1 and 0 = a  b  c + d
⇒ a  b  c + 1 = 0 (ii)
From (i) and (ii), we get
a = 3,b = 4,c = 0 d = 1
Q.9. Find the point at which the local maximum or local minimum value of the function
occurs. Also find the local minimum and for maximum values of the function.
Ans. We have y = f (x) = sin^{4 }x + cos^{4} x
= 4 cos x sin x (cos^{2} x  sin^{2} x)
= 2 sin 2x cos 2x =  sin 4x
Over the given interval is defined for all x , hence for local maximum or minimum value Where we have used the fact that.
Thus is the only point where local maximum or minimum value of the function occurs. Here we can use either first derivative test or the second derivative test to check for local maxima or minimum. Using the second derivative test, we obtain
Thus at a local minimum occurs. The local minimum value is
Note: Students should check using the first derivative test that a local minimum indeed occurs at x =
Q.10. If A > 0,B > 0 then find the maximum value of tan A + tan B.
Ans. We have A + B =
Let, z = tan A+tan B
F or maximum value of because which implies that x = tan A > 0. It can be earily checked that Hence, z is maximum for x = For this value of x, z =
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